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Second Edition
Water Chemistry
Mark M. Benjamin University of Washington
For information about this book, contact: Waveland Press, Inc. 4180 IL Route 83, Suite 101 Long Grove, IL 60047-9580 (847) 634-0081
[email protected] www.waveland.com
Front cover photo of water processing plant by Michael K. Stenstrom. Index prepared by Clive Pyne Book Indexing Services 38 Inglewood Place Ottawa, Ontario K1Y 4C7 Canada
Copyright © 2015, 2002 by Mark M. Benjamin 10-digit ISBN 1-4786-2308-X 13-digit ISBN 978-1-4786-2308-3 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means without permission in writing from the publisher. Printed in the United States of America 7
6
5
4
3
2
DEDICATION
To those who have nurtured me
Arthur and Hannah Benjamin Doc and Hune Smith Judith and Mara Benjamin
iii
Contents
PREFACE
i
ACKNOWLEDGEMENT
iv
1
BASICS OF AQUATIC CHEMISTRY 1 1.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 THE STRUCTURE OF WATER MOLECULES AND INTERACTIONS AMONG THEM . . . . . . . . . . . . . . . . . 3 1.3 THE BEHAVIOR OF SOLUTES IN WATER; DISSOLUTION OF SALTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.4 COMMON APPROACHES FOR EXPRESSING CONCENTRATIONS IN ENVIRONMENTAL CHEMISTRY . . . . . . . . . . . . . . . 10 1.5 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.6 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2
CHEMICAL REACTIVITY, REACTIONS, AND EQUILIBRIUM 2.1 OVERVIEW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 CHARACTERIZING CHEMICAL REACTIVITY . . . . . . . . . 2.3 PREDICTING ACTIVITY COEFFICIENTS FROM KNOWLEDGE OF THE SOLUTION COMPOSITION . . . . . . . . . . . . . . . 2.4 THE ACTIVITY AS AN INTENSIVE PROPERTY; THE ACTIVITY OF SOLIDS DISPERSED IN WATER . . . . . . . . . 2.5 MODELS OF CHEMICAL EQUILIBRIUM . . . . . . . . . . . . 2.6 EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT 2.7 COMBINING CHEMICAL REACTIONS . . . . . . . . . . . . . . v
37 38 38 49 56 57 68 69
2.8 2.9
SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 73
3
REACTION KINETICS AND THE KINETICS-BASED INTERPRETATION OF EQUILIBRIUM 77 3.1 A MOLECULAR-LEVEL PICTURE OF AN ELEMENTARY REACTION AND THE FACTORS THAT AFFECT ITS RATE . . 78 3.2 EFFECT OF TEMPERATURE ON REACTION RATE CONSTANTS 83 3.3 THE KINETICS OF SOME IMPORTANT CATEGORIES OF ENVIRONMENTAL CHEMICAL REACTIONS . . . . . . . . . . . . 84 3.4 KINETICS OF ELEMENTARY CHEMICAL REACTIONS . . . . 86 3.5 REACTION REVERSIBILITY AND THE DEFINITION OF THE EQUILIBRIUM CONSTANT . . . . . . . . . . . . . . . . . . . . 88 3.6 EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT 91 3.7 COMBINING CHEMICAL REACTIONS: KINETICS AND EQUILIBRIUM CONSTANTS OF NONELEMENTARY REACTIONS . 92 3.8 EXPERIMENTAL EVALUATION OF REACTION KINETICS . . 97 3.9 RATE-LIMITING STEPS AND SOME CLASSICAL, MODEL REACTION PATHWAYS . . . . . . . . . . . . . . . . . . . . . . . 104 3.10 HETEROGENEOUS (PHASE-TRANSFER) REACTIONS . . . . . 113 3.11 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 3.12 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
4
POTENTIALS, ENERGY, AND FORCES: INTERPRETING CHANGES IN PHYSICAL/CHEMICAL SYSTEMS 131 4.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.2 POTENTIAL ENERGY, POTENTIALS, AND FORCES . . . . . . 134 4.3 OVERVIEW OF POTENTIAL AND POTENTIAL ENERGY IN CHEMICAL SYSTEMS . . . . . . . . . . . . . . . . 143 4.4 ENTROPY AND GIBBS ENERGY . . . . . . . . . . . . . . . . . 144 4.5 THE CHEMICAL POTENTIAL AND ITS USE TO PREDICT CHEMICAL CHANGE . . . . . . . . . . . . . . . . 148 4.6 OTHER MOLAR THERMODYNAMIC QUANTITIES . . . . . . 150 4.7 QUANTIFYING Gi , Hi , and Si . . . . . . . . . . . . . . . . . . . . 151 4.8 THE DEPENDENCE OF Gi ON SYSTEM COMPOSITION, TEMPERATURE, AND PRESSURE . . . . . . . . . . . . . . . . 158 4.9 COMPARISON OF Gi OF DIFFERENT SPECIES . . . . . . . . . 166 4.10 THE DEPENDENCE OF Gi ON PRESSURE AND TEMPERATURE . . . . . . . . . . . . . . . . . . . . . . . . 168 4.11 COMPARING THE MAGNITUDE OF FACTORS THAT AFFECT Gi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 vi
4.12 CHEMICAL POTENTIAL AS THE DRIVING FORCE FOR CHEMICAL REACTIONS . . . . . . . . . . . . . . 179 4.13 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 4.14 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 5
ACIDS AND BASES, PART 1: ACID/BASE SPECIATION AND EXACT SOLUTIONS TO ACID/BASE PROBLEMS 215 5.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 217 5.2 THE DISSOCIATION OF WATER; Kw . . . . . . . . . . . . . . . 219 5.3 THE STRUCTURE OF ACIDS AND BASES . . . . . . . . . . . . 221 5.4 STRONG AND WEAK ACIDS, K a , AND CONJUGATE ACID/BASE PAIRS . . . . . . . . . . . . . . . . . . . . . . . . . . 222 5.5 A BRIEF DIVERSION: THE STRUCTURE OF SOME IMPORTANT ORGANIC ACIDS AND BASES . . . . . . . . . . . 229 5.6 COMBINING ACIDITY REACTIONS AND THE DEFINITION OF BASICITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 5.7 ACID/BASE SPECIATION AS A FUNCTION OF pH; DIAGRAMS OF LOG C VERSUS pH . . . . . . . . . . . . . . . . . . . . . . . 238 5.8 LOG C-pH DIAGRAMS FOR MONOPROTIC ACIDS . . . . . . . 243 5.9 EFFECTS OF NONIDEAL SOLUTE BEHAVIOR ON LOG C-pH DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 5.10 LOG C-pH DIAGRAMS FOR MULTIPROTIC ACIDS . . . . . . . 249 5.11 DETERMINING SPECIES AND RELEVANT EQUATIONS FOR SOLVING EQUILIBRIUM PROBLEMS . . . . . . . . . . . . . . 256 5.12 NUMERICAL APPROACHES FOR SOLVING ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 5.13 THE pH OF SOLUTIONS CONTAINING ONLY WATER AND STRONG ACIDS OR BASES . . . . . . . . . . . . . . . . . . . . 266 5.14 THE pH OF IDEAL SOLUTIONS CONTAINING WEAK ACIDS AND BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 5.15 A SIMPLE SPREADSHEET ANALYSIS FOR DETERMINING ACID/BASE SPECIATION . . . . . . . . . . . . . . . . . . . . . . 274 5.16 THE pH AND SPECIATION OF SYSTEMS CONTAINING MULTIPROTIC WEAK ACIDS AND BASES . . . . . . . . . . . 279 5.17 ACID/BASE EQUILIBRIA IN NONIDEAL SOLUTIONS . . . . . 280 5.18 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 5.19 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
vii
6
ACIDS AND BASES, PART 2: USE OF Log C-pH DIAGRAMS AND THE TOTH EQUATION 6.1 GRAPHICAL SOLUTIONS FOR SETS OF SIMULTANEOUS EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 USING LOG C-pH DIAGRAMS TO SOLVE WEAK ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 THE MASS BALANCE ON H (THE TOTH EQUATION AND PROTON CONDITION) . . . . . . . . . . . . . . . . . . . . . . . 6.4 DEFINING THE COMPOSITION OF THE SYSTEM: COMPONENTS, SPECIES, AND THE SYSTEM TABLEAU . . . 6.5 EXPRESSING INPUT AND EQUILIBRIUM CONCENTRATIONS IN TERMS OF COMPONENTS; COMPONENT MASS BALANCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 IDENTIFYING DOMINANT SPECIES . . . . . . . . . . . . . . . 6.7 COMPARISON OF APPROACHES FOR SOLVING ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 APPENDIX 6A: ARITHMETIC OPERATIONS AND LOG-LOG DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
8
SOFTWARE FOR SOLVING CHEMICAL EQUILIBRIUM PROBLEMS 7.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 THE COMPUTER ALGORITHM FOR SOLVING A SIMPLE ACID/BASE PROBLEM . . . . . . . . . . . . . . . . . . . . . . . 7.3 SYSTEMS IN WHICH THE EQUILIBRIUM ACTIVITIES OF ONE OR MORE SPECIES ARE KNOWN . . . . . . . . . . . . . . . . 7.4 USING CHEMICAL EQUILIBRIUM SOFTWARE TO DEVELOP LOG C−pH DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . 7.5 MODIFYING THE COMPONENT AND SPECIES DATABASES IN VISUAL MINTEQ . . . . . . . . . . . . . . . . . . . . . . . . 7.6 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TITRATIONS AND BUFFERS 8.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . 8.2 REACTIONS OCCURRING DURING A TITRATION: QUALITATIVE CONSIDERATIONS . . . . . . . . . . . . . . 8.3 QUANTITATIVE INTERPRETATION OF TITRATION DATA 8.4 USING VISUAL MINTEQ TO SIMULATE TITRATIONS . . . viii
291 292 293 308 310
316 327 341 342 343 345 353 354 355 363 368 373 375 375
377 . . 378 . . 379 . . 384 . . 393
8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 9
TITRATIONS WITH WEAK ACIDS OR BASES . . . . . . . . . . THE EFFECT OF THE ACIDITY OF WATER ON TITRATION CURVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TITRATION OF SOLUTIONS CONTAINING UNKNOWN ACIDS AND BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TITRATION EQUIVALENCE POINTS . . . . . . . . . . . . . . . ALKALINITY AND ACIDITY . . . . . . . . . . . . . . . . . . . ACID/BASE TITRATIONS OF NATURAL ORGANIC MATTER . BUFFERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . APPENDIX 8A: DETERMINING ALKALINITY FACTORS AND INSERTING THEM INTO THE VISUAL MINTEQ DATABASE . APPENDIX 8B: DERIVATION OF THE BUFFER INTENSITY OF MULTIPROTIC ACIDS . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
GAS/LIQUID EQUILIBRIUM 9.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 BASIC CONCEPTS AND TERMINOLOGY FOR GAS/LIQUID EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 EFFECT OF GAS/LIQUID EQUILIBRATION ON GAS-PHASE COMPOSITION . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 FACTORS AFFECTING VOLATILITY AND HENRY’S CONSTANT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 HENRY’S LAW AND AQUEOUS-PHASE SPECIATION . . . . . 9.6 THE pH OF SOLUTIONS IN EQUILIBRIUM WITH ACIDIC OR BASIC GASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 GAS-LIQUID EQUILIBRIUM IN VISUAL MINTEQ . . . . . . . 9.8 BUFFERING IN SOLUTIONS EQUILIBRATED WITH WEAK ACID AND WEAK BASE GASES . . . . . . . . . . . . . . . . . 9.9 CO2 DISSOLUTION, ALKALINITY, AND ACIDITY . . . . . . . 9.10 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.11 APPENDIX 9A. THE LIQUID-PHASE ACTIVITY COEFFICIENT IN HENRY’S LAW USING THE RAOULT’S LAW STANDARD STATE CONVENTION . . . . . . 9.12 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
402 405 406 409 413 438 442 453 455 456 457 469 470 471 482 483 490 494 502 508 512 515
516 519
10 CHEMISTRY OF METALS IN AQUEOUS SYSTEMS 1. SOLUTION-PHASE REACTIONS 527 10.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 528 10.2 METAL OXIDATION STATES . . . . . . . . . . . . . . . . . . . 529 ix
10.3 10.4 10.5 10.6
FORMATION AND STRUCTURE OF METAL COMPLEXES COMPLEXES WITH HYDROXIDE IONS . . . . . . . . . . . LOG C-pH DIAGRAMS FOR DISSOLVED METALS . . . . . COMPLEXES WITH LIGANDS OTHER THAN WATER AND HYDROXIDE . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 MIXED LIGAND COMPLEXES AND CHELATING AGENTS 10.8 METAL SPECIATION IN SYSTEMS CONTAINING COMPLEXING LIGANDS . . . . . . . . . . . . . . . . . . . . 10.9 METAL ION BUFFERS . . . . . . . . . . . . . . . . . . . . . 10.10PREDOMINANCE AREA DIAGRAMS . . . . . . . . . . . . . 10.11SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.12PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . 529 . . 530 . . 536 . . 538 . . 539 . . . . .
. . . . .
541 553 556 566 567
2. PRECIPITATION AND DISSOLUTION REACTIONS 11.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 FORMATION OF SOLIDS CONTAINING METAL IONS . . . . . 11.3 THE CHEMICAL ACTIVITY OF PURE SOLIDS . . . . . . . . . 11.4 THE SOLUBILITY PRODUCT . . . . . . . . . . . . . . . . . . . 11.5 PRECIPITATION OF SOLIDS OTHER THAN HYDROXIDES . . 11.6 FORMATION OF NONHYDROXO COMPLEXES IN SYSTEMS WITH SOLIDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 ADDING NEW SOLIDS TO THE VISUAL MINTEQ DATABASE 11.8 DETERMINING WHETHER A SOLID WILL PRECIPITATE UNDER GIVEN CONDITIONS . . . . . . . . . . . . . . . . . . . 11.9 THE GIBBS PHASE RULE . . . . . . . . . . . . . . . . . . . . . 11.10PREDOMINANCE AREA DIAGRAMS CONSIDERING POSSIBLE PRECIPITATION OF SOLIDS . . . . . . . . . . . . . 11.11SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.12PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
575 576 577 578 579 593
12 REDOX CHEMISTRY 12.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 DETERMING OXIDATION NUMBERS . . . . . . . . . . . . . . 12.3 BALANCING REDOX REACTIONS . . . . . . . . . . . . . . . . 12.4 REDOX HALF-REACTIONS . . . . . . . . . . . . . . . . . . . . 12.5 THE ACTIVITY OF FREE ELECTRONS; EQUILIBRIUM CONSTANTS FOR REDOX HALF-REACTIONS . . . . . . . . . 12.6 DEFINITION OF e◦ AND pe◦ . . . . . . . . . . . . . . . . . . . . 12.7 COMPUTING pe FROM SPECIES ACTIVITIES: THE NERNST EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
661 663 667 670 671
11 CHEMISTRY OF METALS IN AQUEOUS SYSTEMS
x
599 604 608 629 641 647 649
674 679 683
12.8 COMBINING REDOX REACTIONS . . . . . . . . . . . . . . . 12.9 REDOX SPECIATION AND LOG C-pe DIAGRAMS . . . . . . 12.10SUMMARY OF SOME KEY SIMILARITIES BETWEEN ACID/BASE AND REDOX SYSTEMS . . . . . . . . . . . . . . 12.11REDOX REACTIONS INVOLVING EXCHANGE OF BOTH ELECTRONS AND PROTONS . . . . . . . . . . . . . . . . . . 12.12COMPUTING EQUILIBRIUM SPECIATION IN REDOX SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.13MODELING REDOX REACTIONS WITH VISUAL MINTEQ . 12.14OXIDATION AND REDUCTION OF WATER . . . . . . . . . . 12.15ENERGY CHANGES ACCOMPANYING REDOX REACTIONS 12.16REDOX TITRATIONS AND THE GEOCHEMICAL REDOX SEQUENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.17pe-pH PREDOMINANCE AREA DIAGRAMS . . . . . . . . . . 12.18REDOX REACTIONS AND ELECTROCHEMISTRY . . . . . . 12.19SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.20PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. 685 . 687 . 689 . 692 . . . .
703 710 719 728
. . . . .
737 740 744 756 757
13 ADSORPTION REACTIONS 13.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 TWO VIEWS OF THE INTERFACE AND ADSORPTION EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 QUANTITATIVE REPRESENTATIONS OF ADSORPTION EQUILIBRIUM: THE ADSORPTION ISOTHERM . . . . . . . . . . . . 13.4 MODELING ADSORPTION WITH VISUAL MINTEQ . . . . . . 13.5 ADSORPTION OF IONS IN THE PRESENCE OF A SURFACE ELECTRICAL POTENTIAL . . . . . . . . . . . . . . . . . . . . . 13.6 MODELING SURFACE-ADSORBATE ELECTRICAL INTERACTIONS IN VISUAL MINTEQ . . . . . . . . . . . . . . 13.7 SURFACE PRECIPITATION . . . . . . . . . . . . . . . . . . . . . 13.8 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
769 770
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774 779 797 812 827 835 840 841
PREFACE TO THE SECOND EDITION In the 14 years since the first edition of Water Chemistry was published, many important advances have been made in the field, including improvements in analytical tools to detect contaminants at ultratrace levels, increasingly sophisticated processes for removing those contaminants or destroying them altogether, and the development of better software for modeling the corresponding reactions and processes. Nevertheless, in my view, the fundamental concepts and tools that should be taught and mastered in an introductory course in aquatic chemistry remain largely the same. This second edition therefore covers largely the same material in the same order as the first edition. That said, however, several significant changes have been made in the presentation, largely in response to feedback I have received from users and my own experience using the book in my classes. These changes can be summarized as follows. The Introduction (Chapter 1) from the first edition has been split into three chapters devoted to distinct topics. The new Chapter 1 contains only the material that is truly introductory and that provides context for what follows — the nature of the water molecule, the types of solutes and the concentrations at which they are typically found in environmental solutions, and units for expressing concentrations. Chapter 2 introduces the concept of, and approaches for, quantifying the reactivity of chemicals. Here, the distinction is made between concentrations and chemical activities, activity coefficients are defined and estimated, and, most importantly, the idea of the chemical equilibrium constant is introduced. Chapter 3 deals with reaction kinetics and has been written almost from scratch. Although a brief, molecular-scale perspective on the factors controlling reaction kinetics was provided in the first edition, several users requested a more comprehensive coverage of the topic, particularly with regard to empirical modeling of kinetics at the macroscopic scale (e.g., reaction orders and rate expressions for nonelementary reactions). I have therefore expanded the discussion to include that material and added corresponding examples. However, I chose to limit the presentation to reaction kinetics per se, and not include any discussion of the impact of reactor hydraulics on the overall extent of conversion. This decision was made both in the interest of brevity and because I believe the latter topic is more appropriately covered in textbooks that focus on environmental processes rather than environmental chemistry. As a transparently self-promotional suggestion, I recommend Water Quality Engineering: Physical/Chemical Treatment Processes1 to anyone looking for an accessible, but thorough discussion of this latter topic. Chapter 4 introduces the thermodynamic perspective on chemical reactivity and has also been extensively revised. The first half of this chapter, which introduces G, 1 Benjamin,
M.M., and Lawler, D.F. (2013) Water Quality Engineering: Physical/Chemical Treatment Processes. John Wiley & Sons, Hoboken, NJ.
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H, and S, has been virtually completely rewritten, with the primary goal of providing readers an intuitive understanding of what those parameters represent. That section includes two key insights that helped me finally resolve some issues that had puzzled me for many years: First, why Gibbs energy can be lost from a system without any form of energy appearing anywhere else (i.e., why the loss of Gibbs energy does not seem to conform to the 1st law of thermodynamics, which enshrines the principle of conservation of energy); and second, why, for a species i in a system at constant temperature, Gi never equals H i − T Si , even though ∆Gi always equals ∆H i − T ∆Si . Although these insights are not essential to correct application of thermodynamic principles to water chemistry problems, I found that they greatly increased my understanding of the relevant phenomena, and I have tried to convey that understanding in the writing. The latter half of the chapter, which focuses on the use of thermodynamic parameters to evaluate the direction of chemical reactions and to determine equilibrium constants, has undergone only minor changes. The traditional bread-and-butter of introductory water chemistry textbooks — the analysis of speciation in aqueous systems at equilibrium — begins with a discussion of acid/base reactions in Chapters 5 and 6. Chapter 5 focuses on the fundamentals of such reactions and the identification and solution of the relevant equations (the mass balances, equilibrium constants, and charge balance equations) using manual and spreadsheet approaches. Log C-pH diagrams are introduced in that chapter as tools for visualizing and unifying the changes in acid/base speciation among conjugate acid/base pairs with vastly different acid strengths. Then, Chapter 6 focuses on the use of those diagrams to aid in the determination of the equilibrium speciation in solutions containing acids and bases. The biggest change in these chapters from the corresponding chapters in the first edition is that, although the proton condition is still introduced as a useful tool for the analysis of acid/base reactions, the tableau method and T OT H equation are now fully explained and used as the primary vehicle for those analyses. This approach has been adopted because of the increased emphasis on chemical equilibrium software, and specifically the use of Visual Minteq, for problem-solving in subsequent chapters. Chapter 7 introduces chemical equilibrium software and has been revised so that it focuses strongly on Visual Minteq (VM), which is the software tool used throughout the remainder of the book. The integration of the software into the text extends even to the inclusion of dozens of screenshots from the program. I made this change because (1) my impression is that VM is the most popular software for such purposes; (2) VM is available without cost and is therefore accessible to all users; and (3) the developer of VM has been very obliging in working with me to add features that improve the program as a tool for teaching water chemistry, in addition to just solving water chemistry problems. I believe that by linking the text to a specific program and showing readers exactly how to use the program, I can help them obtain interesting results and insights while spending less time struggling with the mechanical aspects of using the program. xiv
The remaining chapters have been updated and, as noted, revised to include explicit instructions for and examples of the application of Visual Minteq to solve for equilibrium speciation in various systems. These chapters address, respectively, acid/base titrations (Chapter 8), gas/liquid exchange (Chapter 9), metal complexation (Chapter 10), precipitation reactions (Chapter 11), redox reactions (Chapter 12), and adsorption reactions (Chapter 13). In general, I have received positive feedback about the content and clarity of these chapters as they existed in the first edition. Therefore, other than the focus on VM as the software for analyzing speciation and the fact that the original chapter on metals has been split into separate chapters dealing with reactions purely in solution and those that involve precipitation or dissolution of solids, they have been kept largely intact. The other major change in the book is the addition of many new homework problems at the end of each chapter, including a large number that are more easily solved. This change responds to my own observation, and comments from numerous colleagues, that the homework problems in the first edition were interesting and challenging, but needed to be balanced by more problems that were straightforward applications of the basic content of the chapter. I hope that these changes allow this text to continue to serve a useful role in the training of environmental engineers, and I look forward to and value any feedback that is offered.
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PREFACE TO THE FIRST EDITION OBJECTIVES AND TEXT ORGANIZATION The past three decades have witnessed a virtual explosion in the range of topics gathered under the umbrella of environmental chemistry. Throughout this period, the basic principles of equilibrium chemistry, particularly as they apply to aquatic systems, have served as indispensable tools for understanding the composition of, and direction of change in, environmental systems. This fact, it seems to me, owes as much to the seminal book that established our current paradigm for studying and interpreting the chemistry of aquatic systems as to the centrality of the equilibrium principles themselves. That book, of course, is Aquatic Chemistry, by Stumm and Morgan, first published in 1970 and currently in its third edition. Since the publication of that text, the tools available for solving the equations that define and constrain the equilibrium composition of aquatic systems have been improved significantly, and those tools have been applied to an ever-expanding range of systems, but the basic approach for analyzing the systems has remained largely unaltered. Though it is unarguably the definitive text in the field, Aquatic Chemistry is widely perceived as too advanced for students taking their first course in the subject area, particularly those with little background beyond an introductory course in general chemistry. As a result, over the years, a number of texts have emerged that attempt to convey the key concepts of equilibrium chemistry in a more accessible format. This text follows in that line, covering much of the same material but diverging in a few ways both substantive and stylistic. A brief outline of the text highlights both the similarities and differences. The text starts with an overview of a few simple, well-known physical/chemical concepts: conservation of mass and energy, and the tendency for any system to change toward a more stable (less reactive) condition. In Chapter 1, a good deal of the vocabulary of equilibrium aquatic chemistry is defined, links between chemical parameters and reactivity are introduced, and the kinetic model for chemical equilibrium is developed. Chapter 2 provides a more formalized approach to understanding and predicting chemical change, via the concepts of chemical thermodynamics. The presentation and level of coverage in this chapter, particularly the first half, differ substantially from those in most other texts in this field, and a case can be made that the presentation is beyond what is necessary or appropriate in an introductory course. Frankly, at times I have persuaded myself that this assessment is accurate. However, after deliberation, I always returned to the opinion that if I wanted students to understand how thermodynamics applies to aquatic systems, as opposed to simply understanding how to carry out useful thermodynamic calculations, I had to devote substantial space to the topic. xvii
This decision reflects, in large part, my own frustration at having studied thermodynamics so often without quite seeing the connections among the various pieces. ◦ For instance, the relationships ∆G = ∆H − T ∆H and ∆G = ∆G + RT ln a make it clear that enthalpy and entropy must be related to chemical activity a, but until recently, I did not understand how. Similarly, I could calculate the redox potential of a solution (EH ) and the surface potential on a suspended colloid (ψ), but I was never quite sure if, or how, these two electrical potentials were related. Once I sorted out those issues, I found the insights they provided immensely satisfying, and found that several ideas and principles that I had previously thought disconnected could be interpreted coherently. This cohesiveness is the essential beauty of thermodynamics, and in Chapter 2 I have attempted to convey some of that cohesiveness to students. In Chapters 3 through 5 and 7 through 10, applications of the chemical principles introduced in Chapters 1 and 2 are presented in the context of specific types of chemical reactions. The first such reactions described focus on acid/base chemistry, in a section that comprises Chapters 3 through 5. This section differs from the discussion of acid/base equilibria in other texts in two ways that are significant. First, in presenting an algorithm for solving for the equilibrium pH of solutions prepared with known inputs, I have chosen to introduce both the proton condition and the TOTH equation. In my experience, although students can rapidly master the use of the TOTH equation to solve an acid/base problem, they gain a firmer grasp of the qualitative chemistry and the quantitative analysis of equilibrium solutions by writing out the proton condition table. On the other hand, the TOTH equation provides an excellent introduction to the development of the tableau that is at the core of numerical solutions to such problems. The essential identity of these two equations for characterizing the proton mass balance is emphasized, so that students understand that both equations provide the same information. A second skill that is developed in this section is the ability to predict a priori the dominant acid/base species expected to be present at equilibrium, even when a complicated mixture of acids and bases has been used to prepare a solution. I derived this algorithm almost two decades ago, with a good deal of assistance from Dimitri Spyridakis. It has been very gratifying and more than a little surprising to see how enthusiastically other instructors have adopted the algorithm, now that I have begun to publicize what I had assumed was a widespread approach. Chapter 6 diverges from the preceding and subsequent chapters, being devoted to a presentation of the most common features of some currently available software for solving chemical equilibrium problems. I have emphasized the solution approach taken in the MINEQL family of programs, without tying the discussion to any particular software package. While some instructors may choose to skip this chapter or to have students start using the software packages without going into the solution algorithms, I believe that understanding the basics of those algorithms is valuable, both pedagogically and to ease the learning of the program mechanics when applied xviii
to some important systems that are not covered in the manuals. Chapters 7 through 10 describe, respectively, equilibrium between solutions and a gas phase, reactions of metals in aqueous systems (both complexation and precipitation/dissolution), equilibrium in systems where oxidation-reduction reactions are occurring, and equilibrium between solutions and solid surfaces (adsorption). In each of these chapters, the presentation includes both a formal mathematical analysis of the reactions of interest and a discussion of how those reactions are analyzed by using chemical equilibrium software. All these chapters also refer to the thermodynamic developments in Chapter 2, and the last two chapters rely heavily on that development in the analysis of how the local electrical potential can affect chemical behavior. It is particularly in these latter chapters that, I hope, the detailed discussion of electrical potential and activity coefficients in Chapter 2 pays dividends. A COMMENT ON THE TEXT LENGTH AND A PHILOSOPHY OF INSTRUCTION One of the most difficult parts of writing this text has been finding the right balance between attention to fundamental concepts and problem-solving techniques. In striking that balance, I have been guided by my experience teaching water chemistry courses over the past 20 years, which has convinced me that students want and can handle more fundamentals than most instructors (including myself) have been providing. Ironically, in my opinion, what frustrates these students and sometimes leads them to believe that water chemistry is overwhelmingly difficult is that, as instructors, we have tried too hard to simplify the concepts. Too often, the simplifications we offer provide students the tools to derive correct answers to numerical problems, but only by following algorithms that they do not fully understand. Then, they feel intimidated and lost when faced with a problem for which the algorithm is inapplicable (or worse, they fail to realize that the algorithm is inapplicable and so apply it inappropriately). I have therefore chosen to write longer and more detailed explanations of both the relevant chemistry and mathematics than are found in most other texts. Undeniably, this decision has lengthened the text, perhaps to the dismay of those who are perusing it for the first time. However, the range of topics covered is no greater than in other water chemistry texts, so the added length does not represent an increase in the conceptual material that readers are asked to master. To the contrary, my belief is that the extra explanatory material will actually reduce the time that students need to devote to learning the course content, while simultaneously facilitating a deeper understanding of the subject matter. Having said that, I recognize that water chemistry courses are taught in many different formatsas semester courses, quarter courses, with and without laboratory components, etc.and that many instructors will choose to cover only a only a portion of the text in their courses. In such cases, I believe that a successful course could be taught by omitting coverage of Chapters xix
10, 9, and 2, in that order, depending on the severity of the time constraints. While an understanding of chemical thermodynamics (Chapter 2) is certainly helpful for interpreting all types of chemical reactions, it is more central to discussions of redox and adsorption reactions (Chapters 9 and 10, respectively) than the reactions covered in earlier chapters. Alternatively, an instructor might consider omitting coverage of chemical equilibrium software packages (Chapter 6 and easily identifiable sections of Chapters 7 through 10). One of my goals throughout the writing of the text has been to integrate the material within each chapter and between chapters as seamlessly as possible. The benefits of such integration are self-evident, but the integration does impair any effort to fashion a course based on reading of disparate sections. Therefore, my personal preference is not to respond to time constraints by eliminating coverage of selected, isolated sections of the text. I believe that, in the end, students are better served by reading and mastering Chapters 1 and 3 through 8 in their entirety, than by being exposed to all 10 chapters but feeling unsure about their mastery of any of them. However, I realize that different courses have different objectives, and I hope that instructors will experiment freely with various ways to use the text and provide feedback to me on how well those approaches work.
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ACKNOWLEDGMENTS The first edition of this book, published in 2000, contained some heartfelt acknowledgments that are as true today as they were then, and I am pleased to repeat them here. In the end, this book was written because of the encouragement I received from students who flattered me into believing that I could write about water chemistry in a way that made sense to them. The faculty and students who use the book will be the ultimate judges of whether that flattery was merited. But regardless of the verdict, I owe a debt of gratitude to all the students over the years who have suffered through this process with me and who have challenged and rewarded me so. At the risk of offending the many, I would be remiss if I did not acknowledge by name the few whose support has been so very far above and beyond the call. First and foremost, Paul Anderson has been a friend for lo these twenty years. That he has been such while simultaneously playing the role of student and later colleague, and always that of gentle but firm critic, surely qualifies him for some sort of award. John Ferguson, Bruce Honeyman, Gregory Korshin, Mickey Schurr, John van Benschoten, Ray Simons, and David Waite all contributed generously of their time to help me understand bits of water chemistry that had me confused, and to point out to me portions of the text that needed revision. Jill Nordstrom provided student feedback at a level of detail that no author of a textbook deserves, but every author must dream of. My wife, Judith, has been as a source of support and encouragement throughout the years that I devoted to this project. When it seemed that both of our lives were being dominated by the writing effort, I could be re-energized by my fascination with the subject matter and a sense of making progress toward a lifelong dream. Judith shared neither of these sources of inspiration, yet she has remained steadfast throughout, energized by her love and her willingness to share my dream as her own. For that, I will be forever grateful. Finally I think my daughter Mara for giving me the joy of fatherhood. In preparing this second edition, I received invaluable technical input and tremendous support from several individuals. First and foremost, Dr. Zhenxiao (Nathan) Cai was instrumental in all stages of the effort. Though the phrase is probably overused in book acknowledgments, in this case it is absolutely true that this book could not have been written without his assistance. I value him enormously as a colleague and friend. I also wish to acknowledge my deep debt to Jon-Petter Gustafson, author and maintainer of Visual Minteq c . Aside from his major contribution to our field by developing this software, Jon-Petter has been extremely patient and responsive to my requests for modifications to the program to make it easier to use by those first learning about water chemistry. Though I have never met or even spoken to him, we have for years engaged in a thought-provoking electronic conversation about how xxi
best to convey the ideas underlying water chemistry. I have learned a great deal from this conversation, and I hope I have successfully transferred that learning to the printed page. Paul Anderson, Jim Duncan, David Dzombak, Jong-Yeop Kim, Detlef Knappe, Xiaosong Li, Joel Migdal, Steve Reiber, Phil Singer, and Mike Stenstrom all provided generous assistance through technical feedback, contributions of homework problems, and help with figures and photographs. They have improved the final product greatly, for which I and, I trust, readers thank them sincerely. Portions of several chapters were written simultaneously with the writing of Water Quality Engineering: Physical/Chemical Treatment Processes, and some of the writing has been used directly or in slightly modified forms in both textbooks. My coauthor for that book, Desmond Lawler, provided detailed and thoughtful comments on those sections, for which I am very grateful. I also acknowledge the willingness of John Wiley and Sons to allow those sections to be used in both books. I also gratefully acknowledge the support provided by my the University of Washington throughout my career and by the University of New South Wales during my sabbatical there in 2000 and 2010. Finally, I (again) thank my wife Judith, who has stood by me and supported me for 26 years, during every one of which I promised that one book or another “will be done soon”. Now the books are all done, and she is still by my side, with her strength and humor intact. Amazing!
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ABOUT THE AUTHOR Mark M. Benjamin is a Professor of Civil and Environmental Engineering at the University of Washington, where he has served on the faculty since 1978. He received his undergraduate degree in chemical engineering at Carnegie-Mellon University and his Masters and PhD. Degrees in environmental engineering from Stanford University. Dr. Benjamin has directed and published research on various aspects of water chemistry and water treatment processes. His long-term interests have been in the behavior of metals and their interactions with mineral surfaces, and in the reactions of natural organic matter in water treatment systems. He and his students have received four patents for treatment processes that they have developed. Their publications have won several awards, and four of his students have won awards, given annually, for the best doctoral dissertation in environmental engineering and/or water supply and treatment. In addition to this textbook, Dr. Benjamin co-authored Water Quality Engineering: Physical/Chemical Processes with Dr. Desmond Lawler of the University of Texas. Dr. Benjamin is a member of several professional societies and has served on the board of the Associaton of Environmental Engineering and Science Professors. He is also serves as faculty co-advisor for the University of Washington student chapter of Engineers Without Borders.
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1
BASICS OF AQUATIC CHEMISTRY
1
Contents 1.1 1.2 1.3 1.4 1.5 1.6
1.1
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 2 THE STRUCTURE OF WATER MOLECULES AND INTERACTIONS AMONG THEM . . . . . . . . . . . . . . . . . 3 THE BEHAVIOR OF SOLUTES IN WATER; DISSOLUTION OF SALTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 COMMON APPROACHES FOR EXPRESSING CONCENTRATIONS IN ENVIRONMENTAL CHEMISTRY . . . . . . . . . . . . . . . 10 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
INTRODUCTION
The field of environmental aquatic chemistry, originally focused on the chemistry of impaired water bodies and of water treatment processes, has expanded over the past several decades to encompass studies of the source, distribution, transport, reaction rates, and fate of chemicals in all natural aquatic systems. The primary subject matter of this text is a small but important subset of that field: the prediction of the chemical composition of solutions that have reached chemical equilibrium. Such predictions are useful both because many systems closely approach their equilibrium state within the time frame of interest and because, even in systems that are far from this condition, the direction of chemical change and the expected ultimate composition are of interest. In addition to describing approaches for computing the equilibrium composition of a system, the text provides an introduction to the types of chemicals that are present in natural waters, drinking water, and certain wastewaters, and to the reactions that those chemicals might undergo when the water is used, mixes with other waters, or interacts with soils and sediments. The techniques presented in the text form the basis for analyzing or designing facilities to bring about many desired changes in water quality via water or waste treatment. The focus of the text is on aqueous solutions of environmental significance, which are almost always very dilute: typically, the reactants of interest comprise only 2
10 8 percent to a few percent of all molecules in the system, with the remaining molecules being water. Such systems include most natural waters (lakes, rivers, oceans, groundwater), domestic and industrial wastewaters, runoff from residential or agricultural areas, and many other solutions. The text emphasizes the determination of the equilibrium chemical speciation in a system, where speciation refers to the distribution of the chemicals among their various forms. For instance, in anaerobic (oxygen-free) environments, dissolved sulfur can be present as hydrogen sulfide (H2 S), but in aerobic environments it is usually converted 1 to sulfate ion (SO2– 4 ) . Hydrogen sulfide is a weak acid that reacts strongly with metals to form solid precipitates. It can also escape from solution into the gas phase, generating unpleasant odors at extremely low concentrations and creating significant health risks at higher (but still very low) concentrations. By contrast, sulfate ion is a very weak base that cannot exist as a gas, does not react significantly with most metals, and is generally innocuous as a health hazard. Thus, knowledge of the speciation of the sulfur is essential if we are to understand its behavior and its impact on water quality, or to design engineering processes to remove it from solution. Similar comparisons could be made regarding the properties of various forms of chlorine, chromium, arsenic, and many other chemicals of environmental importance. The calculations emphasized in the text allow us to determine the conditions that favor formation of one chemical form of an element over another. Concentrations of some species of interest in a variety of solutions are presented in Table 1.1. In some solutions, contaminant concentrations of a few micrograms per liter or less can be a cause of major concern, especially for carcinogenic compounds and some heavy metals. While a contaminant concentration of, say, 50 µg/L might be high relative to the threshold for some ecological effect or a regulatory limit, it is important to recognize how phenomenally dilute such a species truly is. For instance, a concentration of 10 µg/L (10 ppb) is equivalent to 70 people out of the world population of 7 billion. Imagine viewing the earth from a spaceship and being told that there are 70 very dangerous criminals scattered around the globe, and that you must find and capture 35 or more of them in a few minutes. This impossible-sounding task is equivalent to what water treatment engineers are asked to do, day in and day out, with respect to certain contaminants.
1.2
THE STRUCTURE OF WATER MOLECULES AND INTERACTIONS AMONG THEM
The shape of a water molecule can be represented roughly by an equilateral tetrahedron (a four-sided, triangular pyramid), with the oxygen in the center and hydrogen 1 An ion is any charged chemical species. Ions carrying positive charges are called cations, and those that carry negative charges are called anions.
3
Table 1.1 Concentrations of some important dissolved constituents in a few natural water bodies and other solutions of interest
Savannah River Ca2+ Mg2+ Na+ Cl– SO2– 4 Alk(a) DOC(b) pH
Cu Zn Cr As Pb Ni Cd pH
mg/L mg/L mg/L mg/L mg/L mg/L as CaCO3 mg/L pH units
mg/L mg/L mg/L mg/L mg/L mg/L mg/L pH units
4 1 12 9 7 23 3 7.0
Mississippi Rain Water River near Great Salt at Menlo Clinton, Lake Park, CA IA 41 19 9 13 49 161 6.3 8.6
0.8 1.2 9.4 17 7.6 3.3 — 5.5
241 7200 83,600 140,000 16,400 206 27 7.4
Mean Seawater 408 1280 10,800 19,400 2710 120 1.0 7.9
Electroplating Rinse Waters(c)
Pretreatment Req’t for Discharge to POTW(d)
Water Quality Standards(e)
MCL(f) for Drinking Water
3 2 3 — 0.7 3 0.5 5–12
3 5 2.75 1 2 2.5 0.5 5.5–12.0
0.009 0.117 0.027 0.148 0.0025 0.052 0.0021 6.5–8.5
1.3 — 0.1 0.05 0.015 0.1 0.005 —
(a) Alkalinity, expressed as milligrams per liter as CaCO3 . This way of expressing concentration is explained later in this chapter. For many natural waters, the alkalinity can be interpreted as the approximate value of the bicarbonate (HCO–3 ) concentration, when both are expressed as equivalents per liter. (b) Dissolved organic carbon. (c) Values shown are in the normal range, but these waters have highly variable composition, depending on the electroplating operation being carried out. (d) A POTW is a publicly owned treatment works; values shown are for King Co., WA. (e) Chronic water quality criteria for protection of aquatic life in ambient water, assuming hardness of 100 mg/L as CaCO3 . Values shown for arsenic and chromium are for these metals in the +3 oxidation state. (f) MCL is the maximum contaminant level allowed by law in the United States.
4
ions at two corners.2 The region between the oxygen and each of the hydrogen ions is occupied by a pair of electrons, one contributed by the hydrogen and one by the oxygen. Since the oxygen atom has six electrons in its outer shell, this arrangement leaves four unshared electrons, which occupy orbitals pointing toward the other two corners of the tetrahedron (two electrons in each orbital). The structure is shown schematically in Figure 1.1.
Figure 1.1 Schematic representation of the structure of a water molecule and the charge distribution around it. The two hydrogen ions and two sets of unshared electron pairs reside at the four corners of a tetrahedron, with the oxygen ion in the center. Two other pairs of electrons are shared — one pair between the oxygen and each hydrogen ion. (Image created by Xiaosong Li.)
Although the electrons forming the hydrogen-oxygen bonds are shared by the two atoms, they are not shared equally: the oxygen attracts them more than the hydrogen does. As a result, although water molecules are electrically neutral overall (i.e., they are not ions), they do have local regions of finite charge. Specifically, a positive charge of about 0.24 esu (electrostatic unit, the charge on a single electron) resides near each hydrogen atom, and a negative charge of about 0.48 esu resides on the side of the oxygen opposite the hydrogen atoms. Because of this asymmetric distribution of charge, the molecule is referred to as being polar. Hydrogen atoms have only one electron and one proton. Because that single electron usually resides between the H and O atoms in a water molecule, from the perspective of the rest of the solution, each hydrogen ion looks almost like a bare proton. The absence of other electrons associated with the hydrogen allows other molecules to approach the “unprotected” proton quite closely, in fact more closely than they could approach the nucleus of any other positively charged atom. This closeness of approach allows unusually strong electrostatic bonds to form between the proton and negatively charged portions of other molecules. Such bonds, called hydrogen bonds, are particularly important in aqueous solutions, because they can form between the electron-poor hydrogen ions of one water molecule and the electron-rich parts of another (i.e., the orbitals with unshared electrons). As a result, every water molecule 2 The
shape of the water molecule actually corresponds to a slightly distorted equilateral tetrahedron, with the H O H angle being 105 as opposed to 109 in an ideal equilateral tetrahedron.
5
can bond to other water molecules by as many as four hydrogen bonds (Figure 1.2). All four such bonds form in ice; experimental evidence suggests that in liquid water each molecule is hydrogen-bonded to three others, on average, at any instant. These bonds cause water to be more cohesive than almost any other liquid and affect the physical and chemical behavior of water in many other important ways.
Figure 1.2 Hydrogen-bonded water molecules. (Image created by Xiaosong Li.)
1.3
THE BEHAVIOR OF SOLUTES IN WATER; DISSOLUTION OF SALTS
The transfer of any substance into or out of a solution changes the energy distribution in both environments. For instance, when a pure substance A enters solution, hydrogen bonds among water molecules break as the molecules rearrange to create a “hole” for the solute to enter. Bonds that linked A molecules to one another prior to dissolution also break, and A-to-H2 O bonds form. In general, the solutions we will consider are sufficiently dilute that molecules of A are widely dispersed in solution, and A-to-A bonds can be ignored.3 The tendency for molecules of A to dissolve is enhanced by the formation of strong A-to-H2 O bonds and is diminished by strong A-to-A bonds in “undissolved” A (i.e., solid or liquid A). Therefore, other factors being equal, the relative strengths of A-to-A and A-to-H2 O bonds can give a good indication of how soluble molecules of A will be.4 Molecules whose dissolution is favorable are called hydrophilic (waterloving), and those whose dissolution is unfavorable are called hydrophobic (water3 If
A-to-A bonds in solution are strong, molecules of the type An (n = 2 or larger) might form, but the An is represented as a new molecule rather than as a group of individual A molecules. 4 The dissolution process also affects the energy distribution among the molecules in ways that might either favor or oppose the reaction. As a result, one cannot conclude whether dissolution will be favorable based solely on bond formation and breakage. This phenomenon is discussed in Chapter 4.
6
fearing). Note, however, that these terms are qualitative, so the same molecule might be considered hydrophilic in one context and hydrophobic in another, depending on what it is being compared with and what is deemed a favorable amount of dissolution in the specific context of interest. The charge on dissolved ions tends to orient the neighboring water molecules to maximize favorable (+ to ) electrical interactions (Figure 1.3). However, the structured orientation encouraged by these interactions is opposed by the randomizing influence of thermal motion, so the water molecules are not frozen in place, and some transient unfavorable (+ to +, or to ) interactions might occur. Nevertheless, at any instant, the favorable electrostatic interactions far outnumber the unfavorable ones. These interactions stabilize dissolved ions significantly, making all ions hydrophilic, and making it possible for large concentrations of ions to be present in solution.
Figure 1.3 The tendency of water molecules to orient themselves around ions (the groupings on the right side of the figure) facilitates dissolution of salt particles (the grouping on the lower left side). (From Silberberg, M.S. Chemistry: The Molecular Nature of Matter and Change. Copyright c The McGraw-Hill Companies, New York, NY. Reproduced by permission of The McGraw–Hill Companies.)
The charged regions of polar solutes can orient neighboring water molecules in much the same way that ions can, but the strength of the orienting force is lower. As a result, polar molecules are typically less soluble than ions, and their solubility generally increases with increasing polarity. Nonpolar molecules, on the other hand, usually interact only weakly with water molecules and have low solubilities. The only two neutral solutes that are commonly present in environmental waters at concentrations comparable to ionic solutes are carbonic acid (H2 CO3 ) and silicic acid 7
(H2x SiO2+x ). Carbonic acid is generated by the hydration of carbon dioxide (CO2 ), which enters the water either from the atmosphere or due to respiration of organisms living in the water. The hydration reaction can be written as
CO2 + H2 O )* H2 CO3
(1.1)
Silicic acid is formed via a similar reaction, but its source is typically silicate rocks rather than the atmosphere or biological activity. This reaction can be represented as
SiO2 + x H2 O )* H2x SiO2+x
(1.2)
where x is typically 1 or 2 (forming H2 SiO3 or H4 SiO4 , respectively). Both of these species are referred to as silicic acid; fully dehydrated silicic acid (SiO2 (aq), with x = 0) is commonly referred to as ‘silica’. Some ions are so stable when surrounded by water molecules that, upon contact with water, they are likely to break off of any compound they are associated with and become surrounded by water molecules. Sodium (Na+ ) and chloride (Cl– ) are examples of ions that behave in this way. Therefore, compounds of the form Naa X(s) or ZClb (s) tend to dissociate completely whenever they are exposed to water, releasing Na+ and Xa– ions, or Zb+ and Cl– ions, respectively, by reactions such as the following: +H O
2 Naa X ) * a Na+ + Xa–
+H O
2 ZClb ) * Zb+ + b Cl–
The Xa– and Zb+ ions might react with other dissolved species, but the Na+ and Cl are so stable when surrounded by water molecules that they rarely undergo further reaction. Compounds that dissolve and dissociate extensively when they contact water are called salts.5 The concentration ranges of several solutes in terrestrial waters are shown in Figure 1.4. Logically enough, species that are present in relative abundance in the geosphere and are highly stable when they dissolve tend to accumulate in natural waters and to dominate the composition of such systems. This group of species is made up primarily of ions and includes sodium (Na+ ), potassium (K+ ), calcium (Ca2+ ), – magnesium (Mg2+ ), chloride (Cl– ), sulfate (SO2– 4 ), and nitrate (NO3 ). The other major ion in most natural waters is bicarbonate (HCO–3 ) which, although it is more reactive than the other ions listed, is present in relatively high concentrations because of the –
5 Other definitions of salts are based on the crystal form and/or chemical bonding arrangements of the compound in the solid form. The definition given here is commonly used by water chemists.
8
Figure 1.4 The cumulative frequency distribution of solute concentrations in terrestrial waters. (From Davis, S., and de Wiest, R. Hydrogeology, John Wiley & Sons, Inc., New York [1966]. Reprinted by permission of the author.)
continual inputs of carbon dioxide (CO2 ) noted previously from the atmosphere and from biological activity. Most of the H2 CO3 molecules formed as a result of those inputs rapidly release an H+ ion to form bicarbonate ion by the following reaction, which is discussed in detail in later chapters:6
H2 CO3 )* HCO3 + H+
(1.3)
The molecules in a gas phase are separated by much larger distances, on average, than those in aqueous solutions, so electrostatic interactions in gas phases are much weaker. As a result, ions are not stable in gases, and the concentration of ions in any gas phase under normal environmental conditions is exceedingly small. Therefore, throughout this text, the assumption is made that gases are made up entirely of neutral molecules. (Note that, overall, aqueous solutions are electrically neutral, just as gases are. However, gases are neutral because they contain only neutral molecules, whereas solutions are neutral because the total positive charge on dissolved cations is exactly balanced by the total negative charge on dissolved anions.) molecules of interest consist of a “core” anion to which one or more H+ ions can attach. In most cases, when one H+ is attached, the name of the molecule is the name of the core anion with the – prefix ‘bi’. In this case, the core anion is carbonate, CO2– 3 , so HCO3 is named bicarbonate. 6 Many
9
Natural waters have a wide range of compositions, depending on their geography, geology, and topography, as well as anthropogenic inputs. (a) An urban waterway in Seattle. (b) A boat plying the waters of Lake Erie during an algal bloom. (c) A river formed by snowmelt o↵ Mt. Rainier. (a) Mark Benjamin. (b) Peter Essick. (c) Mark Benjamin.
1.4
1.4.1
COMMON APPROACHES FOR EXPRESSING CONCENTRATIONS IN ENVIRONMENTAL CHEMISTRY Mass/Volume and Mass/Mass Units for Solute Concentrations
Concentrations of species in aqueous systems are reported in a wide variety of ways. Clearly, the behavior of a chemical does not depend on the units chosen to represent its concentration, any more than the velocity of an object depends on whether that velocity is given in miles per hour or meters per second. However, it is also clear that certain choices of units are more convenient than others (no one would want to have a speedometer read in millimeters per year). The most common way to report the concentration of a chemical that is dissolved or suspended in water is as mass per volume, with units such as milligrams per 10
liter (mg/L) or micrograms per liter (µg/L). Dimensions of moles per liter (mol/L) (also called molar and designated simply as M) are also widely used for expressing concentrations of dissolved constituents; mol/L dimensions are convenient to use in conjunction with chemical reactions, because ratios of stoichiometric coefficients in chemical reactions correspond to molar ratios of the reacting species.7 In this text, the symbol cA is used to designate the concentration of species A in either mass- or mole-based units, and [A] is used exclusively to indicate the concentration in units of moles per liter. Because chemical concentrations can range over many orders of magnitude, it is frequently convenient to express their values in logarithmic units. By convention, in the chemical literature, a lowercase p preceding a symbol is used to designate the negative, base-10 logarithm ( log10 ) of the value associated with the symbol. When this convention is applied to the concentration of a solute, the units of solute concentration are understood to be moles per liter. Thus, for example, if the concentration of Cu2+ ions in a solution (cCu2+ ) is 2 ⇥ 10 5 mol/L, then log10 cCu2+ = 4.70, and pcCu2+ = 4.70. The p convention is applied not only to solute concentrations, but also to chemical activities and equilibrium constants, both of which are defined subsequently. Sometimes, concentrations are normalized based on the mass of the phase in which they are found, rather than the volume. For instance, the concentration of a dissolved species A might be given as a mass fraction, i.e., the mass of A per unit mass of solution, with units such as milligrams per kilogram or micrograms per kilogram. Concentrations reported using this convention are often given as dimensionless quantities [mg/kg = parts per million by mass (ppmm ), g/100 g = percent, etc.]. This approach is commonly used to express the concentrations of constituents in a solid phase; for example, a concentration of 1 ppmm in a solid phase means that there is 1 mg of the given species per kilogram of solid. A similar approach can be used to express concentrations as mole fractions, i.e., the number of moles of a species as a fraction of the total number of moles of all substances in that phase. By convention, the mole fraction of a species A is represented as xA in solid and liquid phases, and as yA in a gas phase. Most aqueous solutions of environmental interest have a density close to 1000 g/L. In such solutions, concentrations expressed as milligrams per liter or micrograms per liter (i.e., mass/volume) are nearly equivalent to those expressed as parts per million or parts per billion by mass, respectively, e.g., 1
mg A mg A 1 mg A '1 = 6 = 1 ppmm L 1000 g solution 10 mg solution
7 One
mole is Avogadro’s number (6.022 ⇥ 1023 ) of items. One mole of a chemical is therefore 6.022 ⇥ 1023 molecules of that chemical.
11
Solute concentrations are also sometimes reported as mass or moles of solute per unit mass of solvent (rather than per unit mass of the total solution). Concentrations computed using this approach are independent of the density of solution, making the approach particularly useful for describing the preparation of reagents for laboratory use. Concentrations given as moles of solute per 1000 g of solvent are referred to as molal concentrations. Example 1.1 A solution contains 8 µg/L lead (Pb, atomic weight 207). Express this concentration as moles per liter, ppmm , and the dimensionless mole fraction of Pb. Assume the solution has a density of 1 g/mL. Solution The atomic weight of an element indicates the mass (in grams) of one mole of the element. The molar concentration of Pb can therefore be computed by dividing the mass concentration by the atomic weight, as follows: (8 µg/L Pb) 1g/106 µg mol Pb = 3.86 ⇥ 10 8 207 g Pb/mol Pb L The solution‘s density is given as 1 g/mL, or 1000 g/L, so the Pb concentration in ppmm can be computed as [Pb] =
8 ⇥ 10 6 g Pb/L = 0.008 ⇥ 10 103 g solution/L
6
g Pb = 0.008 ppmm g solution
Although the solution might contain a variety of solutes, chances are that the vast majority of the molecules are H2 O. We can therefore assume that the total molar concentration of all species in the solution is approximately the same as in pure water. This concentration is 1000 g H2 O/L mol H2 O total moles = 55.6 ' 55.6 18 g H2 O/mol H2 O L L The mole fraction of Pb in the solution can then be computed as xPb =
3.86 ⇥ 10 8 mol Pb/L = 6.95 ⇥ 10 55.6 total moles/L
Thus, the mole fraction of Pb is 6.95 ⇥ 10
1.4.2
10
mol Pb total moles in solution
10 .
Concentrations Reported as Equivalents Per Liter
Sometimes, particularly in the older environmental engineering literature, concentrations of various solutes are reported in units of equivalents per liter 12
(equiv/L) or milliequivalents per liter (meq/l). The concentration in equiv/L is referred to as the normality, N, of the solute. Such an approach implies that the concentration of the substance of interest is “equivalent” in some way to the concentration of another chemical. Because the basis for the equivalency can be ambiguous, there is at least some risk of misunderstanding the basis for the comparison. In current practice, concentrations are widely reported in equivalents per liter in only three cases. When reporting concentrations of salt ions, the implied unit defining one equivalent is one mole of electric charge (without regard to sign); when referring to acids and bases, it is one mole of H+ ; and when referring to oxidation-reduction reactions (reactions in which electrons transfer from one species to another), it is one mole of electrons [e– ]). The meaning intended in a given application is usually clear from the context, although some ambiguity can arise (e.g., if a charged, acidic species is participating in an oxidation-reduction reaction). In general, it is best to avoid using these units or to identify the reference explicitly, e.g., by using the terms charge equivalents, H+ equivalents, or electron (e– ) equivalents. Example 1.2 A number of chlorine-containing chemicals are used to disinfect water. Two such chemicals are hypochlorous acid (HOCl, MW 52.5) and chlorine dioxide (ClO2 , MW 67.5). The relevant reactions in which electrons combine with these species are shown below.8 Express doses of 5 mg/L HOCl and 5 mg/L ClO2 in meq/L. HOCl + H+ + 2e )* Cl + H2 O ClO2 + 4H+ + 5e )* Cl + 2H2 O Solution According to the given reactions, one mole of HOCl corresponds to two electron equivalents, and one mole of ClO2 corresponds to five electron equivalents. The calculations for expressing 5 mg/L of HOCl and ClO2 in meq/L are therefore: ✓ ◆✓ ◆ ⌘ mg HOCl 1 mmol HOCl ⇣ meq meq 5 2 = 0.19 L 52.5 mg HOCl mmol HOCl L ✓ ◆✓ ◆✓ ◆ mg ClO2 1 mmol ClO2 meq meq 5 5 = 0.37 L 67.5 mg ClO2 mmol ClO2 L
One common use of charge equivalents is to test the completeness and accuracy of an analysis of the ionic composition of a water sample, via a charge balance. Theoretical considerations suggest that the concentrations of positive and negative 8 The
reactions shown indicate the maximum number of electrons with which each species can combine. Under typical water treatment conditions, ClO2 often combines with only one electron, to form ClO–2 .
13
charge in any solution should equal one another. The discrepancy between these two values is therefore an indicator of the minimum error associated with the analysis. (The discrepancy represents the minimum error, because there could be other problems with the analysis that are not reflected in the charge balance.) The error in the charge balance (ErrorCB ) can be quantified as the ratio of the difference in the absolute values of the cationic and anionic charge to the sum of those same terms; that is, Â Ni
ErrorCB =
 Ni
cations
anions
 Ni +  Ni
cations
(1.4)
anions
If ErrorCB is less than about 5%, the charge balance is considered to be satisfied; if the analysis fails this test, one would have to conclude that one or more of the chemical analyses is in error, or that some significant ion(s) have been overlooked. Generally, analyses become more accurate as the concentration increases, so the criterion for a satisfactory charge balance might be made stricter for salty waters and more lenient for those with lower ionic concentrations.
Example 1.3 A groundwater at pH 7.5 is reported to have the following ionic composition; only ions present at concentrations > 2 mg/L are shown. Compare the total cationic and total anionic charge. Do the data satisfy the criterion of < 5% error for an acceptable charge balance? Ion Ca
ci (mg/L)
2+
90
Cl–
122
HCO–3 2+
317
Mg
48
Na+
120
NO–3 SO2– 4
11.1 190
Solution The given information about the ionic composition is repeated in the following table, with additional columns for the molecular weight of each ion (MWi ), its charge (zi ), and its normality (Ni ). 14
Ion
ci (mg/L)
MWi
zi
Ni (meq/L) Cations
Anions
2+
90
40
2
–
Cl
122
35.5
1
3.4
HCO–3 2+
317
61
1
5.2
Mg
48
24.3
2
4.0
Na+
120
23
1
5.2
NO–3 SO2– 4
11.1
62
1
0.2
190
96
2
4.0
Ca
SUM:
4.5
13.7
12.8
The error in the charge balance can be computed from Equation (1.4) as  Ni
ErrorCB =
cations
 Ni
anions
 Ni +  Ni
cations
=
anions
|13.7 12.8| = 0.036 < 0.05 13.7 + 12.8
Thus, the analysis meets the criterion.
1.4.3
Gas-Phase Concentrations
Like concentrations in solution, concentrations in gas phases are often reported in mass-per-volume units. However, because gases are so much less dense than water, typical units for dilute constituents are micrograms per cubic meter, rather than milligrams per liter or micrograms per liter. In some cases, the concentration of a gaseous species is given as its partial pressure or its volume fraction. These units of concentration can be understood by recognizing that, under environmental conditions, all gases behave very nearly as ideal gases. For such gases, the pressure, volume, and temperature are interrelated by the ideal gas law, two forms of which are shown in Equations (1.5) and (1.6): Pi =
nG,i RT = cG,i RT Vtot
(1.5)
nG,i RT Ptot
(1.6)
Vi =
where: Pi and Vi are the partial pressure and partial volume of species i, respectively; nG,i is the number of moles of species i in the gas phase; Ptot and Vtot are the total pressure and volume of the gas phase, respectively; 15
cG,i is the gas-phase concentration of i in moles per unit volume; R is the universal gas constant; and T is the absolute temperature. The value of R in a few common sets of units is 8.314 J/(mol·K), 0.0821 L–atm/(mol·K), and 1.986 cal/(mol·K). The partial pressure of i can be defined as the pressure that i would exert if it were present in a hypothetical gas phase that was at the same temperature and contained the same concentration of i as the real gas, but if all other species were removed. Similarly, the partial volume of i is the volume that i would occupy if all other species were removed from the gas phase and the total system pressure and temperature were held constant. Partial pressure is expressed in the usual units of pressure, e.g., atmospheres (atm), kilopascals (kPa), bars, or pounds per square inch (psi).9 According to Equation (1.5), the partial pressure is not a direct measure of the concentration of i, since a given concentration of i is associated with different Pi values at different temperatures. However, if the temperature is specified, the partial pressure can be used as a surrogate for the concentration of i in the gas phase. Use of partial pressure in this way is common in environmental chemistry, often with the implicit assumption that the temperature is 25 C. The sums of Pi , Vi , and ni for all gases in the system equal Ptot , Vtot , and nG,tot , respectively. Ptot = Â Pi
(1.7a)
Vtot = Â Vi
(1.7b)
nG,tot = Â nG,i
(1.7c)
Combining these identities with Equations (1.5) and (1.6) yields the following relationships for any constituent in an ideal gas: nG,i Pi Vi = = = yi Ptot Vtot nG,tot
(1.8)
Equation (1.8) indicates that the volume fraction of i (i.e., Vi /Vtot ) is the same as the mole fraction (yi ). Thus, like the mole fraction, the volume fraction can be considered a measure of concentration if the system temperature is known and constant. Volume fractions of dilute gaseous constituents are typically expressed in parts per million by volume (ppmv ).10 9 The
SI unit for pressure is the pascal (Pa); 105 Pa = 100 kPa = 1.0 bar = 0.987 atm. should be clear that 1 ppmm and 1 ppmV can represent quite different concentrations. Although it would be convenient if subscripts were consistently used with these terms to avoid confusion, that is rarely done. Rather, both ppmm and ppmV are commonly designated simply as ppm, with the implicit understanding that the units are mass-based (1 ppm = 1 mg/kg) when applied to a solid or liquid and volume-based (1 ppm = 1 mL/m3 ) when applied to a gas. 10 It
16
1.4.4
Composite Parameters to Represent Multiple Species Containing the Same Element: Nitrogen Speciation, TOC, and TOX
Aquatic chemistry, more than other fields of chemistry, deals with complex, uncontrolled mixtures of chemicals in samples whose history is poorly known. Under the circumstances, a detailed chemical characterization of the samples is often impossible. Furthermore, in many cases, our interest is in the behavior or effects of groups of chemicals, rather than a single species. Thus, for both fundamental and practical reasons, concentrations of environmentally significant chemicals are frequently reported in ways other than as concentrations of individual chemical species. In this section, a few of those ways of expressing concentration are presented. One approach for describing the collective concentration of several different chemical species, or for comparing the concentrations of several species that contribute to a “pool” of interest, is to report the concentration of a single element that is common to all the species. For instance, nitrogen species commonly found in aquatic environmental samples include molecular nitrogen (N2 ), nitrate ion (NO–3 ), nitrite ion (NO–2 ), ammonium ion (NH+4 ), and nitrogen incorporated into unspecified organic compounds. These species can all be interconverted by microbial action. The concentrations of these species (especially the latter four) are commonly reported in terms of the amount of elemental N that each one contains, rather than in milligrams per liter or moles per liter of the entire species. This approach is useful for comparing the concentrations of the different species and for tracking conversions among them because 1.0 mg of N in any of the species can be converted to exactly 1.0 mg of N in another form. The same 1:1 ratio does not apply if the concentrations are computed for each species in its entirety, because of the different molecular weights of the different species and because N2 and many organic compounds contain more than one N atom per molecule. When this reporting convention is employed, the symbols NO2 –N, NO3 –N, NH4 –N, and org–N, respectively, are used to represent the concentrations of nitrite, nitrate, ammonium, and organic species containing N. For instance, the statement that a solution contains 1.4 mg/L of NO2 –N means that nitrite is present in the solution in an amount sufficient to include 1.4 mg/L of N. This concentration is sometimes conveyed in both writing and speaking by the statement that the concentration of nitrite is 1.4 mg/L as N. The molecular weight of nitrite ion is 46, so the molar and mass concentrations of nitrite ion in such a solution are ✓
mg NO2 –N 1.4 L ✓
10
◆✓
1 mol NO2 –N 14, 000 mg NO2 –N
4 mol
NO2 L
◆ ✓
◆✓
1 mol NO2 mol NO2 –N
46, 000 mg NO2 mol NO2 17
◆
= 4.6
◆
= 10
mg NO2 L
4 mol
NO2 L
If all the nitrite in this solution were converted to nitrate, the NO3 –N concentration in solution would increase by 1.4 mg/L; that is, the increase in NO3 –N is numerically identical to the decrease in NO2 –N. However, since the molecular weight of NO–3 is 62, the increase in the mass concentration of nitrate would be ✓ ◆✓ ◆ mg NO3 –N 62 mg NO3 mg NO3 1.4 = 6.2 L L 14 mg NO3 –N Thus, 1.4 mg/L NO2 –N can be converted to 1.4 mg/L NO3 –N, corresponding to conversion of 4.6 mg/L NO–2 to 6.2 mg/L NO–3 . As noted above, the total concentration of a particular element in a group of different species is also of interest in cases where it is impractical to analyze for each of the species individually. For instance, the overall organic concentration in environmental samples is almost always reported in terms of the total concentration of carbon in all the organic compounds present, because the samples contain a large and disparate collection of organic molecules which would be impossible to identify individually. Such data are usually reported either as the total organic carbon (TOC) concentration of the sample or, if the sample is filtered and only the solution is analyzed, as the dissolved organic carbon (DOC) concentration.11 By analogy, the concentration of total or dissolved inorganic carbon [generally assumed to include only carbonic acid (H2 CO3 ) and bicarbonate (HCO–3 ) and carbonate (CO2– 3 ) ions] is sometimes designated as the TIC or DIC, respectively. Carbon typically accounts for 50±10 percent of the total mass of organic molecules in environmental samples, so the mass-based concentration of all the organic matter is approximately 2⇥DOC. Example 1.4 A natural water has a DOC concentration of 3.5 mg/L. The average composition of the organic molecules in the solution is C30 H33 O18 N. Assuming that all the molecules actually have that composition, compute the molar concentration and mass concentration of the molecules in the sample. (Note that, in truth, the DOC includes molecules with a variety of chemical formulas, so the properties of an “average” molecule should not be interpreted too literally.) Solution The molecular weight of the hypothetical average molecule is (30 ⇥ 12) + (33 ⇥ 1) + (18 ⇥ 16) + (1 ⇥ 14) = 695 Carbon represents a fraction equal to 360/695, or 52 percent, of the mass of the molecule. Since the sample contains 3.5 mg/L DOC, the mass concentration of the dissolved organics is 11 Another
way of characterizing the composite organic content of a sample is to report the amount of oxygen that is consumed when the organic matter is oxidized, either chemically [in the chemical oxygen demand (COD) test] or by a combination of biological and chemical processes [in the biochemical oxygen demand (BOD) test].
18
3.5 mg C/L mg organic molecules = 6.8 0.52 mg C/mg organic molecule L The molar concentration of the hypothetical organic molecules is therefore: 6.8 mg organic molecules/L = 9.7 ⇥ 10 695,000 mg organic molecules/mol
6
mol organics L
A similar situation applies to certain organic contaminants that can form when chlorine-based disinfectants are used to treat drinking water or wastewater. In particular, when either hypochlorous acid (HOCl) or sodium hypochlorite (NaOCl, the active ingredient in bleach) is added to water, it reacts not only with microbes, but also with some of the organic compounds in the water, generating species known as disinfection by-products, or DBPs. Some of these DBPs are carcinogenic, so regulations have been established limiting their allowable concentrations in drinking water. The concentration limit established by law is referred to as the maximum contaminant level (MCL). The DBPs thought to pose the greatest risk to public health are organic compounds into which the halogen atoms chlorine, bromine, and/or iodine have been incorporated.12 The dominant DBPs in most drinking water systems are the trihalomethanes (CHX3 ) and haloacetic acids (Xn H3 n C COOH ), where X can be Cl, Br, I, or a mixture of these elements, and n can be l, 2, or 3. The trihalomethanes and haloacetic acids are commonly designated by their respective acronyms, THMs and HAAs. Currently, in the United States, the MCLs for total THMs and for the sum of five specified HAAs (‘HAA5 ’) are 80 and 60 µg/L, respectively. The total concentration of chlorine, bromine, and iodine incorporated into all the DBPs is also of interest. This parameter is commonly referred to as the total organic halogen (TOX) or the adsorbable organic halogen (AOX) and is reported in units of micrograms per liter as Cl. If a DBP contains bromine or iodine, the contribution of those elements to the TOX is computed by treating them as chlorine atoms, as demonstrated in the following example. Example 1.5 A treated drinking water contains 31 µg/L chloroform (CHCl3 ), 11 µg/L bromoform (CHBr3 ), 8 µg/L dichloroacetic acid (Cl2 HC COOH), and 19 µg/L trichloroacetic acid (Cl3 C COOH). What is the TOX concentration attributable to these four species, expressed in micrograms per liter as Cl? 12 The
brominated and iodinated DBPs can form if chlorine is added to waters that natually contain bromide or iodide ions or iodine [I2 ] molecules. In such a case, the chlorine can convert the other halogens into species that react with and become incorporated into organic molecules.
19
Solution The TOX concentration is the total concentration of halogen incorporated into organic compounds; when the TOX is reported as Cl, the value is computed as if all the halogens were Cl atoms, regardless of their actual identity. The molecular weights of chloroform, bromoform, dichloroacetic acid, and trichloroacetic acid are 119.5, 252.7, 129.0, and 163.5, respectively, so the molar halogen concentrations in these compounds in the given sample are ✓
31 µg CHCl3 /L 119.5 µg CHCl3 /µmol
◆✓
✓
11 µg CHBr3 /L 252.7 µg CHBr3 /µmol
◆✓ ◆ mol halogen µmol halogen 3 = 0.13 mol CHBr3 L
✓
8 µg Cl2 HC-COOH/L 129 µg Cl2 HC-COOH/µmol
◆✓ 2
mol halogen mol Cl2 HC-COOH
✓
19 µg Cl3 C-COOH/L 163.5 µg Cl3 C-COOH/µmol
◆✓ 3
mol halogen mol Cl3 C-COOH
mol halogen 3 mol CHCl3
◆
= 0.78
µmol halogen L
◆
◆
= 0.12
= 0.35
µmol halogen L
µmol halogen L
The total halogen concentration in the four compounds is the sum of the above values, or 1.38 µmol/L. If all of this halogen were Cl, the solution would contain 1.38 µmol/L organic Cl, corresponding to a mass-based Cl concentration of ✓ ◆✓ ◆ µmol Cl 35.5 µg Cl µg Cl 1.38 = 49.0 L µmol Cl L
The concentration of TOX attributable to these species is therefore 49.0 µg/L as Cl. Note that the value of TOX actually measured would include the halogen associated with numerous other species, many of which could not be identified individually.
1.4.5
Species’ Concentrations Represented as the Concentration of a Different Compound: Hardness, Alkalinity, and Oxidant Concentrations
In some cases, the logic behind reporting the concentration of a species as X can be fairly obscure. For instance, calcium ion (Ca2+ ) concentrations are often reported as though the calcium were present as calcium carbonate solid (CaCO3 ), even if no calcium carbonate is actually present in the system. That is, a concentration of 40 mg/L Ca2+ might be reported as 100 mg/L as CaCO3 , meaning that if all the Ca2+ in solution were converted to CaCO3 , the concentration of CaCO3 formed would be 100 20
mg/L. The history that led to this reporting approach is that calcium ion used to be analyzed by adding various reagents to the solution of interest to cause most of the Ca2+ to precipitate as CaCO3 . The mass of CaCO3 solid was then measured, and the Ca2+ concentration in the original solution was inferred. Rather than report the Ca2+ concentration that was inferred, though, people simply reported the concentration of CaCO3 that had precipitated. The collective concentration of several different divalent (doubly charged) ions is also commonly reported as CaCO3 . Specifically, Ca2+ , Mg2+ , and Fe2+ all contribute to the hardness of a water (an indicator of its mineral content and its reactivity with soap and certain other chemicals). In applications where the hardness of the water is of concern, the relative contributions of Ca2+ , Mg2+ , and Fe2+ are usually not particularly important. As a result, the hardness is usually reported as a composite parameter that includes all three contributions. By convention, this value is computed as though all the hardness were contributed by Ca2+ ions. For instance, if a solution contains 10 3 mol/L Ca2+ , 5 ⇥ 10 4 mol/L Mg2+ , and 10 4 mol/L Fe2+ , the total hardness is 1.6 ⇥ 10 3 mol/L. The hardness of this solution might be expressed in units of milligrams per liter as Ca2+ or as CaCO3 as follows: ✓ 1.6 ⇥ 10
3 mol
hardness L
◆✓
1 mol Ca2+ mol hardness
◆✓
◆ 40, 000 mg Ca2+ mol Ca2+ mg hardness = 64 as Ca2+ L
✓ ◆✓ ◆ mg hardness 100 mg CaCO3 mg hardness 2+ 64 as Ca = 160 as CaCO3 2+ L L 40 mg Ca Note that in both of the above expressions, the contributions of Mg2+ and Fe2+ are being reported as the amount of Ca2+ that would contribute an equivalent amount of hardness. In the latter expression, the concentrations of Mg2+ and Fe2+ are first converted to equivalent concentrations of Ca2+ , and then the total equivalent Ca2+ concentration is reported as though it were present as CaCO3 . If any tri- or higher-valent ions contribute significantly to the hardness, their contribution could be calculated based on the amount of calcium that would contribute the same amount of charge (i.e., 10 4 mol/L Fe3+ would contribute 1.5 ⇥ 10 4 mol/L hardness as Ca2+ ). However, such ions almost never make a major contribution, so they can generally be ignored. Although the idea of a mole of hardness is useful for introducing the preceding concepts, that unit is never used in practice; rather, if hardness is not expressed on a mass basis as Ca2+ or as CaCO3 , it is invariably expressed in terms of charge equivalents. Because Ca2+ is divalent, one mole of Ca2+ corresponds to two charge equivalents. The equivalent weight of Ca2+ (the mass of Ca2+ , in grams, that corresponds to one [charge] equivalent) is therefore 21
EWCa2+ =
AWCa2+ 2+
charge equivalents of Ca
per mol
=
40 = 20 2
(1.9)
where AW is the atomic weight. The analogous calculation for CaCO3 (using the molecular weight, MW, instead of the atomic weight) is EWCaCO3 =
MWCaCO3 100 = = 50 charge equivalents of CaCO3 per mol 2
(1.10)
Applying this result to the preceding example calculation, the hardness in the solution containing Ca2+ , Mg2+ , and Fe2+ might be reported as ◆ ⇣ ⌘✓ mg 1 meq meq 160 as CaCO3 = 3.2 (1.11) L 50 mg CaCO3 L Example 1.6 A solution is reported to have a total hardness of 4.5 meq/L. If 80% of the hardness is contributed by Ca2+ and the other 20% by Mg2+ , what are the concentrations of Ca2+ and Mg2+ in mg/L? Solution The contributions of Ca2+ and Mg2+ to the hardness are 80% and 20% of the total, or 3.6 and 0.9 meq/L, respectively. The atomic weight of Mg2+ is 24.3, and it is divalent, so its equivalent weight is 12.15 g. The concentrations of the two ions in mg/L units are therefore ⇣ meq ⌘ ✓ 20 mg Ca2+ ◆ mg cCa2+ = 3.6 = 72 L meq L ⇣ meq ⌘ ✓ 12.15 mg Mg2+ ◆ mg cMg2+ = 0.9 = 10.9 L meq L Another important cumulative parameter that is commonly used to characterize water quality is the alkalinity, which is a measure of the capacity of the solution to neutralize acidic inputs; a corresponding, but slightly less commonly reported value (the acidity) is used to characterize a solution’s capacity to neutralize inputs of bases. Natural aquatic systems and many other solutions of environmental interest contain a variety of species that can neutralize acids or bases. Because it is impractical to analyze for each of these species directly, the composite parameters alkalinity and acidity have been defined to characterize their collective effects, at least with respect to one important aspect of water quality. These parameters are discussed in detail in Chapter 8. 22
Like hardness, alkalinity is often reported either as CaCO3 or in terms of equivalents. In the former case, the meaning is that the real solution contains the same amount of alkalinity as would a hypothetical solution containing the stated concentration of CaCO3 and no other solutes. In the latter case, because alkalinity is inherently related to the acid/base characteristics of the solution, the equivalency is based on the ability of CaCO3 to react with acid. Most readers will already be familiar with the fact that acidity and pH are closely related to the concentration of H+ ions in solution. These relationships are explained in detail in Chapter 5; for the purposes of this discussion, we need only note that one mole of CaCO3 is capable of reacting with two moles of H+ , via the reaction: CaCO3 + 2 H+ )* Ca2+ + CO2 (g) + H2 O
(1.12)
Based on this reaction, the equivalent weight of CaCO3 with respect to acid/base reactions and alkalinity is exactly the same as its equivalent weight with respect to charge and hardness: EWCaCO3 =
H+
MWCaCO3 100 = = 50 equivalents of CaCO3 per mol 2
(1.13)
Example 1.7 A solution is reported to have an alkalinity of 230 mg/L as CaCO3 . Express this value in meq/L. Solution The conversion can be carried out using the equivalent weight of CaCO3 determined in Equation (1.13): ◆ ⇣ ⌘✓ mg 1 meq meq Alk = 230 as CaCO3 = 4.60 L 50 mg CaCO3 L Another scenario in which concentrations are commonly reported in terms of a different species arises when dealing with some oxidation-reduction reactions that are commonly induced in water treatment processes (in particular, disinfection processes). In such cases, the concentration of the oxidant (the chemical that combines with electrons in the reaction) is often expressed in terms of the concentration of Cl2 that could, in theory, combine with the same number of electrons. In such cases, the units used to report the concentration are typically mg/L as Cl2 . The calculation of this Cl2 concentration is based on Reaction (1.14):
Cl2 + 2 e
! 2 Cl 23
(1.14)
Example 1.8 In Example 1.2, doses of 5 mg/L HOCl and 5 mg/L ClO2 were expressed in units of meq/L. Express those same doses in mg/L as Cl2 . Do the same for a dose of 5 mg/L ozone (O3 ), assuming that each molecule of ozone ultimately combines with six electrons, as follows: O3 + 3 H+ + 6 e
! 3 OH
Solution Reaction (1.14) indicates that each mole of Cl2 can combine with two moles of electrons; i.e., one mole of Cl2 corresponds to two (electron) equivalents. Therefore, since the molecular weight of Cl2 is 71.0, its equivalent weight is EWCl2 =
MWCl2
e equivalents of Cl2 per mol
=
71.0 = 35.5 2
In Example 1.2, we determined that doses of 5 mg/L HOCl and ClO2 correspond to 0.19 and 0.37 meq/L, respectively, so those doses can be expressed in mg/L as Cl2 as follows: ✓ ◆ mg HOCl ⇣ meq ⌘ mg Cl2 mg 5 = 0.19 35.5 = 6.75 as Cl2 L L meq L 5
✓ ◆ mg ClO2 ⇣ meq ⌘ mg Cl2 mg = 0.37 35.5 = 13.1 as Cl2 L L meq L
The equivalent weight of ozone is EWO3 =
MWO3
e equivalents of O3 per mol
=
48.0 = 8.0 6
A dose of 5 mg/L O3 can therefore be expressed in mg/L as Cl2 as follows: ✓ ◆✓ ◆ mg O3 1 meq mg Cl2 mg 5 35.5 = 22.2 as Cl2 L 8 mg O3 meq L
1.4.6
Composite Parameters Characterizing the Total Solute Concentration: TDS and Ionic Strength
Three composite parameters are commonly reported as overall measures of the total solute or total ionic content of a sample — the total dissolved solids concentration, the ionic strength, and the electrical conductivity. The total dissolved solids (TDS) 24
concentration (sometimes called the filterable residue) includes the majority of both the ionic and nonionic dissolved species. It is determined by heating a sample of the water at a temperature slightly greater than 100 C until only a dry residue remains. During the drying step, essentially all the gases dissolved in the water [primarily oxygen (O2 ), nitrogen (N2 ), and carbon dioxide (CO2 )] escape. In addition to the inorganic carbon that is present in the initial solution as dissolved CO2 gas, all of the carbonic acid (H2 CO3 ) and some of the bicarbonate ions are converted to CO2 (g) by the following reactions and are then removed when the solution is heated. H2 CO3 )* H2 O + CO2 (g) 2HCO3 )* H2 O + CO2 (g) + CO23
(1.15) (1.16)
Very few other substances that are likely to be dissolved form gases at 100 C, so they remain in the residue. These species include virtually all salt ions and most neutral inorganic and organic solutes. Most of the water molecules also evaporate, although some might remain firmly attached to salts or other solids in the residue. In most natural waters, the majority of the residue is inorganic salts, so the TDS can be thought of as a rough measure of the ionic content of the original solution. Example 1.9 Using the data in Table 1.1, estimate the TDS in rain water at Menlo Park, CA. Assume that the alkalinity of the water is all contributed by bicarbonate ions (HCO–3 ). That is, assume that the concentration of HCO–3 in the water equals the reported value of Alk, when both terms are expressed in meq/L. Assume also that, when the solution is heated to evaporate the water as part of the TDS analysis, all the HCO–3 undergoes Reaction (1.16). The H2 O and CO2 generated by this reaction are volatilized (i.e., transferred to the gas phase), but the CO2– 3 remains as part of the dry solid residual, along with all the other ions in the original solution. Solution The concentrations of most of the dissolved ions in the rain water are given explicitly in Table 1.1, but that of HCO–3 must be evaluated based on the assumption about its relationship to Alk, as follows: ⇣ mg ⌘✓ 1 meq Alk ◆✓ meq HCO ◆✓ mmol HCO ◆ 3 3 cHCO = 3.3 as CaCO3 1 1 3 L 50 mg as CaCO3 meq Alk meq HCO3 = 0.066
mmolHCO3 L
Reaction (1.16) indicates that, for every two moles of HCO–3 in the rain water, one mole of CO2– 3 ends up in the solid residual and thereby contributes to the TDS. Thus, the ultimate contribution of the HCO–3 to the TDS is 25
✓
mmolHCO3 cHCO ,TDS = 0.066 3 L
◆
1mmol CO23 2 mmol HCO3
!
60 mg mmol CO23
!
= 1.98
mg L
The TDS of the rain water is therefore TDS =
Â
all i in residual
ci = cCa2+ + cMg2+ + cNa+ + cCl + cSO2 + cCO2 4
= (0.8 + 1.2 + 9.4 + 17 + 7.6 + 1.98)
3
mg mg = 38.0 L L
The second measure of the overall solute concentration that we will consider — the ionic strength, commonly designated I or µ — accounts only for the ions in the solution; that is, it completely excludes uncharged solutes. The focus on ionic species arises both because ions represent the vast majority of the solutes in most environmental solutions and because ions generate local electrical fields that can have significant effects on the behavior of neighboring ions and dipoles (like H2 O). The ionic strength is a composite measure of the concentration and charge on the ions in the solution. To take into account the observation that the strength of certain chemical interactions varies as the square of the ionic charge, the ionic strength is defined as follows: I=
1 ci z2i 2 all ions
(1.17)
where ci is the concentration of species i in moles per liter and zi is the charge on species i. If zi is taken to be dimensionless, I has units of moles per liter. However, I is often reported without units, i.e., as though it were dimensionless. Note that the ionic strength is a composite property that characterizes the conditions in a solution; individual ions do not have distinct ionic strengths. As is explained shortly, the ionic strength is a key parameter controlling the reactivity of dissolved ions (and, to a lesser extent, neutral solutes). Example 1.10 Compute the ionic strength of solutions containing 0.05 mol/L NaCl, 0.025 mol/L Na2 SO4 , or 0.025 mol/L MgSO4 as the only salts. All these salts dissociate completely, so the resulting solutions contain only Na+ , Cl– , Mg2+ , and SO2– 4 in addition to H2 O. The concentrations of the salts have been chosen such that each solution contains 0.05 mol/L of positive and negative charges.
26
Solution The ionic strengths of the various solutions can be computed using Equation (1.17). Note that two moles of Na+ enter solution per mole of Na2 SO4 dissolving. i ⇤ 1h 1⇥ INaCl = cNa+ z2Na+ + cCl z2Cl = (0.05 M) (+1)2 + (0.05 M) ( 1)2 2 2 = 0.05 M i 1h i 1h cNa+ z2Na+ + cSO2 z2SO2 = (0.05 M) (+1)2 + (0.025 M) ( 2)2 4 2 2 4 = 0.075 M
INa2 SO4 =
i 1h i 1h cMg2+ z2Mg2+ + cSO2 z2SO2 = (0.025 M) (+2)2 + (0.025 M) ( 2)2 4 2 2 4 = 0.100 M
IMgSO4 =
The ionic strengths of the solutions vary by a factor of 2, even though all the solutions contain the same total concentrations of positive and negative charge.
Like the ionic strength, the electrical conductivity, k, of a solution (also called the specific conductance or simply the conductivity) reflects only its ionic composition. The conductivity is measured by imposing an electrical field on the solution, causing the dissolved cations and anions to migrate toward the oppositely charged electrodes and thereby establishing a flow of current (I).13 This current is directly proportional to the cross-sectional area (A) through which the ions flow and to the strength of the electrical field, which equals the difference in the electrical potential (y) between the electrodes divided by their separation (l) . The conductivity is defined as the proportionality constant in this relationship, i.e.: I=k k=
y A l
l I l 1 = Ay AR
(1.18a) (1.18b)
where the final expression in Equation (1.18b) reflects the fact that, by Ohm’s law, the ratio I/y is the inverse of the electrical resistance (R). If I is expressed in amperes, y in volts, and l and A in cm and cm2 , respectively, k has units of ampere/(volt·cm), or 1/(ohm·cm); one (ohm·cm) 1 is defined as one Siemen (S). According to Equation (1.18b), the conductivity of a solution can be determined by measuring the electrical resistance of the solution in a cell with known geometry 13 Note
that the letter I is conventionally used to represent both ionic strength [in Equation (1.17)] and electrical current, but the two parameters are completely unrelated.
27
(l and A). In practice, the geometry of the cell is usually not known exactly and other characteristics of the cell can have a small effect on the measured resistance, so a cell constant (C) is defined that accounts for the l/A ratio and these other factors. The conductivity is then represented as 1 (1.19) R The value of C for a given cell can be determined by measuring the resistance of a standard solution of known conductivity, and this value of C can then be applied when the same cell is used to analyze other solutions. Although each ion in a solution contributes to the conductivity, not all ions contribute equally: ions that have a higher charge carry more current than those with lower charge and, for ions of equal charge, those that move through the solution easily (typically, the smaller ones) carry more of the current than those whose movement is impeded. And, of course, the amount of current carried by any ionic species i increases when the concentration of i increases. To separate the effects of an ion’s mobility from the effects attributable to its charge and concentration, a value called the equivalent ionic conductance (li ) has been defined as the conductivity contributed by that ion (ki ) per unit normality of the ion (Ni ) in the solution, i.e., k =C
ki ki = Ni ci zi ki = li ci zi li =
(1.20a) (1.20b)
The conductivities of the individual ions are assumed to be additive to yield the conductivity of the whole solution, so k = Â ki = Â li ci zi i
(1.21)
i
The value of li in the limiting case of an infinitely dilute solution (i.e., one in which each ion interacts almost exclusively with water molecules and not with other solutes) is designated li,0 . Such values have been measured for many ions; those for some major ions in environmental solutions are shown in Table 1.2. In real solutions, li for any ion is invariably less than li,0 , because the mobility of i is reduced due to interactions with species other than water molecules. The l0 values of common solutes other than H+ and OH– all fall in the range of ⇠50-80 S/(equiv/cm3 ). The unusually large values of l0 for H+ and OH– reflect the fact that these species can, in effect, be transported through the solution by transfer of H+ ions from one water molecule to another. This migration pathway is not available to other ions that are not inherently part of water molecules. However, H+ and OH– 28
Table 1.2 Equivalent ionic conductance at 25 C and infinite dilution [S/(equiv/cm3 )](a)
Cations +
H Na+ K+ NH+4 Ca2+ Mg+ Ba2+ (a) Values
l0
Anions –
349.8 50.1 73.5 73.6 59.5 53.1 63.7
OH HCO–3 Cl– NO–3 CH3 COO– SO2– 4 H2 PO–4
l0 198.3 44.5 76.4 71.5 40.9 80.0 43.0
from Marcus, Y. Ion Properties, Marcel-Dekker, New York (1997)
are typically present at very low concentrations, so despite their relatively large l0 values, they do not contribute significantly to k except in solutions at high or low pH. For any solution, we can define an average value of l as lavg ⌘
 li ci zi i
 ci zi
(1.22)
i
Substitution of this expression into Equation (1.21) yields k = lavg  ci zi = lavg Ntot
(1.23)
i
where Ntot is the total normality of all dissolved ions. Based on this approximation, the conductivity is sometimes used as an indicator of the total ionic content of a solution. This usage is fairly reliable when estimating the ionic content of natural waters or comparing solutions with similar mixtures of ions (e.g., to track the mixing of fresh water with salty water in an estuary or the evaporation of water from a river in an arid climate), but it is less so when comparing solutions of widely different compositions. Note that the conductivity of a solution increases significantly with temperature, so it is important to make such comparisons only if the measurements are at the same temperature or are otherwise adjusted to account for the effect of temperature.
1.5
SUMMARY
Environmental aquatic chemistry comprises an enormous range of topics, many of which relate to the prediction or analysis of the speciation of chemicals in aquatic 29
systems. The focus of this textbook in on the speciation of chemical systems, and in particular aqueous solutions, that have reached chemical equilibrium. In any aqueous environmental solution, the vast majority of the molecules are water molecules. Nevertheless, the relatively low concentrations of other molecules determine the quality of the water and its usefulness for various applications. This chapter has introduced the basic nomenclature used to describe chemical concentrations and some of the factors that control the solubility of various species in aqueous solutions. Water molecules are dipoles, meaning that although they are neutral overall, a net positive charge tends to reside in one part of the molecule and a net negative charge in another part. As a result, water molecules can form hydrogen bonds with one another, causing the liquid to be unusually cohesive. The polarity of water molecules also allows them to bond to other polar molecules and to dissolved ions, stabilizing those species in solution and making the species hydrophilic and soluble. By contrast, potential solutes that are uncharged and nonpolar form weaker bonds with water molecules and are therefore hydrophobic and relatively insoluble. The concentration of a species A (designated here as either cA or [A]) can be expressed in a variety of units, including mass and moles per unit volume, partial pressures (for gases at a fixed temperature), mass fractions, and mole fractions. However, in environmental systems, we must often deal with solutions of unknown history that contain many unknown chemicals, and it is frequently difficult or impossible to determine the exact chemical identity of many of the dissolved constituents. In addition, we are often interested in the concentrations of collections of species as much as or more than in the concentration of each individual species. For these reasons, it is common to characterize the solution composition in terms of collective parameters such as the dissolved organic carbon (DOC) concentration, the hardness, or the conductivity. These parameters are sometimes expressed in ways that are unconventional in other fields of chemistry (e.g., as CaCO3 ). However, with a little effort and experience, these units become as easy to use as more conventional ones. Having defined the essential language needed to discuss environmental solutions, we next consider the factors that control the reactivity of the species in such systems, and then proceed to explore some of the most important types of environmenta chemical reactions.
1.6
PROBLEMS
1. An aqueous solution is prepared by adding 110 mg of calcium chloride (CaCl2 ) and 50 mg of calcium sulfate (CaSO4 ) to 500 mL of water. The solution pH is 8.0 [H+ ] = 10 8 M, [OH ] = 10 6 M . (a) Assuming that the salts dissociate completely, express the concentration of each of the dissociation products (Ca2+ , Cl– , and SO2– 4 ) in mg/L, millimolar, and normal (N) units. 30
(b) If the total molar concentration of all species in the solution is the same as in pure water, what are the mole fractions of H+ , OH– , Ca2+ , Cl– , and SO2– 4 , and H2 O? 2. A solution contains 45 mg/L Ba2+ . A consultant has suggested that the Ba2+ could be removed from solution by converting it to solid barium sulfate [BaSO4 (s)]. To accomplish this, she proposes adding enough Na2 SO4 to the solution to convert all the Ba2+ to BaSO4 (s), plus an extra 5 mg/L SO2– 4 as a safety factor. What dose (mg/L) of Na2 SO4 should be added, and what would the concentration of solids be, in ppm, if essentially all the Ba2+ precipitated? 3. The concentration of silver in seawater is 50 parts per trillion, and the total volume of seawater in the world’s oceans is approximately 1370 ⇥ 106 km3 . The density of seawater is 1.025 kg/L. (a) Determine the total mass (kg) of silver in the oceans of the world. (b) What volume of seawater would have to be processed to recover one kilogram of silver if the extraction process were 100% efficient? 4. The partial pressure of carbon dioxide in air is approximately 10 3.42 atm. In temperate regions, PCO2 in soil gas can be as high as 10 1.5 atm. For each of these two cases, calculate the mass of carbon dioxide (in grams) in 1.0 m3 of gas at T = 25 C. 5. The partial pressures of nitrogen (N2 ) and oxygen (O2 )in air are 0.78 atm and 0.21 atm, respectively. Calculate the number of moles and the mass (mg) of each of these components in 1 L of air at 25 C. 6. Ozone gas (O3 ) is commonly injected into drinking water (also swimming pools and hot tubs) as a disinfectant. Pure ozone decays rapidly, so it cannot be stored for long periods; as a result, it is typically generated onsite by exposing dried air to an electrical discharge that causes gaseous oxygen to be converted to ozone by the reaction: 3 O2 (g) ! 2 O3 (g). Typically, about one-seventh of the oxygen in the gas is converted to ozone in this step. A water treatment plant for a city with a population of 1 million people must produce on the order of 400,000 m3 per day of clean water. If ozone is injected into the water at a dose of 1 mg/L, what volume of air (1.0 atm total pressure, 15 C, 21% O2 ) must be processed per day to provide the ozone? 7. The discharge of chromate ions (CrO2– 4 ) to sewers or natural waters is of concern because of both its ecological impacts and its effects on human health if the receiving water is later used as a drinking water source. One way in which chromate can be removed from solution is by its reaction with ferrous 31
ions (Fe2+ ) to form a mixture of chromic hydroxide and ferric hydroxide solids [Cr(OH)3 (s) and Fe(OH)3 (s), respectively], which can then be filtered out of the water. The overall reaction can be represented as CrO24 + 3 Fe2+ + 8 H2 O ! Cr(OH)3 (s) + 3 Fe(OH)3 (s) + 4 H+ How much particulate matter would be generated daily by this process at a facility that treats 60 m3 /h of a waste stream containing 4.0 mg/L Cr, if the treatment reduces the Cr concentration enough to meet a discharge limit of 0.1 mg/L Cr? 8. A solution contains 65 mg/L Ca2+ as CaCO3 and 85 mg/L alkalinity as CaCO3 . What are the Ca2+ concentration and the alkalinity in meq/L? What is the Ca2+ concentration in molar units? 9.
(a) Express the concentration of Ca2+ in the Mississippi River near Clinton, IA (composition shown in Table 1.1) in mol/L Ca2+ and in mg/L as CaCO3 . (b) Compute the ionic strength of the river water. Assume that the organic molecules are 50% C by mass, have an average molecular weight of 550, and carry an average charge of 1. Also assume that the number of equiv/L of alkalinity can be treated as the number of mol/L of HCO–3 .
10. The following analysis has been reported for a sample of contaminated groundwater obtained near an aluminum production facility: pH Na+ K+ Ca2+ Mg2+
Cl– F– DIC SO2– 4 OH–
= 11.2 = 6, 500 mg/L = 2, 230 mg/L = 3, 400 mg/L = 1, 350 mg/L
= 1, 100 mg/L = 4, 350 mg/L = 1, 740 mg/L (as C) = 3, 940 mg/L = 10 2.8 mol/L
(a) Check the charge balance for this analysis. Does the total cationic charge (in equiv/L) equal the total anionic charge to within ±5% (the typical target for an acceptable charge balance)? (Note: at pH 11.2, the inorganic carbon is predominantly in the form CO2– 3 .) (b) Calculate the TDS of the solution in mg/L. Because most of the DIC is present as CO2– 3 , a negligible portion is released as CO2 (g) during the TDS analysis. (c) Calculate the ionic strength of the solution. (d) Calculate the hardness of the solution in mg/L as CaCO3 . 32
11. The West Virginia Water Research Institute reported the following water quality measurements for the Monongahela River monitoring station at Mile 89 for August 2010. pH = 7.79 ALK = 59.9 mg/L as CaCO3 Na+ = 44.7 mg/L Ca2+ = 39.8 mg/L
Mg2+ = 10.6 mg/L Cl = 15.9 mg/L SO24 = 146 mg/L
(a) Determine the concentration of dissolved inorganic carbon (DIC) in the water in mol/L and in mg/L, assuming that HCO–3 is the only significant contributor to ALK and DIC. (b) Check the cation–anion balance for this analysis. (c) Calculate the ionic strength of the solution. (d) What is the alkalinity in meq/L, if all the alkalinity is contributed by HCO–3 ions? 12. A natural water sample has been analyzed and reported to have the following solute concentrations. [SO24 ] = 10
TDS = 300 mg/L Alkalinity = 1.5 ⇥ 10 +
[Na ] = 2.5 ⇥ 10 2+
[Ca ] = 10
3
3
3
equiv/L
M
3
M
[Cl ] = 0.5 ⇥ 10
3
M
pH = 8.2
M
(a) Check the charge balance, assuming that all the alkalinity is contributed by HCO–3 . If the charge is not balanced, determine the minimum additional concentration of sodium or chloride ions (depending on the sign of the imbalance) that would have to be present to eliminate the charge imbalance. (b) Compare the TDS contributions of solutes that were analyzed to the reported TDS concentration. Would inclusion of the Na+ or Cl– ion concentration determined in part (a) resolve any discrepancy between the measured TDS and the analyses for individual ions? 13. A river contains 8 mg/L DOC in molecules whose average composition is C10 H15 O4 N. (a) What is the mass fraction of C in the organic molecules? (b) What are the mass fraction and mole fraction of these molecules in the solution? 33
14. When organic matter with an average chemical formula of C5 H9 O3 N is consumed by microorganisms under aerobic conditions, it undergoes the following oxidation reaction: C5 H9 O3 N + 5 O2 + 2 H2 O ! 5 HCO3 + 4 H+ + NH+ 4 If 10 mg/L DOC is consumed, how much HCO–3 is generated? Express your answer in mg/L DIC, meq/L, and mg/L as CaCO3 . 15. What is the ionic strength of a solution containing only sodium benzoate (NaC7 H5 O2 ) and water, if the DOC concentration in the solution is 15 mg/L? Assume that, when the sodium benzoate dissolves, it dissociates completely into sodium and benzoate ions (Na+ and C7 H5 O–2 , respectively), and that neither ion undergoes significant subsequent reactions. 16. A solution contains 30 mg/L NH+4 and 5 mg/L NO–3 . (a) Express these concentrations as mg/L NH4 N and NO3 N, respectively. Which species makes a larger contribution to the total N concentration in the water? (b) In biological wastewater treatment processes, conditions are sometimes established that are conducive to microbial conversion of NH+4 to NO–3 (nitrification) in an aerobic step, and of NO–3 to N2 (denitrification) in a subsequent anaerobic step. What volume of N2 (g) would be generated per liter of water treated if the original solution underwent complete nitrification and denitrification? Assume that the gas phase is at 25 C and 1 atm total pressure. 17. The following data summarize the concentration changes of various dissolved nitrogen species in a biological wastewater treatment process. The species listed as “organic matter” has an average composition of C30 H33 O18 N. Determine the change in total dissolved nitrogen (TDN) concentration in the process. Species NH+4 NH3 NO–3 NO–2 Organic matter
Influent conc’n (mg/L) 28 1.8 0.2 0.1 265
Effluent conc’n (mg/L) 4.5 0.4 95 0.0 21
18. Potassium permanganate (KMnO4 ) is sometimes added to water as it enters a drinking water treatment plant to assist in the removal of Fe2+ and Mn2+ 34
ions. The KMnO4 oxidizes those cations by acquiring electrons according to the following reaction: KMnO4 + 4 H+ + 3 e
! K+ + MnO2 (s) + 2 H2 O
(a) What is the (electron) equivalent weight of Mn, based on this reaction? (b) Calculate the mass (g) of KMnO4 in 2 L of a 0.15 M KMnO4 solution. (c) If 1.0 mL of 0.15 M KMnO4 is added to each liter of the drinking water, and all the KMnO4 undergoes the reaction shown, what concentration of MnO2 (s) will be generated by the reaction, in ppm? 19. You wish to add 5 mg/L NaOCl as Cl2 to a solution in a disinfection test, and you have a stock solution (household bleach) that contains 5% NaOCl by weight. Assuming that the density of the stock solution is 1.0 g/mL, how many milliliters of bleach should you add to each liter of test solution? The key reaction that NaOCl undergoes is NaOCl + 2 H+ + 2 e )* Na+ + H2 O + Cl 20. Calculate the normality and equivalent weight of the following species based on the reactions shown. Assume that the species concentration is 10 3 mol/L and that equivalency is based on charge in part (a), acid/base behavior in part (b), and oxidation/reduction behavior in parts (c) through (e). (a) Na+ in: NaCl )* Na+ + Cl– (b) Ac– in: HAc )* H+ + Ac (c) Pb2+ in: Pb2+ + 2 e– )* Pb(s) (d) NO–3 in: NO–3 + 2 H+ + 2 e– )* NO–2 + H2 O (e) NO–3 in: NO–3 + 10 H+ + 8 e– )* NH+4 + 3 H2 O 21. After disinfection of drinking water with chlorine, the following concentrations of halogenated disinfection byproducts were found. How much TOX (µg/L as Cl) and how much TOC (mg C/L) are contributed by these compounds? The chemical formula of dichloroacetonitrile is Cl2 HCCN. Species Chloroform Bromodichloromethane Dibromochloromethane Bromoform Chloroacetic acid
Conc’n (µg/L) 43 19 10 5 10 35
Species Dichloroacetic acid Trichloroacetic acid Bromoacetic acid Dibromoacetic acid Dichloroacetonitrile
Conc’n (µg/L) 20 20 8 15 8
22. Estimate the specific conductance at 25 C of a solution containing 700 mg/L KCl, assuming that the KCl dissociates completely. Compare your result with the estimated specific conductance of a solution of CaSO4 that has the same total concentration of cationic and anionic charge as the KCl solution. 23. Consult some books on analytical chemistry to learn about two of the following analytical methods, and prepare a half-page summary of each technique. Indicate the most common units in which results of these analyses are reported. Cite the sources of your information. (a) atomic absorption (AA) spectroscopy for measurement of metal concentrations in water; (b) inductively coupled plasma (ICP) emission spectroscopy for measurement of metal concentrations in water; (c) selective ion electrodes for measurement of concentrations of ions such as Cu2+ and Cl– ; (d) potentiometric analysis for H+ (pH probe). 24. Consult the APHA/AWWA/WEF reference book Standard Methods for the Examination of Water and Wastewater to learn about standard methods available for analysis of chloride (Cl– ) or selenium (Se). List each method and give the Standard Methods identification number and name for the method. Prepare a brief summary (⇠150 words) of each method, explaining how it works.
36
2
CHEMICAL REACTIVITY, REACTIONS, AND EQUILIBRIUM
37
Contents 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
2.1
OVERVIEW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHARACTERIZING CHEMICAL REACTIVITY . . . . . . . . . PREDICTING ACTIVITY COEFFICIENTS FROM KNOWLEDGE OF THE SOLUTION COMPOSITION . . . . . . . . . . . . . . . THE ACTIVITY AS AN INTENSIVE PROPERTY; THE ACTIVITY OF SOLIDS DISPERSED IN WATER . . . . . . . . . MODELS OF CHEMICAL EQUILIBRIUM . . . . . . . . . . . . EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT COMBINING CHEMICAL REACTIONS . . . . . . . . . . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 38 49 56 57 68 69 71 73
OVERVIEW
As noted in Chapter 1, this text focuses on the analysis of chemical equilibrium, a stable condition characterized by the absence of any net driving force for a chemical reaction to occur. In this chapter, we explore the factors that control the reactivity of various species in chemical systems, the way in which chemical reactivity is quantified, and the conditions that must be satisfied for chemical equilibrium to be achieved.
2.2 2.2.1
CHARACTERIZING CHEMICAL REACTIVITY Factors Affecting the Reactivity of Chemicals
We begin the chapter by developing an approach for quantifying the tendency for a chemical change to occur that causes the concentration of a species to decline at a location of interest. The change might be associated with a chemical reaction that consumes the species, or it might be driven by chemically-induced transport (e.g., 38
diffusion of the species in response to a concentration gradient). In either case, the net effect is that the species disappears from the original location, either because it has been converted to a different species or because it has moved elsewhere. As we will see, a single parameter can be used to characterize this tendency, regardless of which type of change we are considering. This parameter is called the chemical activity (or simply the activity), and, for a species A, is designated here by either aA or {A}.1 Consider, for example, the tendency for a species A to participate in the elementary reaction A + B )* C.2 In this reaction, collisions between molecules of A and B cause those molecules to disappear and a molecule of C to appear. Since the probability of a collision between a molecule of A and a molecule of B is proportional to their respective concentrations (in mass or moles per unit volume), the rate of A B collisions (and hence the rate of reaction) is expected to be proportional to the concentrations of A and B, all other factors being equal. For instance, if the concentration of B is fixed and the concentration of A doubles, we expect twice as many A B collisions in a given volume of solution per unit time, causing A and B to disappear at twice the original rate. This argument applies regardless of whether the reaction occurs in a liquid, gas, or solid phase, so we infer that the tendency for any constituent to participate in an elementary reaction is proportional to its concentration in the phase where the reaction occurs. Often, we are interested in processes in which some constituents cross a phase boundary (solution to solid, gas to solution, etc.). A little thought suggests that the likelihood that a given type of molecule will leave (disappear from) a phase by crossing the boundary is proportional to that molecule’s concentration in the phase that it is leaving. For example, in a system with a fixed amount (i.e., area) of gas/liquid interface, doubling the dissolved concentration of a constituent doubles the frequency with which molecules of the constituent strike the interface and therefore doubles the rate at which they leave solution and enter the gas.3 The same reasoning applies if the dissolved constituent is “removed” from solution by reacting at the gas/liquid interface with a gas-phase species, even if the products of the reaction remain in solution. Thus, the tendency for a species to disappear from a phase is expected to be proportional to its concentration in that phase, regardless of whether the process in1 Different authors use different types of brackets to distinguish between the concentration and activ-
ity of A. For instance, [A], (A), and {A} might each be used to represent concentration in some writings and activity in others. In this text, braces { } are used to designate activities, and square brackets [ ] are used to represent concentrations. 2 An elementary reaction is one that occurs in a single step as a result of a collision between the reactants. 3 As suggested by the wording, the likelihood of cross-boundary transfers also depends on the amount of interfacial area in the system. However, that dependency is generally considered to be a physical characteristic of the system and is handled separately.
39
volves a homogeneous (one-phase) or heterogeneous (multi-phase) reaction, or simply reflects migration from one phase to another without a chemical reaction. Similar thought experiments suggest that this relationship between chemical change and concentration applies to other processes as well. While the concentration of a species is a dominant factor controlling its activity, other factors such as temperature, pressure, and the overall composition of the phase also play a role. For instance, increasing the temperature increases the likelihood of reactions in solution, because as temperature increases, the molecules themselves have more kinetic energy, and in addition the water becomes less viscous, providing less resistance to molecular movement. Furthermore, even if temperature, pressure, and the concentrations of A and B are the same in two solutions, the frequency of A B collisions might not be identical in the two solutions if the other (nonreacting) components of the two solutions differ. For instance, if one solution contains species that surround molecules of A and shield them from molecules of B, the likelihood of a collision between A and B will be lower than in another solution lacking those shielding molecules. Thus, we expect the activity of a given species to be affected by system-specific physical and chemical factors in addition to the species’ concentration. Throughout this text, we will be interested in both the concentration and the activity of chemical species, and we will use these parameters regularly in various kinds of equations. As has been noted, these two quantities are related to one another, but they are not identical. Roughly speaking, concentrations describe the amounts of various species that are present in a system, while activities characterize the species’ behavior. The most important equations in which concentrations are used are mass balances and charge balances, and the most important ones in which activities are used are ‘mass action’ expressions like equilibrium constant expressions. These equations are described in much more detail in subsequent chapters, but it is worthwhile to begin developing this basic understanding of the distinction between concentrations and activities immediately.
2.2.2
Quantifying Chemical Activity: The Standard State
The preceding section describes the main factors that are expected to affect the chemical activity of a species. To quantify that activity, an arbitrary, but well-characterized system containing the species is selected as a baseline, and the species is assigned an activity of 1.0 in that system. The activity of the species in any other system is then compared with that in the baseline system and assigned a corresponding value; e.g., if the species is half as reactive in the system of interest as in the standard system, its activity is 0.5. Note that, by definition, activities are always dimensionless, since they are ratios of the actual reactivity of a substance to the reactivity of that substance in the baseline system. The complete set of conditions describing the baseline system 40
is called the standard state. Throughout the text, standard state conditions are designated by a superscript ; for example, the concentration of A in the standard state is designated cA . Because activity is related to both the concentration of a species and its environment (i.e., the temperature, pressure, and overall composition of the phase), all these factors must be clearly specified when defining the standard state. Historically, the concentration and environmental components of the standard state have been dealt with separately, with the “standard environment” being referred to as the reference state. Thus, a species is said to be in its standard state if it is present both at its standard concentration and in its reference state: ⇢ ⇢ ⇢ Standard Standard Reference State = + State Concentration Environmental Conditions 8 9 ⇢ ⇢ < Standard Pressure = Standard Standard = + Standard Temperature State Concentration : ; Standard Composition Concentration in the Standard State
In environmental systems, the most common choices for standard state concentrations are as follows: • For dissolved solutes: a concentration of 1.0 mol/L. • For bulk liquids and solids: the concentration of the species when it is present as a pure substance (mole fraction = 1.0) at 25 C and 1 atm total pressure. • For gases: the concentration of a pure gas (mole fraction = 1.0) that is behaving in accordance with the ideal gas law at 25 C and 1 atm total pressure. Based on Equation (1.5), this concentration is 40.87 mol/m3 . Environmental Conditions in the Standard State: The Reference State
The reference state for a species A establishes the baseline chemical environment for that species. As noted above, the temperature and pressure normally chosen for the reference state are the same for all chemicals (25 C and 1 atm, respectively). Choices for the chemical composition in the reference state are, however, generally different for different types of constituents. For bulk solids and liquids, the reference state composition is usually defined as the pure substance. Thus, for example, the reference state for water corresponds to the environment in pure water at 25 C and 1 atm. If a solute were dissolved in the water, the mole fraction of H2 O molecules would be <1.0; i.e., the solute would cause the concentration of H2 O to be less than its standard-state concentration. In 41
addition, the solute would change the environment in which the water molecules find themselves, and this change could make each individual water molecule either more or less reactive than it would be in the absence of the solute. Both of these factors would affect the activity of the water. The reference state for gas molecules is usually chosen to be slightly different from the environment in pure, real gases. The ideal gas law represents molecules as hard spheres that occupy no space, engage in perfectly elastic collisions, and have no attraction for one another. No real gas has these properties, but under virtually all conditions of environmental interest, real gases do behave very much like such ideal gases. The reference state for gases is typically chosen to be a state in which these properties apply. Thus, for example, the standard state for oxygen gas is a hypothetical gas containing only O2 molecules (i.e., a mole fraction of 1.0) at 25 C and 1.0 atm, in which each O2 molecule behaves not as molecules actually do in such a system, but rather as they would be expected to behave according to the ideal gas law. The most common choice for the reference state of dissolved species is also based on a hypothetical, idealized situation. Specifically, the reference state is defined as a solution in which each molecule of the solute interacts only with solvent (water) molecules, and not with other constituents. This hypothetical condition corresponds to what would actually occur in an infinitely dilute solution, so it is commonly referred to as the infinite dilution reference state. Thus, for example, the standard state for dissolved zinc ions (Zn2+ ) is a solution at 25 C and 1 atm total pressure, containing 1.0 mol/L Zn2+ (the standard state concentration), with each Zn2+ ion behaving as it would if it were the only ion in solution (the reference state environment). In a real solution containing 1.0 mol/L Zn2+ , zinc ions might interact with one another and with the anions that entered the solution with the Zn2+ ions. As a result, that real solution might behave quite differently from the hypothetical standard-state solution. The most common definitions for the standard state for species in various phases are summarized in Table 2.1. The preceding description makes it clear that standard states need not be real or attainable conditions. For instance, it is not possible to prepare a solution in which the Zn2+ concentration is 1.0 M and simultaneously ensure that every Zn2+ ion behaves as it would if it were the only ion in solution. Nevertheless, despite the impossibility of preparing such a solution, we can imagine ways to infer what the reactivity of Zn2+ ions would be in such a solution. For instance, we could measure the reactivity of successively more dilute real solutions containing Zn2+ , and then extrapolate to a hypothetical solution containing, say, one molecule of Zn2+ per liter. If we multiplied the reactivity of Zn2+ in that (virtually) infinitely dilute solution by Avogadro’s number (6.022 ⇥ 1023 ), the result would be the hypothetical reactivity of a solution containing 1.0 mol/L Zn2+ , with each molecule of Zn2+ behaving as though the so42
Table 2.1 Common choices for the standard state conditions in environmental systems
Standard Conc’n Solid Liquid Gas Solute
Concentration in pure solid Concentration in pure liquid Concentration in ideal gas 1.0 mol/L
Standard state Reference state conditions Temperature Pressure Other 25 C
1 atm
Pure solid
25 C
1 atm
Pure liquid
25 C
1 atm
Ideal gas behavior
25 C
1 atm
Infinite dilution
lution were infinitely dilute; i.e., it would be the reactivity of Zn2+ in its standard state. Return now to the idea that the chemical activity of a species can be defined as the ratio of its reactivity in the system of interest to that in the standard state. Based on this idea, if we determined the reactivity of Zn2+ in a real solution to be 6.022 ⇥ 1020 times as great as that in a hypothetical solution containing one molecule of Zn2+ per liter and no other solutes, we could calculate the activity of Zn2+ in the real solution as follows: aZn2+ = =
Reactivity of Zn2+ in real system Reactivity of Zn2+ in its standard state 6.022 ⇥ 1020 a⇤Zn2+ 6.022 ⇥ 1023 a⇤Zn2+
= 10
3
where a⇤Zn2+ is the reactivity of Zn2+ in the hypothetical, infinitely dilute (1 molecule Zn2+ per L) solution. This result implies that, although the real solution may not actually have a concentration of 6.022 ⇥ 1020 molecules of Zn2+ per liter, and the interactions of Zn2+ molecules with others might not be identical to those that would occur in an infinitely dilute solution, the Zn2+ in the real solution behaves exactly as it would in a solution where those two conditions were met.
2.2.3
Systems in Which Environmental Conditions Differ from the Reference State; The Activity Coefficient
The ratio of the reactivity per molecule or per mole of A in a real system to the reactivity per molecule or per mole in the reference (or standard) state is called the activity coefficient and is designated gA . Thus 43
gA =
reactivity per molecule (or per mole) of A in the real system reactivity per molecule (or per mole) of A in the reference state
⌘ reactivity of A in the real system concentration of A in the real system ⌘ =⇣ reactivity of A in the reference state concentration of A in the reference state
⇣
(2.1b)
⇣
(2.1c)
⌘ reactivity of A in the real system reactivity of A in the reference state ⌘ =⇣ concentration of A in the real system concentration of A in the reference state gA =
(2.1a)
aA cA /cA
(2.1d)
Equation (2.1d) indicates that the activity coefficient is dimensionless, being the ratio of two dimensionless terms. Consider again a hypothetical solution in which the activity of dissolved Zn2+ is 10 3 , and assume that the concentration of Zn2+ in the system is 1.5 ⇥ 10 3 mol/L. If each molecule of Zn2+ in the solution behaved as it would in its reference state, the activity of Zn2+ would be 1.5 ⇥ 10 3 . The data therefore suggest that, on average, each molecule of Zn2+ in the real solution has a reactivity only two-thirds as great as that of a Zn2+ molecule in the reference state, i.e.: g Zn2+ =
aZn2+ 10 3 = = 0.667 cZn2+ /cZn2+ 1.5x10 3 M/1.0 M
Rearrangement of this equation yields the following classic expression for the activity of any species A: aA = gA
cA cA
(2.2)
Based on Equation (2.2), the reactivity of A is identical in all of the following solutions. cA = 0.16 mol/L cA = 0.40 mol/L cA = 0.12 mol/L cA = 1.00 mol/L
gA = 1.00 gA = 0.40 gA = 1.33 gA = 0.16
aA = 0.16 aA = 0.16 aA = 0.16 aA = 0.16
The wide range of activity coefficients of these hypothetical solutions suggests that the solutions have very different compositions and/or environmental conditions. Nevertheless, the equality of aA in the solutions indicates that A is equally reactive in 44
all of them. Thus, the activity provides a consistent way to compare the behavior of a given species in disparate systems. Furthermore, based on the first set of conditions, the reactivity of A in all the solutions is equivalent to that in a solution containing 0.16 mol/L in the reference environment. For this reason, the activity of A is sometimes characterized as the “equivalent concentration” of A in the reference state.
Example 2.1 The concentration and activity of glucose (C6 H12 O6 , MW 180) dissolved in an aqueous solution are determined to be 3.0 g/L and 0.015, respectively. What is the activity coefficient of glucose in the solution, and how can this value be interpreted in terms of the reactivity of glucose molecules compared to that in an infinitely dilute solution? Solution The molecular weight of glucose is given as 180, so its molar concentration in the solution is 3.0/180, or 0.0167 M. Also, by definition, the glucose concentration in the standard state is 1.0 M. Substituting these values and the known activity of glucose into Equation (2.1d), we obtain g C6 H12 O6 =
aC6 H12 O6 cC6 H12 O6 /cC H 6
=
12 O6
0.015 = 0.90 0.0167 M/1.0 M
The result can be interpreted to mean that each molecule of glucose in the real solution is only 90 percent as reactive as it would be in an infinitely dilute solution.
Since the concentration of any species A in the standard state always has a numerical value of 1.0 (1.0 mol/L for dissolved species, 1.0 atm for gases, and a mole fraction of 1.0 for solids and liquids), Equation (2.2) is frequently simplified as follows: cA aA = g A = g A cA (2.3) 1.0 Although activities are frequently represented as in Equation (2.3), this representation can lead to some confusion about dimensions. As has been noted, activities and activity coefficients are both dimensionless quantities. However, when applied in conjunction with a dimensionless value for gA , Equation (2.3) gives the impression that aqueous phase activities have dimensions of concentration. In reality, dimensions only appear to apply to the activity in Equation (2.3) because the fact that the dimensions of cA have been canceled (by normalizing the concentration in the real system to the concentration in the standard state) is hidden. For this reason, Equation (2.2) is the preferred way of expressing the a-c-g relationship. 45
Given that the activity coefficient describes the reactivity of an average molecule in the actual system compared to the reactivity of such a molecule in the reference state, the activity coefficient is bound to be close to 1.0 if the actual chemical environment is quite similar to the reference state. As a result, activity coefficients for most solutes are near 1.0 in solutions that contain very low TDS concentrations (i.e., solutions that are very dilute). Correspondingly, to the extent that the chemical environment in the real system differs from the reference condition, the activity coefficient will differ from a value of 1.0. A species that behaves in the real system exactly as it would in the reference state is said to behave ideally, and a solution in which all species behave as they would in their respective reference states is called an ideal solution. The use of the term ideal in such a circumstance and the idea that activity coefficients “correct” for the nonideality of real solutions are unfortunate. They can mislead one into thinking either that the activity coefficient is a fudge factor, or that activity coefficients far from 1.0 reflect extreme or highly unusual conditions. This is not at all the case. Ideality as used in this context simply means conformity to the arbitrary reference state. To make the point in an exaggerated way, one could choose the reference state for a dissolved substance as a solution containing 10 mol/L dissolved NaCl at the critical point of water. In that case, dilute solutions at normal temperatures and pressures would have activity coefficients very far from 1.0, reflecting the fact that the ions in those solutions behave very differently from ions in the “ideal” reference state.
2.2.4
Alternative Choices for the Standard State Conditions
The standard state conditions for solutes described in Table 2.1 (1.0 mol/L concentration, infinite dilution reference state) are the ones that are most commonly used for dilute aquatic systems, but not the only ones. For organic solutes that can exist as a separate, pure liquid (e.g., benzene), the pure compound in its liquid form is sometimes used to define both the concentration and the environment in the standard state. In other words, even though the system of interest might be a dilute solution of benzene in water, the behavior of the benzene molecules is described by comparison with their behavior in pure liquid benzene. When this is done, the concentration is usually calculated on a mole fraction rather than a molar scale, and the activity coefficients are typically much greater than 1.0. Example 2.2 Experimental evidence suggests that, at 25 C, benzene has the same tendency to enter the gas phase (i.e., it has the same reactivity) when it is dissolved in water at a concentration of 800 mg/L as when it is present as a pure organic liquid. Using the mole fraction concentration scale and a reference state of pure liquid benzene, de46
termine the benzene concentration and activity coefficient in a solution of 800 mg/L benzene in water. Solution If we choose pure liquid benzene as the standard state, the activity of benzene in that form is 1.0 by definition. Furthermore, because dissolved benzene in the aqueous solution of interest has the same tendency to react (as indicated by its tendency to enter the gas phase) as pure liquid benzene, the activity of the dissolved benzene must be 1.0 as well. To compute the mole fraction of benzene in the aqueous solution, we need to know the total molar concentration of all constituents in the aqueous solution. Since the solution is made up overwhelmingly of water molecules, we can assume that its total molar concentration is approximately the same as in pure water (55.6 mol/L), so x benz =
=
moles benzene per liter of solution moles of all species per liter of solution (800 mg/L) (1 mol/78,000 mg) = 1.85 ⇥ 10 55.6 mol/L
4
Knowing both the activity and the concentration of the dissolved benzene, we can find its activity coefficient as g benz =
abenz 1.0 = = 5420 x benz /x benz 1.85 ⇥ 10 4 /1.0
The large value of g benz in the preceding example implies that benzene molecules are far more reactive in water than in pure liquid benzene. This result can be understood qualitatively by recognizing that benzene molecules are nonpolar and are very “comfortable” when surrounded by other nonpolar molecules, as they are in the pure organic liquid (the selected reference state). On the other hand, when present as a dilute solute in water, the benzene molecules are surrounded by polar molecules of water, with which they have much less favorable interactions (i.e., they are very ‘uncomfortable’ in the aqueous environment, hence hydrophobic). This discomfort destabilizes the benzene molecules and increases their tendency to react. As noted previously, the large deviation of the benzene molecules from “ideal” behavior does not imply that they are behaving abnormally or that the solution is anomalous in any way; it simply reflects the very different chemical environment in the aqueous solution from that in the pure organic liquid that was chosen as the reference state. Yet another choice for the reference state of solutes, frequently used by marine chemists, is infinite dilution of the solute in a solution that contains the major species that are present in typical ocean water. According to this convention, for instance, the reference state for Zn2+ in seawater would be a solution in which Zn2+ is infinitely 47
– dilute, but Na+ , Ca2+ , Mg2+ , Cl– , SO2– 4 , and HCO3 are present at their normal concentrations in seawater. In this case, infinite dilution of Zn2+ implies that zinc ions interact with H2 O and the six other ions listed, but not with other Zn2+ ions or other minor ions that might be present in the system. Note that there is no need to define the standard state identically for all solutes in a system; all that is needed is for the standard state of each species to be defined unambiguously. For example, it is perfectly acceptable to use pure liquid benzene as the reference state for dissolved benzene and infinite dilution as the reference state for Zn2+ in the same solution.
Example 2.3 The activity coefficient of strontium ions (Sr2+ ) in seawater is approximately 0.26 when based on a reference state of infinite dilution in pure water, and approximately 1.0 when based on a reference state of infinite dilution in major ion seawater. (a) What would gSr2+ be if Sr2+ were present at nearly infinite dilution in pure water, and if the reference state were defined as major ion seawater? (b) Is Sr2+ an ideal solute in seawater? Solution (a) To begin, we designate the activity coefficient of Sr2+ as g yx , where x identifies the solution in which the Sr2+ is dissolved, and y identifies the solution chosen as the reference state. Then, using ‘fw’ and ‘sw’ to designate infinitely dilute fresh water and major ion seawater, respectively, the question can be reworded as follows: Given sw that g fw sw is 0.26, what is the value of g fw ? 2+ The fact that g fw ions in seawater are only 26% as resw = 0.26 means that Sr active as the same ions would be in infinitely dilute fresh water. The inverse of that statement is that Sr2+ ions in infinitely dilute fresh water are 1/0.26, or 3.85 times as reactive as in major ion seawater. This latter value, i.e., the ratio of the activity of a Sr2+ ion in infinitely dilute fresh water to that in major ion seawater, is, by definition, sw g sw fw ; i.e., it is the term we are seeking. Thus, g fw in an infinitely dilute solution is 3.85. The same argument can be made somewhat more formally by applying Equation (2.1a) to the various solutions of interest. Designating the reactivity per mole of Sr2+ by R, we can apply Equation (2.1a) as follows: g fw sw =
Rsw,real reactivity per mole of Sr2+ in sw = = 0.26 2+ Rfw,ref reactivity per mole of Sr in the fw reference state
g sw sw =
reactivity per mole of Sr2+ in sw Rsw,real = = 1.0 2+ Rsw,ref reactivity per mole of Sr in the sw reference state 48
In addition, because the real, fresh water system is infinitely dilute, we know that g fw fw =
reactivity per mole of Sr2+ in fw Rfw,real = = 1.0 2+ Rfw,ref reactivity per mole of Sr in the fw reference state
We can then manipulate the preceding equations to obtain an expression for the activity coefficient for Sr2+ in the fresh water based on major ion seawater as the reference state: reactivity per mole of Sr2+ in fw reactivity per mole of Sr2+ in the sw reference state ✓ ◆✓ ◆ Rsw,real Rfw,real Rfw,real (1.0) (1.0) Rsw,ref Rfw,ref g sw = = = 3.85 fw = 0.26 Rsw,ref Rsw,real Rfw,ref (b) An ideal solute is one that behaves in a real solution exactly as it would in the reference solution. Thus, in seawater, Sr2+ behaves as an ideal solute if the reference state is major ion seawater, but it is quite nonideal if the reference state is pure, fresh water. g sw fw =
To summarize, the chemical activity of a substance describes its tendency to undergo chemical change, taking into account factors related to the substance’s concentration and its physical and chemical environment. Use of the term activity to describe this synthesis is apt; the value of the activity indicates how active the species is in the system, in terms of its tendency to participate in processes such as chemical reactions, diffusion away from a location, or transfer into a different phase. Changing either the concentration of the species or its chemical environment can alter its activity. While the various standard and reference states used to quantify activity can be daunting at first, these choices are in fact nothing more than a set of simplifying conventions. By adopting them, we can describe the reactivity of any species in a real system simply by stating its activity, in lieu of stating all the details about the system composition and environmental conditions.
2.3
PREDICTING ACTIVITY COEFFICIENTS FROM KNOWLEDGE OF THE SOLUTION COMPOSITION
Under some circumstances, activity coefficients of dissolved species can be estimated from theory. The ionic strength, which was introduced in Chapter 1 [Equation (1.17)], turns out to be a key parameter in this calculation. The basis for the relationship between ionic strength and activity coefficients is that dissolved ions attract 49
oppositely charged ions and repel like-charged ions. Therefore, each cation in a solution is surrounded by a diffuse cloud of other molecules that include anions at a somewhat higher concentration than in the bulk solution, cations at a somewhat lower concentration than in the bulk solution, and water molecules that are mostly oriented to maximize favorable electrostatic interactions. This situation is illustrated schematically in Figure 2.1. An analogous but inverse description applies to the environment surrounding dissolved anions. As noted earlier, the kinetic energy of the molecules prevents them from remaining in place, so ions and water molecules are constantly entering and exiting one another’s sphere of influence.
Figure 2.1 The arrangement of water molecules and ions around a dissolved ion. Although most of the molecules around a given ion are water, oppositely charged ions are selectively attracted to that environment. (From Silberberg, M.S. Chemistry: The Molecular Nature of Matter and Change. Copyright c 2000 The McGraw-Hill Companies, New York, NY. Reproduced by permission of The McGraw–Hill Companies.)
The activity coefficient characterizes the difference between the chemical environment in the system of interest and that in the reference state. Thus, if the reference state for an ion is infinite dilution in pure water, the activity coefficient of the ion reflects the fact that in the real system, the species surrounding the ion include other solutes as well as water molecules. Most of the ions in this region are oppositely charged to the ion of interest and therefore have a greater tendency to remain close to the central ion than do water molecules. These relatively tightly held, oppositely charged ions shield the central ion from interactions with other dissolved species and reduce its activity coefficient. The more highly charged the central ion and the other ions in solution, the greater the attraction and the shielding effect, and the lower the activity coefficient. Similarly, for a given central ion, the higher the concentration of other ions in solution, the greater the shielding and the lower the activity coefficient. 50
In 1923, Debye and H¨uckel independently derived a mathematical model to predict the value of the activity coefficient of symmetric electrolytes (salts in which the magnitude of the charge is the same on the cation as on the anion) in aqueous solutions, for a reference state of infinite dilution in pure water. The derivation can be developed fairly simply using principles of physical chemistry.4 The resulting ¨ equation, known as the Debye–Huckel limiting law, proved to be quite accurate for solutions with low ionic strength. In the Debye–H¨uckel derivation, ions are treated as point charges that can approach infinitely close to one another. Eliminating this assumption and taking ionic ¨ size into account yields another equation known as the extended Debye–Huckel equation, which is useful for predicting ion activity coefficients in significantly more concentrated solutions. The extended Debye–H¨uckel equation is strictly valid only if the ion of interest and the shielding ions are all of the same ionic size, but the equation is commonly used even if that restriction is not satisfied. Davies proposed adding a term to the extended Debye–H¨uckel equation to improve the fit between the equation and experimental observations. Later, Pitzer and coworkers proposed incorporating additional terms to account for interactions among specific pairs of ions, thereby effectively incorporating information about both the size and properties of individual ions into the analysis. Still more terms intended to account for three-way interactions among ions and for other factors have been proposed as well. The resulting equations have been used successfully to estimate activity coefficients in seawater and even more concentrated ionic solutions.5 The equations discussed in the preceding paragraphs are collected in Table 2.2a, and the size parameters for a few ions for use in the extended Debye-H¨uckel equation are given in Table 2.2b. Predicted values of activity coefficients of a few ions are shown as a function of ionic strength in Figure 2.2. Note that, in all the equations, the activity coefficient of an ion depends strongly on that ion’s charge, and the variable that characterizes the ionic composition of the solution is the ionic strength. One interesting phenomenon that has been observed experimentally and that can be modeled with some success by the specific interaction model is that the activity coefficients of cations increase dramatically with increasing ionic strength when the ionic strength exceeds approximately 1 mol/L. Part of the explanation for this phenomenon is the substantial reduction in the concentration of free water molecules in solution, because so many of those molecules are held in the hydration sphere surrounding the ions, but other factors might also be involved. The activity coefficient of OH– follows a trend similar to that of cations, but for most anions, g either declines steadily or increases only slightly as ionic strength increases above 1 mol/L. 4 See, for example, Morel, F.M.M. and Hering J.G. Principles and Applications of Aquatic Chemistry, Wiley-Interscience, New York (1993) or Bockris, J.O’M. and Reddy, A.K.N. Modern Electrochemistry, vol. 1, Plenum Press, New York (1998). 5 A comparison of all these models for estimating g is available in Langmuir, D. Aqueous Environmental Geochemistry, Prentice-Hall, Upper Saddle River, NJ (1997).
51
Table 2.2a Various equations for predicting ion activity coefficients in aqueous solutions, based on a reference state of infinite dilution in pure water
Name and Equation Debye-H¨uckel Limiting Law
Notes and Approximate Range of Applicability
log10 (g D-H ) = Az2 I 1/2
(2.4)
A = 1.82 ⇥ 106 (eT ) 3/2 , where e is the dielectric constant of the medium; for water at 25 C, A = 0.51; z = ionic charge; applicable at I < 0.005 M.
(2.5)
a = ion size parameter (see Table 2.2b), B = 50.3(eT ) 1/2 ; for water at 25 C, B = 0.33. Appropriate in solutions where one salt dominates ionic strength; applicable at I < 0.1 M.
(2.6)
Applicable at I < 0.5 M.
(2.7)
Bi j = specific interaction term between ions i and j; m j = molality (mol/kg solution) of j; applicable at I < 1 M; additional terms can extend range to higher ionic strengths.(a)
Extended Debye-H¨uckel Equation I 1/2 log10 (g Ext.D-H ) = Az2 1 + BaI 1/2 52
Davies Equation log10 (g Davies ) = Az2
(
I 1/2 1 + I 1/2
0.3 I
)
Specific Interaction Model (Pitzer Equation) log10 (g Pitzer ) = log10 (g Ext.D-H ) + Â Bi j I m j j
(a) See Pitzer, K.S. (1983) “Thermodynamics of electrolytes. V. e↵ects of higher-order electrostatic terms.” J. Solution Chem. 4, 249-265, and Millero, F. (1983) “The estimation of the pK⇤HA of acids in seawater using the Pitzer equations .”Geochim. Cosmochim. Acta 47, 2121-2129.
˚ Table 2.2b Values of ion size parameter (in Angstroms) for use in the extended Debye-H¨uckel equation Cations a = 3 : Ag+ , K+ , NH+4 a = 4 : Na+ a = 9 : H+
Anions a = 3 : Cl– , ClO–4 , HS– , I– , NO–3 , OH– a = 4 : CH3 COO– , HCO–3
Doubly charged ions
a = 5 : Ba2+ , Pb2+ , Sr2+ a = 6 : Ca2+ , Cu2+ , Fe2+ , Mn2+ a = 8 : Be2+ , Mg2+
2– a = 4 : HPO2– 4 , SO4 2– a = 5 : CO3
Triply charged ions
a = 9 : Al3+ , Ce3+ , Fe3+ , La3+
a = 4 : PO3– 4
Singly charged ions
A consequence of the assumption that ionic interactions are the only factor that affects activity coefficients is that uncharged species are always predicted to behave ideally when the Debye–H¨uckel or extended Debye–H¨uckel equation is used; i.e., they have activity coefficients of 1.0. Empirically, activity coefficients of neutral species do depend on ionic strength, but the dependence is quite weak (gneut is usually in the range 1.00 ± 0.05 for neutral solutes in natural fresh waters). This dependence is commonly modeled as log gneut = kneut I (2.8) where kneut is positive for most species, but negative for a few. The error associated with assuming gneut = 1.0 is smaller than that associated with other uncertainties in analyzing most systems so, unless otherwise indicated, the assumption that neutral solutes behave ideally is made throughout this text.
The evaporation of water increases the concentration of salt ions in the Dead Sea to the point where the water cannot hold all the ions in solution. In response, ‘salt pillars’ form in the water. (Photo: Richard Nowitz.) 53
Figure 2.2 (a) Activity coefficients of various ions according to the extended Debye–H¨uckel law, based on the infinite dilution reference state. (b) The activity coefficient of Ca2+ in a solution prepared by dissolution of CaCl2 , according to three models. The specific ion interaction (SIT) model (of Pitzer) is the most complex of the three shown, and it fits the experimental data best. However, at low ionic strengths, the other two equations yield values close to those predicted by the SIT. (From Langmuir, D. Aqueous Environmental Chemistry, c 1997, p.135. Reprinted by permission of Pearson Education, Inc., Saddle River, NJ.)
54
Example 2.4 A natural fresh water is modeled as a solution of 10 3 M NaCl, 10 3 M NaHCO3 , and 10 4 M CaCl2 . Assume that when these salts dissolve, they dissociate to Na+ , Cl– , Ca2+ , and HCO–3 and then do not react further. (a) Compute the ionic strength of the solution. (b) What is the activity of each dissolved ion if the infinite dilution reference state is used, and if infinite dilution is in fact a reasonable approximation for the solute concentrations? What are the predicted values of the activity coefficients according to the extended Debye–H¨uckel equation? (c) Calculate the activities of all the ions if the NaCl concentration is increased by two orders of magnitude to 0.1 M to model an estuarine water. Because the ionic strength is higher after the NaCl is added, use the Davies equation to calculate activity coefficients. Solution (a) After dissociation, the concentrations of the various ions and their contributions to the ionic strength are as follows: Ion Na+ Cl– Ca2+ HCO–3
Conc’n (M) 2.0 ⇥ 10 3 1.2 ⇥ 10 3 1.0 ⇥ 10 4 1.0 ⇥ 10 3
0.5 c i z 2i 1.0 ⇥ 10 6.0 ⇥ 10 2.0 ⇥ 10 5.0 ⇥ 10 2.3 ⇥ 10
Sum
The ionic strength of the solution is thus 2.3 ⇥ 10
3
3 4 4 4 3
M.
(b) By definition, if the reference state is infinite dilution and the infinite dilution approximation applies, then g = 1.0. In that case, the activity of each ion equals its molar concentration; that is, {Na+ } = 2.0 ⇥ 10 3 , {Cl– }= 1.2 ⇥ 10 3 , {Ca2+ }= 1.0 ⇥ 10 4 , {HCO–3 }= 1.0 ⇥ 10 3 . The activity coefficients computed according to the extended Debye–H¨uckel equation and the corresponding ion activities computed as gi ci /ci are summarized in the following table. Ion Na+ Cl– Ca2+ HCO–3
log g 0.0230 0.0230 0.0920 0.0230
Size Parameter a 4 3 6 4 55
g 0.948 0.948 0.809 0.948
Activity 1.90 ⇥ 10 1.14 ⇥ 10 8.09 ⇥ 10 9.48 ⇥ 10
3 3 5 4
(c) The ionic strength of the model estuarine water can be computed as in part (a), except that the new Na+ and Cl– concentrations are 0.102 and 0.1012 M, respectively, yielding a new ionic strength of 0.1022 M. Substituting this value into the Davies equation yields the values shown in the following table for the activity coefficients and activities of the ions. Note that the activities of Ca2+ and HCO–3 decrease by 55% and 18%, respectively, when the NaCl concentration is increased, even though the concentrations of these two ions are not altered. Ion Na+ Cl– Ca2+ HCO–3
2.4
log g 0.108 0.108 0.432 0.108
g 0.780 0.780 0.370 0.780
Activity 7.96 ⇥ 10 7.89 ⇥ 10 3.70 ⇥ 10 7.80 ⇥ 10
2 2 5 4
THE ACTIVITY AS AN INTENSIVE PROPERTY; THE ACTIVITY OF SOLIDS DISPERSED IN WATER
Simply increasing the amount of a given phase present without changing its composition, temperature, or pressure does not change the activity of any chemical in that phase. Thus, the activity of Ca2+ in a 10 2 M solution of CaCl2 is the same regardless of whether there is 1 L or 1000 L of solution in the system. Changing the concentration of CaCl2 would change the activity of dissolved Ca2+ , and adding a different chemical to solution might do so as well, but just changing the amount of solution present would not. The same principle applies if the phase of interest is a solid or gas. If, for example, a pure solid is present in a system at 1.0 atm total pressure and 25 C, the activity of the solid is 1.0 no matter how much of it is present. In thermodynamic jargon, properties that characterize a phase and that are independent of how much of the phase is present are called intensive properties. The constancy of the activity of pure solids, regardless of the amount of solid present, can become confusing when one is dealing with systems in which a solid is dispersed in a solution. For instance, copper might be removed from an industrial wastewater by precipitation of solid copper hydroxide, Cu(OH)2 (s). Assuming the solid that formed was pure Cu(OH)2 (s), the mole fraction of Cu(OH)2 (s) in the solid would be 1.0. Therefore, if T = 25 C and P = 1.0 atm, the activity of Cu(OH)2 (s) would be 1.0, regardless of how much solid was present. We may wish to quantify how much solid Cu(OH)2 (s) is suspended per liter of water and refer to that value as the concentration of Cu(OH)2 (s) in the water. How56
ever, such a concentration refers to the amount of one phase dispersed in another and is fundamentally different from the concentration of a substance dissolved in another. In the former case, we are just talking about two bulk substances sharing a region of space, whereas in the latter we are describing chemical mixing on the molecular level in a single phase, with all molecules of the solute able, in theory, to interact with all molecules of the solvent. The use of the same term—concentration—to refer to both dispersed and dissolved substances often confuses students and leads to the error of treating the activity of the solid as being proportional to its concentration in the solid/solution mixture. Be careful to avoid this error! Example 2.5 Hydrated aluminum sulfate, or alum [chemical formula Al2 (SO4 )3 · 14 H2 O, MW 594], is commonly added to water during treatment operations. When this salt dissolves, essentially all of the Al3+ released to solution precipitates as the solid Al(OH)3 (s), often collecting other small particles and removing other contaminants from solution as it does so. The Al(OH)3 (s) and associated contaminants are then removed from the purified water, typically by settling and/or filtration. If a dose of 36 mg/L alum is added to water at a drinking water treatment plant, what are the concentration and activity of the Al(OH)3 (s) that forms? Solution The atomic weight of Al is 27, and each mole of alum contains two moles of Al, so the amount of Al3+ dosed to the water is: ◆✓ ◆✓ ◆✓ ◆ ✓ mg alum 1 mmol alum 2 mmol Al 27 mg Al mg Al 36 = 3.27 L 594 mg alum mmol alum mmol Al L The molecular weight of Al(OH)3 (s) is 78. Assuming that all of the added Al precipitates as this solid, the concentration of solid formed is: ✓ ◆✓ ◆ 78 mg Al(OH)3 (s) mg Al(OH)3 (s) mg Al 3.27 = 9.45 L 27 mg Al L The solid is dispersed throughout the solution, but it is a separate phase, not dissolved in the water. Since it is a pure solid, its activity is 1.0, independent of its concentration.
2.5
MODELS OF CHEMICAL EQUILIBRIUM
Having established both a qualitative understanding of the reactivity of individual species and a way to quantify that property, we next turn our attention to the interactions and relationships among species. In particular, we develop tools to predict 57
whether a potential reaction among various species will actually take place, the direction that any such reaction will proceed, and how much reaction will occur before equilibrium is achieved and the chemical changes cease. Two conceptual models are widely employed to describe chemical equilibrium. In one model, equilibrium is envisioned as a dynamic steady state in which the reaction is proceeding at equal rates in the forward and reverse directions. As a result, no net destruction of reactants or formation of products occurs, and on a macroscopic level, the system behaves as though the molecules are not reacting at all. The second model focuses more on the concept of chemical energy and can be explained by analogy with a rock perched on a hillside. In the absence of other forces, such a system has a natural tendency to change until it reaches its condition of minimum gravitational potential energy, which is achieved when the rock comes to rest at the lowest available elevation. Similarly, chemicals can be thought of as having a chemical potential energy and as being thermodynamically driven to undergo changes (i.e., to react) that cause the total chemical potential energy of the system to be minimized; once the minimum energy condition is attained, the system is at equilibrium, and no further (macroscopic) change occurs. Two important points that are basic to the study of equilibrium chemistry fit well with the rock-on-a-hill analogy. The first is that the equilibrium condition is independent of previous history and of the pathway between the initial and final states. Thus, the equilibrium location of the rock is at the lowest elevation where it can reside, regardless of where it has been previously or what route it follows to reach the lower elevation. Similarly, a chemical system with a given initial composition can be expected to form the same products (at equilibrium) regardless of how the reactants were formed, the order in which they were added to the system, how they are mixed, etc. Whereas the preceding point allows us to predict the equilibrium condition based strictly on knowledge of the initial conditions, a second truism limits the applicability of these predictions: Even though we might be able to predict the ultimate equilibrium conditions with confidence, we cannot predict how long the system will take to attain equilibrium. As is the case with rocks, chemicals can sometimes remain in nonequilibrium states for geologic times. Indeed, all plants and animals are thermodynamically unstable in an atmosphere that contains oxygen; if they reached chemical equilibrium, they (we!) would all be oxidized to carbon dioxide and water. Fortunately, under normal environmental conditions, that reaction proceeds at a negligible rate. How closely a particular reaction or system approaches chemical equilibrium depends on the relative rates of chemical processes (i.e., chemical reactions) and physical processes (bulk flow, mixing, diffusion, etc.) in the system. Consider, for example, an initial mixture of ice at 0 C, water at 20 C, and overlying air at 25 C, and assume that, if the system reached equilibrium, all the ice would melt and the 58
water and air would stabilize at 10 C. Whether the system actually reaches equilibrium depends on the rates of heat transfer and the amount of time that the various components are in contact. In the limit, if the three phases are in contact for a very short time and mixing is minimal, almost no ice melts, the water and air temperatures change negligibly, and the system remains far from equilibrium. On the other hand, after long periods of contact, the system reaches equilibrium. In addition to contact time, the mixing intensity in the system and the physical form of the constituents (whether the ice is present as a solid block or thin shavings, whether the water is present only as a bulk fluid or also as a mist) affect the rate of approach to equilibrium. Similarly, if a chemical reaction in an aquatic system is rapid compared to the time that the reactants spend in the system, the reaction will reach equilibrium by the time the water exits the system. On the other hand, if the reaction is slow compared to the time available, the materials exiting the system will not have progressed significantly toward equilibrium and will be minimally modified from their input conditions. Because this text focuses almost exclusively on the determination of the equilibrium condition in aquatic systems, the calculations described are most directly applicable to systems in which the reactions are rapid compared to the time available. However, even in systems that do not attain equilibrium, the calculations are often useful as indicators of the ultimate equilibrium condition toward which the system tends. The two views of equilibrium described above – the dynamic molecular kinetic model and the more static thermodynamic model – are equally valid. Neither is better than the other in an absolute sense, and either or both models can be used in a given situation, depending on which approach is more useful. If the available data are accurate, both models must predict the same final results, and these results must be experimentally valid. The kinetics-based model of chemical reactions and the relationship of kinetics to chemical equilibrium is presented in Chapter 3, and chemical equilibrium is analyzed from an energy-based (i.e., thermodynamic) perspective in Chapter 4. While of those chapters can provide important insights into the nature of chemical reactions and equilibrium, it is possible to use the key outcomes of those chapters without a full understanding of how those outcomes are derived. Both to preview the major results presented in those chapters for readers who intend to study them, and to allow other readers to gain the maximum benefit from subsequent chapters without studying Chapters 3 and 4, the remainder of this chapter introduces the essential ideas and equations needed to predict speciation in equilibrated systems. By far the most important parameter used to test for and characterize chemical equilibrium is the equilibrium constant, K. This constant is defined as the ratio, at equilibrium, of the activities of the reaction products to those of the reactants, with each term raised to a power equal to its stoichiometric coefficient in the reaction. That is, for any reaction aA + bB )* cC + dD, 59
{C}c {D}d {A}a {B}b
eq
⌘K
(2.9)
Equation (2.9) indicates that, whenever the reaction aA + bB )* cC + dD is at equilibrium, the ratio {C}c {D}d /{A}a {B}b equals K; if the actual value of this ratio is not K, the reaction is not at equilibrium, and it will proceed in either the forward or reverse direction. Note that it makes no sense to speak of a single species being in or out of equilibrium; the concept of equilibrium applies only to a complete reaction. Because K is a ratio of activities and activities are always dimensionless, K is dimensionless as well. In some situations, it is useful to compute a ratio corresponding to K, but to use concentrations rather than activities in the calculation. This practice is most common when the reaction involves transfer of a molecule between solution and either a gas phase or the surface of a solid. Although this concentration ratio is sometimes referred to as an equilibrium constant, that designation is not really accurate: equilibrium constants must be ratios of activities and must be dimensionless. Concentration ratios that characterize the equilibrium state of a system are more correctly referred to as distribution coefficients or partition coefficients. Example 2.6 The reaction shown below is extremely important in water and waste-water treatment because HOCl (hypochlorous acid) is the dominant disinfecting agent when water is chlorinated; hypochlorite ion, OCl– , is a much weaker disinfectant. HOCl + H2 O )* H3 O+ + OCl
(2.10)
To maximize disinfection, it is desirable to adjust solution conditions so that most of the OCl in the system is present as HOCl rather than OCl– . The concentration of H3 O+ (hydronium) ions is a fundamental property of any aqueous solution, since H3 O+ (and hydroxide, OH– ) ions can form by the dissociation of water (2 H2 O )* H3 O+ + OH– ).6 The quantity log{H3 O+ } is designated the pH of the solution and is the most widely used indicator of the acid/base balance in solution. The equilibrium constant for Reaction (2.10) at 25 C is 2.9 ⇥ 10 8 . Compute the equilibrium ratio of HOCl activity to OCl– activity in solutions at pH 6, 7, and 8. Assume that all the solutes behave ideally (g i = 1.0 in all cases) and that the water can be approximated as being in its standard state (pure water), so that its activity is 1.0. 6 Most readers will have previously seen the products of the water dissociation reaction written as H+ and OH– , rather than as H3 O+ and OH– . As explained in Chapter 5, the symbol H+ is really a shorthand way of representing a group of chemicals, of which H3 O+ is the major one. For the purposes of this example, it is useful to represent this group of chemicals as H3 O+ .
60
Solution From the equilibrium constant expression we can write {H3 O+ } OCl {H2 O} {HOCl} {HOCl} OCl
eq
= K = 2.9 ⇥ 10
{H3 O+ } 2.9 ⇥ 10 8 {H2 O}
= eq
8
eq
Assigning a value of 1.0 for the activity of water, {H2 O}, the preceding relationship indicates that, at equilibrium, the ratio {HOCl}/{OCl– } depends only on {H3 O+ }. For instance, in an equilibrium solution at pH 7.0 (i.e., one with {H3 O+ }= 10 7 ): {HOCl} 10 7 = = 3.45 OCl (2.9 x 10 8 )(1.0) Thus, in such a solution, the activity of HOCl is 3.45 times that of OCl– . If g HOCl = g OCl = 1.0, the total concentration of OCl (T OT OCl) is apportioned in the same ratio (i.e., 3.45 HOCl molecules for every one OCl– molecule), so a fraction equal to 3.45/4.45, or 78%, of T OT OCl is present in the form that is more effective as a disinfectant. By adding a little acid or base to the solution we can change the pH to 6 or 8 (i.e., we can increase {H3 O+ } to 10 6 or decrease it to 10 8 ), respectively. T OT OCl would then be apportioned as follows: Case I:
{H3 O+ } = 10
6
(pH = 6.0)
{HOCl} 10 = OCl 2.9 ⇥ 10
6 8
[HOCl] ⇤ = 34.5 = ⇥ OCl (1.0)
[HOCl] 34.5 = = 97% present as HOCl T OT OCl 35.5 Case II:
{H3 O+ } = 10
8
(pH = 8.0)
{HOCl} 10 = OCl 2.9 ⇥ 10
8 8
[HOCl] ⇤ = 0.345 = ⇥ OCl (1.0)
[HOCl] 0.345 = = 26% present as HOCl T OT OCl 1.345 Based on the distribution of T OT OCl between HOCl and OCl– and the knowledge that HOCl is the more effective disinfectant, we conclude that, for a given value of T OT OCl in solution, disinfection will be much more effective at pH = 6.0 than at pH = 8.0, and this is in fact observed. 61
Liquid chlorine to be used for disinfection is often delivered to water treatment plants in pressurized tanks (left). It is then dissolved in a sidestream and added to water flowing through long, narrow tanks like the one on the right. This flow pattern minimizes the dilution of the incoming water by water that has been in the tank longer (and from which the disinfectant has been depleted). The speciation of the Cl is critical. One mg/L Cl in the form of HOCl is toxic to most microorganisms. However, the oceans can support rich microbial life even though they contain 17,000 mg/L of the nontoxic ion Cl– . (Left: Russ Esser; Right: M.K. Stenstrom.)
Example 2.7 One gram of solid silver chloride, AgCl(s), is added to 1.0 L of major-ion seawater containing 0.48 M Cl– and having an ionic strength of 0.56 M. (These values are slightly lower than those that would be computed using the seawater data in Table 1.1, for reasons that are explained in subsequent chapters.) The solid begins to dissolve according to the following reaction, which has an equilibrium constant of K = 10 9.75 : AgCl(s) )* Ag+ + Cl Estimate the activity of Ag+ and the mass of AgCl(s) that has dissolved per liter of solution when the reaction reaches equilibrium. Assume that the Ag+ and Cl– ions that dissolve do not participate in any other reactions. Use the Davies equation to estimate activity coefficients of the solutes (even though the ionic strength is slightly higher than the range in which the Davies equation is normally used). Solution The molecular weight of AgCl(s) is 143, so the initial molar concentration of the solid in the suspension is (l g/L)/(143 g/mol) = 7.0 ⇥ 10 3 M. Given that the initial 62
ionic strength of the solution is 0.56 M, and that even if all the AgCl(s) dissolved, the Ag+ and Cl– ions that entered solution would increase the ionic strength by only 0.007 M, we can assume the final ionic strength equals the initial value. Using the Davies equation [Equation (2.6)] in conjunction with the given ionic strength, we find the activity coefficients of both Ag+ and Cl– to be
2
log10 g Ag+ = log10 g Cl = Az =
2
(0.51) (1)
g Ag+ = g Cl = 10
(
(
I 1/2 1 + I 1/2
0.561/2 1 + 0.561/2
0.133
0.3 I
)
)
0.3 (0.56)
= 0.133
= 0.74
Designating the concentration of AgCl(s) that dissolves as x (mol/L), the molar concentration of dissolved Ag+ once equilibrium is attained will be x, that of Cl– will be 0.48 + x, and that of AgCl(s) will be 0.007 x. The activities of Ag+ and Cl– at that time will depend on their concentrations and activity coefficients (the activity of each ion will be given by g i [i]/[i] ), but the activity of the solid will be 1.0 regardless of how much solid is present. Furthermore, the standard state concentration for both solutes (i.e., [i] ) is 1.0 mol/L. We can therefore substitute values into the equilibrium constant expression for the dissolution reaction, as follows: ⇥ +⇤ ! ⇥ ⇤! Ag Cl ⇤ g Ag+ ⇥ + ⇤ g Cl ⇥ Ag Cl Ag+ Cl K= = {AgCl(s)} {AgCl(s)} ✓ ◆✓ ◆ x mol/L [0.48 + x] mol/L 0.74 0.74 1.0 mol/L 1.0 mol/L 9.75 10 = 1.0 = (0.74)2 (x) (0.48 + x) The preceding expression can be expanded and solved using the quadratic equation. However, noting that the initial system contains 0.48 M Cl– and only 0.007 M AgCl(s), and that only a portion of the solid dissolves, we can be certain that dissolution of the solid will have a negligible effect on the Cl– concentration (i.e., x ⌧ 0.48). Making that assumption, the expression simplifies to 10
9.75
⇡ (0.74)2 (x) (0.48) = 0.26x
x = 6.77 ⇥ 10
63
10
= 10
9.17
We conclude that an exceedingly small amount of the solid dissolves, and that the concentration of Ag+ at equilibrium will be 6.77 ⇥ 10 10 mol/L. The activity of Ag+ at equilibrium is g Ag+ [Ag+ ]/[Ag+ ] , or 5.01 ⇥ 10 10 . Recall that the numerical values of ai , g i , and ci for any species depend on the choice of the standard state conditions for that species. As a result, because the equilibrium constant is a ratio of chemical activities, its value depends on the choices of standard state conditions for all the species participating in the reaction. However, once the standard state conditions are chosen, the equilibrium constant for the reaction is fully established. For example, the equilibrium constant relating the activities of H+ , OH– , and H2 O is commonly designated Kw and is given by Kw =
{H+ } OH H2 O
(2.11)
If the infinite dilution reference state and the usual standard-state concentration (1.0 mol/L) are used for solutes, and the pure-liquid reference state is used for H2 O, the value of Kw at 25 C is 10 14.0 . This value applies in fresh water, salt water, or any other aqueous solution at 25 C, as long as the activities of H+ and OH– are determined using the specified reference states. If a different reference state (e.g., major ion seawater) were chosen for H+ and OH– , Equation (2.11) would still apply generically, but the numerical values of the solute activities and of Kw would be different. That is, the f orm of the equilibrium constant is independent of the standard states being used, but the numerical values for the terms in the equilibrium constant expression do depend on the conditions in those states. Unless otherwise indicated, tabulated equilibrium constants in this text and virtually all other compilations are based on the conventional standard state concentrations described previously and the reference state conditions summarized in Table 2.1. In contrast to the values of ai , g i , and ci , the actual concentrations of solutes in a given solution (i.e., ci values) are unambiguous, experimentally accessible quantities that do not depend on the reference state conditions. This fact allows us to determine the change in K for a given change in reference states, as demonstrated in the following example. Example 2.8 The activity coefficients of the major ions in seawater have been studied extensively. When evaluated using infinite dilution in fresh water as the reference state, values of g for H3 O+ , H2 CO3 , and HCO–3 in seawater are 0.59, 1.14, and 0.57, respectively. The equilibrium constant shown below is based on the infinite dilution reference state for the solutes. Compute the value of the equilibrium constant for the same 64
reaction using major ion seawater as the reference state. Assume that the mole fraction and activity coefficient of H2 O in both fresh water and seawater are ⇠1.0, so the activity of H2 O is ⇠1.0 in both solutions. H2 CO3 + H2 O )* H3 O+ + HCO–3
K1 = 10
6.35
Solution To determine the value of K1 when major ion seawater is used as the reference state, we first use the given information to determine the concentration ratio of the reaction products to the reactants in seawater. Using a superscript to indicate the reference state and a subscript to indicate the actual solution being characterized, and ‘fw’ and ‘sw’ to indicate fresh water (assumed to be infinitely dilute) and seawater, respectively, we can write K1fw = 10
6.35
=
fw {H3 O+ }fw x {HCO3 }x fw {H2 CO3 }fw x {H2 O}x
(2.12)
where the subscript x is used to indicate that the expression applies in any solution. For the current analysis, because we are given information about the activity coefficients in seawater, we will apply the expression to such a solution (i.e., in this case, x = sw). Inserting the known value of 1.0 for {H2 O}fw sw , and expanding the expressions for the activities of the solutes yields ⇥ ⇤ ! ⇥ ⇤ ! + H O HCO 3 3 sw sw g fw g fw ⇤ ⇤fw, H3 O+ ,sw ⇥ HCO3 ,sw ⇥ + fw, H O HCO 3 3 K1fw = ⇥ ⇤ ! H2 CO3 sw g fw ⇤fw, (1.0) H2 CO3 ,sw ⇥ H2 CO3 =
fw g fw H O+ ,sw g HCO 3
g fw H CO 2
3 ,sw
3
,sw
⇥ ⇤ ⇥ ⇤ H3 O+ sw HCO3 sw ⇥ ⇤ H2 CO3 sw
(2.13)
where the standard state concentrations have been eliminated because they all have values of 1.0 mol/L, and the solute concentrations are, by convention, expressed in mol/L units. Note that, consistent with the preceding discussion, it is necessary to specify both the reference state and the identity of the actual solution to quantify activity coefficients; by contrast, solute concentrations are fully established by specifying just the identity of the actual solution, and the standard state concentrations and equilibrium constant are fully established by specifying just the reference states. Values for the activity coefficients that appear in Equation (2.13) and for K1fw are given in the problem statement, so we can compute the solute concentration ratio in seawater as 65
⇥
⇤ ⇥ ⇤ g fw H3 O+ sw HCO3 sw H CO3 ,sw fw ⇥ ⇤ = K1 fw 2 fw H2 CO3 sw g H O+ ,sw g HCO 3
= 10
6.35
3
,sw
1.14 = 10 (0.59)(0.57)
5.82
(2.14)
If major ion seawater, rather than fresh water, is used to define the standard states, the equilibrium constant expression is sw {H3 O+ }sw x {HCO3 }x sw {H2 CO3 }sw x {H2 O}x
K1sw =
(2.15)
Once again letting x = sw, and making the same substitutions and rearrangements as we applied previously to the K1fw expression, we obtain
K1sw =
=
⇤ ! ⇥ ⇤ ! + H O HCO 3 sw sw ⇥ 3 ⇤sw, ⇥ ⇤ g sw g sw H3 O+ ,sw HCO3 ,sw HCO sw, H3 O+ 3 ⇥ ⇤ ! H CO 2 3 sw ⇤sw, (1.0) g sw H2 CO3 ,sw ⇥ H2 CO3 ⇥
sw g sw H O+ ,sw g HCO 3
3
g sw H CO
,sw
3 ,sw
2
⇥
⇤ ⇥ ⇤ H3 O+ sw HCO3 sw ⇥ ⇤ H2 CO3 sw
(2.16)
In this case, because the conditions in both the reference state and the actual solution correspond to those in seawater, the activity coefficients are all 1.0. The terms in the concentration ratio are the actual concentrations of the three species in seawater. These are real, measurable values, independent of the choice of standard states, so the concentration ratio in Equation (2.16) is the same as the ratio computed in Equation (2.14) and equals 10 5.82 . We therefore know all the values on the right side of Equation (2.16) and can compute K1sw as K1sw
=
sw g sw H O+ ,sw g HCO 3
g sw H CO 2
=
3
,sw
3 ,sw
(1.0) (1.0) 10 1.0
5.82
⇥
⇤ ⇥ ⇤ H3 O+ sw HCO3 sw ⇥ ⇤ H2 CO3 sw
= 10
5.82
The calculations and result are consistent with the point made earlier that a change in the choice of reference states changes the values of the solute activities and the equilibrium constant, but not the values of the solute concentrations. 66
The ratio of the activities of the products to those of the reactants (i.e., the ratio {C} {D}/{A} {B} for a reaction A + B )* C + D) is important not only for reactions that are at equilibrium, but also under nonequilibrium conditions. This ratio is called the activity quotient and is commonly represented by the symbol Q. Like K, Q is a ratio of dimensionless quantities (activities) and is therefore dimensionless. When a reaction is at equilibrium, the activity quotient equals the equilibrium constant; i.e., Q = K. If Q 6= K, the reaction is not at equilibrium and will proceed in a direction that causes Q to approach K. For example, if Q > K, products are converted to reactants. This conversion decreases the numerator and increases the denominator of Q, so that its value decreases. Similarly, if Q < K, reactants are converted to products, causing Q to increase. In either case, the reaction causes Q to approach K until, at equilibrium, they are equal. The ratio Q/K is thus an indicator of how far a reaction is from equilibrium. Because Q/K can vary over many orders of magnitude and equilibrium is reached when Q/K = 1, the extent of disequilibrium is conveniently quantified as log(Q/K), with a negative value implying that the forward reaction is favored, and a positive value implying that the reverse reaction is favored. These relationships are summarized in Table 2.3. Table 2.3 Relationships among Q, K, and the direction of chemical reactions
Q
log(Q/K) < 0
Not at equilibrium
Reactants ! Products
Q>K
log(Q/K) > 0
Not at equilibrium
Reactants
Q=K
log(Q/K) = 0
At equilibrium
No net reaction
Products
Example 2.9 At a particular instant, the activities of H3 O+ , HOCl, and OCl– in a solution are 10 7 , 10 3 , and 10 4 , respectively. Determine whether Reaction (2.10) is at equilibrium and, if not, whether the concentration of HOCl in solution will increase or decrease as the reaction proceeds. Solution At the given activities of H3 O+ , HOCl, and OCl– , the activity quotient is Q=
H3 O+ OCl 10 7.0 10 4.0 = = 10 {HOCl} {H2 O} 10 3.0 (1.0)
8.0
Since K = 10 7.53 , Q < K. The reaction is therefore not at equilibrium, and we expect reactants to be converted to products. This process causes Q to increase and 67
will continue until {HOCl} has decreased and {H3 O+ } and {OCl– } have increased sufficiently that equilibrium is attained (i.e., Q = K).
2.6
EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT
The effect of temperature on the equilibrium constant can be derived based on consideration of either the kinetics or the thermodynamics of the reactions, as is illustrated in Chapters 3 and 4, respectively. The resulting relationship is known as the van’t Hoff equation, which can be expressed as follows: K|T2 DH r ln = K|T1 R
✓
1 T1
1 T2
◆
(2.17)
Example 2.10 The reaction for the dissociation of water is shown below as it is conventionally written. The equilibrium constant for this reaction, Kw , is 1.00 ⇥ 10 14 at 25 C, and the standard molar enthalpy of the reaction is 55.81 kJ/mol. Compute the equilibrium constant for the reaction at 4 C and determine the activity of OH– in equilibrium solutions at 25 C and at 4 C, if {H+ } = 10 7.0 in both solutions. H2 O )* H+ + OH Solution Substituting the given values into Equation (2.17), we find ✓ ◆✓ ◆ Kw |4 C 55.81 kJ/mol 1 1 ln = = 1.71 Kw |25 C 298 K 277 K 8.314 ⇥ 10 3 kJ/mol-K Kw |4
C
= exp ( 1.71) Kw |25
C
= (0.18) 1.0 ⇥ 10
14
= 1.8 ⇥ 10
15
Thus, lowering the temperature substantially decreases the tendency for water molecules to dissociate. The equilibrium constant for the water dissociation reaction H+ OH . Since the activity of H2 O is 1.0 (because its mole fraction is Kw = {H2 O} is very nearly 1.0), and the activity of H+ is given as 10 7.0 , the activity of OH– is Kw . Therefore, at 25 C, {OH– } is 1.0 ⇥ 10 7 , and at 4 C it is 1.8 ⇥ 10 8 . 10 7.0
68
2.7
COMBINING CHEMICAL REACTIONS
Overall, observable reactions are often the net result of combinations of reactions that proceed in parallel (if the same reactant participates in more than one reaction) and/or in series (if the product of one reaction serves as the reactant in another). For an overall reaction to be at equilibrium, each of the reactions that contribute to it must be at equilibrium. We can use this principle to derive the value of K for the overall reaction. Consider, for example, the combination of reactions shown in Figure 2.3, in which species A participates in parallel reactions (R1 and R2), and C and E participate in reactions in series. The sum of the three reactions yields a net, overall reaction in which A and B are consumed, D and F are generated, and C and E do not appear at all:
(2.18) Figure 2.3 Example combination of parallel and series reactions.
The product of the equilibrium constants for the three reactions is KR1 KR2 KR3 =
{C} {D} {A}2
eq
{E} {A} {B}
{F}2 eq {C} {E}
= eq
{D} {F}2
{A}3 {B}
(2.19) eq
The resulting expression for the overall reaction has the form of an equilibrium constant; i.e., it is a ratio with the activities of the products in the numerator, those of the reactants in the denominator, and with each term raised to a power equal to its stoichiometric coefficient in the reaction. Equation (2.19) indicates that this ratio must equal a constant (computed as KR1 KR2 KR3 ) if the constituent reactions, and therefore the overall reaction, are at equilibrium. We therefore infer that the given ratio is indeed the equilibrium constant for the overall reaction, i.e.: Koverall =
{D} {F}2
{A}3 {B}
= KR1 KR2 KR3
(2.20)
eq
Because the example includes both parallel and series reactions, we can generalize the result to any such combination, no matter how complex. That is, we conclude that if an overall reaction can be formulated as the sum of any number of other reactions, the equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual reactions. That is, 69
Koverall = ’ Ki
If Roverall = Â Ri , then:
(2.21)
i
i
Similar reasoning shows that if a reaction is reversed or doubled, the equilibrium constant is, respectively, the inverse or the square of the original equilibrium constant. For example, for the reverse of the overall reaction considered above, we can write: K
reverse D+2F ) * 3A+B
Kreverse =
{A}3 {B} 2
{D} {F}
=
(2.22)
1
(2.23)
Kforward
Example 2.11 Equilibrium constants are known for the following three reactions, each of which is written for the transfer of one H+ ion to a water molecule. Reaction I:
H3 PO4 + H2 O )* H2 PO–4 + H3 O+
Ka1 = 10
2.148
Reaction II:
+ H2 PO–4 + H2 O )* HPO2– 4 + H3 O
Ka2 = 10
7.198
Reaction III:
3– + HPO2– 4 + H2 O )* PO4 + H3 O
Ka3 = 10
12.375
These reactions are examples of acid dissociation reactions, for which the equilibrium constants are commonly designated Ka1 , Ka2 , and Ka3 . The starting material in this sequence is phosphoric acid (H3 PO4 ), a common additive in foods and soft drinks, and the product of the third reaction is phosphate ion (PO3– 4 ), a common constituent of cleaning agents. Furthermore, all forms of PO4 -P (i.e., all the PO4 -containing species in these reactions) contribute to the pool of PO4 available for microbial growth in aquatic systems and wastewater treatment plants. Find the equilibrium constant for the following overall reaction: H3 PO4 + 3 H2 O )* 3 H3 O+ + PO34 Solution The overall reaction of interest can be generated as the sum of the three given reactions: 2 3 + ⇠⇠ + HPO2 + 3 H O )* H ⇠ ⇠⇠ H3 PO4 + ⇠ H2⇠ PO ⇠2 PO4 + HPO4 + PO4 + 3 H3 O 2 4 4
The equilibrium constant for the overall reaction is therefore the product of the K values for the three constituent reactions: Koverall = Ka1 Ka2 Ka3 = 10
2.148
10
70
7.198
10
12.375
= 10
21.721
Example 2.12 Given the equilibrium constants for the acid dissociation reactions designated as Reactions I and II below, find pK for Reaction III. Reaction I: Reaction II: Reaction III:
HOCl + H2 O )* OCl– + H3 O+ + HCO–3 + H2 O )* CO2– 3 + H3 O – – HOCl + CO2– 3 )* OCl + HCO3
Ka,HOCl = 10 7.53 Ka,HCO = 10 10.33 3 K =?
Solution Reaction III can be generated by adding Reaction I to the reverse of Reaction II: HOCl + H2 O )* H3 O+ + OCl– + – + [CO2– 3 + H3 O )* HCO3 + H2 O]
– – HOCl + CO2– 3 )* OCl + HCO3
K = Ka,HOCl = 10
7.53
K = 1/Ka,HCO = 1/10
10.33
3
K=
Ka,HOCl 10 = Ka,HCO2 10
7.53 10.33
= 102.80
3
The equilibrium constant for Reaction III is 102.80 ; pK for that reaction is therefore log10 10+2.80 = 2.80. In addition to demonstrating how equilibrium constants for composite reactions can be derived from K values for the constituent reactions, the preceding analysis and examples make the important point that, although the underlying concepts of chemical activities and equilibrium constants relate to chemical principles and reactions, the values associated with those terms are simply numbers. Thus, in the preceding examples, the value of Koverall is a certain number that can be determined from relationships among other numbers (KR1 , KR2 , KR3 ). These numbers can be manipulated using any valid mathematical operation, including addition, multiplication, taking square roots, etc. The mathematical manipulations and numerical results are independent of any meaning that we ascribe to the parameters, i.e., of whether we interpret the results in a chemical context or simply as values of the unspecified variables x, y, and z.
2.8
SUMMARY
‘Reactivity’ is defined very broadly here, encompassing the tendency for a chemical to participate in both conventional chemical reactions in which chemical bonds are broken or forged and other processes that cause chemicals to disappear from their original location, such as diffusion or transfer into another phase. The tendency for all such changes to occur can be quantified by a well-defined quantity known as the 71
chemical activity, whose value depends on the species’ concentration in the phase of interest, the composition of the phase, temperature, and pressure. The chemical activity of a species A (designated aA or {A}) can be represented as the product of two terms: the concentration of A relative to a standard concentration (cA /cA ), and an activity coefficient (gA ). If A is a solute, cA is usually defined to be 1.0 mol/L. The activity coefficient is defined to have a value of 1.0 under certain, specified conditions (the reference state conditions). If the environmental conditions in the system of interest conform to the reference state conditions, the activity coefficient is 1.0, and the species is said to behave ideally. In other solutions, the activity coefficient might be different from 1.0, indicating that the molecular interactions are different from those in the reference state. If A is a solute and cA is chosen to be 1.0 mol/L, the (dimensionless) activity of A has the same value as its concentration in moles per liter. Although the standard-state concentration and the reference-state conditions can be chosen arbitrarily, once those are defined, gA and aA are fixed. Activity coefficients of ions and, to a lesser extent, of neutral solutes depend on the ionic composition of the solution. Some simple equations have been developed to predict activity coefficients as a function of ionic strength. In general, the larger the charge on the ion of interest and the higher the ionic strength of the solution, the lower the activity coefficient. At ionic strengths greater than approximately 1.0 mol/L, the activity coefficients of many cations begin to increase with increasing ionic strength. The chemical activities of solvents (e.g., water), gases, and solids can also be represented by a product of the form gA cA /cA . However, for these substances, cA is often expressed as a mole fraction, and cA corresponds to a mole fraction of 1.0, i.e., to the pure substance. Also, for these substances, the reference state is usually defined in such a way that aA is close to 1.0 under normal environmental conditions. When these conventions are used, the value of aA is typically close to the mole fraction of A in the phase in which it is found if A is a solvent or a solid, and close to the partial pressure of A (when pressure is expressed in atm) if A is a gas. The chemical activity is an intrinsic thermodynamic property, meaning that its value does not depend on the amount of material in the phase of interest. As a result, the activity of a solid is the same no matter how much of that solid is present. When a chemical reaction is at equilibrium, there is no net driving force for conversion of reactants to products, or vice versa. Conceptually, equilibrium can be understood as either a state in which the reaction is proceding in the forward and reverse directions at equal rates or as the the state in which the chemical potential energy of the system has been minimized. The ratio of the activities of products to reactants for a reaction is called the reaction quotient, Q. The value of Q for a given reaction must always be the same once the reaction has reached equilibrium, although that value can be achieved via many different combinations of activities for the individual reacting species. The 72
value of Q at equilibrium is called the equilibrium constant, K. For reactions that are not at equilibrium, Q does not equal K, and the reaction proceeds in the direction that causes Q to approach K. The ratio Q/K [or log(Q/K)] is an indicator of the extent of disequilibrium. The equilibrium constant varies with temperature according to the van’t Hoff equation [Equation (2.17)]. The equilibrium constant for an overall reaction that can be written as the sum of other reactions equals the product of the K values of the reactions in that summation. The derivation of this relationship emphasizes that chemical activities and equilibrium constants are simply numbers that can be manipulated according to the rules of algebra, independent of the fact that they have chemical significance.
2.9
PROBLEMS
– 1. Compute the activities of Ca2+ , SO2– 4 , and Cl in the Mississippi River near Clinton, IA, using the Davies equation to estimate activity coefficients. Make the same assumptions about the composition of the organic matter as were made in Problem 9 of Chapter 1.
2. Explain briefly why we always assume that activity coefficients are 1.0 for pure solids, regardless of the ionic strength of the solution in which they are found. 3. Water in a reservoir at 20 C has a pH of 7.7 and contains the following species. Species Na+ K+ Ca2+ Mg2+
Conc’n (mg/L)
Species SO2– 4 Cl– NO–3 F– HCO–3
85 5.6 38 24
Conc’n (mg/L) 21 65 4.0 1.6 183
(a) Check the charge balance for this analysis by calculating the ratio of cation to anion equivalents per liter. As noted in Chapter 1, such an analysis is typically considered acceptable if the error in the charge balance, computed with Equation (1.4), is less than 5%. Is this analysis in the acceptable range? (b) Estimate the activities of Na+ , Ca2+ , and HCO–3 , using an appropriate equation to compute the activity coefficients. 4. An aqueous solution is prepared by dissolving 100 mg/L of CaCl2 and 230 mg/L of Na2 SO4 in water. Assuming that these salts dissociate completely, and that the dissociated ions do not undergo any additional reactions, calculate 73
(a) the molar concentration of each ionic species; (b) the ionic strength of the solution; (c) the activity coefficient for each species (with the Davies equation); (d) the activity of each species 2– 6 5. What are p[SO2– 4 ] and p{SO4 } in solutions of 8.5 ⇥ 10 M and 4.5 ⇥ 10 Na2 SO4 , assuming that the salts dissociate completely?
3M
6. When the activity of Na+ in an estuary is evaluated using the conventional standard state conditions (molar concentration of 1.0, chemical environment corresponding to infinite dilution in pure water), aNa+ is estimated to be 0.015. Would this value increase, decrease, or remain the same if the reference state environment were changed to infinite dilution in major-ion seawater? Explain your reasoning. 7. A solution of 0.01 M CaCl2 and 10 5 M CuCl2 is prepared. Both of these salts dissociate completely when dissolved in water. If the concentration of CaCl2 is then doubled, how much will the activities of Ca2+ , Cl– , and Cu2+ change? 8. Sodium chloride is added to pure water to increase its ionic strength from essentially zero to 0.1 mol/L. The solution remains at 25 and pH 7.0 throughout the process. How many moles of water molecules dissociate per liter of solution when the salt is added? By what fraction does the concentration of H+ increase? By what fraction does the activity of H+ increase? 9. One gram of solid calcium sulfate [CaSO4 (s)] is added to 1.0 L of pure water. Some of the solid then dissolves according to the following reaction, which has an equilibrium constant of K = 10 4.85 : CaSO4 (s) )* Ca2+ + SO24 (a) What are the concentration (mol/L) and activity of CaSO4 (s) before any dissolution occurs? (b) What is the activity of CaSO4 (s) after enough dissolution has occurred so that the reaction reaches equilibrium, assuming that not all of the solid dissolves? (c) Compute the concentrations of Ca2+ and SO2– 4 in mol/L and in mg/L at equilibrium, assuming that gCa2+ = gSO2 = gCaSO4 (s) = 1.0. 4
(d) Estimate the ionic strength of the solution based on the result of part (c). Then compute gCa2+ and gSO2 with the Extended Debye-H¨uckel equation, 4 and make a new estimate of the concentration of each ion in the equilibrated solution. Iterate between calculating the concentrations of Ca2+ 74
and SO2– 4 and the corresponding ionic strength and activity coefficients until each calculation converges. What fraction of the original solid is dissolved at equilibrium? 10. The reactions for the dissolution of limestone (CaCO3 (s)) and for the dissociation of bicarbonate ion (HCO–3 ) to carbonate ion (CO2– 3 ) are as follows: CaCO3 (s) )* Ca2+ + CO23
K = 10
8.3
HCO3 )* H+ + CO23
K = 10
10.33
These reactions have reached equilibrium in a solution at pH 9.33 that contains 10 2 mol/L CaCO3 (s) and 10 3.3 mol/L Ca2+ . (a) What is the HCO–3 concentration in the solution, assuming that activity coefficients for all species are 1.0? (b) Would you expect the Ca2+ concentration to increase, decrease, or not change if the system changed in the following ways? For this part of the problem, do not assume that activity coefficients are 1.0. In each case, explain your reasoning briefly (1-2 sentences). For this part of the question, consider only the effect on the first reaction shown above. (i) The concentration of CaCO3 (s) is doubled. (ii) The ionic strength of the solution increases, e.g., by addition of NaCl. (iii) The pH is lowered slightly, without significantly altering the ionic strength. 11. Having become thoroughly frustrated with the concept of moles, a group of reformers decides to defy convention and carry out all calculations using a new set of definitions for the standard state. They choose the standard state concentration to be 1 mg/L for all solutes and 1000 g/L for water, but they retain the conventional reference state conditions of infinite dilution for solutes and pure liquid water for H2 O. The equilibrium constant for the following reaction using the conventional standard state definitions is 1011.32 . Compute the equilibrium constant using the revised conventions. Include dimensions, if appropriate. H2 CO3 + 2 OH )* CO23 + 2 H2 O
75
76
3
REACTION KINETICS AND THE KINETICS-BASED INTERPRETATION OF EQUILIBRIUM
77
Contents
3.1
A MOLECULAR-LEVEL PICTURE OF AN ELEMENTARY REACTION AND THE FACTORS THAT AFFECT ITS RATE . . 78 3.2 EFFECT OF TEMPERATURE ON REACTION RATE CONSTANTS 83 3.3 THE KINETICS OF SOME IMPORTANT CATEGORIES OF ENVIRONMENTAL CHEMICAL REACTIONS . . . . . . . . . . . . 84 3.4 KINETICS OF ELEMENTARY CHEMICAL REACTIONS . . . . 86 3.5 REACTION REVERSIBILITY AND THE DEFINITION OF THE EQUILIBRIUM CONSTANT . . . . . . . . . . . . . . . . . . . . 88 3.6 EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT 91 3.7 COMBINING CHEMICAL REACTIONS: KINETICS AND EQUILIBRIUM CONSTANTS OF NONELEMENTARY REACTIONS . 92 3.8 EXPERIMENTAL EVALUATION OF REACTION KINETICS . . 97 3.9 RATE-LIMITING STEPS AND SOME CLASSICAL, MODEL REACTION PATHWAYS . . . . . . . . . . . . . . . . . . . . . . . 104 3.10 HETEROGENEOUS (PHASE-TRANSFER) REACTIONS . . . . . 113 3.11 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 3.12 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
3.1
A MOLECULAR-LEVEL PICTURE OF AN ELEMENTARY REACTION AND THE FACTORS THAT AFFECT ITS RATE
As was explained in Chapter 2, chemical equilibrium is a stable condition characterized by the absence of any net driving force for a chemical reaction to occur, and one way of understanding this condition is as a state in which the forward and reverse
78
reactions are proceeding at equal rates. In this chapter, we explore reaction kinetics in greater detail.1 Chemical reactions happen when reactant molecules collide with sufficient energy to break existing bonds and rearrange into a different configuration. Therefore, a complete description of the process requires knowledge about the frequency of collisions, the energy involved in the collisions, the energy required for molecular rearrangement, and the relative stability of the various configurations possible. Often, the net reaction that is observable at a macroscopic scale is actually the result of several reactions. That is, when two reactants are mixed, they might form a new species, which then reacts with others to form different species, which in turn might react with yet others. Eventually, a stable suite of products is formed. In some cases, the individual steps are slow enough to be readily identified and studied. In others, though, the intermediate steps occur very quickly, and we might not even be aware of their existence without the assistance of sophisticated analytical instrumentation. The step-by-step sequence that leads to the overall reaction is referred to as the reaction mechanism or reaction pathway. Each step in the reaction pathway occurs as the result of individual, sufficiently energetic collisions among the reactants for that step and is called an elementary reaction. Theory suggests that collisions among more than two molecules are rare, so that any overall reaction involving three or more reactants is virtually certain to be nonelementary. A hypothetical sequence of elementary reactions that leads to an overall, nonelementary reaction is shown below. The designation 2 (X + C )* Y + Z) means that the elementary reaction is between one X and one C molecule, but that two such elementary reactions occur each time the overall sequence proceeds. Sequence of Elementary Reactions:
A + B )* 2 X 2 (X + C )* Y + Z) 2 (Y + Z )* D)
Overall Reaction:
A + B + 2 C )* 2 D
In this sequence, species X, Y, and Z are intermediates that are created and then destroyed by the elementary reactions, so the concentrations of these species do not change when the overall reaction occurs. An elementary reaction is initiated when two reactant molecules approach one another, causing the bonds in each to become strained. This strain increases the po1 The
contents of this chapter draw heavily on the discussion in Chapter 3 of Benjamin, M.M. and Lawler, D.F. Water Quality Engineering: Physical/Chemical Treatment Processes, J. Wiley & Sons, Hoboken, NJ (2013).
79
tential energy stored in the bonds, much like compressing a spring increases the potential energy stored in it. The intensity of the strain and the amount of energy stored in the bonds increase dramatically with decreasing separation of the molecules. Since energy is conserved during this interaction, the increased bond energy must be balanced by a decrease in energy elsewhere in the system. In this case, most of the energy used to strain the bonds comes from molecular kinetic energy — the molecules slow down as the distance between them decreases. In the absence of other factors, the molecules would eventually stop their mutual approach and begin moving away from one another, thereby relieving the strain and converting the bond energy back into kinetic energy. Although the molecules might indeed separate without undergoing any long-term changes, it is also possible that, as the molecular structures adjust to the strain, new bonds will begin to form. At some critical point, the original bonds may be sufficiently distorted, and the new bonds may form to a sufficient extent, that the strain can be relieved just as easily by rearrangements that form product molecules as by a reversal of the process and re-formation of the original reactants. At this point, the molecules are not identifiable as either reactants or products, but are an intermediate species of negligible stability, analogous to a ball perched at the crest of a hill with equal likelihood of returning along the path of its climb or proceeding down the other side. The amount of energy necessary to bring molecules from far apart (no interaction) to the critical point is called the activation energy E ⇤ , and the process of reaching the critical condition is referred to as overcoming the activation energy barrier. Catalysts operate by providing an alternative path by which a reaction can occur. Specifically, they allow reactants to be converted to products via a route that has a lower activation energy than the route that is taken in their absence. Extending the analogy to a ball on a hill, a catalyst might be viewed as providing an alternative path for the ball to follow that does not require quite so much of a climb before arriving at a point on the downhill slope. In some cases, catalysts can increase the rate of an overall reaction by many orders of magnitude. However, catalysts cannot alter the overall energy change associated with the reaction or the ultimate mix of products that forms. Although the preceding discussion focuses on the changes in bond energy within the reacting molecules as they approach one another, the amount of energy in the molecules changes in a myriad of other ways as well during this process (e.g., the bond strength between the reactants and their nonreacting neighbors also changes). The sum of the energy terms that can change as the reaction proceeds is sometimes referred to colloquially as the heat content of the molecules, or more formally as a thermodynamic parameter known as the enthalpy, H. The enthalpy per mole of a substance i (the molar enthalpy of i) is represented as H i , and the enthalpy change accompanying the complete reaction, when normalized to the amount of material 80
reacting, is represented as DH r . The enthalpy and its relationship to other molecular properties are discussed in detail in Chapter 4; here, we simply introduce the term and use it without further explanation, understanding it as a composite measure of various components of molecular energy that can change as a reaction proceeds. In addition to the energy of the colliding molecules, their orientation when they collide and the distribution of energy within the molecules can affect the likelihood that a reaction will occur. For example, if the molecules are not spherically symmetric, only a fraction of all collisions can cause the bonds to distort in a way that leads to the formation of product, even if the collisions involve an amount of energy greater than the activation energy. Thus, the overall rate of reaction depends on the frequency with which reactant molecules collide, the likelihood that colliding molecules are properly oriented and have sufficient kinetic energy to overcome the activation barrier, and the rate at which the activated species is converted to products. The most widely accepted mathematical model for chemical reaction kinetics is called the activated complex or transition state model. In this model, the passage from reactants to products is viewed as a continuum, with the reactant molecules being converted to products by proceeding along a path on the “energy landscape” described above. Molecules that have acquired an amount of energy equal to or greater than the activation energy are called activated complexes. For a generic elementary reaction between A and B, the activated complexes are commonly represented as AB⇤ ; that is, the complete elementary reaction is A + B )* AB⇤ ! P. The arrows in this representation signify that, as they ‘climb’ the activation barrier, molecules of A and B might reverse their approach and revert to separate molecules, but once they overcome that barrier, they proceed to form P. A schematic of the key energy relationships during the transition from reactants through the activated complex to products according to this model is shown in Figure 3.1 for a system in which the reactants and products are all in their standard states.2 Figure 3.1 includes three curves — one for the enthalpy (H) of the molecules, one for the negative product of the system temperature and the entropy (S) of the molecules, and one for their Gibbs energy (G). Like the enthalpy, the entropy and Gibbs energy are thermodynamic parameters that are explained in detail in Chapter 4, and we use them here without further definition. These three parameters are related by DG = DH T DS (3.1) For this example reaction, the products have more enthalpy than the reactants, 2 The preceding discussion might give the impression that the reacting molecules engage in only one
collision as they climb the activation barrier. In truth, the activated complex model suggests that, as the reactant molecules approach one another, each of them participates in many collisions that collectively increase its energy content. Simultaneously, the mutual approach of the reactants strains the bonds and converts some of the kinetic energy acquired in the collisions into chemical energy. The frequent nonreactive collisions are essential for the molecules at all chemical energy levels to be at the same temperature, which is a central assumption of the activated complex model.
81
so energy must be added to keep the system at constant temperature as the reaction proceeds. Such a reaction is referred to as endothermic. A reaction in which the products have less enthalpy than the reactants is referred to as exothermic. In the activated complex model, the activation energy is associated with the change in Gibbs energy between the reactants and the activated complexes. Note that the magnitude of the activation energy required to induce the reaction is independent of the change in Gibbs energy accompanying the overall reaction.3
Figure 3.1 Schematic representation of the transition from reactants to products according to the activated complex model. H, G, and S are the enthalpy, Gibbs energy, and entropy of the system at any point in the process (i.e., the sum of those values for all the species that participate in the reaction), and DHr , DGr , and DSr are the corresponding changes for the overall reaction.
The key result of the activated complex model is that the rate at which A and B cross the activation barrier to form P is 3 Diagrams
like Figure 3.1 are intended to convey a combination of qualitative and quantitative information, and it is important to understand which parts of the diagram fall into each category. Because the properties of activated complexes are not measurable (indeed, the very existence of such species is a theoretical construct), one cannot know the value of their enthalpy, entropy, or Gibbs energy. As a result, any representation of the magnitude of the activation barrier is conceptual and, at best, semiquantitative. On the other hand, the reactants and products of the overall reaction are real species whose thermodynamic properties are unambiguous. Thus, even though the heights of the activation barriers for the forward and reverse reactions cannot be known with certainty, the difference in those two heights is measureable.
82
rP =
kB T ⇤ K cA cB h
= kcA cB
(3.2a) (3.2b)
where kB and h are the Boltzmann and Planck constants, respectively; K ⇤ is a temperature-dependent value that relates the Gibbs energy of the activated complexes to that of the reactants; and k is called the reaction rate constant. K ⇤ increases with increasing temperature and decreasing activation energy (E ⇤ ). Therefore, Equation (3.2a) indicates that all elementary reactions proceed at rates that are proportional to the concentrations of the reacting species, increase with increasing temperature, and decrease with increasing activation energy.
3.2
EFFECT OF TEMPERATURE ON REACTION RATE CONSTANTS
The earliest successful attempt to describe the dependence of reaction rate constants on temperature was by Arrhenius, who proposed the relationship ✓ ◆ EAr k = kAr exp (3.3) RT where
k kAr EAr R T
is the reaction rate constant is a reaction-dependent constant with the same units as k; kAr is sometimes called the frequency factor is a constant, with dimensions of energy per mole is the universal gas constant, in appropriate units so that the argument of the exponential is dimensionless; and is absolute temperature.
Taking the logarithm of both sides of Equation (3.3) yields EAr 1 (3.4) R T Equation (3.4) indicates that a plot of ln k versus 1/T is expected to be a straight line with slope EAr /R and intercept ln kAr , as shown in Figure 3.2. Arrhenius’s result was empirical, and it is widely applied to characterize the temperature dependence of rate constants for both elementary and nonelementary reactions. In an effort to blend this empirical relationship with the more theoretical activated complex model, EAr is commonly referred to as an activation energy and interpreted (not entirely correctly) as the enthalpy required to form the activated complexes from the reactants. ln k = ln kAr
83
Figure 3.2 Characteristic plot of the rate constant versus inverse absolute temperature, from which the values of kAr and EAr can be computed; ln kAr is found by extrapolating the straight line to the hypothetical condition 1/T = 0.
3.3
THE KINETICS OF SOME IMPORTANT CATEGORIES OF ENVIRONMENTAL CHEMICAL REACTIONS
Except for Chapters 4 and 7 (which are devoted to chemical thermodynamics and a discussion of the software used to solve chemical equilibrium problems, respectively), each of the remaining chapters of this text focuses on a particular category of chemical reactions. It is useful to develop a sense of the kinetics of each of these categories of reactions, and thereby to identify those that typically proceed rapidly enough to reach equilibrium in the aquatic systems of interest. One such generalization that is almost universally valid is that acid/base reactions (in which an H+ ion is transferred from one dissolved molecule to another) are very fast; these reactions are discussed in Chapters 5, 6, and 8. The only major exception to this generalization is that acid/base reactions of large polymers are sometimes slow if the acquisition or release of the H+ ion is accompanied by a significant change in the polymer’s conformation (e.g., coiling or uncoiling of the molecule). Another broad group of reactions can be characterized generically as follows: A (H2 O)m + B (H2 O)n )* AB (H2 O)p + (m + n p)H2 O
(3.5)
In this reaction, two types of molecules (A and B) that are each initially surrounded by water molecules come together to form a new species (AB) that is also surrounded by water molecules, but in most cases fewer than the number required to surround the original molecules. This type of reaction occurs commonly between dissolved 84
metal ions (e.g., Zn2+ ) and dissolved anions (e.g., SO2– 4 ) or neutral molecules (e.g., NH3 ). A simple way to think about such a reaction is that a weak bond exists between each of the reacting species and the surrounding water molecules. When the two species approach one another, one of the water molecules adjacent to each of them is released, and a bond between the two nonwater species forms in its place. These kinds of reactions are described in more detail in Chapter 10. Like acid/base reactions, they tend to proceed very quickly in aqueous solutions at normal temperatures, and they therefore typically reach equilibrium in all systems of interest. Again, a few exceptions exist. For example, the activation energy for stripping away water molecules that surround Cr3+ ions is unusually large, so reactions like (3.5) are slow if A or B is Cr3+ . The only major category of aqueous-phase reactions that are frequently slow compared to the amount of time available are those in which electrons are transferred from one molecule to another. Such reactions are called oxidation/ reduction or ‘redox’ reactions; these reactions are the central topic of Chapter 12. Not all redox reactions in solution are slow; some, such as the reaction of hydrogen sulfide (H2 S) with hypochlorous acid (HOCl) proceed very quickly. The rates of many others depend strongly on the pH. For example, as shown in Figure 3.3a, the reaction of dissolved ferrous iron (a term referring collectively to various chemical species containing iron (Fe) atoms with a +2 charge) with dissolved oxygen is very slow in acidic solutions but proceeds rapidly if the solution is made alkaline. The reaction rates of many organic compounds with HOCl and OCl– also have a complex dependence on pH, as exemplified by tetracycline in Figure 3.3b. On the other hand, the redox reactions of many organic molecules with oxygen are extremely slow at any pH, so that, in the absence of catalysts (e.g., enzymes), the reactants can coexist and barely react at all over periods of centuries. The rates of many important environmental redox reactions have been studied extensively and are discussed in detail in many books and journal articles.4 The other broad category of relatively slow reactions that involve aqueous solutions are those in which a species crosses a phase boundary (heterogeneous reactions). This category includes transfer across gas/liquid and solid/liquid boundaries when, for instance, carbon dioxide [CO2 (g)] dissolves into a solution or diffuses out of it (described in Chapter 9), or solids such as calcium carbonate [CaCO3 (s)] or iron hydroxide [Fe(OH)3 (s)] precipitate or dissolve (Chapter 11). In these cases, the actual transfer from one phase to the other usually occurs rapidly, so that the two phases are very nearly equilibrated right at the interface. However, concomitant processes that occur adjacent to the interface are often slow, and they can place significant limitations on the rate of the overall reaction. Such processes include diffusion of dissolved species from the bulk solution through the liquid boundary layer and, in the case of 4 See,
for example, Schwarzenbach, R., Gschwend, P., and Imboden, D. Environmental Organic Chemistry, 2nd ed., J. Wiley & Sons, Hoboken, NJ (2003), or Stumm, W., and Morgan, J.J. Aquatic Chemistry, 3rd ed., J. Wiley & Sons, New York (1996).
85
precipitation or dissolution reactions, the formation or breaking of bonds between the reacting ions and the solid structure. More detailed discussions of these processes is provided in Section 3.10 and in other texts.5
Figure 3.3 pH–dependence of some important environmental reaction rates. (a) Apparent (observed) first-order reaction rate constant for oxidation of ferrous iron by dissolved oxygen, in a solution equilibrated with pure O2 (g) at 1 atm total pressure. Open circles are from Singer and Stumm (1970), and filled circles from Millero et al. (1987). From Benjamin, M.M. and Lawler, D.F. Water Quality Engineering: Physical/Chemical Processes, John Wiley & Sons, Hoboken, NJ (2013). (b) Apparent second-order rate constant for oxidation of tetracycline by OCl species (HOCl and OCl– ). After Wang, et al. (2011).6
3.4
KINETICS OF ELEMENTARY CHEMICAL REACTIONS
We begin the quantitative analysis of chemical reaction kinetics by defining some nomenclature and describing generic approaches for characterizing the rates of elementary reactions. The kinetics of nonelementary reactions are discussed subsequently. Although the chemical activity is the most direct measure of the tendency for a molecule to react, chemical reaction rates are often written using concentrations as 5 See, for example, Clark, M.M. Transport Modeling for Environmental Engineers and Scientists, J. Wiley & Sons, Hoboken, NJ (2009), or Benjamin, M.M., and Lawler, D.F. Water Quality Engineering: Physical/ Chemical Processes, J. Wiley & Sons, Hoboken, NJ (2013). 6 Singer, P.C., and Stumm, W. (1970) “The solubility of ferrous iron in carbonate-bearing waters.” Science 167, 1121-1123; Millero, F.J., Sotolongo, S., and Izaguirre, M. (1987) “The oxidation kinetics of Fe(II) in seawater.” Geochim. Cosmochim. Acta 51, 793-801; Wang, P., He, Y.L, and Huang, C-H. (2011) “Reactions of tetracycline antibiotics with chlorine dioxide and free chlorine.” Water Research 45, 1838-1846.
86
the key parameters. That is, for an elementary reaction between A and B (Equation 3.6), the reaction rate might be written as shown in either Equation (3.7a) or (3.7b): A + B )* products
(3.6)
rf = kf0 {A} {B}
(3.7a)
rf = kf [A] [B]
(3.7b)
where rf is the rate of the forward reaction, expressed as the mass or number of moles of A or B reacting per unit volume per unit time (e.g., mol/L-s), and kf0 and kf are both referred to as the forward reaction rate constant (kf0 when the rate is computed based on reactant activities, and kf when it is computed based on concentrations). The rate constant accounts for all factors that affect the reaction rate, other than the concentrations or activities of the reactants. The primary such factors are the activation energy and temperature, although ionic strength also plays a role for reactions in solution. Because chemical activities are dimensionless, kf0 has the same units as rf (e.g., mol/L-s). On the other hand, for a two-molecule (bimolecular) reaction, kf has units corresponding to rf divided by concentration squared (e.g., L/mol·s). The two forms of the rate constant are related by: kf = kf0
g Ag B [A] [B]
(3.8)
where [i] (with i = A or B) is the concentration of i in its standard state. For a reaction requiring a collision between two A molecules, B can be replaced by A in the reaction and rate expression, yielding 2A )* products
(3.9)
rf = kf0 {A}2 = kf [A]2
(3.10)
Generalizing this observation, we infer that the rate of any elementary reaction is given by the product of a reaction rate constant and the concentrations or activities of the reactants, each raised to a power equal to its stoichiometric coefficient. The exponent on the concentration or activity term in a rate expression is called the order of the reaction with respect to that species, and the sum of these exponents is called the order of the overall reaction. Thus, for example, Reaction (3.6) is said to be first order with respect to A, first order with respect to B, and second order overall. In Reactions (3.6) and (3.9), both molecules involved in the collision are destroyed by the reaction. However, it is also possible for an elementary reaction to cause only a single molecule to disappear. For example, a molecule might be unstable and susceptible to a rearrangement of its bonds that converts it to one or more different species, but the reaction might proceed only when the molecule acquires 87
sufficient energy via a collision with a second molecule in the system. In such a case, the reaction and rate expression might be written as follows:
A + X )* products + X rf =
kf0 {A} {X}
= kf [A] [X]
(3.11) (3.12)
where X represents a composite of all the molecules in the system (because any of those molecules might collide with A and provide the activation energy needed for the reaction). In aqueous solutions, water molecules account for the overwhelming majority of X, and both [X] and {X} are essentially identical in all such solutions. As a result, the rate can be expressed as the product of the activity or concentration of A and an effective rate constant that incorporates the activity or concentration of X: 0 {A} = kf,eff [A] rf = kf,eff (3.13)
0 where kf,eff = kf0 {X} and kf,eff = kf [X]. If the activation energy for a reaction is very small, the reaction occurs virtually every time that the reactant molecules collide. The overall rate of such reactions therefore approaches the collision rate, which in turn is limited by the rate at which reactants can diffuse toward one another. Reactions that proceed at rates close to these maximum predicted values are said to be diffusion-controlled. For bimolecular elementary reactions between uncharged solutes in aqueous solutions, diffusioncontrolled rate constants at room temperature are on the order of 1010 (mol/L) 1 s 1 ; this rate constant can be up to about an order of magnitude larger if the reaction is between two ions that have opposite charges, and up to about two orders of magnitude smaller if it is between ions that have like charges. Although the reaction rates of some acid-base reactions approach those estimated based on diffusion control, the vast majority of reactions in aquatic systems have activation energy barriers that are much higher than the barrier imposed by diffusion; the corresponding reaction rates are anywhere from a few orders of magnitude to tens of orders of magnitude slower than the diffusion-controlled rates. In such cases, the energy required for bond rearrangement provides the dominant impediment to the progress of the reaction, and the reaction rate is said to be chemically controlled.
3.5
REACTION REVERSIBILITY AND THE DEFINITION OF THE EQUILIBRIUM CONSTANT
A fundamental tenet of chemical reaction theory is that all reactions are reversible: if a collision between certain reactants can generate certain products, a collision between those products can regenerate the reactants. In some cases, the reverse reaction is equally or more likely than the forward reaction, whereas in others it is extremely 88
unlikely (e.g., if an elementary, bimolecular reaction generates three molecules as products, the likelihood of the reverse reaction is very small). Regardless of its likelihood, though, the reverse reaction is possible in principle. Such reversibility is intimately tied to the idea of chemical equilibrium because, as noted previously, equilibrium can be described as a condition in which a reaction is proceeding in the forward and reverse directions at equal rates. We next derive the quantitative link between these two important concepts using the following generic example reaction, which we assume to be elementary in both directions: k
f A+B ) * C+D
kr
(3.14)
where kr is the rate constant for the reverse reaction. In any system containing all the species that participate in a given reaction, each species is constantly being both created and destroyed. Thus, for example, the net rate at which Reaction (3.14) generates species C, written in terms of activity-based rate constants, is: rC,net = rate of C production =
kf0 {A}{B}
rate of C destruction
kr0 {C}{D}
(3.15a) (3.15b)
At equilibrium, rC,net must be zero, so: kf0 {A}eq {B}eq = kr0 {C}eq {D}eq
(3.16a)
kf0 {C} {D} = {A} {B} kr0
(3.16b)
eq
⌘K
we can recognize K as the equilibrium constant that was defined in Chapter 2. Recall also from Chapter 2 that the reaction quotient, Q, is defined by the same ratio of activities as defines K, except that Q is not constrained to equilibrium conditions. That is Q = K for a reaction at equilibrium, but that equality does not apply under other conditions. By combining the definition of Q with the result shown in Equation (3.16b), we can now demonstrate that an elementary reaction that is not at equilibrium will always proceed toward equilibrium (i.e., in the direction that causes Q to approach K). {C} {D} Q kr0 {C} {D} rr {A} {B} = = = K kf0 /kr0 kf0 {A} {B} rf
(3.17)
This result indicates that, if Q > K, the reverse reaction proceeds more rapidly than the forward reaction, causing a net conversion of products to reactants. This conversion decreases the numerator and increases the denominator of Q, so that its 89
value decreases. Similarly, if Q < K, reactants will be converted to products, causing Q to increase. In either case, the reaction causes Q to approach K until, at equilibrium, they are equal. The ratio Q/K is thus an indicator of how far a reaction is from equilibrium. Because Q/K can vary over many orders of magnitude and equilibrium is reached when Q/K = 1, the extent of disequilibrium is conveniently quantified as log(Q/K), with a negative value implying that the forward reaction is favored, and a positive value implying that the reverse reaction is favored. If Q ⌧ K, the forward reaction proceeds much more rapidly than the reverse reaction, and if Q K, the opposite is true. The net rate of the reaction, rnet , is rf rr , so if log Q/K is smaller than approximately 1.5 (i.e., Q/K < 0.03), rnet ⇡ rf , and if log Q/K is larger than approximately +1.5 (i.e., Q/K > 30), rnet ⇡ rr . Under either of these conditions, the reaction is said to be irreversible, meaning that the reaction is proceeding so much faster in one direction than the other that only the faster reaction need be considered when evaluating the net reaction rate. Many reactions of interest in environmental systems meet this criterion. These ideas relating Q/K to the direction and rate of the reaction are summarized in Table 3.1, which is an expanded version of Table 2.3. Table 3.1 Relationships among Q, K, and the direction and reversibility of chemical reactions
Q
log(Q/K) < 0
Not at equilibrium Reactants ! Products
Forward rate > Reverse rate
Q⌧K
log(Q/K) ⌧ 0
Not at equilibrium Reactants ! Products ⇠ Irreversible
Forward rate
Q>K
log(Q/K) > 0
Not at equilibrium Reactants Products
Forward rate < Reverse rate
Q
log(Q/K)
Not at equilibrium Reactants Products ⇠ Irreversible
Forward rate ⌧ Reverse rate
At equilibrium No net reaction
Forward rate = Reverse rate
K
Q=K
0
log(Q/K) = 0
Reverse rate
Example 3.1 In Example 2.9, we determined that if the activities of H3 O+ , HOCl, and OCl– in a solution were 10 7 , 10 3 , and 10 4 , respectively, HOCl would dissociate into H+ and OCl– as equilibrium was approached. Assuming that the association and 90
dissociation reactions are elementary, would it be reasonable to treat the reaction as irreversible under these conditions? Solution In the solution to Example 2.9, we determined that the activity quotient for HOCl dissociation under the given conditions was Q = 10 8.0 . The equilibrium constant for the reaction was given as K = 10 7.53 , so Q/K = 10 0.47 = 0.34. The forward reaction is therefore proceeding ⇠3.0 (i.e., 1/0.34) times as fast as the reverse reaction. This difference in rates is insufficient to justify ignoring the slower reaction, so it would not be reasonable to consider the reaction irreversible under the given conditions.
3.6
EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT
According to Equation (3.16b), the equilibrium constant for a reaction that is elementary in both directions equals the ratio of the forward to the reverse, activity-based rate constants. Writing that equality, but using Equation (3.3) to substitute for both kf and kr (which we now write as kf0 and kr0 , since we are applying the equation to elementary reactions), we obtain ✓ ◆ EAr,f ✓ ◆ kAr,f exp kf0 kAr,f EAr,f EAr,r RT ✓ ◆= K= 0 = exp (3.18a) EAr,r kr kAr,r RT kAr,r exp RT ✓ ◆ kAr,f DEAr = exp (3.18b) kAr,r RT Empirical studies of the effect of temperature on equilibrium constants have shown that, over the range of temperatures typically of interest in environmental systems, both the ratio kAr,f /kAr,r and the difference EAr,f EAr,r are approximately constant, so the ratio of the equilibrium constant for a given reaction at two different temperatures can be expressed as ✓ ◆ K|T2 DEAr 1 1 = exp (3.19) K|T1 R T1 T2 ln
K|T2 K|T1
=
✓
DEAr R
◆✓
91
1 T1
1 T2
◆
(3.20)
Recall that the empirical Arrhenius model for reaction kinetics could be linked, at least qualitatively, to the changes in well-defined thermodynamic parameters in the activated complex model. The thermodynamically based analysis of the effect of temperature on equilibrium constants is presented in Chapter 4 and yields an equation almost identical to Equation (3.20), except that EAr is replaced by the normalized (molar) enthalpy of the reaction when all the reactants and products are in their standard states (DH r ). Thus, the thermodynamic equation analogous to Equation (3.20), known as the van’t Hoff equation, is ln
K|T2 K|T1
DH r = R
✓
1 T1
1 T2
◆
(3.21)
Example 3.2 The reaction for the dissociation of water is shown below as it is conventionally written. The equilibrium constant for this reaction, Kw , is 1.00 ⇥ 10 14 at 25 C, and the standard molar enthalpy of the reaction is 55.81 kJ/mol. Compute the equilibrium constant for the reaction at 4 C and determine the activity of OH– in equilibrium solutions at 25 C and at 4 C, if {H+ } = 10 7.0 in both solutions. H2 O )* H+ + OH
Solution Substituting the given values into Equation (3.21), we find ✓ ◆✓ ◆ Kw |4 C 55.81 kJ/mol 1 1 ln = = 1.71 Kw |25 C 298 K 277 K 8.314 ⇥ 10 3 kJ/mol-K Kw |4
C
= exp ( 1.71) Kw |25
C
= (0.18) 1.0 ⇥ 10
14
= 1.8 ⇥ 10
15
Thus, lowering the temperature substantially decreases the tendency for water molecules to dissociate. The equilibrium constant for the water dissociation reaction H+ OH is Kw = . Since the activity of H2 O is 1.0 (because its mole fraction {H2 O} is very nearly 1.0), and the activity of H+ is given as 10 7.0 , the activity of OH– is Kw . Therefore, at 25 C, {OH– } is 1.0 ⇥ 10 7 , and at 4 C it is 1.8 ⇥ 10 8 . 10 7.0
3.7
COMBINING CHEMICAL REACTIONS: KINETICS AND EQUILIBRIUM CONSTANTS OF NONELEMENTARY REACTIONS
While the description at the beginning of this chapter provided a framework for understanding chemical reactions at the molecular level, in most cases we do not have 92
the information necessary to interpret reactions at that level of detail. Furthermore, as explained previously, most overall reactions are considerably more complicated than the preceding description suggests, because they represent the net result of several elementary reactions. Fortunately, even without knowing the molecular-level details of reactions, it is possible to make some general observations about their tendency to occur and the stable endpoint (i.e., the equilibrium condition) toward which they progress. This section describes ways of thinking about and analyzing the kinetics and equilibrium of nonelementary reactions.
3.7.1
Deriving Equilibrium Constants of Nonelementary Reactions
Nonelementary reactions are generated by combinations of elementary reactions that might proceed in parallel (if the same reactant participates in more than one reaction) and/or in series (if the product of one reaction serves as the reactant in another). We saw in Chapter 2 that the equilibrium constant for an overall reaction that can be represented as the sum of two or more other reactions equals the product of the equilibrium constants of the contributing reactions. Once again considering Reaction (2.18) (repeated below) as an example, and now specifying that the three reactions that combine to generate that overall reaction are elementary, we can combine the equilibrium constant we derived for that reaction [Equation (2.20)] with the relationship derived here between K and the rate constants for elementary reactions to obtain Equation (3.22): 3 A + B )* D + 2 F (2.18) Koverall =
{D} {F}2
{A}3 {B}
= KR1 KR2 KR3 = eq
0 0 0 kf,R1 kf,R2 kf,R3 0 0 0 kr,R1 kr,R2 kr,R3
(3.22)
The final equality in Equation (3.22) indicates that Koverall for the nonelementary Reaction (2.18) can be computed as the ratio of the product of the forward rate constants to the product of the reverse rate constants for the constituent reactions. However, the numerator and denominator of this expression cannot be interpreted as forward and reverse rate constants for the overall reaction; i.e., one cannot express 0 0 0 {A}3 {B}, nor the rate of the the rate of the forward overall reaction as kf,R1 kf,R2 kf,R3 2 0 k0 k0 reverse reaction as kr,R1 r,R2 r,R3 {D} {F} . This point can be demonstrated by considering the rate at which the products of the overall reaction are generated. Based on the reaction scheme, the rate of creation 0 {C} {E}. Thus, increasing the concentrations of A and B of F at any instant is kf,R3 would not necessarily have any immediate effect on the rate of formation of F (e.g., 0 0 if kf,R1 and kf,R2 were both very small, the added A and B would not be converted to C and E for a long time). Since increasing the activities of A and B might or might not have any immediate effect on the rate of formation of D and F, it is clear that the 93
0 0 0 {A}3 {B}; a rate of the overall forward reaction cannot be computed as kf,R1 kf,R2 kf,R3 similar argument applies to the reverse reaction. Generalizing this result, we conclude that, unlike the case for elementary reactions, the forward and reverse rates of nonelementary reactions cannot be expressed by any simple or universal relationship involving a rate constant and the activities of the reacting species. Rather, detailed understanding of the reaction steps and their individual rates is required to derive the rate expression for nonelementary reactions. Nevertheless, for both elementary and nonelementary reactions, equilibrium is defined by equal rates of the overall forward and reverse reactions (i.e., by rnet = 0), and for both groups, the extent of disequilibrium can be quantified as log(Q/K). Thus, regardless of whether a reaction is elementary or not, values of Q/K that are much larger or much smaller than 1 indicate that the reaction is approximately irreversible.
Example 3.3 The conversion of Fe2+ to Fe3+ by reaction with dissolved oxygen can be represented by the following reaction, which has an equilibrium constant of 2.95 ⇥ 108 . Fe2+ + 0.25 O2 (aq) + H+ )* Fe3+ + 0.5 H2 O A solution at pH 6.8 contains 2.0 mg/L Fe2+ , 0.01 mg/L Fe3+ , and 0.1 mg/L O2 (aq). If all solutes behave ideally, in which direction will the reaction proceed? Is it reasonable to treat the reaction as irreversible? Solution The molar concentrations of Fe2+ , Fe3+ , and O2 (aq) are ⇥ 2+ ⇤ Fe = ⇥ 3+ ⇤ Fe = [O2 (aq)] =
2.0 mg/L = 3.57 ⇥ 10 56, 000 mg/mol
5 mol
0.01 mg/L = 1.79 ⇥ 10 56, 000 mg/mol
7 mol
0.1 mg/L = 3.13 ⇥ 10 32, 000 mg/mol
6 mol
L L L
The assumption of ideal solute behavior means that the activity of each solute can be equated with its molar concentration. In addition, the activity of H+ is, by definition, 10 pH , and we will assume that the activity of water is ⇠1.0. The reaction quotient for the given conditions is therefore Q=
=
Fe3+ {H2 O}0.5
{Fe2+ } {O2 (aq)}0.25 H+ 1.79 ⇥ 10
3.57 ⇥ 10
5
7
(1.0)0.5
3.13 ⇥ 10 94
6 0.25
10
6.8
= 7.50 ⇥ 105
The reaction quotient is less than the equilibrium constant, indicating that the reaction will proceed by conversion of the reactants to the products. Furthermore, the ratio Q/K is only 0.0025, meaning that the reaction is far from equilibrium, so it can be considered irreversible under the specified conditions.
3.7.2
Kinetics of Nonelementary Reactions
Detailed kinetic information is rarely available for the elementary reactions that combine to generate overall reactions of environmental interest. The overall rate expressions are therefore usually developed empirically, with a goal of characterizing the dependence of the reaction rate on the parameters that have the greatest influence over the range of conditions likely to be encountered. Typically, the approach is to hypothesize the general form of the rate expression and then conduct experiments to test that hypothesis. Using rate expressions for elementary reactions as models, the rate expressions that are initially hypothesized for nonelementary reactions are often “power law” equations that include the concentrations or activities of the reactants raised to various powers. However, unlike for elementary reactions, the exponents do not necessarily equal the reactants’ stoichiometric coefficients. Thus, for example, a rate expression of the following form might be postulated for the net rate of formation of A by Reaction (2.18): rA,net = kf,A {A}a {B}b + kr,A {D}d {F} f
(3.23)
where a, b, d, and f (i.e., the order of the reaction with respect to A, B, D, and F, respectively) are treated as empirical constants that must be determined from experiments and that have no fundamental relationship to the reaction stoichiometry. Note that, although the exponents in Equation (3.23) are decoupled from the reaction stoichiometry, the rates of destruction or formation of the reacting chemicals must still conform to the stoichiometry. Thus, for example, the reactants and products other than A must be generated at rates that are related to rA,net by rA,net = 3rB,net = 3rD,net =
3 rF,net 2
(3.24)
If we write an expression analogous to Equation (3.24), but for the rate of reaction of B, and then divide Equation (3.24) by that expression, we obtain rB,net = kf,B {A}a {B}b + kr,B {D}d {F} f
(3.25)
kf,A {A}a {B}b + kr,A {D}d {F} f
(3.26)
rA,net = rB,net
kf,B {A}a {B}b + kr,B {D}d {F} f 95
However, Equation (3.24) indicates that rA,net /rB,net equals 3. Combining that observation with Equation (3.26), we conclude that kf,A = 3kf,B and kr,A = 3kr,B . This result illustrates that, if a reaction includes species with stoichiometric coefficients other than 1, the reaction rate and the rate constant must be referenced to a particular species to avoid ambiguity. For example, a statement that “the forward rate constant for Reaction (2.18) is xyz” is ambiguous, because such a statement does not make clear whether the specified value is kf,A or kf,B [i.e., whether it applies to Equation (3.23) or (3.25)]. Equations (3.23), (3.25), and analogous expressions for rD,net and rF,net all refer to the same reaction. It is therefore both possible and convenient to identify a single rate expression that applies to all of the reactants and products identically. To do this, we define one mole of stoichiometric reaction as conversion of reactants to products in amounts that correspond to one mole times the corresponding stoichiometric coefficients. For example, for Reaction (2.18), one mole of stoichiometric reaction occurs when 3 moles of A and one mole of B are converted to one mole of D and two moles of F. The net rate of the reaction could then be written as follows, without reference to any particular species: r(2.18) = kf {A}a {B}b
kr {D}d {F} f
(3.27)
In such a case, a single set of kf and kr values apply, because the rate expression describes the number of moles of stoichiometric reaction per unit volume of solution per unit time, not the rate of appearance or disappearance of any individual species. The rate of reaction with respect to each species can then be computed as ri = ni r(2.18)
(3.28)
where ni is the stoichiometric coefficient for i in the reaction. Example 3.4 We found in Example 3.3 that, in a solution that has pH= 6.8 and contains 2.0 mg/L Fe2+ , 0.01 mg/L Fe3+ , and 0.1 mg/L O2 (aq), the reaction converting ferrous ions and dissolved oxygen to ferric ions is out of equilibrium, with Q < K. (a) Express rO2 (aq) for this reaction in terms of rFe2+ . (b) If 90% of the available dissolved oxygen is consumed in the reaction, how many moles of stoichiometric reaction have occurred?
96
Solution (a) According to the reaction stoichiometry shown in Example 3.3, 0.25 mole of dissolved oxygen reacts per mole of Fe2+ reacting, so 1 rO2 (aq) = rFe2+ 4 (b) For the given reaction, one mole of stoichiometric reaction corresponds to conversion of 1 mole of Fe2+ , 0.25 mole of O2 (aq), and 1 mole of H+ to 1 mole of Fe3+ and 0.5 mole of H2 O. The solution described in Example 3.3 contained 3.13 ⇥ 10 6 mol/L O2 (aq), so consumption of 90% of this oxygen corresponds to ◆✓ ◆ ✓ mol stoichiometric reaction 6 mol O2 (0.90) 3.13 ⇥ 10 1 L 0.25 mol O2 = 1.13 ⇥ 10
3.8
5 mol
stoichiometric reaction L
EXPERIMENTAL EVALUATION OF REACTION KINETICS
In theory, the coefficients in a reaction rate expression could be evaluated by investigating the rate under a wide variety of conditions (i.e., reactant concentrations) and testing whether the data fit a particular hypothesized rate expression. However, using this approach to simultaneously evaluate the rate constants and the orders of the reaction with respect to all the relevant species would likely require extensive experimentation and complex data analysis. Therefore, simplified approaches have been developed that allow evaluation of a single parameter at a time. The most common such simplifications involve limiting the experimental conditions to ranges where the reaction can be considered irreversible and the concentrations of all but one of the constituents remain approximately constant throughout the experiment. In such a scenario, if the reaction rate was hypothesized to follow a power law, the rate expression could be approximated as rA = kn [A]n
(3.29)
where A is the one reactant whose concentration changes significantly during the experiment. When this approach is used, the concentration of A is typically analyzed in a batch reactor as a function of time, and the data are analyzed using one of two methods. 97
In the integral method, a mass balance on the reactant is written and integrated. Assuming that the only process that alters the concentration of A is the reaction, the rate of change of the concentration of A (i.e., d [A]/dt ) can be equated with the reaction rate, so7 d [A] = rA dt Z[A]t
[A]0
d [A] = rA
Zt
(3.30)
dt
(3.31)
0
Next, a guess can be made for the order of the reaction with respect to A, and the corresponding rate expression can be substituted into Equation (3.31). For instance, if we guess that the reaction is first order with respect to A, we can substitute k1 [A] for rA and integrate the equation to obtain Zt
Z[A]t
d [A] = k1 [A]
ln
[A]t = k1 t [A]0
(3.33a)
[A]t = [A]0 exp ( k1t)
(3.33b)
[A]0
dt
(3.32)
0
The two forms of Equation (3.33) show that, if the guessed rate expression is correct, the concentration of A will decline exponentially over time, and a plot of ln [A]t /[A]0 versus t will be linear with a slope of k1 , as shown in Figure 3.4. There is, of course, no guarantee that the hypothesized rate expression will accurately reproduce the experimental observations. If it does not, we would conclude that the rate expression is not first order in A, make an alternative guess, and repeat the process. Expressions for the concentration of A over time for any nth -order irreversible reaction proceeding in a batch reactor are summarized in Table 3.2. While first- and second-order kinetics are commonly reported, reactions can be 7 Formally,
d[A]/dt is the rate of change of the concentration of A in the reactor. This rate depends not only on the reaction rate, but also on all other processes that bring A into, or take it out of, the reactor (e.g., advection, transfer from the gas phase, etc.). The term rA , on the other hand, refers solely to the rate of change of the concentration of A in the reactor due to chemical reactions. Therefore, rA equals d[A]/dt if and only if the only process affecting the concentration of A is the reaction. For a thorough discussion of mass balances in aquatic systems and ways to evaluate rate expressions using data from batch and other types of reactors, see Benjamin, M.M. and Lawler, D.F. Water Quality Engineering: Physical/Chemical Processes, John Wiley & Sons, Hoboken, NJ (2013).
98
Figure 3.4 Characteristic profiles for decay of a reactant undergoing an irreversible, firstorder reaction in a batch reactor.
of other (and sometimes even fractional) orders. In such cases, guessing the correct value of n is difficult, if not impossible. However, it is sometimes possible to determine both n and kn by an analysis of the time required for a given fraction of the reactant to disappear. For example, by substituting [A]t = 0.5[A]0 into either of the integrated rate expressions in the final row of Table 3.2, we can solve for the amount of time needed for the concentration of A to decline by 50%; this value of t is referred to as the half-time of the reaction or the half-life of A, and is designated t1/2 . The result is 8 n 1 2 1 > > [A]1 n if n 6= 1 (3.34a) > > < kn (n 1) 0 t1/2 = > > > ln 2 > : if n = 1 (3.34b) k1 lnt1/2 = ln
2n 1 kn (n
1 + (1 1)
n) ln [A]0
if n 6= 1
(3.35)
These results indicate that, if a reaction proceeds according to the rate law rA = kn [A]n , a plot of ln t1/2 versus ln [A]0 will yield a straight line with a slope of 1 n. Thus, the reaction order with respect to A can be determined from the slope of such a plot without guessing the value of n in advance. If n 6= 1, the y-intercept of such a 2n 1 1 plot equals ln , so once n is determined from the slope, kn can be evaluated kn (n 1) from the intercept. If n = 1, Equation (3.34b) indicates that t1/2 is independent of the initial concentration of A, and k can be evaluated as (ln 2)/t1/2 . 99
100
Any n 6= 1
Two
One
Zero
rA =
d [A] = kn [A]n dt
d [A] rA = = k0 dt d [A] rA = = k1 [A] dt d [A] rA = = k2 [A]2 dt
Reaction Differential form order d [A] = k0 Z
Z
dt
Z
dcA = kn cnA
Z
dt
d [A] = k1 dt [A] Z Z d [A] = k dt 2 [A]2
Z
Z
Integral form [A]t
[A]0 [A]t 1 1 k2t = [A]t [A]0 ⌘ 1 ⇣ 1 n knt = [A]t [A]10 n n 1 k1t = ln
k0t = [A]0
Integrated expression for kit
Rate Expression
k0t [A]t = [A]0 exp( k1t) ✓ ◆ 1 1 [A]t = + k2 t [A]0 ⇣ ⌘ 11n [A]t = [A]10 n + (n 1) knt
[A]t = [A]0
Integrated expression for [A]t
Table 3.2 Expressions for the concentration of reactant A as it undergoes a reaction with rate expression rA = kn [A]n in a batch reactor
An alternative approach that can be used to characterize the rate expression without postulating a value for n is known as the differential method. This method relies on the following logarithmic transformation of Equation (3.29): ln ( rA ) = ln kn + n ln [A]
(3.36)
Equation (3.36) indicates that, if the reaction is nth order with respect to A, a plot of ln ( rA ) versus ln [A] should be linear, with slope n and intercept ln kn . Therefore, if the reaction rate is determined experimentally at several values of [A] and the data are plotted as indicated by Equation (3.36), a linear plot implies that the reaction is indeed nth order in A and allows n and kn to be evaluated; if the plot is not linear, the conclusion is that the dependence of the rate on A cannot be represented by a simple power-law expression. The rate expressions developed above are for irreversible reactions in which A is the only species whose concentration changes significantly during the reaction. The conditions that must be met for the reaction to be approximately irreversible have been described previously (viz., it must be far from equilibrium). The approaches for maintaining approximately fixed concentrations of other species include buffering the solution composition and starting with a concentration of the target constituent that is much less than that of the other reactants. The former approach is often used to maintain a constant concentration of H+ in solutions, while the latter is used in most other circumstances.8 A study that investigated the reaction of bisphenol A (BPA) with hypochlorous acid illustrates both approaches.9 BPA is used in the production of various plastics and can leach from those plastics into water. It has been found in trace concentrations in natural waters, drinking water, wastewater, and landfill leachate, and is of concern because it can interfere with the endocrine systems of animals by mimicking the biological activity of estrogen. The structure of BPA is shown in Figure 3.5. When BPA is contacted with HOCl, chlorine attacks the aromatic rings to form a variety of reaction products (some of which are of equal or more health concern than unreacted BPA). If the reaction proceeded to equilibrium under conditions typically encountered in disinfection processes, essentially all of the BPA would be destroyed. Thus, as long as any BPA remains, the reaction is far from equilibrium and can be considered irreversible. In the study, batch experiments were first conducted with initial concentrations of 1 µmol/L total BPA (T OT BPA) and 38 µmol/L total OCl (T OT OCl). It was anticipated that up to four HOCl molecules would react per BPA molecule. As a result, 8 Buffering
is achieved by adding chemicals to the system that rapidly replenish certain species if they are depleted, or consume those species if they are generated. Approaches for buffering the H+ activity (and therefore the pH) are described in Chapter 8. 9 Gallard, H., Leclercq, A., and Crou´e, J P. (2004) “Chlorination of bisphenol A: Kinetics and byproducts formation.” Chemosphere 56, 465-473.
101
Figure 3.5 Molecular structure of BPA
even if all the BPA reacted, T OT OCl would decline by only about 10%; T OT OCl could therefore be treated as approximately constant under the experimental conditions. The speciation of both T OT BPA and T OT OCl is sensitive to pH (as shown for OCl in Example 2.6), so the solutions were buffered to maintain a constant pH during each experiment. Several experiments were conducted at each pH of interest. To model the reaction kinetics, the authors postulated the power-law rate expression rBPA = kf [T OT BPA]a [T OT OCl]b (3.37) However, based on the assumption that [T OT OCl] was approximately constant, this expression could be simplified to rBPA = kobs [T OT BPA]a
(3.38)
where the ‘observed’ first-order rate constant, kobs , equals kf [T OT OCl]b . Some of the experimental results are plotted as ln ([T OT BPA]t /[T OT BPA]0 ) vs. t in Figure 3.6. The linearity of the plot at each pH indicates that a = 1; i.e., the reaction is first order in TOTBPA. Therefore, as indicated by Equation (3.33a) and Figure 3.4, kobs can be evaluated as the negative of the slope; e.g., kobs, pH 6.75 = slopepH 6.75 = 1.43 ⇥ 10
3
s
1
(3.39)
Experiments at other TOTOCl values (still at the same pH, and still with TOTOCl much greater than TOTBPA) were then conducted to assess the dependence of the reaction rate on TOTOCl. In each case, a plot of ln ([T OT BPA]t /[T OT BPA]0 ) vs. t was linear, and kobs was determined from the slope. These kobs values were then plotted against TOTOCl, as shown in the inset in Figure 3.6. The linearity of this plot indicates that kobs is directly proportional to TOTOCl, implying that the reaction is first order with respect to TOTOCl and allowing kf to be evaluated from the slope. Based on all the experimental results, the overall reaction rate at pH 6.75 was reported as L rBPA,pH 6.75 = 36 [T OT BPA] [T OT OCl] (3.40) mol-s 102
Figure 3.6 Kinetics of the loss of TOTBPA from aqueous solutions initially containing 1.0 µmol/L TOTBPA and 38 µmol/L TOTOCl, with the pH buffered at various values. Inset is for experiments at various initial TOTOCl concentrations at pH 6.75. From Gallard et al. (2004) (see footnote 9).
Example 3.5 Plot the expected TOTBPA as a function of time in a batch reactor for initial concentrations of 0.1 µmol/L TOTBPA and 15 µmol/L TOTOCl at pH 7.5. Gallard et al. (2004) estimated kf to be 130 L/(mol·s) at this pH. How long would be required to destroy 99% of the BPA in this solution? Solution As in the study described previously, the initial value of TOTOCl is much larger than that of TOTBPA, so TOTOCl can be considered constant, and the BPA destruction reaction can be considered pseudo-first order: ✓ ◆ L 6 mol [T OT BPA] 15 ⇥ 10 = 1.95⇥10 3 s 1 [T OT BPA] rBPA,pH7.5 ⇡ 130 mol·s L The BPA concentration therefore declines exponentially, consistent with Equation (3.33b): ⇥ ⇤ [T OT BPA] = [T OT BPA]0 exp 1.95 ⇥ 10 3 s 1 t (3.41a) ⇥ ⇤ = [T OT BPA]0 exp 0.117 min 1 t (3.41b) This trend is plotted in Figure 3.7.
103
Figure 3.7 Decay of TOTBPA by reaction with chlorine at pH 7.5, under the example conditions.
The time required for 99% destruction of TOTBPA can be found by rearranging Equation (3.33a) and setting [T OT BPA] equal to 0.01[T OT BPA]0 : ln
[T OT BPA] = ln 0.01 = [T OT BPA]0 t=
3.9
ln 0.01 0.117 min
1
0.117 min
1
t
= 39.4 min
RATE-LIMITING STEPS AND SOME CLASSICAL, MODEL REACTION PATHWAYS
To complete this introduction to the kinetics of nonelementary reactions, we consider two important subclasses of such reactions – sequential irreversible reactions, and a sequence consisting of one reversible and one irreversible reaction. When irreversible reactions proceed in series, the process can be compared to flow of water through a series of tanks, with a valve controlling the resistance to flow between any two tanks. Figure 3.8a shows such an arrangement when the first tank is filled and its outlet is closed. When the valve is opened and flow commences, water accumulates in all the tanks, but to different depths, with water building up preferentially in the tanks whose valves provide the greatest resistance (Figure 3.8b). Later, the tanks upstream of the tank with the highest resistance drain completely, but water remains 104
in that tank and, to a lesser extent, in the downstream tanks (Figure 3.8c). If the difference in resistance of the valves was even greater than in the illustration, at some point essentially all the water remaining in the system would be in the tank with the high-resistance outlet valve, and the rate at which water passed through the whole network would be controlled by the rate at which it flowed out of that tank. That is, increasing or reducing the resistance of that step would affect the rate at which water appeared at the outlet, whereas altering the resistance of other steps would have almost no effect.
Figure 3.8 Water flowing through a series of tanks. (a) Initial condition, with the outlet from the first tank closed; (b) A short time later, with all valves open, but with the resistance imposed by the valve downstream of the third tank much larger than that of any of the other valves; (c) Later, after the first two tanks have drained.
Similarly, irreversible chemical reactions can be thought of as having a resistance that must be overcome for reactants to be converted to products. If an overall reaction comprises a series of such steps, the step with the greatest resistance exerts the greatest control over the rate of the overall reaction; this step is referred to as the rate-limiting step or rate-limiting reaction. The resistance associated with a chemical reaction is more complex than that of a valve, because it can change in response to a change in the concentrations of the reactants (analogous to the resistance of a valve depending on the depth of water in the upstream tank). The easiest way to quantify this resistance is in terms of a characteristic time, defined as the time for a given amount of reaction to occur under the conditions of interest. The longer the characteristic time, the greater the resistance to the reaction. One common definition for the characteristic time of an irreversible reaction that is nth order in a single constituent is the time that would be required for the constituent to be completely destroyed in a batch reactor if the reaction rate 105
stayed at its initial value continuously. Based on the equations in Table 3.2, this characteristic time can be computed as tchar =
1 kn [A]n0
(3.42)
1
Like the half-time, tchar for first-order reactions is independent of the concentration of A, but tchar for reactions of any other order depends on the initial concentration of the reactant. The idea of the rate-limiting step can be demonstrated for a hypothetical sequence of a second-order reaction followed by a first order reaction, with both reactions irreversible: A !B !P rA = kA [A]2 ;
rB = kA [A]2
kB [B] ;
rP = kB [B]
(3.43)
Note that the first reaction in the sequence is second-order in A, even though the stoichiometric coefficient of A is 1; because the rate expression does not conform to the stoichiometry, we infer that the reaction is not elementary. Assume that kA is 500 (mol/L) 1 min 1 , kB is 0.01 min 1 , and that we wish to predict the reaction progress in a batch reactor that initially contains 10 3 mol/L A and no B or P. The rate of conversion of A to B is independent of whether or not B participates in subsequent reactions, so the decline in the concentration of A conforms to the equation in Table 3.2 for a second-order irreversible reaction. Changes in the concentration of B, on the other hand, reflect a balance between its formation from A and its conversion to P. Because the reaction is taking place in a batch reactor, d [B] /dt can be equated with rB , so d [B] = kA [A]2 kB [B] (3.44) dt Similarly, the rate of formation of P at any instant can be found from the mass balance on P. Because P can be formed only by destruction of B and does not participate in any other reactions, this mass balance is simply d [P] = kB [B] dt
(3.45)
The results from the simultaneous solution of the three mass balance equations in conjunction with the specified initial condition are shown in Figure 3.9. The analysis indicates that A is depleted quickly and that B builds up to approximately 0.8 mmol/L within 15 min, before decaying to P over the next few hours. According to Equation (3.42), the characteristic times of the two reactions can be computed as 1/(kA [A]0 ) and 1/kB , with values of 2 min and 100 min, respectively. These characteristic times suggest that the second reaction imposes more resistance 106
Figure 3.9 Concentration versus time profiles for the example, sequential reactions.
and is therefore rate-limiting. Applying this idea, we can make the approximation that A is converted to B instantaneously, simplifying the overall reaction to the single reaction B ! P, with [B]0 = 0.001 mol/L. The predicted concentration of P when that approximation is applied is shown in Figure 3.9 by the curve labeled P⇤ . The curve closely follows that obtained from the more complete analysis, suggesting that, by identifying the rate-limiting step, we can significantly simplify the evaluation of the reaction kinetics without much loss of accuracy. Note that, in an analysis like the preceding one, the characteristic reaction time for each step is calculated assuming that that reaction step is rate-limiting. This assumption implies that essentially all of the initial reactant has been converted to the reactant for the step being investigated. As a result, the appropriate values to use for the initial concentrations in the calculation are not the actual initial concentrations of the reactants, but rather the hypothetical concentrations that would exist if all the upstream reactions had proceeded to completion, and all the reactants were “detained” just upstream of the step being evaluated. The benefits of carrying out such calculations and identifying the rate-limiting step increase dramatically with increasing complexity of the overall reaction. Example 3.6 The widely prescribed anti-bacterial agent ciprofloxacin (‘cipro’) is incompletely metabolized by humans and is therefore routinely found at low concentrations in domestic wastewater. Dodd et al. (2005)10 reported that the reaction of this compound 10 Dodd,
M.C., Shah A.D., von Gunten, U., and Huang, C.H. (2005) “Interactions of fluoroquinolone antibacterial agents with aqueous chlorine: Reaction kinetics, mechanisms, and transformation pathways.” Environ. Sci. Technol. 39, 7065-7076.
107
with HOCl can be modeled as occurring in two steps, comprising incorporation of a chlorine atom into an intermediate species and subsequent cleavage of that species to release small organic molecules and a Cl– ion. RI z }| { Cipro + HOCl )* Intermed-Cl )* Organic Products + Cl | {z } RII
Both reactions are first-order with respect to each of the reacting species and, at pH 5.5, the rates of the reactions are consistent with the following expressions: rRI = 6.68 ⇥ 104 M rRII = 3.11 ⇥ 10
1 4
s
s
1 1
[Cipro] [HOCl]
[Intermed-Cl]
A solution at pH 5.5 containing 1⇥10 6 M Cipro is dosed with 2⇥10 5 M HOCl. Predict the concentrations of the intermediate and the product for reaction times up to 10, 000 s (about 2.8 hours), if RI and RII are the only reactions occurring. Which step is rate limiting? Solution The reaction of Cipro is first order with respect to Cipro and HOCl, and second order overall. We can develop a spreadsheet to compute the rate of this reaction at any instant, assume that that reaction rate applies over a short time step Dt, and then compute the decline in the concentrations of the two reactants as the product rRI Dt. A new rate can then be determined for the next time step based on the concentrations at the end of the first time step, and the process can be repeated for the duration of interest. Alternatively, if we wish to simplify the calculations, we can assume that the concentration of HOCl is approximately constant throughout the reaction, since its initial concentration is 20 times that of the Cipro. A similar process can be carried out to compute the changes in the concentrations of the intermediate species and the products, except that in these cases the changes during any time step are (rRI rRII ) Dt and rRII Dt, respectively. The results of the calculations are shown in Figure 3.10. Essentially all the Cipro disappears within the first 5 seconds, but the intermediate reacts much more slowly. As a result, after the reaction has proceeded for 5 s, the concentration of the intermediate is very close to the initial concentration of Cipro, i.e., 10 6 M. It then decays gradually over the next few hours, and the product accumulates. Clearly, in this case, reaction RII is rate limiting, and the time required for reaction RI to proceed is virtually irrelevant if the primary interest is the generation of the product.
108
Figure 3.10 Concentrations of the reactants, intermediate, and products of the reaction of Cipro with HOCl under conditions described in the problem statement. (a) Changes during the first minute of reaction; (b) Longer-term changes.
Another reaction sequence that has been widely applied to model nonelementary reactions was first developed for enzyme-based reactions. This sequence begins with a reversible, rapidly equilibrating reaction between an enzyme (E) and a substrate (S) to generate an intermediate species (ES⇤ ). The intermediate is then irreversibly converted to the product (P) by a reaction that is first order in ES⇤ , re-generating the enzyme in the process. This sequence can be represented as K
S + E )* ES⇤
k1
! P+E
(3.46)
When this reaction model is applied, the reacting species are invariably assumed to behave ideally, so the equilibrium constant for the first reaction (K) is numerically 109
equal to the concentration ratio of the products to the reactants; we represent this concentration ratio as [K]. In addition, because the enzyme is not consumed, its total concentration (T OT E, the sum of [E] and [ES⇤ ]) remains constant as S is converted to P, so we can write [K] =
[ES⇤ ] [ES⇤ ] = [E] [S] ([T OT E] [ES⇤ ]) [S] [ES⇤ ] =
[T OT E] [K] [S] 1 + [K] [S]
(3.47) (3.48)
Inserting this expression for ES⇤ into the rate expression for formation of P yields rP = k1 [ES⇤ ] = k1
[T OT E] [K] [S] 1 + [K] [S]
(3.49)
In the microbiological literature, it is common to rewrite Equation (3.49) in terms of a new parameter, Km , equal to the inverse of [K]. That is, rP = k1
[T OT E] (1/Km ) [S] [T OT E] [S] = k1 1 + (1/Km ) [S] Km + [S]
(3.50)
Finally, the concentration of ES⇤ in the solution at any time is assumed to be much less than the initial concentration of S, so that essentially all the S present initially either remains in that form or has been converted to P. In such a case, the rate of formation of P can be equated with the rate of disappearance of S, so rS = k1
[T OT E] [S] Km + [S]
(3.51)
The dependence of rP on [S] for such a reaction and the concentration changes over time in a batch system are shown in Figure 3.11. In systems where the substrate is being consumed by a microbial culture (as opposed to an abiotic system with purified enzymes), the enzyme concentration is typically assumed to be proportional to the biomass concentration, [X]. In such cases, Equation (3.51) is commonly rewritten as follows: rS kmax [S] = [X] Km + [S]
(3.52)
where kmax = k1 [T OT E] / [X]. The left side of Equation (3.52) represents the rate of substrate consumption per unit amount of biomass present and is known as the specific rate of substrate utilization. Inspection of the right side of the equation shows that, when [S] Km , the specific rate of substrate utilization is maximized and equals kmax . Also, when [S] = Km , the specific rate of substrate utilization is 0.5kmax ; for this reason, Km is called the half-saturation constant. 110
Figure 3.11 Classic kinetics of enzyme-based reactions, based on the sequence shown in Reaction (3.46). (a) Product formation rate as a function of substrate concentration; (b) Changes in concentration in a batch system. The parameters used in these simulations are Km = 1/K = 8.3 ⇥ 10 4 mol/L; k1 = 3 min 1 ; T OT E = 10 6 mol/L; in part (b), [S]0 = 10 3 mol/L.
Both Equations (3.51) and (3.52) are known as the Michaelis-Menten equation. When Michaelis-Menten kinetics apply, the rate of the overall reaction is first order in T OT E or X under all conditions, but the order with respect to S varies from approximately zero (when [S] Km ) to approximately one (when [S] ⌧ Km ). Because the equilibration of S and E with ES⇤ is assumed to be rapid and therefore to impose negligible resistance on the progress of the reaction, the conversion of ES⇤ to P is always rate-limiting. 111
Example 3.7 The rate of an enzyme-mediated reaction is consistent with the Michaelis-Menten equation, with Km = 3 ⇥ 10 3 M, [T OT E]= 10 6 M, and k1 = 85 min 1 . (a) How long would it take to convert 99% of the initial substrate S to product P in a batch reactor if the initial concentration of S was 10 2 M? (b) Develop a curve showing the expected value of [S] during the first four hours of reaction, for the system in part (a).
Solution (a) We wish to determine the time required to reduce the substrate concentration from 10 2 M to 10 4 M. Because the reaction is taking place in a batch reactor, we can substitute d[S]/dt for rS in Equation (3.51). Making that substitution and integrating the resulting expression, we obtain d[S] [T OT E] [S] = k1 dt Km + [S] [S(t)] Z
[S(0)]
[S(t)] Z
[S(0)]
Km + [S] d[S] = k1 [T OT E] [S]
d[S] Km + [S]
Zt
dt
0
[S(t)] Z
d[S] = k1 [T OT E]t
[S(0)]
[S(t)] + {[S(t)] [S(0)]} = k1 [T OT E]t [S(0)] ⇢ 1 [S(t)] Km ln + {[S(t)] [S(0)]} k1 [T OT E] [S(0)]
Km ln t=
Substituting the known values yields ⇢ 1 10 4 M 3 t= 3 ⇥ 10 M ln + 10 4 M 10 2 M 85 min 1 (10 6 M)
10 2 M
= 279 min
(b) We can develop the [S] vs. t curve for the system by substituting a range of values for [S] in the expression derived in part (a) and solving for the corresponding values of t. The resulting plot is shown below.
112
The preceding discussion of irreversible reactions in series and Michaelis-Menten kinetics describes only two of a virtually infinite variety of possible combinations of elementary reactions that lead to the overall reactions observed at a macroscopic scale. The particular examples presented have been chosen not because they are more plausible than other combinations, but because they are relatively simple to develop, they lead to interesting patterns of change in constituent concentrations over time, and they have been found to characterize many overall reactions in important practical systems. Regardless of the reaction scheme or rate expression that is ultimately adopted to describe an overall reaction, the procedure is as described here – collect experimental data, postulate a rate expression, and iterate on the functional form of that expression until it reproduces the data satisfactorily.
3.10
HETEROGENEOUS (PHASE-TRANSFER) REACTIONS
Reactions in which solutes enter a solution by transferring from a gas phase or by dissolution of a solid, or in which solutes exit the solution by the reverse processes, play a central role in determining the composition of water bodies and the rates of many geochemical cycles. When the kinetics of these phase-transfer reactions are modeled, the physical transport of the constituents up to or away from the interface (i.e., solute transport) is typically dealt with separately from the step in which they actually cross the interface and enter the other phase. Two models for the kinetics of these reactions are widely used in environmental systems, differing with respect to whether solute transport or phase transfer is assumed to be the rate-limiting step. Both models are discussed in this section. 113
The model based on phase transfer as the rate-limiting step is applied primarily to precipitation and dissolution of solids, in which case phase transfer requires the formation or breaking of a bond with the solid. In this model, transport between the bulk solution and the interface is assumed to proceed very rapidly (e.g., due to intense mixing), so the composition of the solution is assumed to be uniform from the bulk liquid right up to the interface. We next develop the key equations for this model using an example system in which particles of SiO2 (s) are added to an intensely mixed, ideal solution that initially contains no dissolved silica. In such a system, a driving force would exist for SiO2 molecules to be released from the surface and enter the solution. Assume, for now, that the rate-limiting elementary reaction in the dissolution process is collision of a water molecule with a molecule of SiO2 at the surface of the solid, causing the surface molecule [which we will designate as SiO2 (surf )] to enter solution as H2 SiO3 (aq). When the SiO2 (surf ) molecule dissolves, the SiO2 (s) molecule “below” it in the solid phase is converted to an SiO2 (surf ) molecule, so the overall reaction can be represented as follows:11 SiO2 (surf) + H2 O )* H2 SiO3 (aq) SiO2 (s) )* SiO2 (surf) SiO2 (surf) + SiO2 (s) + H2 O )* SiO2 (surf) + H2 SiO3 (aq)
(3.53) (3.54) (3.55)
Although the reaction does not affect the concentration of SiO2 (surf ) in the system (because one such molecule is covered up and a new one is generated), we show that species on both the reactant and product sides of the reaction to emphasize that it does participate in the reaction (and because its concentration is a key parameter in an equation that is derived shortly). However, because Reaction (3.54) is an immediate and direct consequence of Reaction (3.53), the rate of the overall reaction equals the rate of just Reaction (3.53), so we can write: rf = kf [SiO2 (surf)](H2 O) = kf [SiO2 (surf)]
(3.56a) (3.56b)
where [SiO2 (surf )] is the molar concentration of SiO2 (surf ) molecules in the system (i.e., with units of moles of SiO2 (surf ) per liter of solution). This concentration can be expressed as the product of the concentration of SiO2 in the surface layer (sSiO2 (surf) , with units like moles of SiO2 (surf ) per m2 ) and the concentration of surface area in the system, ([ASiO2 (s) ], m2 per liter of solution): h i [SiO2 (surf)] = sSiO2 (surf) ASiO2 (s) (3.57) 11 Throughout this section, the parameters r
designation, all refer to Equation (3.55).
f , rr , rnet , kf , kr , and K, when shown without any additional
114
Substituting this expression into Equation (3.56b), we obtain h i rf = kf sSiO2 (surf) ASiO2 (s)
(3.58)
Equation (3.58) suggests that the rate of solid dissolution will be proportional to the total surface area of SiO2 (s) particles dispersed in the solution. For example, it suggests that SiO2 (s) will enter solution much more rapidly if the solid is present in the form of many 0.1-mm particles than if it is present as a single 1.0-mm particle, even if both systems contain the same mass of SiO2 (s) per liter of solution. Essentially the same logic applies to the precipitation reaction [i.e., the reverse of Reaction (3.55)]. That process can be envisioned as being initiated by a collision of a dissolved H2 SiO3 (aq) with a surface molecule. The collision causes the H2 SiO3 to split apart, with the SiO2 portion attaching to the solid (and thereby covering up a surface molecule), and the remaining portion (an H2 O molecule) returning to solution. The reaction rate is therefore h i rr = kr (H2 SiO3 (aq))sSiO2 (sur f ) ASiO2 (s) (3.59) where the assumption that the solution behaves ideally allows us to use either the molar concentration or activity of H2 SiO3 (aq) in the rate expression. The net solid dissolution rate is rnet = rf
rr
h i = kf sSiO2 (sur f ) ASiO2 (s) = kf
h i kr (H2 SiO3 (aq))sSiO2 (sur f ) ASiO2 (s) h i kr (H2 SiO3 (aq)) sSiO2 (sur f ) ASiO2 (s) (3.60)
At equilibrium, the forward and reverse reaction rates are equal, and the net reaction rate is zero. Therefore, the equilibrium activity of dissolved H2 SiO3 is (H2 SiO3 (aq))eq =
kf kr
(3.61)
Substituting this result into Equation (3.60) shows that, like first-order reactions that occur entirely in one phase, phase-transfer reactions proceed at rates that are proportional to the extent of disequilibrium. However, in phase-transfer reactions, the rate is also proportional to the concentration of interfacial area in the system:
r = kr
⇢
kf kr
h i (H2 SiO3 (aq)) sSiO2 (sur f ) ASiO2 (s)
= kr (H2 SiO3 (aq))eq
h i (H2 SiO3 (aq)) sSiO2 (sur f ) ASiO2 (s) 115
(3.62)
Equation (3.61) demonstrates that the concentration of H2 SiO3 (aq) in equilibrium with the solid is independent of the concentration of either bulk solid ([SiO2 (s)]) or SiO2 surface species ([SiO2 (surf )]). Thus, although the rate of the phase transfer reaction depends on how much surface area is available to react, the equilibrium condition does not. The preceding discussion relied on an analysis of the reaction between SiO2 molecules at the solid surface and those in solution, and did not directly address those in the bulk solid. However, we can incorporate the activity of the bulk solid into the analysis by writing the equilibrium constant for Reaction (3.55), as follows: (H2 SiO3 (aq))eq kf /kr = {SiO2 (s)}(H2 O) SiO2 (s) ⇢ kf /kr {SiO2 (s)} = = constant K(3.55)
K=
(3.63a) (3.63b)
Equation (3.63b) indicates that the activity of SiO2 (s) is the same in any system where it is present. In particular, its activity is independent of the amount of solid present, consistent with the discussion in Chapter 2 about the activity of any pure solid. The numerical value of the fixed activity of SiO2 (s) is not established by Equation (3.63b) and depends on the choice of the standard state conditions. However, as explained in Chapter 2, the universal convention is to make choose the standard state in a way that causes the constant value of the solid activity to be 1.0. A second model for the rate of phase-transfer reactions is based on the assumption that equilibrium is reached very rapidly at the interface, and that the rate of solute transfer between the two phases depends primarily on the rate at which the transferable species migrate through the boundary layer between the interface and the bulk fluid phase.12 This migration is assumed to proceed by molecular diffusion and therefore to be characterized by Ficks law: dci (3.64) dx where Ji,x is the flux (mass transported per unit area of interface per unit time) of species i in the x direction, Di is the diffusion coefficient of i, and dci /dx is the local concentration gradient of i.13 Ji,x = Di
12 When this model is applied to transfer between solution and a gas phase, boundary layers exist on both sides of the interface, but if the resistance to transport is much greater in one boundary layer than in the other, the overall kinetics can be modeled considering only the high-resistance layer. We make that assumption and consider only a single boundary layer in this analysis. Some discussion of systems where both boundary layers contribute to the overall resistance is provided at the end of this section. 13 Fick’s law actually predicts that the flux is proportional to the gradient in the activity, not the concentration, of i. However, it is almost universally written in terms of the concentration gradient, implicitly assuming ideal behavior of the species.
116
The earliest model of this process assumed that the fluid in the boundary layer was stagnant, that ci at the interface (ci,int ) was controlled by the rapid equilibration with the other phase at that location, and that ci at the outer edge of the boundary layer equaled that in the bulk fluid (ci,b ). In that case, if the concentrations at the edges of the boundary layer change slowly compared to the time frame needed for diffusion through the layer, a steady-state concentration profile develops in the boundary layer, in which the concentration changes linearly across the layer (i.e., the concentration gradient in the boundary layer is constant). In that case, Equation (3.62) indicates that the flux of i is constant through the layer. Defining x to be positive in the direction from the interface to the bulk fluid, the flux into the fluid is Ji,x = Di
ci,b
ci,int dBL
= Di
ci,int ci,b dBL
(3.65)
where dBL is the thickness of the boundary layer, which is expected to depend on factors like the geometry of the interface and the intensity of fluid mixing in the system. Equation (3.65) is commonly written as follows: Ji,x = kmt,i (ci,int
ci,b )
(3.66)
where kmt,i equals Di /dBL and is referred to as the mass transfer coefficient for i in the boundary layer. Subsequent experiments and modeling conducted after the stagnant-layer model was developed suggested that boundary layers in systems with any significant amount of mixing are actually characterized by continuous exchange of fluid packets between the boundary layer and the bulk fluid. Like the stagnant-layer model, models for the kinetics of phase-transfer reactions that take this exchange into account assume that equilibrium is achieved rapidly right at the interface and that transport through the boundary layer occurs by molecular diffusion. However, in the latter models, packets of fluid are envisioned to depart the boundary layer before a linear concentration gradient develops. Rather, while those packets are in the boundary layer, the concentration profile is curvilinear, with the intensity of the curvature gradually decreasing the longer the packet spends in the layer before being swept back into the bulk solution. A schematic illustrating the gradually changing shape of the concentration profile is shown in Figure 3.12. The mathematics of models that take fluid exchange between the boundary layer and the bulk solution into account are more complex than those for the stagnantlayer model and are not critical for the current discussion. The key outcome is that, as in the simpler model, the flux through the boundary layer can be expressed as the product of a mass transfer coefficient and the concentration difference across the boundary layer. However, in this model, the mass transfer coefficient is predicted to be given by 117
Figure 3.12 The changing concentration profile of a dissolved species in a packet in the liquid boundary layer adjacent to a gas/liquid or solid/liquid interface. In this hypothetical system, species are migrating from the nonaqueous phase into solution, so the concentration declines from the interface to the bulk solution phase. If the packet stays in the boundary layer long enough, a linear profile develops, as predicted by the stagnant-layer model. (After Benjamin and Lawler, 2013 [see footnote 1]).
kmt,i =
r
" ⇢ • Di 1 + 2 Â exp pt j=1
2 j2 dBL Dit
#
(3.67)
where t is the time that the packet spends at the interface. Now, consider how we would apply model for transport-limited phase transfer reactions to the SiO2 (s) dissolution reaction that we evaluated previously. In this scenario, the interfacial concentration of H2 SiO3 (aq) would be the fixed value in equilibrium with the solid. Inserting that value into Equation (3.66), we obtain n o JH2 SiO3 (aq) = kmt,H2 SiO3 (aq) H2 SiO3 (aq) eq H2 SiO3 (aq) (3.68) The rate at which H2 SiO3 (aq) enters the solution is the product of the flux into solution and the concentration of interfacial area, so h i r = JH2 SiO3 (aq) ASiO2 (aq) n oh i = kmt,H2 SiO3 (aq) H2 SiO3 (aq) eq H2 SiO3 (aq) ASiO2 (aq) (3.69)
Comparing Equations (3.61) and (3.69), we see that both models predict that the rate of dissolution is directly proportional to the extent of disequilibrium between the two phases and to the volume-based concentration of interfacial area. Where the 118
models differ is in the interpretation of the coefficients that appear in the governing equations. Equation (3.61) includes two such coefficients - the rate constant of the precipitation reaction and the concentration of SiO2 (surf ) molecules in the surface layer - whereas Equation (3.69) includes just one - the mass transfer coefficient for i in the boundary layer [defined either as Di /dBL or by the expression in Equation (3.67)]. Modeling of gas-transfer processes almost always assumes that transport through the liquid and/or gas boundary layer (as opposed to transfer across the gas/liquid interface) is rate-limiting. Transport through the two individual layers is linked by the assumption of equilibrium at the interface and by the fact that the flux out of either phase must equal the flux into the other. When this assumption and constraint are combined with two applications of an equation like Equation (3.69) (one written for each boundary layer), the following expressions are obtained for the flux and the overall transfer rate of a species from the bulk gas to the bulk aqueous solution, assuming ideal solute behavior:14 ⇣ ⌘ Ji = KL,i c⇤i(aq) ci(aq) (3.70) ⇣ rgt,i = KL,i aL c⇤i(aq)
ci(aq)
⌘
(3.71)
In these equations, KL,i is the overall gas transfer coefficient, with dimensions of length per time (e.g., cm/s); c⇤i(aq) is the hypothetical concentration of dissolved i that would be in equilibrium with the bulk gas phase; ci(aq) is the concentration of dissolved i that is actually present in the solution; and aL is the volume-based concentration of interfacial area in the system.15 KL,i is related to the mass transfer coefficients through the individual boundary layers and the gas/liquid equilibrium constant by KL,i =
kL,i kG,i Keq kL,i + kG,i Keq
(3.72)
where Keq is the equilibrium constant for dissolution of gaseous i into the solution. Equation (3.73) indicates that, if kL,i ⌧ kG,i Keq , then kL,i ⇡ kL,i ; i.e., the overall gas transfer coefficient is approximately equal to the mass transfer coefficient through the liquid boundary layer. This approximation implies that most of the resistance to gas transfer resides in the liquid boundary layer, and it applies to environmental 14 The derivation of Equation (3.70) as well as a more thorough discussion of both the conceptual basis
and the mathematics of models that account for fluid exchange between a bulk phase and a boundary layer are provided in Chapter 5 of Water Quality Engineering: Physical/Chemical Treatment Processes by Benjamin, M.M., andhLawler, D.F., i J. Wiley & Sons, Hoboken, NJ (2013) 15 a L
is analogous to ASiO2 (s)(aq)
(aq)
in Equation (3.57); by convention, this parameter is always
represented as a or aL in the gas transfer literature.
119
gas transfer systems (especially for species that are only slightly soluble in solution, like oxygen). In cases where kL,i kG,i Keq , the overall gas transfer coefficient is approximately equal to kG,i Keq ; this situation can arise for highly soluble gases (like ammonia), and it implies that most of the resistance to gas transfer is in the gas-phase boundary layer. Example 3.8 Wastewater is being treated in a pond that is aerated by a fountain aerator that sprays droplets with an average diameter of 0.4 cm into the air. The pond water contains 1.5 mg/L dissolved oxygen, the liquid-phase mass transfer coefficient is 0.02 cm/s, and almost all the resistance to gas transfer is in the liquid boundary layer. What will the dissolved oxygen concentration be just before a droplet enters the pond after spending 5 s in the air, if the equilibrium dissolved oxygen concentration is 10.5 mg/L? Solution While the droplet is in the air, it can be treated as a batch reactor, so rgt,O2 (aq) = dcO2 (aq) /dt. Also, because the gas transfer resistance is almost all in the liquid boundary layer, KL ⇡ kL . The ratio of interfacial area to liquid volume in the drop is aL =
pd 2 6 6 = = = 15 cm 3 pd /6 d 0.4 cm
1
We can insert these values and the other, given information into Equation (3.72) to find dcgt,O2 (aq) dt
Z
c⇤O
⇣ ln c⇤O
dcgt,O2 (aq)
2 (aq)
2 (aq)
cO2 (aq)
cO2 (aq)
⌘
5s 0
⇣ = KL,O2 (aq) aL c⇤O = KL,O2 (aq) aL
Z
= KL,O2 (aq) aLt
2 (aq)
dt 5s 0
mg cO2 (aq) |5 s ⇣ cm ⌘ L ln = 0.02 15 cm mg mg s 10.5 1.5 L L 10.5
cO2 (aq) |5 s = 8.5
120
mg L
cO2 (aq)
1
(5 s)
⌘
3.11
SUMMARY
All observable chemical reactions are either elementary reactions – that is, they occur in a single step via a collision among the reacting molecules – or combinations of elementary reactions. Elementary reactions require an initial input of energy to destabilize existing chemical bonds, after which those bonds might re-form or break and allow new bonds to form. The minimum energy input needed to destabilize the bonds and allow reactants to be converted to products is called the activation energy of the reaction. The lower the activation energy, the faster a reaction proceeds. In addition, all such reactions proceed at a rate that is proportional to the activities of the reacting species. However, it is common to write the rate expression in terms of the concentrations of the reactants, thereby incorporating the activity coefficients and standard state concentrations into the rate constants. Nonelementary overall reactions are the net result of a group of elementary reactions that proceed in parallel and/or series. The rates of all reactions increase with increasing temperature. This relationship can be quantified by the Arrhenius equation. When the Arrhenius equation is applied to both the forward and reverse directions of a reaction, the result is the van’t Hoff equation, in which the key parameter is the standard molar enthalpy of reaction (DH r ). In principle, all chemical reactions are reversible. The net effect of the simultaneous forward and reverse reactions is that all reactions proceed toward equilibrium, a condition that is characterized by the absence of any net production of either reactants or products by the reaction. Such a scenario can be interpreted as the result of a balance in which the forward and reverse reactions proceed at equal rates, or a condition of minimum chemical potential energy. The ratio of the activities of products to reactants for a reaction is called the reaction quotient, Q. Based on the recognition that the forward and reverse reaction rates must be equal for a reaction at equilibrium, the value of Q for a given reaction must always be the same once the reaction has reached equilibrium, although that value can be achieved via many different combinations of activities for the individual reacting species. The value of Q at equilibrium is called the equilibrium constant, K. For elementary reactions, K equals the ratio of the forward to the reverse activitybased rate constants; for nonelementary reactions, this latter relationship does not apply, but K can still be computed as the ratio, at equilibrium, of the activities of the products to those of the reactants. For reactions that are not at equilibrium, Q does not equal K, and the reaction proceeds in the direction that causes Q to approach K. Any overall reaction can be derived as the sum of a group of other (elementary or nonelementary) reactions; the equilibrium constant for the overall reaction equals the product of the K values of the reactions in that summation. The derivation of this relationship emphasizes that chemical activities and equilibrium constants are simply 121
numbers that can be manipulated according to the rules of algebra, independent of the fact that they have chemical significance. The ratio Q/K [or log(Q/K)] is an indicator of the extent of disequilibrium. If log(Q/K) is greater than ⇠1.5, the reaction is proceeding much faster in one direction than the other, and the overall reaction rate can be computed considering only the faster reaction. In such cases, the reaction is effectively irreversible. The rates of nonelementary reactions can depend in complex ways on the reactant concentrations and the rate constants for the constituent elementary reactions. As a result, the rate expressions for such reactions cannot be predicted without detailed knowledge of the reaction pathway and the rates of the individual steps, information that is frequently unavailable for reactions of environmental importance. In these cases, rate expressions for the overall reactions are derived empirically, often by studying the reaction kinetics under simplified conditions where the reaction is effectively irreversible and the rate depends on the concentration of a single constituent. When reactions proceed in series, one reaction often provides the majority of the resistance, and that reaction is referred to as the rate-limiting step in the overall reaction. If the resistance associated with that step is much larger than the resistance of the rest of the sequence, the kinetics of the overall reaction can be closely approximated by considering the kinetics of that step alone.
3.12
PROBLEMS
1. An elementary reaction A )* B has an equilibrium constant of 10, and a forward rate constant of 1.0 d 1 . (a) Plot the concentrations of A and B as a function of time in a batch reactor that initially contains 10 3 M each of A and B. Show the data from the initial condition until the concentrations are changing at an instantaneous rate of less than 10 5 M per day. (b) Repeat part (a) if the initial concentrations are 2 ⇥ 10
3M
A and no B.
(c) What are the molar concentrations of A and B at equilibrium in part (b)?
2. The reaction 2 A )* B + C is elementary in both directions and has an equilibrium constant of 75. In a batch solution that contains 10 mmol/L A and no B or C, the initial rate of reaction generates 0.3 mmol B per liter per minute. What is the net rate of formation of B under conditions where the concentrations of A, B, and C are 2, 10, and 10 mmol/L, respectively? 3. A reversible reaction that is elementary in both directions has the following stoichiometry: A + B )* 2 P 122
Hydroxide ion has been found to catalyze (i.e., to increase the rate of) the forward reaction. How would the following parameters change when the pH is increased? Briefly explain your reasoning. (a) The activation energy for the forward reaction. (b) The rate constant for the reverse reaction. 4. Two reversible, elementary reactions proceed in the following sequence: A + B )* 2 C )* D + E The rate constants for the reactions are as follows, where ‘1’ and ‘2’ refer to the first and second reaction, respectively: k1,f = 0.04 M
1
·s
1
k1,r = 0.01 M
1
·s
1
k2,f = 0.10 M
1
·s
1
k2,r = 0.10 M
1
·s
1
(a) If a solution initially contains 60 mmol/L of A, 40 mmol/L of B, and no C, D, or E, what will the distribution of species be at equilibrium? (Hint: the equilibrium solution composition comprises five unknowns [the concentrations of the five species], which can be determined by writing and solving five independent equations involving those unknowns. The available equations include mass balances that characterize the stoichiometry of the reactions, and equilibrium constant expressions. Try to write and solve five independent equations that characterize the system.) (b) What is the equilibrium constant for the reaction 0.5A + 0.5B )* C? 5. A network of elementary reactions is shown below, along with rate constants for the forward (f) and reverse (r) of each reaction. Note that species A is consumed in reactions 1 and 2 and is then generated in reaction 3. Similarly, species D is generated in reaction 2 and then consumed in reaction 3. The rate constants apply when the species’ activities (not their concentrations) are used in the reaction rate expressions. The units of ‘mol’ refer to moles of stoichiometric reaction. A + B )* C + H2 O 2 A + C )* D D + H2 O )* E + A
kf,1 = 6.6 ⇥ 10 kf,2 = 4.0 ⇥ 10 kf,3 = 2 ⇥ 10
7
3
mol/L·s,
5
mol/L·s,
mol/L·s,
kr,1 = 3 ⇥ 10 kr,2 = 1 ⇥ 10
11 9
mol/L·s
mol/L·s
kr,3 = 0.10 mol/L·s
(a) Write the overall reaction corresponding to the sum of these elementary reactions and determine the equilibrium constant that applies to that reaction. 123
(b) A system that contains all the species that participate in these reactions has reached equilibrium. Do you think any of the elementary reactions can be treated as irreversible under these conditions? Why or why not? 6. The data below summarize the loss of a solute from water due to a chemical reaction in a batch reactor. Determine whether the data are best fit by a zeroorder, first-order, or second-order rate expression, and report the rate constant with proper units. Time (h) 0 0.5 1.0 2.0 3.0 5.0
Conc’n (mg/L) 100 27.0 11.1 6.5 3.7 2.3
7. The following data were collected in a laboratory batch experiment investigating the degradation of a contaminant in a water treatment process. Are the data consistent with an nth -order reaction rate? If so, what are the values of n and the rate constant? [Hint: Derive an equation that approximates the given c vs. t data, and then use that equation to estimate the half-time for several different initial concentrations. Then, apply Equation (3.35) to see if the data fit an nth -order rate expression.] Time (min) 0 5 10 15 20 25 30
Conc’n (µM) 250 220 200 170 155 145 130
Time (min) 35 40 45 50 55 60
Conc’n (µM) 105 100 90 75 70 60
8. The radionuclide 32 P has a half-life of 14.3 days. How long would an aqueous waste containing 1.0 mg/L of this nuclide have to be stored to reduce the concentration to 0.03 mg/L through radioactive decay? 9. The following reaction is elementary but is reported to be “pseudo-first-order” with a rate constant of 10 1 s 1 . CO2 (aq) + H2 O ! H2 CO3 124
Explain why the reaction is pseudo-first-order, even though it requires a collision between two molecules. Present an appropriate equation to support your explanation. 10. The decomposition of ozone in water fits the general rate equation ⇥ ⇤a ⇥ ⇤b rO3 (aq) = kd O3 (aq) OH
Staehelin and Hoigne (1982)16 reported the data in the following table for the decrease in the O3 (aq) concentration over time at pH=10 in batch reactors. Expt 1*
Expt 2
Expt 3
[O3 ]
Time
[O3 ]
Time
[O3 ]
Time
0 15 30 45 60 75 90
12.00 8.76 6.36 5.04 4.08 3.24 2.16
0 10 30 40 50 60 70 80 90
3.00 2.49 1.77 1.50 1.23 1.02 0.90 0.75 0.63
0 10 20 30 40 50 60 80
0.330 0.261 0.225 0.180 0.159 0.123 0.114 0.075
* In all experiments, the units for [O3 ] and time are µM and s, respectively.
(a) Determine the order of the reaction with respect to O3 (aq). (b) Write the rate equation and evaluate kd , assuming b = 1. (c) Still assuming that b = 1, estimate the pseudo-first-order rate constant for ozone decay at fixed pH values of 8.0, 9.0, and 10.0. What is the half-life of ozone in water at each of these pH’s? 11. Consider the following irreversible reaction for conversion of chlorophenol (C6 H4 OHCl, or ‘CP’) to dioxin [(C6 H3 ClO)2 ]. 2 C6 H4 OHCl + 0.5 O2 (aq) ! (C6 H3 ClO)2 + H2 O rCP = k[CP]2 [O2 (aq)]0.5 If k = 10 L1.5 /(mol1.5 · s), how long would be required to form 10 9 M dioxin in a batch reactor if the initial solution contained 10 7 M chlorophenol and [O2 (aq)] was maintained at 2.5 ⇥ 10 4 M?
16 Staehelin,
J., and Hoigne, J. (1982) “Decomposition of ozone in water: Rate of initiation by hydroxide ions and hydrogen peroxide.” Environ. Sci. Technol. 16, 676-681.
125
12. At pH below about 3.5, ferrous iron (Fe2+ ) oxidizes in streams according to the following overall reaction: Fe2+ + 0.25 O2 (aq) + H+ )* Fe3+ + 0.5 H2 O The rate law for the abiotic oxidation of ferrous iron under these conditions and at 20 C is ⇥ ⇤ rFe2+ = k Fe2+ PO2 where PO2 is the partial pressure of gas-phase oxygen that is in equilibrium with the solution, and k = 10 3.2 atm 1 ·d 1 .
Calculate the time required for the concentration of Fe2+ to be reduced to 1% of its initial concentration in a batch reactor where the solution is in continuous equilibrium with the atmosphere. 13. When chlorine is added to water containing ammonia, chloramine compounds (NHx Cl3 x , with x=0, 1, or 2) form. Monochloramine (NH2 Cl) can split (‘disproportionate’) into dichloramine and ammonia according to the following reaction and rate expression: 2 NH2 Cl ! NHCl2 + NH3
rNH2 Cl = k(NH2 Cl)2
If k = 20 M 1 ·s 1 , what will the monochloramine concentration be 10 min after chlorine is added, if the concentration immediately after dosing is 0.1 mM and no more NH2 Cl forms thereafter? 14. Hypochlorous acid (HOCl) can react with ferrous iron (Fe2+ ) to produce ferric iron (Fe3+ ) via the following reaction: 2 Fe2+ + HOCl ! 2 Fe3+ + Cl– + OH– Two experiments are carried out with solutions that initially contain 0.01 M Fe2+ , no Fe3+ , and 10 3 M HOCl. In one, conducted at 10 C, the concentration of Fe3+ generated in 1 minute is 1.0 ⇥ 10 5 M. When the temperature is increased to 20 C, 2.4 ⇥ 10 5 M Fe3+ is generated in one minute. Assuming the rate expression has the following form, calculate the rate constant at each temperature and estimate the activation energy for the reaction. ⇥ ⇤2 rFe3+ = k Fe2+ [HOCl]
15. The hydrolysis of pyrophosphate can be described by the reaction H2 PO27 + H2 O ! 2 H2 PO4
If the pH is held constant, the reaction is pseudo-first order with respect to H2 PO2– 7 . 126
(a) If, at a given pH, the half-life of pyrophosphate is 140 h at 75 C and 13 h at 100 C, what is EAr for the reaction? (b) Estimate the time required for 50% hydrolysis of pyrophosphate in a solution at the same pH as in part (a), but at 20 C. 16. In the presence of dissolved oxygen, microorganisms can oxidize nitrite ions (NO–2 ) to nitrate ions (NO–3 ) via the following reaction: NO2 + 0.5 O2 (aq) ! NO3 (a) This reaction is occurring in a batch bioreactor at a rate that is first order with respect to both NO–2 and O2 (aq), with k2 = 3 ⇥ 104 L/mol·d. How much time is required for the nitrite concentration to be reduced ˙ if O2 is resupplied continuously to maintain the from 0.3 mM to 0.1mM O2 (aq) concentration at 4 mg/L? (b) Repeat part (a), but assume that O2 is not resupplied, so that the O2 (aq) concentration is initially 4 mg/L but declines as the reaction proceeds. 17. The stoichiometry of the reaction of hydrogen sulfide with dissolved oxygen is given by H2 S(aq) + 2 O2 (aq) ! SO24 + 2 H+
Millero et al. (1987)17 reported that, at pH < 7 and 25 C, this reaction is first order with respect to both H2 S and O2 , with a rate constant of k = 11 L/mol·h. Consider a solution containing 10 5 M H2 S(aq) and a constant dissolved oxygen concentration of 0.25 mM (8.0 mg/L). If the given reaction is the only way that sulfide is lost from the system, how much dissolved sulfide will remain after one day at pH 5? (As is explained in Chapter 5, at pH 5.0, H2 S(aq) accounts for virtually all the dissolved sulfide in the system.)
18. In the biochemical oxygen demand (BOD) test, a sample of water is inoculated with microorganisms and then incubated with essential nutrients to determine how much dissolved oxygen is consumed by the constituents in the sample under conditions favorable for microbial growth. The rate of oxygen consumption is typically modeled as first-order with respect to the concentration of dissolved oxygen that could be consumed by reaction with other constituents present in the solution; this parameter is conventionally represented as L. Thus: rO2 = k1 L A BOD bottle is filled with 270 mL of treated wastewater plus 30 mL of ‘dilution water’ containing the bacterial seed and nutrients. The constituents in 17 Millero,
F.J., Hubinger, S., Fernandez, M., and Garnett, S. (1987) “Oxidation of H2 S in seawater as a function of temperature, pH, and ionic strength.” Environ. Sci. Technol. 21, 439-443.
127
the treated wastewater have the potential to consume a maximum of 15 mg/L of oxygen. (This value is known as the ultimate BOD of the sample, L0 .) How long can the solution be incubated before the dissolved oxygen concentration decreases to 1.0 mg/L if the constituents in the dilution water consume almost no oxygen, the mixture of sample and dilution water initially contains 10.5 mg/L dissolved oxygen, and the rate constant for oxygen consumption is 0.45 d 1? 19. N-nitrosodimethylamine (NDMA) is a suspected carcinogen that can form via reactions between dimethylamine (DMA) and monochloramine (NH2 Cl). DMA is a precursor for the formation of many industrial chemicals and can enter water supplies as a contaminant in polymers that are used in water treatment. NH2 Cl is frequently generated intentionally in water supplies during water treatment operations, because it can disinfect water without forming chlorinated disinfection byproducts (DBPs). Choi and Valentine (2002)18 suggested that the relevant elementary reactions and the corresponding rate constants are as shown below.
Figure 3.13 Reaction pathway for formation of NDMA from DMA and monochloramine.
Reaction
1 · s 1)
Rate Constant (M
DMA + NH2 Cl )* DMCA + NH3 DMA + NH2 Cl ! UDMH
UDMH + NH2 Cl ! NDMA
kf = 1.4 ⇥ 10 1 , kr = 5.83 ⇥ 10 1.28 ⇥ 10 1.11 ⇥ 10
3
3 1
Predict the concentrations of all species that participate in these reactions as a function of time for 500 hours in a batch experiment with initial concentrations 18 Choi,J., and Valentine, R.L. (2002) “Formation of N-nitrosodimethylamine (NDMA) from reaction
of monochloramine: a new disinfection by-product.” Water Research 36, 817-824.
128
of 10 5 M DMA, 10 4 M NH2 Cl, 2 ⇥ 10 5 M NH3 , and zero for the other species shown. Explain the concentration trends qualitatively. 20. The Chick-Watson model characterizes the rate of inactivation of microorganisms by disinfectants. The model indicates that, for a given type of organism, disinfectant, pH, and temperature, ln(Nt /N0 ) = Lcnt where Nt is the concentration of living organisms at time t, c is the concentration of disinfectant, and L and n are constants. (These constants are sometimes referred to, respectively, as the coefficient of specific lethality and the coefficient of dilution.) The value of n is often approximately 1.0. The following values of L were reported by Scarpino et al. (1974)19 for inactivation of E. coli and poliovirus by HOCl and OCl– (the active ingredients in bleach) when c is in mg/L as Cl2 and t is in seconds. Disinfectant HOCl OCl–
E. coli bacteria 1.64 0.088
Poliovirus 1 0.042 0.20
(a) For a water at pH 7.5 that contains 10 4.5 M T OT OCl, calculate the concentrations of HOCl and OCl– in mol/L and in mg/L as Cl2 . (Hint: use the approach shown in Example 2.6 to determine the relative concentrations of HOCl and OCl– .) (b) Plot log(N/N0 ) vs. time (in seconds) for E.coli and for Poliovirus 1, for the solution described in part (a). Assume that n = 1, the disinfectant concentrations are approximately constant and that the effects of HOCl and OCl– are additive. How long is required to reduce the population of each organism by a factor of one million? 21. Fine particles suspended in water can collide as a result of differential settling rates, fluid mixing, and/or Brownian motion. Under certain conditions, these collisions cause the particles to aggregate into larger particles which are more likely than the original particles to settle out of the water column. By itself, the aggregation process reduces the number concentration of suspended particles (i.e., number of particles per unit volume of solution), but not the mass concentration. 19 Scarpino, P.V., Lucas, M., Dahling, D.R., Berg, G., and Chang, S.L. “Effectiveness of hypochlorous
acid and hypochlorite ion in destruction of viruses and bacteria.” Chapter 15 in Chemistry of Water Supply, Treatment, and Distribution. A.J. Rubin, ed. Ann Arbor Science (Ann Arbor, MI) (1974).
129
Aggregation is sometimes modeled as an irreversible rate process with second order dependence on the number concentration of suspended particles. Consider a suspension that initially contains 108 particles per liter and in which the rate constant for aggregation is 5 ⇥ 10 10 L/particle·s. (a) Determine the number concentration of particles remaining in suspension after 30 min. (b) How long would be required for the particle concentration to decline to 105 /L? 22. In the early sanitary engineering literature, there was a great deal of interest in the effect of temperature on the rates of biodegradation and oxygen transfer in aquatic systems, where the range of temperatures is limited to around 20 ± 20 C (293±20 K). Based on that work, an empirical model was developed suggesting that kT2 = q T2 T1 kT1 where q is an empirical constant. Typically, the value of k at T1 = 20 C was determined in the laboratory, and its value at other temperatures was estimated from the preceding equation. For T in C or Kelvins, commonly cited values of q were 1.047 for utilization of oxygen by microorganisms (BOD consumption) and 1.016 for oxygen transfer from the atmosphere to water. Compare the expressions for the temperature dependence of these two reactions to the Arrhenius equation and estimate EAr for each reaction.
130
4
POTENTIALS, ENERGY, AND FORCES:
INTERPRETING CHANGES IN PHYSICAL/CHEMICAL SYSTEMS
131
Contents 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14
4.1
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . POTENTIAL ENERGY, POTENTIALS, AND FORCES . . . OVERVIEW OF POTENTIAL AND POTENTIAL ENERGY IN CHEMICAL SYSTEMS . . . . . . . . . . . . . ENTROPY AND GIBBS ENERGY . . . . . . . . . . . . . . THE CHEMICAL POTENTIAL AND ITS USE TO PREDICT CHEMICAL CHANGE . . . . . . . . . . . . . OTHER MOLAR THERMODYNAMIC QUANTITIES . . . QUANTIFYING Gi , Hi , and Si . . . . . . . . . . . . . . . . . THE DEPENDENCE OF Gi ON SYSTEM COMPOSITION, TEMPERATURE, AND PRESSURE . . . . . . . . . . . . . COMPARISON OF Gi OF DIFFERENT SPECIES . . . . . . THE DEPENDENCE OF Gi ON PRESSURE AND TEMPERATURE . . . . . . . . . . . . . . . . . . . . . COMPARING THE MAGNITUDE OF FACTORS THAT AFFECT Gi . . . . . . . . . . . . . . . . . . . . . . . . . . . CHEMICAL POTENTIAL AS THE DRIVING FORCE FOR CHEMICAL REACTIONS . . . . . . . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . 132 . . . 134 . . . 143 . . . 144 . . . 148 . . . 150 . . . 151 . . . 158 . . . 166 . . . 168 . . . 178 . . . 179 . . . 204 . . . 206
INTRODUCTION
This chapter introduces the thermodynamic perspective on changes in chemical systems. This perspective has at its core the principle that underpins our understanding of all physical and chemical changes in the world — the second law of thermodynamics. A concise statement of the second law, some form of which all readers will certainly have encountered previously, is that any physical or chemical change in any system occurs if and only if it increases the total entropy of the universe. 132
Historically, the concept of entropy was first developed from the empirical evidence that, in systems isolated from their surroundings, certain types of spontaneous change always proceed in the same direction: hot objects transfer heat to cool ones, odors distribute themselves throughout a gas phase rather than stay localized, water ‘seeks its own level’, etc. In a truly prodigious intellectual feat based primarily on their studies of thermomechanical systems (systems that interact with their surroundings via heat exchange and mechanical work), engineers of the 19th century inferred the form and significance of the entropy function. The key insight from this effort is often expressed by some variation of the statement that isolated systems always change in a direction that decreases their ability to do useful work. Although “useful work” might sound imprecise and excessively anthropocentric, it actually relates to entropy and is a legitimate thermodynamic quantity. Later, in an equally impressive effort, physical chemists demonstrated that entropy could be explained and quantified via a statistical analysis of the number of ways that energy can be distributed in a system. The statistical analysis provided a satisfying explanation as to why universal entropy is always increasing and also extended the entropy concept to non-thermomechanical systems (e.g., systems in which the chemical composition can change). It is from the statistical analysis that the popular characterization of entropy as a measure of the “disorderliness” of a system derives, and from that, the characterization of the second law as postulating that the disorderliness of the universe is always increasing.1 Notwithstanding some excellent efforts to explain entropy in accessible terms, most people still find it easier to grasp analyses of the possible changes in a system in terms of potential energy (PE). PE is not as universal as entropy as a tool for predicting or explaining changes in the world, but it is an acceptable surrogate for entropy when the system of interest conforms to certain restrictions. For example, in certain systems where fluids can move around, but the temperature (T ), pressure (P), fluid velocity, and chemical composition remain constant, the gravitational potential energy can serve as a surrogate for entropy. In such systems, any change that decreases the gravitational potential energy of the system (e.g., downhill flow of the fluid) will increase the entropy of the universe and will therefore be favorable. In this text, our interest is primarily in environmental systems in which the temperature (T ) and pressure (P) remain approximately constant, but the chemical makeup of the system can change.2 In such cases, changes in universal entropy are closely related to changes in a function known as the Gibbs free energy, or simply the Gibbs 1 When used to explain entropy, ‘disorderliness’ is often presented in terms of the distribution of the mass in a system. However, entropy is inherently an energy-based idea, and the mass-based explanation is incomplete, at best. For example, precipitation of solids and coalescence of oil droplets both increase the orderliness of the available mass, but they can nevertheless occur spontaneously via processes that increase universal entropy. Explanations of entropy based on the disorderliness of the available energy, rather than mass, are more complete but, unfortunately, also more complex. 2 In fact, a host of other parameters, such as the gravitational potential energy, are also assumed to be constant. The constancy of these other parameters is taken for granted and is not usually mentioned explicitly.
133
energy, G, of the system. The reason that G is so useful is that, even though the value of G depends only on the properties of the system, changes in G at constant T and P implicitly account for changes in the entropy of both the system and the surroundings. As a result, the criterion for a spontaneous process can be expressed in terms of the change in G in the system rather than the change in the entropy of the whole universe. Derivations of the Gibbs energy function starting from first principles (the three fundamental laws of thermodynamics) are presented in virtually all texts dealing with thermodynamics and/or physical chemistry, and readers are directed to those texts for the corresponding details. Here, an overview of those derivations is provided, but the emphasis is on the conceptual meaning of the parameters and relationships, especially as they apply to aquatic chemistry. We begin with a brief review of some more familiar forms of potential energy, both because those forms of energy sometimes contribute to the Gibbs energy and because the discussion lays the foundation for understanding the other, less familiar forms of energy that we consider subsequently.
4.2 4.2.1
POTENTIAL ENERGY, POTENTIALS, AND FORCES Gravitational Potential Energy
Formally, the gravitational potential energy of an object i (PEgrav,i ) can be computed from its mass (mi ), its elevation (hi ), and the earth’s gravitational constant (g), as follows: Z
dPEgrav,i = mi g dhi
(4.1a)
dPEgrav,i = mi g
(4.1b)
Z
dhi
PEgrav,i + PE0grav,i = mi g hi + h0i
(4.1c)
where the two terms with the designation 0 are constants of integration. The values of hi and PEgrav,i in Equation (4.1) are not uniquely defined; they can only be quantified relative to arbitrarily assigned datum values. To specify these datum values, we first establish a scale for height by assigning h = 0 to an arbitrary elevation, and then choose a value of h at which PEgrav,i is defined to be zero. Designating this elevation as h⇤ , we can apply Equation (4.1c) at h⇤ to obtain 0 + PE0grav,i = mi g(h⇤ + h0i )
(4.2)
Subtracting Equation (4.2) from Equation (4.1c), we find PEgrav,i = mi g (hi
h⇤ )
= mi ghi + PE⇤grav,i 134
(4.3a) (4.3b)
where PE⇤grav,i = mi gh⇤ . Based on Equation (4.3b), PE⇤grav,i can be defined as the potential energy of i when it is at elevation hi = 0. Usually, the most convenient choice is to assign PEgrav,i = 0 when hi = 0 (i.e., to assign h⇤ = 0). In that case, PE⇤grav,i = 0, and Equation (4.3b) simplifies to PEgrav,i = mi ghi (4.4) This choice is made so commonly that Equation (4.4) is frequently presented as the defining equation for PEgrav,i . Formally, however, the datum conditions for h and PEgrav are independent, and it is perfectly acceptable to assign PEgrav a value of zero at some nonzero value of h. The relationships among several of the parameters defined in the preceding paragraphs are shown graphically in Figure 4.1.
Figure 4.1 Relationships among h, h⇤ , PEgrav,i , and PE⇤grav,i , as defined in the text. The solid line applies for an arbitrary choice of h⇤ (in this case, h⇤ < 0), and the broken line applies for the common choice of h⇤ = 0.
The total gravitational potential energy of a system, PEgrav,sys , is the sum of the potential energies of all the objects in the system. Applying this principle in conjunction with Equations (4.3a) and (4.3b), we find 0 1 PEgrav,sys =
Â
PEgrav,i =
all i in system
Â
all i in system
mi g (hi
h⇤ ) = @
Â
all i in system
mi ghi A + PE⇤grav,sys
(4.5)
where PE⇤grav,sys = Â PE⇤grav,i . Note that h⇤ and PE⇤grav,sys drop out when we calculate the change in PEgrav,sys between any two states for a system with constant total mass. That is, 135
0
DPEgrav,sys = D @
Â
all i in system
1
0
PEgrav,i A = D @
Â
all i in system
1
mi ghi A
(4.6)
Once the datum locations where h = 0 and where PEgrav,i = 0 have been chosen, the values of PEgrav,i and PEgrav,sys depend only on the current distribution of objects in the system (i.e., the state of the system), regardless of how it got to that condition. Correspondingly, DPEgrav,sys between two states of the system depends only on the characteristics of those states and not on the way the transition from State 1 to State 2 is carried out. For this reason, PEgrav is referred to as a state function or state variable. It is important to recognize that, when an object is raised or lowered in a system, DPEgrav,sys is not the total energy input required to cause the movement, but only the portion of that energy that is stored in the object as a result of the change in elevation. During such a move, some energy is unavoidably expended to overcome frictional resistance. Unlike DPEgrav,sys , this energy depends on the details of the move; different “pathways” require different inputs of energy to overcome friction, and one cannot know the magnitude of this input just by knowing the two endpoints. In the limit of negligible frictional resistance, the total energy required for the move approaches DPEgrav,sys . Thus, DPEgrav,sys can be identified as the theoretical, minimum amount of energy that must be expended to carry out the operation. The preceding discussion focused on a change in h for objects of fixed mass. However, PEgrav,sys can also change by adding mass to or removing mass from the system. The increment in the total gravitational potential energy of the system per unit mass added when a differential amount of mass, dm, is added at elevation h is called the gravitational potential at h; we will represent this quantity as PEgrav |h . Formally, PEgrav |h can be computed as PEgrav |h ⌘
∂ PEgrav,sys ∂ m|h ∂ Â mi g (hi
=
all i in system
∂ m|h
(4.7a) h⇤ ) (4.7b)
Equation (4.7b) can be simplified and understood by considering a specific example. Assume that we wish to evaluate PEgrav at an elevation, h1 , in a system that currently contains two objects — one at elevation h2 with mass m2 , and another at elevation h3 with mass m3 . In other words, we wish to know how the total potential energy of the system changes per unit amount of mass added at h1 . Expanding Equation (4.7b), and considering the mass added at h1 as well as the mass of the objects already at h2 and h3 , we can write 136
PEgrav |h1 = =
∂ [m1 g(h1
h⇤ ) + m2 g(h2 h⇤ ) + m3 g(h3 ∂ m |h1
∂ PE1 ∂ PE2 ∂ PE1 + + ∂ m |h1 ∂ m |h1 ∂ m |h1
h⇤ )]
(4.8a)
(4.8b)
The first term in the summation on the right side of Equation (4.8b) represents the increment in PEgrav at h1 per unit mass added at h1 ; this quantity is g (h1 h⇤ ). The second and third terms in the summation represent the increment in PEgrav at h2 and h3 , respectively, when mass is added at h1 . Clearly, the addition of mass at one elevation has no effect on the the potential energy stored at another location, so these terms are both zero. Equation (4.8b) can therefore be written as PEgrav |h1 = g (h1 = g (h1
h⇤ ) + 0 + 0
(4.9a)
h⇤ )
(4.9b)
Generalizing this result, we can conclude that, at any elevation, the gravitational potential is simply ⇤ PEgrav |h = g (h h⇤ ) = gh + PEgrav (4.10) ⇤
where PEgrav = gh⇤ and can be interpreted as the gravitational potential at h = 0. The key feature that makes the gravitational potential a useful parameter is that it characterizes the properties of the gravitational field independent of the number or types of objects that are in that field. In fact, as shown by the preceding calculation, we can even compute the magnitude of the gravitational potential at a location where no mass currently resides. Furthermore, comparison of the first equality in Equation (4.10) with Equation (4.3a) indicates that PEgrav,i |h = PEgrav |h mi
(4.11)
That is, if an object is present at a given location, the gravitational field interacts with it in such a way that the product of the gravitational potential and the object’s mass yields its gravitational potential energy. As we will see, this separation of potential energy into one strictly system-dependent component and one strictly objectdependent component is also applicable to other forms of potential energy and is particularly useful when applied to chemical systems. Sometimes, even though one datum is widely adopted for reporting the potential energy of an object or the potential at a given location, it is most convenient to measure those parameters relative to a different datum and then adjust the value so that it applies to the conventional datum. For example, one might wish to report an 137
elevation relative to sea level, but find it easier to measure the elevation relative to a local landmark, and then adjust the value based on the elevation of the landmark relative to sea level. Such adjustments are readily made based on the fact that potential energy and potential are state functions. Thanks to this property, the calculation of either parameter can be carried out in two steps as follows: Value in state of interest (state 2) using conventional datum = Value at alternate datum (state 10 ) based on conventional datum + Value in state 2 based on use of state 10 as datum
(4.12)
For example, say we are interested in the gravitational potential at an elevation h2 that is 75 m above street level at a location where street level is 1500 m above sea level. Assume that the conventional datum for both elevation and potential energy is sea level (i.e., both h and PEgrav are assigned values of zero at sea level, which we designate h1 ). Then, if we wish to use street level as the alternate datum, we can apply Equation (4.12) in conjunction with Equation (4.10) to write 0
PEgrav |h2 = PEgrav |h 0 + PEgrav |1h2 1
= g(1500 m
0 m) + g(1575 m
= g(1575 m)
1500 m) (4.13)
where the subscript identifies the location of the potential being reported, and the superscript identifies the datum level if it is different from the conventional one (i.e., the absence of a superscript indicates that the datum is at sea level). PEgrav at the alternate datum (i.e., PEgrav |h 0 ) is the same regardless of the eleva1
tion of interest (h2 ). Therefore, designating this value as PEgrav , we can evaluate the gravitational potential at any elevation h relative to the conventional datum as 0
PEgrav |h = PEgrav + PEgrav |1h = PEgrav + mi gh0
(4.14a) (4.14b)
where h0 is the elevation of interest measured from the alternate datum. Thus, if PEgrav is evaluated once, Equation (4.14) can be used to evaluate PEgrav,h at any elevation simply by measuring the difference between that elevation and the alternate datum. Although the logic and mathematics leading to Equation (4.14) are fairly simple, the underlying approach is broadly applicable and serves as a useful template for other types of potential for which the analysis might be less intuitive. One final relationship of note is that the force acting on an object i in the direction j (Fi, j ) equals the negative gradient of the object’s potential energy in that direction: Fi, j =
dPEi dl j
138
(4.15)
where l j is the distance in direction j, and the minus sign is required because the force is in the direction of decreasing PE. In gravitational systems, we are generally interested in downward movement, so that dl j = dh, and the equation simplifies to dPEgrav,i = dl j
Fgrav,i, j =
d(mi gh) = mi g dh
(4.16)
In other cases, the direction of the force and the direction of movement might not coincide, in which case the more general form [Equation (4.15)] must be used.
4.2.2
Electrical Potential Energy
Consider next the case of a charged particle i in an electrical field. Analogous to the situation for gravitational potential energy, if the particle moves, the change in its electrical potential energy can be quantified as the product of a property of the particle itself (in this case, its charge, si ) and the change in a parameter that characterizes the local electrical environment. This latter parameter is typically represented by the Greek letter Y, so, by analogy with Equation (4.1b), we can write Z
dPEelec,i = si
Z
dYi
(4.17)
Typical units of s and Y are coulombs and volts, respectively. Like PEgrav and h, values of PEelec and Y can be quantified unambiguously once datum levels for those parameters are defined. The key equations analogous to those developed in the preceding section for gravitational systems are then as follows: PEelec,i = si (Yi PEelec,sys =
Â
Y⇤ ) = si Yi + PE⇤elec,i PEelec,i =
all i in system
si (Yi
all i in system
(4.20) ⇤
Y⇤ = Yx + PEelec
PEelec |x = PEelec |x si 0
dPEelec,i dY = si dl j dl j 139
(4.21) (4.22)
PEelec |x = PEelec + PEelec |1x Felec,i, j =
(4.19)
Y⇤ )
∂ s |x
PEelec |x = Yx
Y⇤ )
all i in system
∂ Â si (Yi PEelec |x ⌘
Â
(4.18)
(4.23) (4.24)
where Yx refers to the electrical potential at a generic location x, and Yi refers to the potential affecting a specific object i. In these equations, Y⇤ is the value of Yi at which PEelec,i is defined to be zero. In practice, PEelec,i is virtually always chosen to equal zero at the same location where Yi is defined to be zero, so Y⇤ is zero [by Equation (4.18)]. As a result, PEelec |x = Yx [by Equation (4.21)], and both PEelec |x and Yx are commonly referred to as the electrical potential at x. The electrical potential plays a central role in certain oxidation/reduction and adsorption reactions, which are covered in Chapters 12 and 13, respectively. Example 4.1 A 1.0-mg particle with a positive charge of 10 5 coulomb (C) is in a region where the electrical potential varies in the vertical (h) direction according to the equation Yh = 30 mV·cm/h. Thus, the potential is 30 mV at h = 1 cm, 3 mV at h = 10 cm, and approaches zero as h gets very large. Since the potential increases in the downward direction and the particle is positively charged, the particle experiences increasing electrical repulsion as it moves downward. What is the equilibrium position of the particle if gravity and the electric field are the only two influences on its motion? Solution A One approach for solving this problem is to find the value of h at which the downward force due to gravity is exactly balanced by the upward electrical force, so the net force acting on the particle is zero. This equilibrium criterion can be expressed as Net force = Fgrav + Felec = 0 (4.25) In the scenario of interest, the direction of movement (l j ) is the vertical direction (h), so dl j /dh = 1. The gravitational and electrical forces are therefore given by Fgrav = mg Felec = s
=
dl j = dh
1 ⇥ 10 3 g
dY d = s dh dh 10
5
C
✓
✓
⇣
30 mV·cm h
30 mV·cm h2
cm ⌘ (1.0) = 0.98 dyn s2 ◆
980
◆✓
10
4 dyn·cm
mV·C
The term 104 dyn·cm/(mV·C) is a unit conversion identity.
140
◆
=
3.0 dyn·cm h2
Substituting these expressions into Equation (4.25), we find that, at equilibrium, 0 = Fgrav + Felec = 0.98 dyn +
3.0 dyn·cm2 h2
h = 1.74 cm We conclude that the particle will come to rest at a height 1.74 cm above the datum. Solution B A second approach that is only slightly different is to recognize that isolated systems change spontaneously in the direction that reduces their total potential energy, and that equilibrium therefore coincides with the condition of minimum potential energy. Using the conventional baselines for PE, so that PEgrav and PEelec are zero when h and Y are zero, respectively, the total potential energy of the particle is PEtot = PEgrav + PEelec = mgh + s Y |h
(4.26)
PEtot is plotted as a function of h in Figure 4.2.
Figure 4.2 Gravitational and electrical potential energy of a particle as a function of distance above the datum plane, for the system in Example 4.1.
The value of h at the minimum of the curve is found by differentiating the total potential energy function and setting it to zero:
141
dY dPEtot = mg + s dh dh ◆ ✓ d 30 mV·cm = mg + s dh h ✓ ◆ ⇣ cm ⌘ 1 dyn 3 = 1 ⇥ 10 g 980 2 + s g · cm/s2 ◆ ⇣ ⌘ ✓ 30 mV · cm ◆ ✓ 5 4 dyn · cm 10 C 10 h2 mV · C
0=
h = 1.74 cm
The result is once again that h = 1.74 cm. Thus, equilibrium can be identified either as the condition where there is no net force on the particle or that where the total potential energy of the system is minimized.
4.2.3
Mechanical Potential Energy
Objects can also store potential energy in a form that we will refer to as ‘mechanical’, in which the potential energy field is generated by pressure, and the parameter on which the pressure acts is volume. The existence of such potential energy is apparent from, for instance, the fact that pressure can push water through a horizontal pipe (i.e., in the absence of a favorable gravitational force); as the water moves, the mechanical potential energy declines (because pressure decreases) as work is done to overcome the friction at the pipe wall. Just as two objects with equal mass have the same gravitational potential energy at a given elevation, and two objects with equal charge have the same electrical potential energy at a given value of Y, two objects with equal volume have the same mechanical potential energy if they are exposed to the same pressure, regardless of their other properties. If the material in the system is incompressible, the mathematics are essentially identical to those shown previously for gravitational and electrical potential energy, yielding the following relationships: Z
PEmech = Vi
Z
dP
PEmech,i = Vi (Pi PEmech |x ⌘
(4.27) P⇤ ) = Vi Pi + PE⇤mech,i
∂ PEmech,sys = P |x ∂V |x
(4.28) ⇤
P⇤ = P |x + PEmech
(4.29)
where P⇤ is the value of P at which PEmech is assigned a value of zero, and PEmech |x is the mechanical potential of the system at location x. The mechanical potential is a key consideration when analyzing the effect of pressure on chemical reactions. 142
The preceding equations apply only to incompressible material because, if the material is compressible (e.g., if i is a gas), its volume changes as a function of pressure, and Equations (4.27) and (4.28) must be written as follows: Z
PEmech =
Z
d (Vi P)
PEmech,i = Vi Pi
Vi⇤ P⇤
(4.30) (4.31)
volume Vi , and Vi⇤
where Pi is the pressure acting on the material with is the volume ⇤ that i would occupy at P = P . In this case, the expression for PEmech |x [analogous to Equation (4.29), but for compressible materials] depends on how the volume at x is increased (i.e., how the ∂V |x is achieved). For a compressible material, volume can be increased either by increasing the amount of the material at x while holding the temperature constant, or by increasing the temperature while holding the amount of material constant. This issue is addressed in greater detail later in the chapter.
4.3
OVERVIEW OF POTENTIAL AND POTENTIAL ENERGY IN CHEMICAL SYSTEMS
An approach similar to that used to analyze gravitational, electrical, and mechanical potential energy can be applied to chemical systems, in which case the ‘objects’ that make up the systems are various chemical species, and the amount of each species in the system is typically expressed in moles. As in the earlier example in which we analyzed the gravitational and electrical potential energy of various possible system states to determine the equilibrium position of a charged object, we can analyze a chemical system to identify conditions that minimize its chemical potential energy and associate those conditions with equilibrium. However, in chemical systems, the possible states correspond to different chemical compositions and distributions rather than different physical locations of an object. The terms that must be included in the calculation of the total chemical potential energy depend, in part, on the constraints that apply to the particular system being analyzed. This is because our ultimate interest is in changes in chemical energy between two states of the system. If system constraints prevent certain energy terms from changing between the two states of interest, those energy terms need not be considered in the analysis, and so they need not be included in the calculation of the chemical potential energy. The most important constraint that we will impose is to restrict our analysis to systems in which no mass crosses the system boundaries (but thermal energy can do so). Such systems are referred to as closed systems. In addition, we note that most reactions of environmental interest occur in systems (natural water bodies, engineered reactors with water in tanks or basins, or the atmosphere) where the temperature and pressure change minimally during or as a result of the reaction. Therefore, when analyzing chemical reactions in a given system, 143
we will usually assume that the system temperature and pressure remain constant. However, we might be interested in a given reaction in several different environments, in each of which the temperature and pressure remain approximately constant over the time frame of interest. For example, we might wish to investigate the likelihood that calcium carbonate will precipitate in a water treatment plant in summer or winter, or near the bottom of a deep lake. In each case, the reaction would take place in an environment where T and P are approximately constant, but the specific values of T and P would be different in the different systems. We therefore will explore the effects of different, but constant, values of T and P on reaction thermodynamics.3
4.4
ENTROPY AND GIBBS ENERGY
We begin our analysis by considering the forms in which energy is stored in molecules. The contributors to this energy include the relatively strong bonds that hold the molecules together, weaker bonds between adjacent molecules, and energy associated with the molecules’ motion (both translation through space and internal motions such as vibration and rotation). These forms of energy are referred to collectively as internal energy, typically represented as U. The sum of the changes in internal energy and mechanical potential energy accompanying any change in a system is defined as the change in enthalpy, H: dH = dU + d (PV )
(4.32a)
DH = DU + D (PV )
(4.32b)
Like the gravitational, electrical, and mechanical potential energy, the internal energy and enthalpy of the system depend only on the current conditions in the system and not on its history, so they are state variables. As is true for other forms of energy discussed previously, changes in internal energy and enthalpy can be assessed unambiguously, but absolute values can be assigned to those parameters only by defining an arbitrary datum level for each. Once such datum levels are established, the three energy terms appearing in Equation (4.32) can be related by H = U + PV + H ⇤ (4.33) where H ⇤ is analogous to the PE⇤ terms in Equations (4.3b), (4.18), and (4.28) and can be interpreted as the value of H when both U and PV are zero; i.e., H ⇤ accounts 3 The phrase “constant temperature and pressure” is widely used to describe the constraint that is ap-
plied in the upcoming analyses. However, a more accurate statement is that the systems are constrained to have the same temperature and pressure at the end of the process as at the beginning; changes in those parameters are allowed as the process is carried out. Because the constraint is so commonly described as “constant temperature and pressure”, and because that phrase is so much more concise than the full explanation, it will be used throughout the chapter. Keep in mind, though, that the “constancy” of T and P refers only to a comparison of the initial and final states of the system.
144
for any differences in the zero-levels chosen for the three other terms in the equation. If the datum levels are chosen to coincide with one another, so that all three parameters are assigned values of zero under the same conditions, H ⇤ = 0, and Equation (4.33) can be written as4 H = U + PV (4.34) Because universal energy must be conserved in any process, a change in either U or PV in a system must be compensated by energy changes elsewhere in the universe. For example, if a chemical reaction converts one group of species to another that has stronger internal bonding, the energy associated with bonding declines.5 The energy that is released when the stronger bonds form might remain in the system as mechanical potential energy (if the system’s pressure or volume increased) or as internal energy (if the system’s temperature increased), or it could be transferred to the surroundings. In the former case, the total enthalpy of the system would remain constant, whereas in the latter it would decline. In this example scenario and most other chemical systems of interest, the only significant changes in energy are associated with changes in internal energy (U) and/or mechanical potential energy (PV ) and are therefore reflected in the enthalpy. The enthalpy is therefore commonly referred to as the energy content or heat content of such systems. The preceding discussion introduces some key energy-related parameters that appear in the first law of thermodynamics, which establishes the principle of energy conservation. However, our ultimate goal is to predict whether certain changes in chemical systems are thermodynamically favorable, and to do that we need to utilize the second law and bring entropy into the discussion. The second law defines entropy as a state variable and posits that any change in the universe is accompanied by an increase in total universal entropy, DStot . It is convenient to separate this total entropy change into two additive portions — one accounting for entropy changes within the system (DSsys ), and the other for entropy changes in the surroundings (DSsurr ): dStot = dSsys + dSsurr
(4.35a)
DStot = DSsys + DSsurr
(4.35b)
The key link between energy and entropy for systems of interest to us is an equation that can be derived by combining the first and second laws. As noted earlier, this derivation is presented in detail in virtually all basic texts on thermodynamics; here, 4 Similar to the case for the datum levels for PE grav , the datum levels for U, P, and H are usually chosen in a way that causes H ⇤ to be zero, so Equation (4.34) almost always applies. As a result, that equation is sometimes presented as the definition of enthalpy. However, Equation (4.33) is a more general form of the relationship among those parameters and is useful as a template for other relationships to be developed shortly. 5 As in other types of systems, increased stability in a chemical system is characterized by a decrease in potential energy. Thus, bond energy declines (and the energy is converted to some other form) when stronger bonds form.
145
only the result is shown. That result states that, for any change in a system at constant temperature and pressure, the entropy gained by the surroundings equals the enthalpy lost by the system divided by the temperature (which, by the assumption of constant T and P, must be the same in the system and surroundings); i.e., DSsurr =
DHsys T
(4.36)
Substituting this relationship into Equation (4.35b) and multiplying through by T , we find T DStot = T DSsys DHsys (4.37) Equation (4.37) achieves the important goal of relating the change in universal entropy to changes in parameters within the system, thereby eliminating the need to measure anything in the surroundings when determining whether a change will occur in the system. Although we could use this equation directly to determine whether a change will occur, the terms on the right side are typically combined into a single parameter to make the mathematics more concise and to link the analysis to the idea of potential energy. This parameter is the change in Gibbs (free) energy of the system, as defined by the following equations: dGsys ⌘ dHsys
DGsys = DHsys
d(T S)sys
(4.38a)
D (T S)sys
(4.38b)
Being a combination of state variables, Gsys is also a state variable. For a change between two different states of the system, both at the same temperature, Equation (4.38b) becomes DG = DH
T DS
(4.39)
where the subscript ‘sys’ has been dropped for conciseness. This convention will be applied in all subsequent equations; i.e., parameters without subscripts refer to conditions within the system. Comparison of Equations (4.37) and (4.39) yields a direct relationship between DG and DStot : DG = T DStot
(4.40)
Because T is always positive, Equation (4.40) indicates that, for systems at constant T and P, an increase in universal entropy is always accompanied by a decrease in Gibbs energy inside the system. The Gibbs energy therefore has the same useful characteristics when considering closed chemical systems at constant temperature and pressure as PEgrav and PEelec do when considering gravitational and electrical systems, respectively. In all cases, the parameters are state variables that are associated strictly with conditions within the system and that always decline when a 146
spontaneous change occurs. It is therefore reasonable to identify G as the chemical potential energy of the system. As is true for other forms of potential energy, it is useful to assign an unambiguous value to the Gibbs energy of a particular system or particular objects (chemical species) in the system. To do that, we can integrate Equation (4.38a) to obtain G⌘H
(T S) + G⇤
(4.41)
where G⇤ is a composite constant of integration that, like analogous terms in previous equations, accounts for differences in the datum conditions (i.e., the “zero-levels”) chosen for G, H, T , and S. While the Gibbs energy does represent potential energy, in the sense that DG is the maximum amount of energy that can (potentially) be converted to useful work (like driving a piston or lifting a weight) when the system undergoes a change of state, it differs from other kinds of PE (e.g., PEgrav , PEelec , or PEmech ) in at least one important way. These other kinds of PE all contribute to the energy content of the system and therefore must be included in an energy analysis based on the first law of thermodynamics. For example, when an object loses gravitational potential energy, the principle of conservation of energy (i.e., the first law) requires that that energy appear in some form somewhere else in the universe; similar statements apply to changes in electrical and mechanical potential energy. Some changes in G (specifically, those that arise from a change in H) do reflect changes in the energy content of the system and must also be accompanied by compensating changes elsewhere in the universe. However, changes in G that arise from a change in S do not have the same requirement. For example, Stot increases and G decreases when two pure gases intermix at constant temperature and pressure, but no energy is gained or lost by the system. The implication is that a system can lose some of its ability to do useful work via internal rearrangements of energy and mass, without actually releasing any of its energy content. As a result, when such a rearrangement occurs, it makes no sense to ask, “Where did the Gibbs energy go?” because the change in G does not represent a loss of energy content from the system; it only represents a loss of the potential to do work. Thus, to summarize, PEgrav , PEelec , and PEmech all contribute directly to the total energy of the universe, which must be conserved. By contrast, G is a composite parameter that has the dimensions of energy and includes energy content (H), but also includes an entropy-related term that is unrelated to energy content.6 As a result, G does not contribute directly to universal energy, and a decline in G need not be accompanied by an increase the energy residing in some other reservoir. 6 For this reason, it is incorrect to think of entropy, or the product T S, as a ‘different kind’ of energy. Entropy can be considered a measure of the distribution of energy in a system, but not the amount of energy.
147
The fact that G is not perfectly analogous to other parameters that we refer to as potential energy does not diminish its usefulness in the analysis of possible changes in chemical systems. As noted, G does share critical features with those parameters, in particular the fact that it is a state variable and that a chemical system at constant T and P is at equilibrium if and only if its Gibbs energy has been minimized. In light of these shared features, and for semantic simplicity, G is referred to throughout the chapter as ‘chemical potential energy’. Nevertheless, when reading this phrase, it is useful to keep in mind the differences between Gibbs energy and other types of potential energy, so as not to apply the analogy in inappropriate ways.
4.5
THE CHEMICAL POTENTIAL AND ITS USE TO PREDICT CHEMICAL CHANGE
As was the case for other forms of potential energy, it is useful to define a new parameter as the partial derivative of the system’s chemical potential energy with respect to the substances that ‘carry’ the energy. In this case, those substances are the chemical species in the system. When the partial derivative is taken with respect to the number of moles of species i, the new parameter is called the chemical potential of i (µchem,i ) or the molar Gibbs energy of i (Gi ) (or, sometimes, the partial molar Gibbs energy of i). Thus: ∂G µchem,i = Gi ⌘ (4.42) ∂ ni The independent variables that affect G include the pressure, temperature, and composition of the system, which is defined by the number of moles (n) of each species present. We can therefore apply the chain rule to write the derivative of G as follows: dG =
=
∂G ∂G ∂G ∂G ∂G dP + dT + dn1 + dn2 + ... + dnk ∂P ∂T ∂ n1 ∂ n2 ∂ nk
(4.43)
∂G ∂G dP + dT + Â Gi dni ∂P ∂T all i
(4.44)
Equation (4.44) expresses the change in G in a system as the sum of separate contributions attributable to changes in pressure, temperature, and the number of moles of each species present. As has been noted, the scenario of most relevance when a chemical reaction occurs in environmental systems is a change in composition at constant T and P, in which case the equation simplifies to dG = Â Gi dni all i
148
(4.45)
By definition, Gi is the increment in G per unit amount of i added to the system, when a differential amount of i is added and T , P, and the amounts of all other species remain constant. Thus, the portion of the system’s Gibbs energy that is attributable to the added i is Gi dni . However, the added i is indistinguishable from all the other i in the system, so the contribution to G of all the i in the system can be identified as Gi ni , which we designate as Gi : Gi ⌘ Gi ni (4.46)
We can interpret Gi as the chemical potential energy associated with (or ‘stored in’) species i in a given system, and Equation (4.46) indicates that this quantity can be computed as the product of the chemical potential of i (a property of the system) and the amount of i in the system (the property on which the chemical potential ‘acts’). The equation thus further emphasizes the analogy between chemical potential and the types of potential discussed previously. The contributions to G from all the species in the system are additive, so: G = Â Gi = Â Gi ni all i
(4.47)
all i
Now consider how G changes if a differential amount of some reaction occurs in a system. Because we are considering only a differential amount of reaction, the system composition changes negligibly, so Gi (being a property of the system) remains approximately constant for each constituent. In that case, Equation (4.47) can be applied to the conditions before and after the reaction to compute the change in G as dG = Â dGi = Â Gi dni (4.48) all i
all i
Equation (4.48) is a key result that we can use to evaluate the likelihood that a given chemical reaction will occur under specified conditions. We will explore this application in detail later in the chapter, but as a preview of that discussion, consider the precipitation of a differential amount of calcium carbonate by the following reaction under conditions where the temperature and pressure remain constant: Ca2+ + CO23 )* CaCO3 (s) According to Equation (4.48), the change in the total system Gibbs energy accompanying the reaction is dG = Â dGi = dGCaCO3 (s) + dGCa2+ + dGCO2 3
= GCaCO3 (s) dnCaCO3 (s) + GCa2+ dnCa2+ + GCO2 dnCO2 3
149
3
(4.49)
By the reaction stoichiometry, dnCaCO3 (s) = dnCa2+ = dnCO2 , so 3
⇣
dG = GCaCO3 (s)
GCa2+
GCO2 3
⌘
dnCaCO3 (s)
(4.50)
Equation (4.50) indicates that if a small amount of CaCO3 (s) precipitates (i.e., if dnCaCO3 (s) is positive), dG will have the same sign as the expression in parentheses. Hence, if that expression is less than zero, dG will be negative, and the reaction will proceed spontaneously. Generalizing this result, we infer that if we knew the Gi values of all the reactants and products of any reaction in a system of interest, we could predict whether the reaction would proceed, based on the sign of a simple expression that contains only those Gi values. We explore ways to quantify Gi shortly.
4.6
OTHER MOLAR THERMODYNAMIC QUANTITIES
Equations analogous to (4.42) can be used to define the molar enthalpy (H i ), molar entropy (Si ), and molar volume (V i ) of i. That is, in all three cases, we can write: Xi ⌘
∂X ∂ ni
(4.51) n j6=i ,P,T
where X is H, S, or V , respectively. Expressions similar to Equation (4.47) also apply to these quantities, i.e., for any of them: X = Â X i ni
(4.52)
all i
Using these definitions and taking the partial derivative with respect to ni of each term in Equation (4.41), we obtain Gi = H i
⇤
T Si + Gi
(4.53)
⇤
⇤
Like the PE terms defined in prior sections, Gi accounts for differences in the ⇤ datum levels chosen for the other terms appearing in the equation. Gi therefore depends on those choices, but not on the conditions in the system being evaluated. As a result, this parameter does not appear in expressions relating changes in Gi , H i , and Si . That is, for a change from state 1 to state 2, ⇣ ⌘ ⇣ ⌘ ⇤ ⇤ DGi = H i T Si + Gi H i T Si + Gi 2
= DH i
D T Si
1
(4.54)
The conventional datum choices made for Gi , H i , and Si are introduced next, followed by an explanation of how these parameters are evaluated under nondatum conditions. 150
4.7 4.7.1
QUANTIFYING Gi , Hi , and Si Datum Choices for Gi , Hi , and Si
A (virtually) universal convention has been adopted for the datum conditions for molar Gibbs energies and molar enthalpies. This convention assigns a value of zero to Gi and H i for all pure elements (mole fraction of 1.0) in their stable phase at 25 C and 1.0 atm, and also to dissolved H+ ions when they are in their standard state (i.e., aH+ = 1.0, at 25 C and 1.0 atm total pressure). For reasons that are explained shortly, the subscript ‘f’ and a superscript are conventionally added to the symbols for the molar Gibbs energy and molar enthalpy of species under these conditions. A few examples of these datum assignments are provided below. Assigned value
Conditions
Gf,O = H f,O ⌘ 0 kJ/mol
pure O2 gas at 25 C, 1 atm
Gf,H = H f,H ⌘ 0 kJ/mol
pure H2 gas at 25 C, 1 atm
Gf,Hg = H f,Hg ⌘ 0 kJ/mol
pure liquid Hg at 25 C under Ptot = 1 atm
Gf,Fe = H f,Fe ⌘ 0 kJ/mol
pure solid Fe at 25 C under Ptot = 1 atm
2 2
2 2
Gf,H+ = H f,H+ ⌘ 0 kJ/mol
1 M H+ in an ideal solution at 25 C, 1 atm
The datum conditions that are used almost universally for Si differ from those for Gi and H i and are based on theoretical considerations that indicate that all pure, ideal crystals must have the same molar entropy at 0 K.7 By convention, Si is assigned a value of zero for such crystals. Because entropy always increases with temperature, Si is finite and positive for all elements (and, in fact, for all species) under standardstate conditions.
4.7.2
Gi , Hi , and Si Under Nondatum Conditions
Because changes in Gibbs energy, enthalpy, and entropy accompanying a change in system conditions can be measured, the values of these parameters under many conditions of interest can be determined experimentally. Consider, for example, a system that initially contains only pure O2 (g) in its standard state. In such a scenario, both the molar Gibbs energy of O2 (g) (GO2 (g) ) and the total Gibbs energy of the system (G) are zero (state 1). If the pressure is then reduced to 0.5 atm at constant temperature (state 2), G will change by a measureable amount, DG. The system will still contain only O2 (g), so all of the system’s Gibbs energy will still be attributable 7 This requirement is a consequence of the 3rd law of thermodynamics (see, e.g., Atkins, P. and dePaula, J. Physical Chemistry, 9th ed., Oxford University Press [2009]).
151
to that species The molar Gibbs energy of O2 (g) in the new condition can therefore be calculated as G2 = G1 + DG = 0 + DG = DG (4.55) GO2 (g) |2 =
G2 nO2 (g)
=
DG
(4.56)
nO2 (g)
Both the numerator and denominator in the final expression in Equation (4.56) are known, so GO2 (g) of pure O2 (g) at 25 and 0.5 atm could be determined from the results of this hypothetical experiment. Analogous experiments can be imagined that would allow the molar Gibbs energy of any element to be determined under many conditions of interest. A similar approach applies to the determination of Gi for compounds. For instance, consider the reaction of hydrogen gas and oxygen gas to form liquid water under conditions where all three species are in their conventional standard states. We can think of this process as being carried out by taking one mole of H2 (g) from a reservoir of pure H2 (g) in its standard state and one-half mole of O2 (g) from a different reservoir in which it is in its standard state, and combining them to generate one mole of H2 O(l), which is placed in yet another reservoir where it is in its standard state. If all three reservoirs are defined to reside within the boundaries of the system, the chemical changes in the system can be represented as follows: 1 H2 (g) | pure, 25
C, Ptot=1 atm
+ 0.5 O2 (g) | pure, 25
C, Ptot=1 atm
! 1 H2 O (l) | pure, 25
C, Ptot =1 atm
(4.57)
In this system, both H2 (g) and O2 (g) are at the datum conditions where their molar Gibbs energies are zero (i.e., they are elements in their normal forms at standard state). The water is, of course, not a pure element, so we do not know its Gibbs energy a priori. However, we can determine its value based on the measureable change in Gibbs energy of the system, DG, when the reaction proceeds, as follows. The overall change in system Gibbs energy, DG, accounts for the loss of Gibbs energy due to the disappearance of H2 (g) and O2 (g) from the system, as well the gain of Gibbs energy associated with the appearance of H2 O(l). However, because both GH2 (g) and GO2 (g) are zero, no Gibbs energy is lost when those elements disappear. The measured DG can therefore be attributed entirely to H2 O(l), and we can determine GH2 O(l) as GH2 O(l) =
DG nH2 O
(4.58)
Similar experiments could be carried out, at least in theory, to form any other compound from its constituent atoms, with all the reactants and products in their standard states. In addition, if experiments were carried out in which the reactants were elements in their standard states, but the products were in nonstandard states, 152
essentially the same analysis would allow us to determine Gi for virtually any compound in any conditions. The assignment of GH+ = 0 when H+ is in its standard state means that a similar analysis can be used to determine Gi for ions. Furthermore, if we measured the enthalpy and entropy changes in these experiments, we could apply the same logic to determine H i and Si as well. The actual experimental conditions used to evaluate Gi , H i , and Si are typically quite different from those described in the preceding examples; details of these approaches are available in texts focusing on experimental physical chemistry. The key point made by the examples is nevertheless valid: once the datum levels for the thermodynamic parameters have been established for elements under some specified conditions, the corresponding values for both elements under other conditions and for compounds and ions under any conditions of interest can be determined from experiments. Based on the preceding discussion, Gi and H i for any species i under any conditions can be interpreted as the increment in Gibbs energy and enthalpy required to form one mole of i under those conditions, starting with only pure elements and H+ ions, all in their standard states. Such values of Gi and H i are referred to as the molar Gibbs energy of formation of i (Gf,i ) and the molar enthalpy of formation of i (H f,i ), respectively, where the word ‘formation’ is shorthand for ’formation from pure elements in their standard states’. As noted previously, the datum at which Si = 0 differs from that for Gi and H i (i.e., it is not elements in their standard states). Therefore, the interpretation given in the preceding paragraph for Gi and H i does not apply to Si . Correspondingly, the value of Si is referred to as the molar entropy of i, but it is not called the entropy of formation. Because the conventional datum conditions have been almost universally adopted, the subscript ‘f’ is often implied but not shown when values of Gf,i and H f,i are reported. Similarly, unless otherwise indicated, it can be assumed that any value reported for Si is based on the conventional datum conditions for that parameter.
4.7.3
An Alternate Datum Convention for Gi , Hi and Si
Recall that, in the discussion of gravitational potential energy, we noted that it was sometimes convenient to measure PE or PE from a datum level different from the conventional one; we could then compute the values of those parameters relative to the conventional datum using Equation (4.12). This approach is used almost universally when reporting values of Gi , H i , and Si , in which case the alternate datum level is the species i in its standard state. That is, we separate these parameters into two additive portions — one associated with the transition from the conventional datum (pure elements in their standard states for G and H, pure solids at 0 K for S) to an intermediate datum of i in its standard state, and a second one associated with the 153
transition from i in its standard state to i in the actual state of interest. As we did previously, we use a superscript to designate the alternate datum, so that the Gibbs energy, enthalpy, and entropy of a species in its standard state, calculated based on the conventional datum assignments, are designated Gi , H i , and Si , respectively. Thus, for example, for the molar Gibbs energy of i in some state 2, we can write Gi,2 = Gi + DGi |2
(4.59)
Values of Gf,i , H f,i , and Si are referred to as the standard molar Gibbs energy of formation, the standard molar enthalpy of formation, and the standard molar entropy of i, respectively. The application of the preceding ideas to Gf,i values of liquid water, CO2 gas, and dissolved H2 CO3 is illustrated in Figure 4.3. These species can all be made by combinations of H, C, and O. The most stable forms of these elements under standard conditions are H2 (g), C(graphite), and O2 (g), so the standard molar Gibbs energy of the elements in these forms is, by definition, zero. The relevant reactions for forming the compounds of interest from these elements are R1: R2: R3:
C(graphite) + O2 (g) )* CO2 (g) 1 H2 (g) + O2 (g) )* H2 O(l) 2 3 C(graphite) + O2 (g) + H2 (g) )* H2 CO3 (aq) 2
(4.60) (4.61) (4.62)
When these reactions proceed, with all the species in their standard states, a substantial amount of Gibbs energy is released in all cases. The standard molar Gibbs energies of formation of CO2 (g), H2 O(l), and H2 CO3 (aq) are therefore all significantly less than zero. In the figure, this result is represented by the arrows labeled R1, R2, and R3, respectively. In environmental systems, liquid water is typically present in a state that approximates its standard state, but CO2 (g) and H2 CO3 (aq) are usually far more dilute than their standard state concentrations. For example, the partial pressure of CO2 (g) in the atmosphere is approximately 10 3.46 atm, and H2 CO3 (aq) concentrations in most natural solutions range from approximately 10 5 to 10 2 M. Values of Gf,i for these latter species could be determined based on a hypothetical starting point of the elements in their standard states, but as has been noted, it is often more convenient to use the species themselves in their standard states as an alternate datum. Gf,i values can then be determined based on the (constant) value of Gf,i at the alternate datum (i.e., Gf,i ) and the DGf,i accompanying the change from the standard state to the actual state of interest. Thus, for example, the molar Gibbs energy of CO2 (g) in the atmosphere is typically evaluated based on independent consideration of the Gibbs energy changes represented by arrows R1 and R4. Similar, two-step sequences for 154
Figure 4.3 Schematic showing the molar Gibbs energy of formation of C(graphite), O2 (g), H2 (g), H2 O(l), CO2 (g), and H2 CO3 (aq) under various standard and nonstandard conditions. Note that the Gibbs energy scale is discontinuous.
evaluating GH2 CO3 (aq),i at activities of 10 3.0 and 10 5.0 are also shown in the figure (arrows R3 + R5 and R3 + R6, respectively). Values of Gf,i , H f,i , and Si of many species are widely available. A few values of Gf,i are compiled in Table 4.1, and a more extensive listing that also includes values of H f,i and Si is provided in Appendix A.1 at the back of the book. Both Gf,i and H f,i are negative for most compounds. A value of Gf,i < 0 implies that formation of i from elements, with all species in their standard states, is accompanied by a decline in Gibbs energy (i.e., it is thermodynamically favorable), and H f,i < 0 implies that the reaction is accompanied by release of thermal energy (i.e., it is exothermic). To illustrate the type of information that can be garnered by considering the changes in G, H, and S accompanying a chemical reaction, consider once again the reaction of hydrogen gas and oxygen gas to form liquid water under standard con155
Table 4.1 Standard molar Gibbs energy of formation of some species of interest a Species Ag+ AgBr(s) AgCl(s) Al3+ Al(s) Br– C(graphite) CO2 (g) H2 CO3 (aq) HCO–3 CO2– 3 (aq) C2 H4 O2 (HAc) C2 H3 O–2 (Ac– ) Ca2+ CaO(s) CaCO3 (calcite) Cl– HOCl OCl– HCrO–4 Cr2 O2– 7 a
Gf , kJ/mol 77.12 96.9 109.8 485.0 0.0 104.0 0.0 394.37 623.2 586.8 527.9 396.6 369.4 553.54 604.4 1128.8 131.3 79.9 36.8 764.8 1301.0
Species Cu(s) Cu2+ Fe(s) Fe2+ Fe3+ H+ H2 (g) H2 (aq) H2 O(l) Hg(l) NH3 (g) NH3 (aq) NH+4 Na+ O2 (g) O2 (aq) H3 PO4 H2 PO–4 HPO2– 4 PO3– 4
Gf , kJ/mol 0.0 65.5 0.0 78.87 15.4 0.0 0.0 17.57 237.18 0.0 16.48 26.57 79.37 240.0 0.0 16.32 1142.6 1130.4 1089.3 1018.8 .
Values are provided in the table only for chemicals whose Gibbs energies are used in examples or problems in this chapter. Values of G f for additional species are provided in Appendix A4.
ditions for all three constituents. The relevant thermodynamic parameter values for these constituents are given in Table A4.2 as follows. Gi (kJ/mol)
H i (kJ/mol)
H2 (g)
0.00
0.00
O2 (g) H2 O(l)
0.00 237.18
0.00 285.83
Si (kJ/mol-K) 130.6 205 69.91
The change in the Gibbs energy of the system when one mole of H2 O is formed, and one mole of H2 (g) and one-half mole of O2 (g) are destroyed, is therefore DG = 1 GH O(l) 2 mol H2 O(l) formed
1 GH2 (g)
= [1 ( 237.18)
1 (0)
156
0.5 GO2 (g) 0.5 (0)]
kJ kJ = 237.18 mol mol
G thus declines by 237.18 kJ for each mole of H2 O formed, indicating that the reaction is favorable. (Keep in mind, though, that the reaction kinetics are not necessarily favorable. This particular reaction proceeds at a negligible rate at room temperature, but it can proceed very rapidly [indeed, explosively] if a catalyst or the energy input from a spark is provided.) Similar calculations can be carried out for the changes in system enthalpy and entropy, yielding DH = 1 H H O(l) 2 mol H2 O(l) formed
1 HH
= [1 ( 285.83) DS = 1 SH O(l) 2 mol H2 O(l) formed = [1 (69.91)
2 (g)
0.5 H O
1 (0)
0.5 (0)]
1 SH
2 (g)
kJ kJ = 285.83 mol mol
0.5 SO
2 (g)
2 (g)
1 (130.6)
0.5 (205)]
J mol·K
J mol·K The results indicate that the formation of water under these conditions is driven by a favorable enthalpy change (enthalpy declines, which lowers G of the system), but is opposed by an unfavorable entropy change (entropy declines, which increases G). The effect of the enthalpy change dominates, as is evident from the net decline in G when both terms are taken into account. Note that, although S declines, that decline accounts only for changes within the system. The entropy change of the surroundings [which is not calculated explicitly above but is related to DH via Equation (4.36)], increases by a larger amount than the decrease in the system entropy, so that the entropy of the universe increases, as it must for a favorable process. ⇤ The tabulated thermodynamic values also allow us to calculate the values of Gi for the three species in the reaction by rearranging Equation (4.53) and applying it to standard-state conditions: = 163.19
⇤
GH
2 (g)
= GH
kJ mol
0
= 38.92
kJ mol
=0
⇤
GO
2 (g)
HH
2 (g)
= GO =0
kJ mol
0
kJ = 61.1 mol
+ T SH
(4.63)
2 (g)
✓
kJ J + (298K) 130.6 mol mol·K
HO
2 (g)
2 (g)
2 (g)
+ T SO
◆✓ 10
3 kJ
J
◆
(4.64)
2 (g)
✓
kJ J + (298K) 205 mol mol·K
157
◆✓ 10
3 kJ
J
◆
⇤
GH
2 O(l)
= GH
HH
2 O(l)
= 237.18 = 69.48
2 O(l)
kJ mol
kJ mol
✓
+ T SH
2 O(l)
285.83
kJ mol
◆
✓
+ (298K) 69.91
J mol·K
◆✓ 10
(4.65) ◆
3 kJ
J
⇤
Because Gi and H i are both zero for all elements, Gi at 25 C for any element or for H+ is simply (298 K)Si . The values of G⇤ for H2 (g) and O2 (g) computed above are consistent with this expression.
4.8 4.8.1
THE DEPENDENCE OF Gi ON SYSTEM COMPOSITION, TEMPERATURE, AND PRESSURE Overview
The preceding sections describe the conventional datum conditions where Gi = 0 for all elements and for H+ ions and explain how it is possible, in principle, to determine Gi for elements or any other chemical species under any conditions via experimental measurements. While the experimental approach is valid in theory, it is clearly impractical to use that approach every time we want to determine Gi under a new set of conditions. Rather, we would like to relate Gi to commonly measured properties of the system and the species of interest via a simple equation, so that we can calculate Gi in a manner analogous to how we use Equation (4.14b) to compute PEgrav,i . This equation, which characterizes the dependence of Gi on species activity and is developed shortly, is Gi = Gi + RT ln ai (4.66) This important result quantifies how the molar Gibbs energy of any species depends on the activity of that species, for a system at the standard-state temperature and pressure. Specifically, it indicates that the Gibbs energy per mole of i increases as the activity of i increases. The implication is that the larger the activity of i in a system, the more energy that is required to increase that activity. Thus, it takes more energy to increase the concentration of a dissolved solute from 0.30 to 0.31 M than to increase it from 0.10 to 0.11 M. Applied to a gas phase, the equation indicates that more energy is required to increase the pressure by a given amount if the initial pressure is high than if it is low (a result that has been experienced by anyone who has ever inflated a bicycle tire!). Furthermore, because the standard-state values of T and P can be chosen arbitrarily, we can always select the temperature and pressure of the system of interest as the standard-state T and P. If we do that, we see that Equation (4.66) must apply in a system at any temperature or pressure, as long as T and P remain constant when ai changes. 158
Using Equation (4.66) to determine the change in Gi when ai changes between any two values at constant T and P, we find DGi = Gi + RT ln ai,2
Gi + RT ln ai,1 = RT ln
ai,2 ai,1
(4.67)
The conventional choice for the standard-state temperature is 25 C and, for this temperature, the product RT equals 2.48 kJ/mol. Substituting this value into Equations (4.66) and (4.67) and replacing ln ai by 2.303 log ai yields, at 25 C:8 ✓ ◆ kJ Gi = Gi + 5.71 log ai (4.68) mol ✓ ◆ ai,2 kJ DGi = 5.71 log (4.69) mol ai,1 In the following section, Equation (4.66) is derived from a combination of conceptual and mathematical arguments. While the derivation provides valuable insight into the thermodynamic basis of chemical activities (in particular, the relationship of a species’ activity to its enthalpy and entropy), Equation (4.66) alone is sufficient for computing Gi from commonly available information about the state of an environmental system and determining the thermodynamic likelihood that a chemical reaction will occur. Readers whose primary interest is in such applications can skip the derivation (up to Example 4.2) without compromising their ability to master the material in the rest of the chapter.
4.8.2
The Dependence of Gi on System Composition — Contributing Factors
As discussed previously, the Gibbs energy includes contributions from the potential energy associated with molecular motion and chemical bonds within individual molecules (intramolecular potential energy, which we will designate Gintra ); chemical and electrical interactions among different molecules (intermolecular potential energy, Ginter ); system pressure and volume (mechanical potential energy, Gmech ); and an entropy-related term (GS ). These contributions are additive, so, for a species i, Gi = Gintra,i + Ginter,i + Gmech,i + GS,i (4.70) Based on these definitions, the sum of Ginter and Gintra can be identifed as the internal energy of i (Ui ), and the sum of those terms and Gmech can be identified as the enthalpy of i (Hi ). 8 Equations
(4.67) and (4.69) implicitly assume that Gi is the same in states 1 and 2. In general, this requirement is met only if i is in the same phase in both states, so these equations do not apply if the change from state 1 to state 2 involves a phase transfer (e.g., dissolution of O2 (g) into a solution to generate O2 (aq)).
159
We can write each term on the right side of Equation (4.70) as the sum of its value when i is in its standard state (Gi ) and the change in Gi when the system is modified from that state. As a reminder that the changes of interest are computed relative to the standard state, we designate them as D Gi , i.e.: Gi = Gintra,i + D Gintra,i + Ginter,i + D Ginter,i + Gmech,i + D Gmech,i + GS,i + D GS,i
(4.71)
By definition, all the D Gi terms in Equation (4.71) equal zero when i is in its standard state. Combining the four contributions that apply to the standard state, we can rewrite Equation (4.71) as Gi = Gi + D Gintra,i + D Ginter,i + D Gmech,i + D GS,i
(4.72)
where Gi ⌘ Gintra,i + Ginter,i + Gmech,i + GS,i ; Gi is the standard molar Gibbs energy
of i, as defined previously. We next evaluate D Gi in systems where the composition, but not the temperature or pressure, differs from that in the standard state. We designate such a change as D Gi ; that is, comp
Gi |T
,P
= Gi + D Gi
(4.73)
comp
where the subscripts T and P have been added on the left to reinforce that the equation applies only in systems where the temperature and pressure are constant at their standard-state values. We consider how changes in T and P affect Gi subsequently. D Gintra,i . The intramolecular Gibbs energy per mole of i, Gintra,i , accounts comp
for energy that is internal to the molecule (internal bonding, kinetic energy, etc.). We expect this term to depend on the identity of the molecule and the system temperature, but not on the system composition. As a result, D Gintra,i = 0. comp
D Ginter,i . The intermolecular Gibbs energy of i, Ginter,i , accounts for interac-
comp
tions between i and other molecules in the system, so we expect Ginter,i to change when the system composition changes. However, we also expect this change to be system-specific and not easily predicted. We therefore acknowledge that D Ginter,i comp
is likely to be nonzero, but we defer any attempt to quantify its value for now. D Gmech,i . For a system at constant pressure, Gmech,i can be computed as comp
Gmech,i =
∂ Gmech ∂ (PV ) ∂V = =P = PV i ∂ ni ∂ ni ∂ ni
(4.74)
where V i is the molar volume of i, as defined in Equation (4.51). Two limiting cases for Gmech,i are of interest. First, if i is an ideal gas, we can substitute for V i from the ideal gas law, yielding 160
RT = RT (4.75) P Thus, for an ideal gas, Gmech,i depends only on temperature, so D Gmech,i = 0. Gmech,i(g) = PV i(g) = P
comp
Second, if i is in a condensed phase (an incompressible liquid such as water, or a solid), V i is approximately independent of the system composition. That is, addition of a small increment of, say, sodium chloride to an aqueous solution is assumed to change the solution volume by approximately the same amount regardless of the solution composition. Therefore, for a constituent in a condensed phase, Gmech,i(aq or s) = PV i(aq or s) ⇡ constant
(4.76)
D Gmech,i(aq or s) ⇡ 0
(4.77)
comp
This approximation is considered acceptable for all species in condensed phases under normal environmental conditions and will be applied throughout the remainder of the text. Thus, Gmech,i is exactly (for ideal gases) or approximately (for constituents of solutions and solids) constant for any species i in any system at constant temperature and pressure. Correspondingly, for all species, D Gmech,i ⇡ 0. comp
D GS,i . Based on Equation (4.53), it is clear that the “entropy-related” contri-
comp
bution to Gi is
T Si , so, for a change at constant pressure and temperature, D GS,i = T D Si
comp
(4.78)
comp
The term D Si characterizes the change in the molar entropy of i when the comp
system composition changes at constant T and P. We can express this term in a more convenient form by applying one of the key outcomes of the statistical analysis of entropy. That analysis demonstrates that Si can be expressed as the sum of two independent components – one reflecting the number of ways that energy can be distributed within individual molecules of i, and the other reflecting the number of ways that molecules of i can be distributed among other molecules in the system. The former component depends only on the identity of the molecules and the temperature, independent of the system composition, while the latter depends only on the mole fraction of i (xi ), independent of its identity and the system temperature. Only the latter of these two components is relevant for the calculation of D Si . Furthermore, comp
the analysis leads to the simple result that the composition-dependent portion of the molar entropy of i equals R ln xi , where R is the universal gas constant. As a result, D Si can be written as comp
D Si = R ln xi + R ln xi = R ln
comp
161
xi xi
(4.79)
Substituting this result into Equation (4.78), we find: D GS,i = RT ln
comp
4.8.3
xi xi
(4.80)
The Overall Dependence of Gi on System Composition at Constant T and P
Inserting the results for the various D Gi terms into Equation (4.72) yields comp
Gi
T ,P
= Gi + D Gintra,i + D Ginter,i + D Gmech,i + D GS,i comp
comp
comp
= Gi + 0 + D Ginter,i + 0 + RT ln comp
= Gi + D Ginter,i + RT ln comp
comp
xi xi
xi xi
(4.81)
Under conditions where the molar density (i.e., total number of moles per liter) of a phase is constant, the mole fraction of each species (xi ) is directly proportional to its concentration (ci ). This condition is met exactly for ideal gases, and it is a good assumption for any condensed phase under normal environmental conditions. As a result, we can rewrite Equation (4.81) as Gi
P ,T
= Gi + D Ginter,i + RT ln comp
ci ci
(4.82)
Because Gi is a constant, independent of the system composition, the dependence of Gi on the composition is fully captured by the final two terms on the right side of Equation (4.82). This dependence includes one component ( D Ginter,i ) that comp
accounts for the chemical interactions of i with other species and therefore depends ci strongly on their identities, and another (RT ln ) that accounts for the relative ci amounts of i and other species, independent of their identities. We can combine these two terms by rewriting the expression for D Ginter,i so that it has the form comp
RT ln X. When we do that, Equation (4.82) becomes ✓ ◆ D Ginter,i ci Gi P ,T = Gi + RT ln exp + RT ln RT ci 02
0
= Gi + RT ln @4exp @ 162
D Ginter,i
comp
RT
13
1
A5 ci A ci
(4.83)
Equation (4.83) indicates that, at standard T and P, the difference in the molar Gibbs energy of a species between its standard state and a state with a nonstandard 2 0 composition 13is determined by the argument of the logarithmic term, i.e., G D inter,i 4exp @ comp A5 ci . This expression has a value of 1.0 in the standard state RT ci
(because, in that state, D Ginter,i = 0, and ci = ci ). Furthermore, an increase in the comp
value of the expression leads to an increase in Gi , corresponding to a decrease in the stability of i and therefore an increase in its reactivity. Recall that in Chapter 2 we defined the chemical activity of i, ai , as an indicator of the reactivity of i, with a value of 1.0 in the standard state and higher values indicating increased reactivity. We did not quantify ai in any other way at the time. We did, however, establish that ai could be expressed as the product of an activity coefficient (gi ) and the ratio of the concentration in the system of interest to that in the standard state (ci /ci ). All of these properties align perfectly with those of the logarithmic argument in Equation (4.83), suggesting that that argument can provide a fundamental, thermodynamically-based interpretaton of ai . We could choose to define the relationship between ai and the logarithmic argument in Equation (4.83) in many different ways. However, the simplest approach is just to equate those terms, as follows: 2 0 13 D Ginter,i comp A5 ci ai ⌘ 4exp @ (4.84) RT ci Substitution of Equation (4.84) into Equation (4.83) yields the classical expression relating Gi to ai that was presented earlier: Gi = Gi + RT ln ai
(4.66)
By rewriting ai in Equation (4.84) in terms of concentrations and the activity coefficient, it becomes clear that gi is a direct measure of the molar potential energy associated with the interactions of i with surrounding molecules, that is, of Ginter,i : 2 0 13 D Ginter,i ci comp A5 ci ai = g i = 4exp @ (4.85) ci RT ci 0
g i = exp @
D Ginter,i
comp
RT
1 A
(4.86)
Ginter,i has been introduced here as a conceptual aid and is not a conventionally recognized thermodynamic parameter. However, because we have found that, at 163
constant T and P, DGintra,i and DGmech,i are both zero, we can add those terms to the numerator of the fraction in Equation (4.86) without altering its value. Then, recalling that the sum of intramolecular, intermolecular, and mechanical potential energy is the enthalpy, we see that gi is directly related to D H i : comp
0
gi = exp @
D Gintra,i + D Ginter,i + D Gmech,i
comp
comp
comp
RT
1
0
D Hi
1
A = exp @ comp A RT
(4.87)
In addition, by applying the approximation that xi /xi ⇡ ci /ci to Equation (4.80) and rearranging that equation, we see that the concentration ratio appearing in the expression for ai accounts for the effect of entropy on Gi : 0 1 D GS,i ci comp A = exp @ (4.88) ci RT
Thus, to summarize the key results of the preceding sections, Equations (4.66) through (4.68) allow us to determine the change in the molar Gibbs energy of a species when its activity changes, regardless of whether the change in ai is induced by a change in ci or gi . Equations (4.85) through (4.88) then provide a thermodynamic (i.e., energy-based) explanation for the relationships among, ci , gi , ai , and the reactivity of the species, thus closing the loop that connects all of these concepts. Example 4.2 The value of GCa2+ is 553.54 kJ/mol. Compute GCa2+ in two solutions at 25 C, each containing 10 3 mol/L Ca2+ but different concentrations of other solutes. In one solution, gCa2+ is approximately 1.0, and in the other, the ionic strength is large enough (⇠ 0.04 M) to reduce gCa2+ to 0.5. Solution The molar Gibbs energy of Ca2+ in each solution can be computed from Equation (4.66), as follows: ! cCa2+ GCa2+ = GCa2+ + RT ln aCa2+ = GCa2+ + RT ln gCa2+ cCa2+ Substituting the given information into this expression, we find that GCa2+ is 570.66 kJ/mol and 572.38 kJ/mol in the solutions with activity coefficients of 1.0 and 0.5, respectively. The chemical potential energy is lower in the system with the higher ionic strength, meaning the Ca2+ is more stable (less reactive) in that system. The independent effects of the Ca2+ concentration and the activity coefficient on GCa2+ are shown schematically in Figure 4.4. 164
Figure 4.4 Contributions to the change in GCa2+ in the example solutions.
Example 4.3 Methane (CH4 ) gas can be generated by anaerobic decomposition of organic matter, both in natural environments and in engineered processes for the treatment of wastewater ⇣ ⌘ and organic solids. The standard molar Gibbs energy of⇣methane ⌘gas GCH4 (g) is 50.75 kJ/mol, and that of methane dissolved in water GCH4 (aq) is 34.39 kJ/mol. If a gas phase and a solution that both contain methane are in contact with one another, and GCH4 is different in the two phases, then the total Gibbs energy of the system could decline by transfer of methane out of the phase with the higher GCH4 and into the phase with the lower one. Therefore, for the two phases to be in equilibrium (i.e., for there to be no driving force for transfer of CH4 ), GCH4 must be the same in both phases. A solution at 25 C contains 7 ⇥ 10 4 M dissolved methane. What must the partial pressure of methane be in the gas phase for the two phases to be in equilibrium? Since methane is a neutral molecule, assume that the activity coefficient of dissolved methane is 1.0. Solution Because T = 25 , the expressions for GCH4 in both phases are given by Equation (4.68). Writing and expanding the expression for GCH4 (aq), we find
165
✓
◆ kJ GCH4 (aq) = GCH (aq) + 5.71 log aCH4 (aq) 4 mol ( ) ✓ ◆ cCH4 (aq) kJ = GCH (aq) + 5.71 log gCH4 (aq) 4 mol cCH (aq) 4
The standard-state concentration for methane in solution is 1.0 mol/L, and we are assuming that gCH4 (aq) = 1.0. We can therefore substitute known values for all the parameters on the right side of this equation to obtain ✓ ◆ ⇢ kJ kJ 7 ⇥ 10 4 M GCH4 (aq) = 34.39 + 5.71 log (1.0) mol mol 1.0 M kJ = 52.40 mol We can carry out essentially the same calculation for GCH4 (g), taking into account that the standard-state pressure in the gas phase is 1.0 atm and that we always assume that gases behave ideally at normal environmental T and P, so gCH4 (g) = 1.0. ✓ ◆ kJ GCH4 (g) = GCH (g) + 5.71 log aCH4 (g) 4 mol ( ) ✓ ◆ PCH4 (g) kJ = GCH (g) + 5.71 log gCH4 (g) 4 mol PCH (g) 4 ✓ ◆ ⇢ P kJ kJ CH4 (g) = 50.75 + 5.71 log (1.0) mol mol 1.0 atm Then, equating the values of GCH4 in the two phases to satisfy the equilibrium criterion, we can solve for PCH4 (g) as follows: ✓ ◆ ⇢ PCH4 (g) kJ kJ kJ 52.40 = 50.75 + 5.71 log (1.0) mol mol mol 1.0 atm PCH4 (g) = 0.51 atm
Thus, gas-phase methane would be in equilibrium with the dissolved methane in this system if and only if the methane partial pressure was 0.51 atm.
4.9
COMPARISON OF Gi OF DIFFERENT SPECIES
Because Gi is zero for all elements, Equation (4.66) indicates that they all have the identical relationship between Gi and ai . This relationship is plotted in Figure 4.5. 166
Figure 4.5 The molar Gibbs energy of formation of all elements at 25 C, when present in the same phase that they are in under standard conditions (e.g., gas for O2 , H2 , N2 ; liquid for Hg; solid for Fe, Cu, Zn). The two graphs show the same relationship, part (a) with an arithmetic scale for the abscissa and part (b) with a logarithmic scale. In accord with Equation (4.68), Gi increases by 5.71 kJ/mol for each factor of 10 increase in ai (at T = 25 C).
Equation (4.66) also applies to compounds and solutes, and Figure 4.6 shows the relationship between Gi and ai for a few solutes at 25 C. Each of the species represented in the plot has a unique, nonzero value of Gi , corresponding to the amount of Gibbs energy that is acquired (if Gi > 0) or released (if Gi < 0) when it is formed from the constituent elements, with all the chemicals in their standard states. However, the change in Gi for a given change in log ai (indicated by the slope of the lines) is the same for all species, because the molar Gibbs energy of all species depends on activity in the same way, as given by Equation (4.66). 167
Figure 4.6 The molar Gibbs energy of formation of various compounds at 25 C. The abscissa is the same as in Figure 4.5b, but the ordinate is compressed in order to show the data for compounds with widely varying Gi values (as indicated by the values of Gi at an activity of 1.0). The relationship for elements (from Figure 4.5b) is shown for reference. For all species, Gi increases by 5.71 kJ/mol for each 10⇥ increase in ai .
4.10
THE DEPENDENCE OF Gi ON PRESSURE AND TEMPERATURE
Equation (4.66) is applicable regardless of what temperature and pressure are chosen as the standard-state values, but tabulated values of Gi are widely available only for the conventional standard-state conditions of 25 C and 1.0 atm. Since we are often interested in systems that have T and P values that are constant but different from those in the tabulations, it is important to know how Gi (or, in general, Gi ) depends on those parameters. Like the case for the dependence of Gi on ai , useful insights can be acquired by understanding the derivations of the Gi vs. P and Gi vs. T functions, but those insights are not essential to correct application of the relationships. Recognizing that different readers have different learning goals and that different courses have different time constraints, the key results of the derivations are presented in Table 4.2. The derivations of those equations follow, but can be skipped without impeding the understanding of the rest of the chapter. Readers who choose not to follow the derivations should review the example problems in the following section to see how the equations are applied and then proceed to Section 4.11. In Table 4.2, Equations (4.91) through (4.93) all express the same relationship, and all are referred to as the Gibbs-Helmholtz equation. These equations are widely represented without including the term G⇤i (or, equivalently, based on an assumption 168
Table 4.2 Dependence of Gi on pressure and temperature
Dependence of Gi on P at constant T For species whose molar volume (V i ) is approximately constant (all species in liquid or solid phases, for normal environmental conditions): DGi = V i DP
(4.89)
For gaseous species (assumed to behave as ideal gases): DGi = RT ln
P2 P1
(4.90)
Dependence of Gi on T at constant P For any species: ⇤
d Gi /T = dT
Gi,T2
T2 = Gi,T1 T1
DGi = Gi,T2 Gi,T1 =
✓
T2 T1
H i + Gi T2
(4.91)
ZT2
H i + G⇤i dT T2
(4.92)
ZT2
(4.93)
T2
T1
◆
1 Gi,T1
T2
T1
H i + G⇤i dT T2
that G⇤i = 0). However, as has been noted, G⇤i is nonzero when conventional datum conditions are adopted. As a result, leaving that parameter out of the equation yields incorrect results for the effect of temperature on Gi . The most common application of the Gibbs-Helmholtz equation is to assess how temperature affects the equilibrium constant for chemical reactions, in which case the error from ignoring G⇤i for the reactants exactly balances the error from ignoring it for the products, and the result for the predicted effect of T on Keq is correct. Nevertheless, there are times when one wishes to know the effect of T on H i for a single species, and to carry out those calculations correctly, the Gibbs-Helmholtz equation must include G⇤i in the way shown in Table 4.2.
4.10.1
Derivation of the Dependence of Gi on P
To derive the dependence of Gi on P, we first write an expression for dG in terms of traditional thermodynamic parameters, as follows: 169
dG = dU + d (PV )
d (T S)
= dU + PdV +V dP
T dS
SdT
(4.94)
Comparing this expression with Equation (4.44), and equating the dependence of G on P and T in the two expressions, we see that ∂ G/∂ P = V and ∂ G/∂ T = S. Substituting those equalities into Equation (4.44) yields dG = V dP
SdT + Â Gi dni
(4.95)
all i
Yet another expression for dG can be derived by differentiating Equation (4.47): dG = Â Gi dni + Â ni dGi all i
(4.96)
all i
Equating the two preceding expressions yields the Gibbs-Duhem equation:
 ni dGi = V dP
SdT
(4.97)
all i
Or, for a system with constant T and P:
 ni dGi = 0
(4.98)
all i
The Gibbs-Duhem equation indicates that any change in the molar Gibbs energy of one species in a system at constant T and P must lead to a change in the molar Gibbs energies of at least some other species, and it establishes an exact relationship among these changes. The most obvious application of this result is that if one species (i) is added to a system, so that ai and Gi increase, the molar Gibbs energies of at least some other species ( j) must decrease. These decreases occur primarily because those species are diluted by the addition of i; in addition, some changes in G j might occur because interactions with the added i alter their activity coefficients. Writing V and S in Equation (4.97) in terms of the contributions from individual species, and then applying the result to any single species, we obtain ! !
 ni dGi =  niV i
dP
ni dGi = niV i dP
ni Si dT
all i
all i
dGi = V i dP
 ni Si
dT
(4.99)
all i
Si dT 170
(4.100)
Equation (4.100) relates changes in Gi to changes in pressure and temperature. It indicates, for example, that if the system pressure is increased at constant temperature, and if the molar volume of a species i (V i ) is constant over the pressure range of interest, Gi changes in proportion to DP: Z
dGi = V i
Z
dP
DGi = V i DP
(4.101) (4.89)
As noted previously, V i is approximately constant for most species in solid and liquid phases in environmental systems. If V i varies with pressure (e.g., if i is a gaseous species), that variation must be incorporated in the integration. For example, for an ideal gas, Z
dGi =
Z
V i dP =
DGi = RT ln
P2 P1
Z
RT dP P
(4.102) (4.90)
Example 4.4 A reverse osmosis system for purifying water is operated under steady conditions, in which the salty water on the feed side of the membrane contains 0.25 M NaCl, and the purified water that has passed through the membrane (the ‘permeate’ solution) contains 0.004 M NaCl. In both solutions, the NaCl is fully dissociated into Na+ and Cl– ions. The feed-side pressure is 70 atm, the permeate is at atmospheric pressure, and both parts of the system are at 25 C. Assume that the molar volume of water is 0.018 L/mol in both solutions, corresponding to a molar density of 55.56 mol/L. What are the direction and magnitude of the change in G of the system when 1 L of water passes from the high- to the low-pressure side? Solution The transfer of water in this process can be separated into two steps, each of which occurs at constant temperature and pressure. In the first, water is removed from the pressurized, saltier, feed (fd) side of the system, and in the second, it is added to the less salty, low-pressure, permeate (pm) side. Because the temperature is 25 C, the molar Gibbs energy of water in each location is given by Equation (4.68): ✓ ◆ kJ GH2 O = GH O + 5.71 log aH2 O 2 mol
The activity of water equals its mole fraction. Because the NaCl is fully dissociated, the feed solution contains 0.25 M each of Na+ and Cl– , and the permeate contains 0.004 M of each ion, so 55.56 mol/L aH2 O,fd = xH2 O,fd = = 0.991 (55.56 + 0.25 + 0.25) mol/L 171
aH2 O,pm = xH2 O,pm =
55.56 mol/L = 0.9999 ⇡ 1.00 (55.56 + 0.004 + 0.004) mol/L
To find GH2 O,fd , we first account for the fact that the feed is not pure water by using Equation (4.68) to determine GH2 O at 25 C and 1.0 atm total pressure in a solution with xH2 O (and therefore aH2 O ) equal to 0.991. The tabulated value of GH O for the conventional standard-state conditions of T = 25 C and P = 1 atm is 2 237.18 kJ/mol, so ✓ ◆ kJ GH2 O | x=0.991, = GH O + 5.71 log (0.991) 2 mol 1 atm,25 C kJ = ( 237.18 0.022) mol kJ = 237.202 mol Next, we apply Equation (4.89) and the identity that 1 L·atm = 0.101 kJ to determine how GH2 O changes in a solution with a fixed composition (in this case, the composition of the feed) when the pressure is increased from 1 to 70 atm. GH2 O |
x=0.991, 70 atm,25 C
GH2 O |
x=0.991, 1 atm,25 C
= V H2 O DP ✓ ◆ L = 0.018 ([70 mol kJ = 0.126 mol
✓
kJ 1] atm) 0.101 L · atm
◆
Combining the results of the two preceding calculations, we find the molar Gibbs energy of water in the feed, at a pressure of 70 atm, to be GH2 O,fd = GH2 O | = GH2 O |
x=0.991, 70 atm,25 C x=0.991, 1 atm,25 C
+V H2 O DP
= ( 237.202 + 0.126) = 237.076
kJ mol
kJ mol
The permeate is at the standard-state temperature and pressure, and it is essentially pure water, so GH2 O,pm equals GH O . The change in Gibbs energy when 1.0 L 2 of water passes from the feed to the permeate is therefore GH2 O,pm
GH2 O,fd = ( 237.18
[ 237.076])
172
kJ kJ = 0.104 mol mol
Because 1 L of water contains 55.56 mol of water, the change in Gibbs energy of the system per liter of water transferred is ✓ ◆ kJ DG = 0.104 (55.56 mol) = 5.78 kJ mol Although the Gibbs energy of the system declines when water transfers from the feed solution to the permeate, this result does not formally prove that the process is thermodynamically favorable, because the water that passes through the membrane undergoes a change in pressure. A decline in a system’s Gibbs energy guarantees that the entropy of the universe will increase, and therefore that the process will be thermodynamically favorable, only if all parts of the system are at the same temperature and pressure in the final state as in the initial state. If these criteria are not met, then some entropy changes occur that are not captured by the Gibbs function, and those unaccounted-for entropy changes could cause a process to be unfavorable even if the Gibbs energy of the system declines. As it happens, the unaccounted-for entropy changes accompanying even a large change in pressure are typically small, so that a decline in the system Gibbs energy usually does indicate that the process is favorable even when the entropy changes associated with the pressure change are ignored. That is the case in this example system; i.e., water does pass from the feed to the permeate side of the membrane. Nevertheless, it is important to keep in mind that the use of the change in the Gibbs energy to determine whether a process is favorable is strictly valid only in systems at constant T and P. Example 4.5 Compare the contributions of the concentration and pressure differences to DGCl across the RO membrane in the system described in the preceding example. Use the Davies equation to estimate the activity coefficient of Cl– in the two solutions. V Cl at 25 C is 0.023 L/mol. Solution Following the same procedure as in the preceding example, we first determine the activity of the species of interest, in this case Cl– , in the feed and permeate. We know the concentration of Cl– in the feed, so to determine aCl ,fd , we need to find its activity coefficient. The ionic strength (in mol/L) and activity coefficient of Cl– in the feed solution can be computed from Equations (1.17) and (2.6), respectively.
Ifd =
i 1 1h 2 2 2 c z = (0.25) (+1) + (0.25) ( 1) = 0.25 i 2Â i 2 173
log gCl
,fd
=
0.501z2Cl
= 0.501 ( 1)2 gCl
,fd
1/2
Ifd
0.3Ifd
1/2
1 + Ifd
[0.25]1/2 1 + [0.25]
! !
0.3 [0.25]
= 0.129
= 0.74
We can then find the activity of Cl– in the feed as ✓ ◆ cCl ,fd 0.25 M aCl ,fd = gCl ,fd = (0.74) = 0.185 cCl 1.0 M
⇣ ⌘ To compute GCl in the feed, we start with the standard-state value GCl and then make appropriate adjustments for both the nonstandard activity of Cl– and the nonstandard pressure in the feed solution. Making both of these adjustments at once, we find ✓ ◆ kJ GCl ,fd = GCl ,fd + 5.71 log aCl ,fd +V Cl ,fd DP mol ✓ ◆ kJ kJ = 131.3 + 5.71 log (0.185) mol mol ✓ ◆ ✓ ◆ L kJ + 0.023 ([70 1] atm) 0.101 mol L · atm kJ = 135.32 mol Analogous calculations indicate that the ionic strength, the activity coefficient, and the activity of Cl– in the permeate are Ipm = 0.004 M, gCl ,pm = 0.93, and aCl ,pm = 0.00374, respectively. The temperature and pressure in the permeate are the same as in the standard state, and the tabulated value of GCl is 131.3 kJ/mol, so we can compute the molar Gibbs energy of Cl– in the permeate from Equation (4.68): ✓ ◆ kJ GCl ,pm = GCl + 5.71 log 3.74 ⇥ 10 3 mol kJ kJ kJ = 131.3 13.86 = 145.16 mol mol mol Finally, we find the change in GCl across the membrane as GCl
,pm
GCl
kJ ,fd = 145.16 mol 174
✓
kJ 135.32 mol
◆
= 9.84
kJ mol
The molar Gibbs energy of the permeate is less than that of the feed, as was the case for water. However, whereas the overall change in Gibbs energy for water was dominated by the pressure difference across the membrane, the concentration difference accounts for 98% of the overall difference in GCl .
4.10.2
Derivation of the Dependence of Gi on T
Like the effect of pressure, the effect of temperature on Gi in a system with fixed composition can be derived from Equation (4.100). Specifically, for such a system at constant pressure:9 ∂ Gi = Si (4.103) ∂ T P, composition Although this equation describes how Gi changes in response to a change in temperature, we will see later in the chapter that the temperature dependence of reaction equilibrium constants is related to the values of Gi /T for the reacting species. As a result, a relationship between Gi /T and T is often more useful than Equation (4.103). Such a relationship can be derived from basic calculus as d Gi /T dT
=
1 dGi T dT
Gi T2
(4.104)
In addition, we can substitute Equation (4.103) into (4.53) and rearrange to solve for dGi /dT . If we restrict the analysis to changes in temperature at constant pressure and composition, the partial derivative in Equation (4.103) becomes a total derivative, and we find Gi = H i
T Si + G⇤i
= Hi + T dGi Gi = dT T
dGi + G⇤i dT
H i + G⇤i T
(4.53) (4.105) (4.106)
Substituting this result into Equation (4.104), we obtain the differential form of the Gibbs-Helmholtz equation [Equation (4.91) in Table 4.2]: 9 Because molar entropies are always positive, Equation (4.103) leads to the counter-intuitive conclusion that the molar Gibbs energy of any species decreases with increasing temperature. According to Equation (4.53), Gi = H i T Si + G⇤i . An input of thermal energy is always required to raise the temperature of i, and that input is reflected in an increase in H i . However, the increase in temperature also increases the product T Si (and has no effect on G⇤i ). The fact that Gi decreases with an increase in T at constant composition and pressure indicates that, under those conditions, the increase in T always causes T Si to increase more than H i does, leading to a net decline in Gi .
175
◆ Gi H i + G⇤i Gi (4.107) T T T2 H i + G⇤i = (4.91) T2 Multiplying both sides of Equation (4.91) by dT and integrating the left side for a change in temperature from T1 to T2 yields d Gi /T 1 = dT T
ZT2
d Gi /T
✓
ZT2
=
T1
T1
Gi |T2
Gi |T1
T2
T1
Z
=
T2 Gi |T2 = Gi |T1 T1
H i + G⇤i dT T2
T2
H i + G⇤i dT T2 ZT2
T1
H i + G⇤i dT T2
(4.92)
Finally, subtracting Gi |T1 from each side and rearranging yields DGi = Gi |T2
Gi |T1 =
✓
T2 T1
◆
1 Gi |T1
T2
ZT2
T1
H i + G⇤i dT T2
(4.93)
Equations (4.92) and (4.93) are two other forms of the Gibbs-Helmholtz equation that were included in Table 4.2. Example 4.6 (a) Determine the change in GH2 O if pure liquid water is heated from 25 C to 35 C. Over this temperature range, H H2 O can be approximated by ✓ ◆ kJ kJ H H2 O = 308.25 + 0.0752 T mol mol · K (b) Carry out the same calculation as in part (a), but assume G⇤H O = 0, as is 2 implied by the conventional representation of the Gibbs-Helmholtz equation. What do you conclude by comparing this result with that in part (a)? Solution (a) We can insert the known information into Equation (4.93) to solve for DGH2 O . DGH2 O =
✓
T2 T1
◆ 1 GH2 O,T1 176
T2
ZT2 H
T1
⇤ H2 O + GH O 2 dT T2
We found G⇤H
2O
earlier in the chapter to be 69.48 kJ/mol (by the identity G⇤i =
Gi H i +T Si ). Substituting this value, the expression for H H2 O , and the given values of T1 and T2 into the integral, we can evaluate the integral as ⇤ ZT2 H H2 O + GH O 2
T2
T1
=
308K Z
298K
=
308K Z
298K
dT
✓ ◆ kJ kJ kJ 308.25 + 0.0752 T + 69.48 mol mol · K mol dT T2 0
1 kJ kJ 238.77 0.0752 B mol + mol · K C @ A dT T2 T
kJ 1 = +238.77 mol 308 K = 0.0235
✓ ◆ 1 kJ 308 K + 0.0752 ln 298 K mol · K 298 K
kJ mol · K
Inserting these results and the tabulated value of GH2 O = 237.18 kJ/mol into Equation (4.93) yields ✓ ◆✓ ◆ ✓ ◆ kJ 308K kJ DGH2 O = 1 237.18 (308K) 0.0235 298K mol mol · K = 0.711
kJ mol
This result is consistent with critical reviews of available experimental data. (b) If G⇤H O were zero and the same calculations were carried as in part (a), the 2 value of the integral in the Gibbs-Helmholtz equation would be 0.0311 kJ/mol, and the value of DGH2 O would be +1.62 kJ/mol. Using this (incorrect) approach, the result is not only quantitatively different from that in part (a), but it suggests that the molar Gibbs energy increases when the temperature increases, which is in the opposite direction from the experimental results, the conclusion inferred from Equation (4.103), and the discussion in footnote 9.
177
4.11
COMPARING THE MAGNITUDE OF FACTORS THAT AFFECT Gi
In the preceding sections, we have developed equations and carried out several example calculations for the effects of concentration, activity coefficients, temperature, and pressure on Gi . The results of these calculations are collected in Table 4.3. Table 4.3 Example effects of ci , gi , T , and P on Gi
Parameter Value State 1 State 2
Example
Species
Parameter
4.2
Ca2+
cCa2+
1.0 mol/L
gCa2+
1.0
0.5
4.4
H2 O
4.5
Cl–
4.6
H2 O
xH2 O P aCl (b) P T
0.991 70 atm 0.186 70 atm 25 C
1.000 1 atm 3.74 ⇥ 10 3 1 atm 35 C mol/L
10
3
mol/L
(a)
DGi
17.12 kJ/mol 1.72 kJ/mol +0.022 kJ/mol 0.126 kJ/mol 9.68 kJ/mol 0.161 kJ/mol 0.711 kJ/mol
(a) Defined as Gi,State 2 Gi,State 1 (kJ/mol); a negative value indicates that transport of the species from State 1 to State 2 is favored if T and P are constant (b) Accounts for changes in both cCl and gCl .
Although the example calculations are for a few different species and therefore are not directly comparable, the magnitudes of the DGi values provide insight into the relative importance of different parameters in terms of their effects on the Gibbs energy. For water, the effect of a 10 C change in temperature outweighs that caused by the presence of a substantial concentration of solutes (0.25 M NaCl) or an enormous change in pressure (70 atm, roughly the pressure under 700 m of water). The reason that the solutes have so little effect is that even at high concentrations, they reduce the mole fraction of water by only a few percent. Thus, we expect even small temperature differences to have a larger effect on the molar Gibbs energy of water than large differences in pressure or solution composition. One important scenario in which differences in pressure and composition dominate the change in GH2 O is in the type of system described in Example 4.4, in which a very high pressure is applied to a solution to drive water through a reverse osmosis membrane. In this case, although the pressure increases GH2 O by only a small amount, that increase in GH2 O is enough to overcome the even smaller difference in GH2 O attributable to the presence of solutes, so the pressure can be used to drive water across a membrane from a salty solution and into a solution that approximates distilled water. 178
The effects of temperature and pressure on Gi for solutes are of the same order of magnitude as for water (although no numerical example was shown for the effect of T ). Thus, as is true for water, Gi for solutes is much more sensitive to temperature than to pressure. However, unlike the case for water, the range in values of ai for a given solute in environmental systems can be several orders of magnitude. This range primarily reflects the wide range of solute concentrations, but variations in solute activity coefficients also play a role. As a result, differences in Gi for solutes between two different environmental solutions are usually dominated by differences in ai . Thus, the pressure difference across the RO membrane accounted for most of the driving force for water transport in Example 4.4, but a negligible portion of the driving force for transport of Cl– in Example 4.5, in which the difference in aCl was so much more important. A sense of the relative magnitudes of these different contributors to Gi for different species can help us understand why species behave as they do in different environments, and also guide us in designing (or choosing not to pursue) certain engineering options for modifying water quality. We return to these considerations after exploring what the Gibbs energies of reactants and products can tell us about the tendency of chemical reactions to occur.
4.12
CHEMICAL POTENTIAL AS THE DRIVING FORCE FOR CHEMICAL REACTIONS
4.12.1
The Driving Force for Chemical Change and the Gibbs Energy of Reaction
The preceding sections of this chapter establish the basis for understanding and quantifying chemical reactivity in terms of Gibbs energy. We have seen that in closed chemical systems at constant temperature and pressure, changes occur if and only if they reduce the total Gibbs energy of the system; correspondingly, equilibrium coincides with the condition of minimum Gibbs energy. As with other types of potential, the negative gradient in G can be interpreted as the driving force for change in the system. For instance, a spatial gradient in Gi can be viewed as a force driving diffusion of i from regions of higher to lower potential (i.e., higher to lower activity). In that case, when the species migrates, the part of the system from which it departs loses more Gibbs energy than is gained by the part to which it moves, so G of the overall system declines. If a chemical reaction can occur, we also expect changes in the state of the system to be in the direction of decreasing G. In these cases, one or more species “depart” the system by reacting, and others enter the system as products of the reaction. The amount of reaction that occurs in such cases can be conveniently quantified in terms of the number of moles of stoichiometric reaction, which was defined in section 179
3.7.2. For the generic reaction aA + bB )* cC + dD, one mole of stoichiometric reaction occurs when a moles of A plus b moles of B are converted to c moles of C plus d moles of D. For instance, for the following reaction, one mole of stoichiometric reaction corresponds to conversion of two moles of H2 and one mole of O2 to two moles of H2 O: 2H2 (g) + O2 (g) $ 2H2 O (l) (4.108) Note that if the same reaction were written with the coefficients divided by 2, then one mole of stoichiometric reaction would correspond to conversion of 1 mol of H2 and 0.5 mol of O2 to 1 mol of H2 O: H2 (g) + 0.5O2 (g) $ H2 O (l)
(4.109)
Therefore, when quantities are expressed in terms of moles of stoichiometric reaction, it is critical to specify the reaction completely, i.e., not only the identities of the reactants and products, but also their stoichiometric coefficients. The number of moles of stoichiometric reaction that occur in a given system is commonly represented as Dh. Thus, for the water formation reaction written as in Reaction (4.108), conversion of 0.1 mol of H2 and 0.05 mol of O2 to 0.1 mol of H2 O corresponds to Dh = 0.05 mol. Extending this relationship to the generic reaction, we see that whenever a reaction proceeds by an amount Dh, the changes in the number of moles of A, B, C, and D in the system are DnA = aDh, DnB = bDh, DnC = +cDh, and DnD = +dDh, or in general, Dni = ni Dh, where ni is the stoichiometric coefficient of i in the reaction (positive for products, negative for reactants). Now consider a system in which A, B, C, and D are each present and a differential amount of reactants is converted to products. Because the activity of each of the constituents remains almost constant, each of the Gi values changes negligibly. The total Gibbs energy of the system before and after the reaction (designated as states 1 and 2, respectively) and the differential change in that value (dG) can therefore be written as G1 = nA,1 GA + nB,1 GB + nC,1 GC + nD,1 GD G2 = nA,2 GA + nB,2 GB + nC,2 GC + nD,2 GD dG = (dnA ) GA + (dnB ) GB + (dnC ) GC + (dnD ) GD
(4.110)
Representing the stoichiometric coefficient of i as ni , we can substitute ni dh for each dni term. Then, dividing through by dh yields an expression for dG normalized to the amount of stoichiometric reaction that has occurred: dG = ( a dh) GA + ( b dh) GB + (c dh) GC + (d dh) GD dG ⌘ DGr = aGA bGB + cGC + dGD dh 180
(4.111) (4.112)
or, more succinctly:
DGr = Â ni Gi
(4.113)
i
The parameter DGr is called the molar Gibbs energy of reaction. As the name implies, DGr is the Gibbs energy change per mole of (stoichiometric) reaction. The definition of DGr as dG/dh indicates that this parameter quantifies a gradient in the potential energy of the system, but along a coordinate that measures the progress of the reaction rather than physical distance. By analogy with gradients of potential energy in other systems, we can interpret DGr as an indicator of the driving force for the reaction. If DGr is evaluated under conditions where all the reactants and products are in their standard states, it is called the standard molar Gibbs energy of reaction, represented as DGr : DGr = a GA
= Â ni Gi
b GB + c GC + d GD
(4.114a) (4.114b)
i
Because Gi values for many compounds have been determined experimentally, the value of DGr for many reactions of interest can be computed from tabulated values. Using Equation (4.66) to expand the Gi term for each constituent in Equation (4.112) and substituting the definition of Gi into the result, we find DGr =
a GA + RT ln {A}
b GB + RT ln {B}
+ c GC + RT ln {C} + d GD + RT ln {D} =
a GA
bGB + cGC + dGD
+ RT ( a ln {A} =DGr + RT ln
b ln {B} + c ln {C} + d ln {D})
{C}c {D}d
{A}a {B}b
(4.115)
(Recall that the designation {i} indicates the activity of and is interchangeable with ai . The braces notation is used here to avoid confusion from using ‘a’ for both activity and the stoichiometric coefficient for reactant A.) The ratio in the logarithmic term in Equation (4.115) is the reaction quotient, Q, so we can rewrite that equation as DGr = DGr + RT ln Q
(4.116)
Converting Equation (4.116) to base–10 logarithms and evaluating it at 25 C yields ✓ ◆ kJ DGr = DGr + 5.71 log Q (4.117) mol 181
Equations (4.116) and (4.117) are critically important results, allowing us to calculate whether a differential amount of conversion of reactants to products will increase or decrease the Gibbs energy of the system, and thus to determine whether the reaction is favorable under the given conditions. Several of the key thermodynamic terms and relationships relevant for determining whether a reaction is thermodynamically favorable are summarized in Table 4.4. Table 4.4 Summary of some thermodynamic terms and relationships Term Description Gi
Molar Gibbs energy of species i
DGr
Standard molar Gibbs energy of reaction
DGr
Molar Gibbs energy of reaction
DGr
Gibbs energy change accompanying a given amount of reaction
Typical Units kJ mol i
Relationship Gi = Gi + RT ln ai
kJ DGr = Â ni Gi mol stoichiometric reaction (ni > 0 for products, ni < 0 for reactants) kJ DGr = Â ni Gi mol stoichiometric reaction kJ
DGr = DGr Dh =
 Gi ni
Dh = Â Gi Dni
(Dh = moles of stoichiometric reaction, Dni = moles of i reacted) Relationship among DG, DH, and T DS
kJ mol
DG = DH
T DS
Example 4.7 Compute the Gibbs energy change when 15 µmol of atmospheric CO2 at a partial pressure of PCO2 = 3 ⇥ 10 4 atm combines with water to form 30 µmo1 of H+ and 11 and 10 4 M, respec15 µmol of CO2– 3 , which are present at concentrations of 10 tively, in a system at 25 C. The reaction and the standard Gibbs energies of formation of the constituents are as follows. 182
CO2 3 ⇥ 10 4 atm
+
H2 O nearly pure
Species: Gi (kJ/mol):
)*
2 H+ 10 11 mol/L
CO2 (g)
H2 O(l)
394.37
237.18
CO2– 3 527.9
CO2– 3 10 4 mol/L
+
H+ 0.00
Assume that the solution is ideal (i.e., that all activity coefficients are 1.0) and that the solution volume is large enough that this amount of reaction does not change the concentrations of dissolved H+ and CO2– 3 significantly. Is the reaction thermodynamically favorable? Solution We first compute the standard Gibbs energy of reaction (DGr ) from Equation (4.114): DGr = 2GH+ + GCO23 GCO2 (g) GH2 O(l) ✓ ◆ ✓ ◆ ✓ ◆ kJ kJ kJ =2 0 +1 527.9 1 394.37 mol mol mol = 103.65
1
✓
kJ 237.18 mol
◆
kJ mol
Then, using Equation (4.117): ✓ ◆ kJ DGr = DGr + 5.71 log Q mol
✓ ◆ 2 10 11 10 4 kJ kJ = 103.65 + 5.71 log mol mol 3 ⇥ 10 4 (1.0) = 24.69
kJ mol
Since the molar Gibbs energy of reaction is negative, the reaction is favorable. For the given stoichiometry, reaction of 15 µmol of CO2 (g) corresponds to 15 µmol of stoichiometric reaction, i.e., Dh = 15 µmol. Therefore, the overall Gibbs energy change is ✓ ◆ kJ DGr = D Gr (Dh) = 24.69 (15 µmol) = 3.70 ⇥ 10 4 kJ mol
183
4.12.2
Enthalpy and Entropy Changes in Chemical Reactions
The molar enthalpy and entropy of reaction can be computed in a manner analogous to the molar Gibbs energy of reaction, i.e.: DH r = Â ni H i
(4.118)
DSr = Â ni Si
(4.119)
By applying Equation (4.54) to each constituent in a reaction, we obtain the following relationship among the molar Gibbs energy, the molar enthalpy, and the molar entropy of a reaction occurring at temperature T : DGr = DH r
T DSr
(4.120)
DGr = DH r
T DSr
(4.121)
or, under standard conditions
Substitution of Equation (4.121) into Equation (4.116) yields DGr = DH r = DH r
T DSr + RT ln Q ✓
= DH r
T DSr + RT ln
T DSr + RT ln ✓
gC
cC cC
cA gA cA
◆c ✓
gD
◆a ✓
cD cD
cB gB cB
◆d
◆b
gCc gDd ccC cdD (cA )a (cB )b + RT ln + RT ln c d gAa gBb caA cbB cC cD
(4.122a)
(4.122b)
(4.122c)
where the final term is zero because all the standard-state concentrations have a numerical value of 1. Like Equation (4.115), Equation (4.122c) expresses the molar Gibbs energy of reaction as the sum of its value when all reactants and products are in their standard states, plus terms that describe how the system composition differs from those standard-state conditions. For a given reaction, any of the terms on the right sides of these equations might dominate the sum. Thus, for example, a favorable reaction (DGr < 0) might be “driven” primarily by: (1) the formation of strong bonds in the product molecules compared to those in the reactants (a large, negative value of DH r ); (2) an increase in the number of ways that energy can be distributed in product molecules compared to reactant molecules (a large, positive value of DSr ); (3) an increase in the strength of intermolecular interactions [which is reflected in a small value of the activity coefficient ratio in Equation (4.122c)]; or (4) a system composition such that more concentrated reactants are converted to more dilute products, leading to a small value of the concentration ratio in Equation (4.122c). 184
4.12.3
The Relationship of the Gibbs Energy of Reaction to the Equilibrium Constant
In Chapter 2, we defined the equilibrium constant for a reaction as the ratio of the chemical activities of the products to those of the reactants in a system that has reached equilibrium, and in Chapter 3 we derived that relationship based on reaction kinetics expressions and the recognition that equilibrium corresponds to the condition where the forward and reverse reactions proceed at equal rates. The principle that, at constant T and P, reactions always proceed in the direction that minimizes the total Gibbs energy of the system allows us to derive the same relationship from thermodynamic considerations, as follows. Equation (4.111) indicates that, for the reaction aA+bB )* cC+dD, the change in Gibbs energy accompanying a differential amount of stoichiometric reaction is dG = aGA dh
bGB dh + cGC dh + dGD dh
(4.111)
When G is at a minimum, dG/dh = 0.10 However, dG/dh is, by definition, DGr [Equation (4.112)], so at equilibrium dG dh
eq
= DGr
eq
=0
(4.123)
According to Equation (4.116), the molar Gibbs energy of any reaction can be computed as DGr = DGr + RT ln Q (4.116) Inserting the condition that DGr = 0 at equilibrium, we obtain 0 = DGr + RT ln Q|eq DGr = RT ln Q|eq
(4.124) (4.125)
The value of Q at equilibrium is, by definition, the equilibrium constant K. Thus, the equilibrium constant can be related to thermodynamic data by DGr = RT ln K K = exp
DGr RT
!
At T = 25 C, Equation (4.126a) becomes ✓ ◆ kJ DGr = 5.71 log K mol 10 The
(4.126a) (4.126b)
(4.127)
condition of minimum G could also be found by differentiation with respect to the number of moles of one of the species, say, dG/dnA = 0, or dG/dnB = 0, etc. Substituting dni = ni dh into any of these equations leads to Equation (4.123).
185
Equations (4.126) and (4.127) establish a link between the thermodynamics of a reaction and its equilibrium constant. Note that, whereas the evaluation of K based on reaction kinetics requires experimental study of the specific reaction of interest, the thermodynamic equations allow K to be evaluated for any hypothetical reaction for which all the Gi values are known. This possibility is often of critical importance, since many reactions of interest are extremely difficult to study experimentally under equilibrium conditions, due to the slow approach to equilibrium and/or difficulties analyzing for all the reactants and products at the accuracy required. Example 4.8 One approach for removing dissolved nickel from solution is to precipitate it as nickel sulfide, NiS(s). The standard molar Gibbs energies of Ni2+ , NiS(s), and HS– are 45.6, 86.2, and 12.0 kJ/mol, respectively. Use these values to compute the equilibrium constant for the following reaction at 25 C: Ni2+ + HS )* NiS(s) + H+ Solution The standard molar Gibbs energy of H+ is zero by definition, so the standard Gibbs energy of the reaction is DGr = GNiS(s) + GH+ = [ 86.2 + 0.0
GNi2+
GHS
( 45.6)
12.0] kJ/mol = 52.6 kJ/mol
Using Equation (4.127), the equilibrium constant for the reaction at 25 C is kJ 52.6 DGr mol = +9.2 K = 109.2 log K = = kJ kJ 5.71 5.71 mol mol Equation (4.126) also provides the possibility of calculating Gi from the equilibrium constant for a reaction involving i if the molar Gibbs energies of all the other species in the reaction are known. For instance, if GH CO were known from prior 2
3
investigations, GHCO could be found based on the experimentally determined equi3 librium constant for the following reaction: H2 CO3 )* H+ + HCO3 Applying Equations (4.114b) and (4.126a), we can write: DGr = GHCO + GH+ 3
And, since GH+ ⌘ 0,
GH2 CO3 = RT ln K
GHCO = RT ln K + GH2 CO3 3
186
(4.128) (4.129)
GHCO could then be determined by inserting the known values for K and GH CO 3 2 3 into this expression. Example 4.9 When air at 25 C containing O2 (g) at a partial pressure of 0.21 atm equilibrates with pure water, the equilibrium concentration of dissolved oxygen (O2 (aq)) is 9.31 mg/L (2.91 ⇥ 10 4 M). Assuming that gO2 (aq) = 1.0, compute GO (aq) . 2
Solution The reaction of interest can be written as O2 (g) )* O2 (aq) Based on the given information, the activities of O2 (aq) and O2 (g) are 2.91 ⇥ 10 and 0.21, respectively. The equilibrium constant for the reaction is therefore K=
{O2 (aq)} 2.91 ⇥ 10 = {O2 (g)} 0.21
4
= 1.39 ⇥ 10
4
3
The temperature is 25 C, so we can use Equation (4.127) to find DGr for the reaction: ✓ ◆ kJ kJ DGr = 5.71 log K = 16.32 mol mol Since GO (g) is zero by definition, GO (aq) can be found by a rearrangement of 2 2 Equation (4.114b): GO2 (aq) = DGr + GO2 (g) kJ kJ = (16.32 + 0) = 16.32 mol mol
4.12.4
The Gibbs Energy of Reaction under Nonequilibrium Conditions
By substituting Equation (4.126a) into Equation (4.116), we obtain the following expression for the molar Gibbs energy of any reaction as a function of the ratio of Q to K, which can be thought of as the extent to which the reaction is out of equilibrium: DGr = DGr + RT ln Q = RT ln K + RT ln Q = RT ln and, at 25 C:
✓ ◆ Q kJ DGr = 5.71 log mol K 187
Q K
(4.130)
(4.131)
Equations (4.130) and (4.131) indicate that if Q > K, DGr is greater than zero. In such a case, converting reactants to products increases G, so the reaction is thermodynamically unfavorable. In fact, the reverse reaction (products to reactants) would proceed spontaneously. On the other hand, if Q < K, DGr is less than zero and the reaction proceeds to form products from reactants. If Q = K, DGr equals zero and the reaction is at equilibrium. These relationships are summarized in Table 4.5. Table 4.5 Summary of relationships between Gibbs energy of reaction and system condition Relationship between standard Gibbs energy of reaction and the equilibrium constant DGr = RT ln K Expressions for the Gibbs energy of reaction under equilibrium or nonequilibrium conditions DGr = DGr + RT ln Q Q K Relationship between sign of Gibbs energy change and direction of reaction DGr = RT ln
DGr < 0: Reaction proceeds to form products DGr > 0: Reaction proceeds to form reactants DGr = 0: Reaction is at equilibrium
Example 4.10 Historically, chromium (Cr) has been used in numerous industrial processes, including leather tanning and metal plating operations. In these applications, if the water is mildly acidic, the Cr is typically present as bichromate (HCrO–4 ) and/or dichromate (Cr2 O2– 7 ) ions. These ions can be interconverted by the reaction 2HCrO4 )* Cr2 O27 + H2 O The standard Gibbs energies of formation of HCrO–4 , Cr2 O2– 7 , and H2 O are 1301, and 237.18 kJ/mol, respectively.
764.8,
(a) Compute the equilibrium constant for the conversion of HCrO–4 to Cr2 O2– 7 by the reaction shown. (b) An aqueous solution is prepared by adding 0.10 mol of HCrO–4 to 1.0 L of pure water. Compute the total Gibbs energy of the solution as a function of the amount of HCrO–4 that is converted to Cr2 O2– 7 . Assume that the molar volumes of H2 O, HCrO–4 , and Cr2 O2– are all equal (1 L per 55.56 mol), so that 7 the solution volume remains constant as the reaction proceeds. Compare the 188
condition of minimum chemical potential energy with the equilibrium condition predicted using the equilibrium constant determined in part (a). Assume all activity coefficients are 1.0. (c) Compute DGr as a function of the amount of HCrO–4 converted to Cr2 O2– 7 . Solution (a) The standard Gibbs energy of the reaction is: DGr = GCr
2 2 O7
+ GH
2O
2GHCrO
4
= [ 1301 + ( 237.1)
2( 764.8)]
kJ kJ = 8.58 mol mol
Applying Equation (4.127), we obtain DGr log K = = kJ 5.71 mol K = 31.8
kJ mol = +1.503 kJ 5.71 mol 8.58
(b) The total Gibbs energy of the solution is the sum of the Gibbs energy contributions of the various constituents: ⇣ ⌘ ⇣ ⌘ G = nH2 O GH O + RT ln aH2 O + nHCrO GHCrO + RT ln aHCrO 2 4 4 ⇣ ⌘4 + nCr O2 GCr O2 + RT ln aCr O2 (4.132) 2 7
2 7
2 7
where ni is the number of moles of i in the system. The ni values for various amounts of reaction can be computed based on the composition of the initial solution (0.1 mol HCrO–4 plus 55.6 mol H2 O) and the reaction stoichiometry. Noting that one mole each of HCrO–4 and H2 O are generated for each mole of HCrO–4 that is consumed, we can write: 0.1 nHCrO 4 nCr O2 = 2 7 2 0.1 nHCrO 4 nH2 O = 55.56 + = 55.56 + nCr O2 2 7 2 The activity of water equals its mole fraction,11 and the activities of the chromium species equal their molar concentrations normalized by the standard-state concentration of 1.0 mol/L, so nH2 O aH2 O = nH2 O + nHCrO + nCr O2 4
11 Although
2 7
the activity of water is very nearly 1.0 regardless of the distribution of Cr species, the number of moles of H2 O in the solution changes slightly when the reaction proceeds. As a result, the contribution of H2 O to G changes, and this change must be taken into account.
189
aHCrO = 4
aCr
2 2 O7
=
nHCrO /V 4
1.0 mol/L nCr O2 /V 2 7
1.0 mol/L
where V is the solution volume. Based on the assumption that 1 L of solution always contains 55.56 moles and the fact that the total number of moles in the system remains constant as the reaction proceeds, V is also constant and is given by ⇣ ⌘ nH2 O + nHCrO + nCr O (55.56 + 0.1 + 0.0) mol 4 2 7 init V= = = 1.0018 L 55.56 mol/L 55.56 mol/L Substitution of the preceding relationships into Equation (4.132) converts that equation into a long but not complex expression for G in terms of a single variable, nHCrO . This relationship is plotted in Figure 4.7a. The point of 4
minimum Gibbs energy (and hence the equilibrium condition) is at HCrO4 = 0.0326 mol, corresponding to nCr O2 = 0.0337 mol and nH2 O = 55.59 mol. 2 7 The activities of the three species and the activity quotient for the reaction are therefore {HCrO4 } = {Cr2 O27 } = {H2 O} =
nHCrO /V 4
cHCrO
=
0.0326 mol/1.0018 L = 0.0325 1.0 mol/L
=
0.0337 mol/1.0018 L = 0.0336 1.0 mol/L
4
nCr c
2 2 O7
/V
Cr2 O27
nH2 O
nH2 O + nHCrO + nCr
2 2 O7
4
=
55.9 mol (55.59 + 0.0326 + 0.337) mol
= 0.9988
Q=
{Cr2 O27 }{H2 O} (0.0336)(0.9988) = = 31.8; (0.0325)2 {HCrO4 }2
log Q = 1.502
The computed value of Q is (essentially) identical to the value of K computed in part (a). That is, the activity ratio indicates that, when the total potential energy of the system is minimized, Q = K, so the reaction is at equilibrium. (c) For any distribution of Cr between its two species, DGr can be computed as DGr = GCr
2 2 O7
+ GH2 O
190
2GHCrO
4
Figure 4.7b shows DGr as a function of the HCrO–4 concentration. Comparison of Figures 4.7a and 4.7b makes it clear that the condition of minimum Gibbs energy corresponds exactly with the condition DGr = 0.
Figure 4.7 The (a) total Gibbs energy and (b) molar Gibbs energy of reaction for conversion of HCrO–4 to Cr2 O2– 7 in the example system.
The relative amounts of reactants and products present in a system are sometimes represented in terms of a coordinate e, called the extent of reaction. This coordinate has a value of 0.0 if the reaction has proceeded to the left (toward reactants) until one of the products is completely depleted, and it has a value of 1.0 if the reaction 191
has proceeded to the right (toward products) until one of the reactants is completely depleted.12 Since a change in e represents some conversion of reactants to products (or vice versa), any change De can also be represented as a corresponding number of moles of stoichiometric reaction, i.e., De = kDh
de = kdh
(4.133)
where the proportionality factor k depends on the total amounts of reactants and products in the system. The difference between h and e is that h is a generic term that applies to a given reaction in any system, whereas e is defined for a specific system containing known total concentrations of the various constituents. According to Equation (4.123), equilibrium is the condition in which dG/dh = 0. Since Equation (4.133) indicates that any change de is proportional to dh, the condition where dG/dh = 0 must also be a condition where dG/de = 0. Example 4.11 The initial concentrations of HOCl and OCl– in a solution are 10 3 M and 10 4 M, respectively, and the value of {H+ } in the solution is fixed at 10 7 . Given the following thermodynamic data, plot DGr versus e for the dissociation of hypochlorous acid in this system, assuming ideal solution behavior. What are the initial and equilibrium values of e? GH+ = 0.00 kJ/mol GHOCl = 79.9 kJ/mol GOCl = 36.8 kJ/mol
Solution The reaction of interest and its standard molar Gibbs energy are HOCl )* H+ + OCl DGr = GH+ + GOCl = 0 + ( 36.8)
GHOCl ( 79.9) = +43.1 kJ/mol
12 The conditions e = 0 and e = 1.0 correspond to the maximum extent that the reaction can proceed toward reactants and products, respectively. In some circumstances, certain constituents are assumed to be replenished as they are depleted and therefore are not considered as potentially limiting reactants. For instance, H+ and OH– are usually assumed to be available in virtually infinite quantities by dissociation of water (H2 O ! H+ + OH– ), so the determination of the condition where e = 0 or e = 1.0 does not consider the possibility that all the H+ or all the OH– can be consumed. In other circumstances, one might assume that another constituent is replenished, depending on the specifics of the system of interest; e.g., oxygen would not be a potentially limiting reactant in a reaction if, in the system of interest, oxygen could enter the solution from the atmosphere.
192
The value of DGr under the specified, nonstandard conditions can be computed from Equation (4.117) as ✓ ◆ H+ OCl kJ kJ DGr = 43.1 + 5.71 log {HOCl} mol mol For this system, the potential limiting conditions are for all the OCl (1.1 ⇥ 10 3 M) to be in the form of HOCl (e = 0.0) or for all of it to be in the form of OCl– (e = 1.0). Thus, e = [HOCl]/1.1 ⇥ 10 3 , and einit is 0.091. Also, since [HOCl] + [OCl ] = 1.1 ⇥ 10 3 , we can express the concentration of OCl– as [OCl ] = 1.1 ⇥ 10 3 (1 e). Formally, e is a concentration-based term. However, since we are assuming that the activity coefficients of the solutes are 1.0, we can substitute the activity of HOCl for its molar concentration in the calculation of e, i.e., e = {HOCl}/1.1 ⇥ 10 3 . Then, substituting the given, fixed value of {H+ } into the expression for DGr and writing {OCl } and {HOCl} in terms of e, we find ✓ ◆ 10 7.0 1.1 ⇥ 10 3 e kJ kJ DGr = 43.1 + 5.71 log mol mol 1.1 ⇥ 10 3 (1 e)
A plot of DGr against e (Figure 4.8) indicates that the equilibrium condition (DGr = 0) is at e = 0.22. Thus, as the solution equilibrates, e increases from 0.091 to 0.22; i.e., HOCl dissociates to form additional H+ and OCl– . The fact that {H+ } remains constant in the solution means that the H+ generated by the dissociation reaction must be consumed by some other reaction taking place in solution. Approaches for maintaining fixed {H+ } in such solutions are discussed in subsequent chapters. Note that, in the system just considered, DGr is a large positive number, indicating that under standard conditions, the driving force strongly favors formation of HOCl from H+ and OCl– . However, under the conditions stated, the net driving force is in the opposite direction. Because the concentrations of most dissolved constituents in environmental solutions are orders of magnitude less than 1 mol/L, they are far from their standard state conditions. As a result, the standard molar Gibbs energy of reaction (DGr ) for a reaction involving these species is often very different from the molar Gibbs energy of reaction under the ambient conditions (DGr ), and the value of DGr is not, in general, a good indicator of the direction in which the reaction will proceed. One commonly encountered situation where DGr often is an acceptable approximation of DGr is in the analysis of oxidation/ reduction reactions, and in particular, reactions describing the aerobic biological degradation of organic substrates. For these reactions, the magnitude of DGr is often so large that adjustments to account for nonstandard-state conditions have only a small effect. In these cases, a fairly widespread practice has developed of approximating DGr under any conditions by 193
Figure 4.8 The molar Gibbs energy of reaction for the conversion of HOCl to OCl– at fixed {H+ } = 10 7.0 as a function of the extent of conversion, e.
its value in a system where all the reactants and products except H+ are in their standard state, and the activity of H+ is 10 7 (pH = 7.0). The DGr values computed in this way are often designated DGr (W), where the W stands for water (the idea being that the standard conditions have been adjusted to apply to pure water, which has a pH of 7.0). Example 4.12 A water source at 25 C contains 10 mg/L of dissolved organic carbon (DOC), of which approximately 20% is in compounds that are relatively easily biodegraded. The water also contains 8 mg/L dissolved oxygen (O2 (aq)) and 10 3 M HCO–3 , and it is at pH 7.5. Estimate the molar Gibbs energy that would be available to microorganisms that consume the easily degraded compounds by the reaction shown below, assuming that those compounds have the elemental composition and Gibbs energy of glucose (C6 H12 O6 ). Compare the result with DGr (W). Standard molar Gibbs energies for glucose, O2 (aq), and HCO–3 are 908.01, 16.32, and 586.8 kJ/mol, respectively. C6 H12 O6 + 6O2 (aq) ! 6HCO3 + 6H+ Solution The standard molar Gibbs energy of H+ is zero by definition, so the standard Gibbs energy of the reaction is DGr = 6GHCO + 6GH+ 3
GC
= {6 ( 586.8) + 6 (0)
6 H12 O6
6GO
1 ( 908.01) 194
2 (aq)
6 (16.40)}
kJ mol
= 2711.2
kJ mol
The molar concentrations of oxygen and the glucose-like, biodegradable DOC in the solution are [O2 (aq)] = ⇥
⇤
8 mg/L = 2.5 ⇥ 10 4 M 32,000 mg/mol
C6 H12 O6 =
✓
2 mg C/L 12,000 mg C/mol C
◆✓
1 mol C6 H12 O6 6 mol C
◆
= 2.78 ⇥ 10 5 M
Assuming that g = 1.0 for all the solutes, the molar Gibbs energy of reaction under the given conditions can be computed as ✓
◆ 10 3 kJ DGr = DGr + 5.71 log mol 2.78 ⇥ 10
6 5
10
7.5 6
2.5 ⇥ 10
4 6
= 2922
kJ mol
The calculation of DGr (W) is essentially identical to the preceding one, except that the pH is assumed to be 7.0, and the activities of all the other species participating in the reaction are assumed to be l.0. Thus, ✓
◆ (1.0)6 10 7.0 kJ DGr (W) = DGr + 5.71 log mol (1.0) (1.0)6
6
= 2951
kJ mol
The difference between DGr and DGr (W) is only ⇠1%, so the approximation that DGr = DGr (W) is acceptable.
4.12.5
The Dependence of G and K on Temperature and Pressure
Equation (4.126a) (DGr = RT ln K) expresses the equilibrium constant for a reaction at standard temperature and pressure as a function of the standard Gibbs energy of the reaction. However, if we wish to determine the equilibrium constant at, say, 50 C in order to model the solution chemistry of a hot spring or some industrial waste stream, or at the elevated pressures extant near the ocean floor or on the pressurized side of reverse osmosis units, knowledge of the standard Gibbs energy of reaction in the conventional standard state (25 C and 1 atm) does not suffice. Rather, we need to know the Gibbs energy of reaction for a system in which the reactants and products are all at unit activity, under the temperature and pressure conditions of interest. In other words, we need to know how DGr varies with temperature and pressure. The concept that DGr varies with temperature and pressure might seem a bit confusing at first; since DGr is defined for a specified set of environmental conditions, the idea of computing a value under different conditions and still referring to it as DGr 195
seems paradoxical. The key to understanding this idea lies in recognizing that even though Equations (4.126a) and (4.126b) apply only under standard state conditions, they are not restricted to any particular choice of the standard state. That is, the equations are equally applicable at 50 C as at 25 C, but to use them at 50 C, we need to define the reference state to be at 50 C, and we need data for DGr at that temperature. Since DGr values are normally available only at 25 C and 1.0 atm, the challenge is to estimate those values under different conditions. We have already developed equations quantifying the effects of T and P on the molar Gibbs energy of individual species [Equations (4.89) through (4.93)]. Computing DGr at some temperature other than 25 C or pressure other than 1.0 atm simply involves applying those equations to all the reactants and products of a reaction, and applying Equation (4.126) to the results. For instance, to assess the dependence of K on T at a given P, we first choose some temperature and the given pressure as the standard-state values for those parameters, and we apply Equation (4.91) to each reactant or product of the reaction under standard conditions: d Gi /T dT
H i + G⇤i T2
=
(4.134)
We then multiply through by the stoichiometric coefficient for each species in the reaction and sum the resulting equations to obtain
d
d ni Gi /T = dT !
 ni Gi /T i
dT
ni H i + G⇤i T2
(4.135)
 ni H i
 ni G⇤i
T2
T2
=
i
i
(4.136)
At any temperature, the sum in the numerator on the left side of Equation (4.136) is DGr /T , and the sum in the numerator of the first term on the right is DH r , so the equation can be rewritten as
 ni G⇤i d DGr /T DH r i = (4.137) dT T2 T2 Recall, however, that G⇤i accounts strictly for differences in the datum conditions chosen for G, H, and S, and that changes in the values of those thermodynamic parameters between two states of the system are independent of the datum conditions (because the same “datum adjustment factor” is applied both before and after the change). As a result, the summation in the final term of Equation (4.137) always equals zero. This point can be made more formally by rearranging the defining equation for G⇤i [Equation (4.53)] as follows: 196
⇤
Gi = Gi
H i + T Si
(4.138)
If we then apply this equation to each of the reactants and products of a reaction of interest, multiplying it in each case by the corresponding stoichiometric coefficient, we obtain
 ni G⇤i =  ni Gi  ni H i +  ni (T Si ) i
i
= DGr
i
(4.139a)
i
DH r + T DSr
(4.139b)
=0
(4.139c)
where the final equality is based on the definition of the Gibbs energy and was presented earlier in Equation (4.120). Applying this result, Equation (4.137) simplifies to d DGr /T DH r = (4.140) dT T2 Finally, substituting for DGr from Equation (4.126a), rearranging, and integrating, we obtain d ( R ln K) = dT ZKT2
KT1
1 d ln K = R
ZT2
T1
DH r T2
(4.141)
DH r dT T2
(4.142)
Even though the molar enthalpies of individual species (H i ) can vary significantly with temperature, the change in the total enthalpy of the system when reactants are converted to products (DH r ) tends to be insensitive to temperature over the range of interest for most environmental systems. As a result, when Equation (4.142) is integrated, DH r is typically treated as a constant equal to the DH r at 25 C, leading to the following simplified result: ln K|T2
DH r,25 ln K|T1 = R
DH r,25 KT ln 2 = KT1 R
C
✓
1 T1
C
ZT2
T1
1 T2
◆
1 dT T2 (4.143)
Equation (4.143) is known as the van’t Hoff equation and can be used to evaluate DH r, 25 C by plotting experimentally determined values of ln K at various temperatures against 1/T . The same equation can then be used to compute K at any other temperature. 197
Example 4.13 The equilibrium constants for the reaction in which bromoform (CHBr3 ) transfers from an aqueous solution to the gas phase [CHBr3 (aq) )* CHBr3 (g)] are 0.0071, 0.017, and 0.028 at 4, 20, and 30 C, respectively. Use the van‘t Hoff equation to estimate the standard enthalpy of the reaction and the equilibrium constant at 15 C. Solution When plotted as ln K versus 1/T (Figure 4.9), the data fit the linear equation 4390 K ln K = + 10.9. Equating the slope with DH r, 25 C /R, we compute T DH r, 25 C to be 36.5 kJ/mol. Inserting a value of T = 15 C (288 K) into the equation of the line, we find that K at that temperature is 0.013.
Figure 4.9 The van’t Hoff relationship for transfer of bromoform from solution to the gas phase.
A similar approach yields an expression for the dependence of K on pressure. For instance, for reactions involving only condensed phases (liquids and solids), when we apply Equation (4.89) to each reactant and product in a reaction at standard state and then sum the terms, we obtain DGr |P
Z
2
dDGr =
ZP2
DV r dP
(4.144)
P1
DGr |P
1
where DV r is the molar volume of reaction (the change in system volume per mole of stoichiometric reaction). If that parameter is approximately independent of pressure, 198
we can write DV r |any P = DV r |1.0 atm ⌘ DV r . In such cases, Equation (4.144) can be rewritten and integrated as follows: K|P
Z
2
d ( RT ln K) = DV r
dP
(4.145)
P1
K|P
ln
ZP2
1
K|P2 K|P1
=
DV r (P2 P1 ) RT
(4.146)
DV r is typically exceedingly small for reactions in condensed phases unless the pressure change of interest is hundreds of atmospheres or more. Therefore, unless one is comparing a reaction at the earth’s surface with one in the deep ocean or in a very high-pressure reactor, the effect of pressure on K can generally be ignored.
4.12.6
The Entropy, Enthalpy, and Gibbs Energy of Composite Reactions
Recall that, in Chapter 2, we considered the following composite reaction in which three H+ ions transfer from phosphoric acid (H3 PO4 ) to water molecules, generating phosphate ion (PO3– 4 ): H3 PO4 + 3 H2 O )* PO34 + 3 H3 O+ This overall reaction can be derived by adding the following three reactions: H3 PO4 + H2 O )* H2 PO4 + H3 O+ H2 PO4 + H2 O )* HPO24 + H3 O+ HPO24 + H2 O )* PO34 + H3 O+ The Gibbs energies of these reactions can be computed as DGr,1 = GH
GH3 PO4
GH2 O
DGr,2 = GHPO2 + GH3 O+
GH
GH2 O
DGr,3 = GPO3 + GH3 O+
GHPO2
2 PO4
+ GH3 O+
4
4
DGr,overall = GPO3 + 3GH3 O+ 4
2 PO4 4
GH3 PO4
GH2 O 3GH2 O
Comparison of these four expressions leads to the simple result that DGoverall = DGr,1 + DGr,2 + DGr,3
199
(4.147)
That is, the molar Gibbs energy of the overall reaction is just the sum of the molar Gibbs energies of the reactions that are added to generate it. Generalizing this result yields DGr,overall = Â DGr,i (4.148) i
where the summation is over the reactions that are added to generate the overall reaction. The same relationship applies, of course, to standard molar Gibbs energies of reaction. Furthermore, application of this result to the case where the direction of a reaction is reversed (i.e., the reactants are replaced by the products, and vice versa) demonstrates that, in such a situation, the magnitude of DGr is unchanged, but its sign is reversed. Values of DSr and DH r for composite reactions can be computed in analogous ways. The result for DGr of a composite reaction is consistent with the derivation in Chapter 2 showing that K for an overall reaction equals ’ Ki , where the Ki values are i
the equilibrium constants of the reactions that are combined to generate the overall reaction. For instance, for the above example reactions involving phosphate, we can write ! DGoverall (4.149) Koverall = exp RT ! DG1 + DG2 + DG3 = exp RT ! ! ! DG1 DG2 DG3 = exp exp exp RT RT RT = K1 K2 K3
(4.150)
Example 4.14 The reaction for the dissociation of acetic acid (CH3 COOH) and the corresponding equilibrium constant are CH3 COOH )* CH3 COO + H+
K = 1.72 ⇥ 10
The standard molar Gibbs energy of acetate ion (CH3 COO– ) is (a) Calculate GCH
3 COOH
5
365.35 kJ/mol.
.
(b) Find DGr and K for the reaction “doubled”, i.e., 2 CH3 COOH )* 2 CH3 COO– + 2 H+ . 200
(c) Write out the form of the equilibrium constant for the reactions in the problem statement and in part (b). Solution (a) Since K = 1.72 ⇥ 10 5 , log K = 4.76. Therefore ✓ ◆ ✓ ◆ kJ kJ kJ DGr = 5.71 log K = 5.71 ( 4.76) = 27.07 mol mol mol DGr equals the difference in the Gi values of the products and reactants, so DGr = DGCH
+ DGH+
3 COO
DGCH
27.07
kJ kJ kJ = 369.4 +0 mol mol mol
GCH
= 396.5 kJ/mol
3 COOH
3 COOH
GCH
3 COOH
(b) Designating the reactions written with coefficients of 1 and 2 as Reactions I and II, respectively, we can write DGr,II = 2DGCH COO + 2DGH+ 2DGCH COOH = 2DGr,I 3 3 ✓ ◆ kJ kJ DGr,II = 2 27.07 = 54.14 mol mol DGi = 9.48 kJ 5.71 mol KII = 3.03x10 10 log KII =
(c) The forms and values of the equilibrium constants in the two cases are KI = 1.74 ⇥ 10 KII = 3.03 ⇥ 10
5
=
10
{CH3 COO }{H+ } {CH3 COOH}
=
{CH3 COO }2 {H+ }2 {CH3 COOH}2
Note that, although the numerical values of DGr and K depend on the stoichiometry, the derived relationship among the variables is the same. That is, if we take the square root of both sides of the equilibrium constant expression for Reaction II, we obtain the result for Reaction I. This outcome emphasizes the fact that we must refer 201
values of DGr and K to a specific reaction stoichiometry. The sentence “The standard Gibbs energy of reaction for dissociation of acetic acid is 27.07 kJ/mol” is, formally, ambiguous unless we specify that the reaction is written for dissociation of one mole of CH3 COOH. In reactions such as the dissociation of acetic acid, the chance of misinterpretation is minimal, since all the coefficients in the reaction are the same and would normally be assumed to be 1. However, in reactions in which the reactants and products have different stoichiometric coefficients, the reactions could reasonably be written with a number of stoichiometries, so the need for clarity is greater.
4.12.7
Dissolution versus Reaction with the Solvent
By definition, the molar Gibbs energy of a reaction is the difference between the molar Gibbs energies of the products and the reactants. If water is a reactant or a product, its Gibbs energy must be considered in the overall change. However, if water is simply the solvent, then it is not viewed as participating in the reaction, so it is not included explicitly in the energy analysis (in such cases, any changes in the energy of the water are embedded in the Gibbs energy values of the solutes). For instance, when gaseous oxygen dissolves in water, the water molecules arrange to allow the oxygen molecules to fit in. However, no individual water molecules react with the oxygen, so the relevant reaction can be written as either Reaction (4.151) or (4.152), but not as Reaction (4.153). In fact, in a formal sense, Reaction (4.153) is not balanced. On the other hand, when gaseous carbon dioxide dissolves, some of the CO2 (aq) molecules react with water molecules to form a new, identifiable species [H2 CO3 , as shown in Reaction (4.154)]. In this case, it is necessary to include the Gibbs energy of water in evaluating the Gibbs energy of reaction. O2 (g)+H2 O )* O2 (aq) + H2 O 2O2 (g) )* O2 (aq) 2O2 (g) + H2 O )* O2 (aq) CO2 (g) + H2 O )* H2 CO3 (aq)
4.12.8
(4.151) (4.152) (4.153) (4.154)
Thermodynamics, Kinetics, and the Activated Complex Model
As explained in Chapter 3, the activated complex model postulates a link between the thermodynamics and kinetics of reactions. According to that model, the rate of an elementary reaction is related to the magnitude of the (Gibbs) energy barrier that must be overcome as the reactants are converted to activated complexes. Gibbs energy is then released when the activated complexes are converted to products. Because products can be converted to reactants by traveling the same path in the opposite 202
direction, the difference in activation energies between the forward and reverse reactions equals the Gibbs energy change accompanying the overall reaction. This concept was illustrated schematically in Figure 3.1. That figure also illustrated the separate contributions of enthalpy and entropy to the change in Gibbs energy. We can now revisit these ideas in light of a deeper understanding of the meaning of the thermodynamic parameters. First, we note that if the reactants and products are all in their standard states, then the changes shown in the figure as DGr , DHr , and T DSr correspond to DGr , DHr , and T DSr , respectively. Furthermore, we now have a clearer understanding of how these terms can be normalized to one mole of stoichiometric reaction and thereby converted into the molar quantities DGr , DH r , and T DSr . By comparing the magnitudes of DH r , and T DSr , we can determine whether the change in enthalpy or entropy makes a greater contribution to the driving force for the reaction (i.e., DGr ) under any specified conditions. Deviations from the standard state can arise from either nonstandard concentrations (which is reflected in the molar entropies of the constituents) or a nonstandard environment (which is reflected in the molar enthalpies). In most cases of interest to us, the concentrations of the reacting species are far more ‘nonstandard’ than the environment is, at least with respect to their impact on Gibbs energy. Therefore, if the curves for H and T S in Figure 3.1 are normalized to one mole of stoichiometric reaction and drawn for a reaction in a standard-state system and a nonstandard-state system of environmental interest, chances are that the curves for DH in the two systems will be very similar to one another, whereas those for T DS will differ substantially. This idea is illustrated in Figure 4.10.
Figure 4.10 Changes in H and T S accompanying a hypothetical elementary reaction under standard- and nonstandard-state conditions.
For the hypothetical systems represented in this figure, the curve for H in the nonstandard system is slightly below that in the system at standard state, indicating 203
that the constituents are slightly more stable in the nonstandard system. Such a situation might arise if the ionic strength of the nonstandard solution caused the reactants and products to have activity coefficients <1. By dividing DHr in each system by the number of moles of stoichiometric reaction occurring, we can convert the values to DH r values. The (approximate) constancy of DHr in the two systems is typical for most reactions under most environmental systems. It is this fact that allows us to approximate DH r as DH r in most applications of the van’t Hoff equation. By contrast, the difference in the T DSr values between the standard and nonstandard systems is large, and much larger than difference in the corresponding DH values. The difference between T DSr and T DSr would typically be caused by vastly different concentrations of the reacting species in the two systems (e.g., 10 4 M vs. 1.0 M for a dissolved species). When both the enthalpy and entropy contributions are taken into account, the Gibbs energy change for the overall hypothetical reaction is negative (DGr < 0), indicating that the reaction is favorable, even though it is unfavorable under standardstate conditions (DGr > 0). As the reaction proceeds, the concentrations of the reactants will decline and those of the products will increase, causing T DSr to decrease and DGr to increase. Eventually, T DSr = DH r , and DGr = 0. At this point, the activation energies of the forward and reverse reactions are equal, so those reactions proceed at equal rates, and the reaction is at equilibrium.
4.13
SUMMARY
This chapter describes chemical reactivity from a thermodynamic perspective. Chemical reactions, like all physical and chemical changes in the world, occur if and only if they increase the total entropy in the system and its surroundings. For the particular case of reactions occurring at constant temperature and pressure (conditions that are met, at least approximately, in many environmental systems), this entropy-based criterion can be replaced by the much simpler criterion that the Gibbs (potential) energy of the system decreases. The Gibbs energy includes components attributable to bonding within and among molecules, mechanical potential energy, and the tendency for energy to distribute itself among all possible arrangements. The Gibbs energy of an individual chemical species in a system equals ni Gi , where ni and Gi are the number of moles of i present and the molar Gibbs energy of i, respectively. Gi is also called the chemical potential of i (µi ). Values of Gi are normally reported based on a scale on which all elements are assigned values of zero when they are in their normal state at 25 C and 1.0 atm total pressure. When reported in this way, the values are called molar Gibbs energies of formation, formally represented as Gf,i but often written simply as Gi . The most important expression relating Gi to system composition at a given temperature and pressure is Equation (4.66): Gi = Gi + RT ln ai 204
(4.66)
In this equation, Gi is the molar Gibbs energy of formation of i under standard-state conditions, and RT ln ai accounts for changes in the molar Gibbs energy caused by a change in system composition. These changes include alterations in the intermolecular interactions, which are quantified by changes in the molar enthalpy of i (DH i ), and in the number of ways that i can be distributed among other species in the system, quantified by changes in the molar entropy (DSi ). The intermolecular interaction quantified by DH i is directly related to the activity coefficient (gi ), while the distribution or DSi term is directly related to the mole fraction (xi ) or concentration (ci ). Equation (4.66) applies at any given, fixed T and P, but the value of Gi changes when those parameters change. The variation of Gi with pressure and temperature can be computed via Equations (4.89), (4.90), (4.92), and (4.93). The change in Gibbs energy accompanying a chemical reaction (the Gibbs energy of reaction) can be computed as the difference between the Gibbs energies of the products and the reactants, for the given reaction conditions. This value is typically normalized to one mole of stoichiometric reaction, defined as the conversion of reactants to products in molar amounts that correspond to the stoichiometric coefficients of the reaction. If the molar Gibbs energy of a reaction is negative, the reaction will proceed by conversion of reactants to products; if it is positive, the reaction will proceed in the opposite direction. The reaction is at equilibrium if and only if the Gibbs energy of reaction is zero. The equilibrium constant for a reaction bears a simple relationship to the molar Gibbs energy of the reaction under standard conditions, as given by Equation (4.126): DGr = RT ln K DGr RT
K = exp
(4.126a)
!
(4.126b)
The molar Gibbs energy of reaction is related to the extent of disequilibrium by Equation (4.130) [or, at 25 C, Equation (4.131)]: Q K
(4.130)
◆ kJ Q log C = 5.71 mol K
(4.131)
DGr = RT ln DGr,25
✓
Although K and DGr can change significantly with temperature, the corresponding change in DH r is generally small. As a result, the effect of temperature on K can be computed by the van’t Hoff equation [Equation (4.143), which assumes that DH r is insensitive to temperature]: DH r,25 KT ln 2 = KT1 R
C
205
✓
1 T1
1 T2
◆
(4.143)
For overall reactions that represent the sum of several other reactions, the molar Gibbs energy (or enthalpy or entropy) of reaction is the sum of the corresponding values of the individual reactions.
4.14
PROBLEMS
1. Calculate the standard Gibbs energy of reaction and the equilibrium constant at 25 C for the following reactions: (a) Fe3+ + 3 OH– )* Fe(OH)3 (s) (b) Mg2+ + OH– )* MgOH+ (c) Ca2+ + SO2– 4 + 2 H2 O )* CaSO4 ·2 H2 O(s)
– + (d) HAsO2– 4 + H )* H2 AsO4
2. Calculate DGr for the dissolution of ferrous hydroxide [Fe(OH)2 (s)] into Fe2+ and OH– ions. If an ideal solution containing 2 mg/L Fe2+ and 10 mg/L of the solid is at pH 7.0, in which direction will the reaction proceed? 3. When a piece of metal is exposed to a strong acid, the metal corrodes and gas bubbles are sometimes formed. These gas bubbles are pure hydrogen (H2 ), and indicate that the partial pressure for H2 (g) in the system is 1.0 atm. For example, if the metal is lead, the reaction would be Pb(s) + 2 H+ )* Pb2+ + H2 (g) The Gibbs energy of this reaction is 9.06 kJ/mol when the pH is 4.0, the Pb2+ concentration is 1 mg/L, and hydrogen bubbles are forming on the surface of the Pb(s). Use this information to find the equilibrium constant for the reaction. 4. Consider systems where the following reactions are occurring under the conditions indicated: + (i) H2 PO–4 )* HPO2– 4 +H ;
[H2 PO4 ] = 10
3
M, [HPO24 ] = 5 ⇥ 10
3
M, pH = 7.0, I = 0 M
(ii) Hg2+ + 2 Cl– )* HgCl2 (aq); [Hg2+ ] = 10
7
M, [HgCl2 (aq)] = 10
6
M, [Cl ] = 0.5 M, I = 0.5 M
(iii) NH3 (g) + H+ )* NH+4 ; PNH3 (g) = 10
6
atm, [NH+ 4 ] = 10 206
6
M, pH = 9.0, I = 0.008 M
(iv) Cd(OH)2 (s) )* Cd2+ + 2 OH– ; [Cd(OH)2 (s)] = [Cd2+ ] = 10
4
M, pH = 6.5, I = 0.05 M
For each reaction: (a) Use thermodynamic data to determine the equilibrium constant, K, at T = 25 C. (b) Would the reaction be more likely to proceed to the right, to the left, or neither, with increasing ionic strength? (c) Calculate the Gibbs energy of reaction in kJ/mol for the conditions specified. Use the Davies equation to calculate activity coefficients. 5.
(a) Determine the standard Gibbs energy of reaction and the equilibrium constant for the following reactions: (i) (ii) (iii) (iv) (v) (vi)
AgBr(s) )* Ag+ + Br– 2– + * PO3– 4 + H ) HPO4 – + HPO2– 4 + H )* H2 PO4 – + * PO3– 4 + 2 H ) H2 PO4 AgBr(s) + Cl– )* AgCl(s) + Br– AgCl(s) + Br– )* AgBr(s) + Cl–
(b) A solution contains 10 2 M Cl– , 10 5 M Ag+ , and 10 6 M AgCl(s) initially. Is the reaction for precipitation of AgCl(s) from Ag+ and Cl– at equilibrium? If not, what is the Gibbs energy of reaction for formation of more AgCl(s) under these conditions? What reaction will occur spontaneously? Find the concentrations of all the species, including the solid, when equilibrium is attained. Assume all activity coefficients are unity. 6. Find DGr and K for the following reaction for conversion of solid calcite to lime and carbon dioxide. GCaO(s) = 604.0 kJ/mol. CaCO3 (calcite) )* CaO(s) + CO2 (g) A dry mixture containing 1 g of each calcite and CaO(s) is in contact with the atmosphere. Does the thermodynamic driving force favor conversion of one of the solids into the other, or are the solids equilibrated with one another? 7.
(a) Use thermodynamic data to determine the equilibrium constants at 25 C for the following reactions. GH SiO = 1260.5 kJ/mol. 3
3
am-SiO2 (s) + 2 H2 O )* H4 SiO4 am-SiO2 (s) + 2 H2 O )* H3 SiO–4 + H+ 207
(b) Both of the above reactions are at equilibrium in an aqueous system at pH= 8.0 and 25 C. What are the concentrations of H4 SiO4 and H3 SiO–4 in the solution, if the solutes behave ideally? (c) Repeat part (b) for pH= 10.0. (d) Considering possible nonideal solute behavior, if the ionic strength of the solution in part (c) were increased at constant pH, would some amSiO2 (s) dissolve, would more am-SiO2 (s) form, or would the amount of am-SiO2 (s) in the system remain steady? 8. The oxidation of Mn2+ by dissolved oxygen can yield the manganese dioxide [MnO2 (s)] mineral pyrolusite, as indicated below: Mn2+ + 0.5 O2 (aq) + H2 O )* MnO2 (s) + 2 H+ Calculate the maximum ‘useful’ energy theoretically available per mole of stoichiometric reaction under the following conditions: [Mn2+ ] = 1.0 ⇥ 10 5 M; 8.0 mg/L dissolved oxygen; pH = 7. Assume that the ionic strength of the water is low. 9. Why is it typically more convenient to predict the direction of a reaction in terms of Gibbs energies of the constituents rather than entropies, even though the most universal statement of the second law of thermodynamics is expressed in terms of entropy? 10. When a gas phase containing ammonia [NH3 (g)] comes in contact with water, some ammonia can dissolve, forming NH3 (aq). The equilibrium constant for the reverse of this reaction (i.e., transfer of dissolved ammonia to the gas phase) is referred to as Henry’s constant for ammonia (HNH3 , not to be confused with the molar enthalpy of ammonia, H NH3 ) and is discussed extensively in Chapter 9. Once in solution, NH3 (aq) can combine with an H+ ion to form ammonium ions (NH+4 ). These reactions can be represented as follows: NH3 (aq) )* NH3 (g) NH3 (aq) + H+ )* NH+ 4
K = HNH3 K = 109.24
A solution at pH 7.0 is in equilibrium with NH3 (g) at a partial pressure of 10 6.0 atm. The pH is then rapidly lowered to 6.0. Assuming that the NH3 /NH+ 4 reaction equilibrates instantaneously, but that the exchange of NH3 between the gas and solution does not, compute the following values. (a) HNH3 . (b) DGr for the dissolution of NH3 (g) prior to the change in pH. 208
(c) The concentrations of NH3 (aq) and NH+4 in the original solution. (d) The concentrations of NH3 (aq) and NH+4 after the pH has been lowered, but before any NH3 has been transferred between the gas and solution. (Hint: Follow the approach used in Example 2.6, where the concentrations of HOCl and OCl– were determined as a function of pH for a solution with known T OT OCl.) (e) DGr for the dissolution of NH3 (g) for the same conditions as in part (d). Will NH3 (g) transfer into or out of the water after the pH change? 11. A resort has a swimming pool that is kept at 25 C and a hot tub that is kept at 40 C. The water in both facilities contains 2 ⇥ 10 4 M total hypochlorite (T OT OCl, the sum of the HOCl and OCl– concentrations) and is adjusted to pH 7.2 aH+ = 10 7.2 . (a) The equilibrium constant for the dissociation of HOCl into H+ and OCl– in the swimming pool is 10 7.53 . What is its value in the hot tub? (b) What is the concentration of HOCl (the more powerful of the two disinfectant species) in each solution, if the solutes behave ideally? 12. Compute K for the dissolution of oxygen from the atmosphere (PO2 = 0.21 atm) into a lake under winter (3 C) and summer (28 C) conditions. The reaction can be represented as: O2 (g) )* O2 (aq). What is the equilibrium concentration of dissolved O2 in mg/L under each condition? 13. Under a given set of conditions, DGr , DGr and Keq for the following reaction are determined to be 21.0 kJ/mol, 19.41 kJ/mol, and 10 3.40 , respectively. 2 Cr3+ + 3 HOCl + 5 H2 O )* 2 CrO24 + 13 H+ + 3 Cl What are the values of those three parameters for the same conditions if the reaction is written such that it involves only one Cr3+ ion (i.e., with all the stoichiometric coefficients divided by 2)? Will the reaction proceed from left to right, from right to left, or will it not proceed in either direction? Explain briefly. 14. The equilibrium constant for the dissociation of hydrogen sulfide (H2 S) to bisulfide ion (HS– ) at 25 C is 10 7.02 . Calculate the equilibrium constant for this reaction from 0 C to 30 C in 5 C increments and plot log K vs. T . Does dissociation of H2 S become more or less favorable as temperature increases? 15. When a solid PQ(s) is added to distilled water at 25 C, 10 5.0 mol/L of the solid dissolves, releasing P+ and Q ions. When the same experiment is carried out at 15 C, only one-half as much dissolves. How much solid will dissolve if it is added to a solution at 30 C that already contains 10 5 M P+ and no Q ? 209
16. Determine the equilibrium constant for dissolution of covellite [CuS(s)] by the following reaction at 5, 25, and 35 C. CuS(s) + H+ )* Cu2+ + HS 17. In the 1990’s, the Boeing Co. developed a process for treating wastewater from electroplating and printed circuit board manufacturing. The central feature of the process was addition of scrap aluminum metal to an acidic solution containing dissolved copper ions. If the solution contained enough fluoride or chloride ions, a reaction proceeded in which the aluminum dissolved and the copper precipitated as metallic Cu(s), which could then be removed from the suspension by settling and/or filtration. The relevant reaction can be written as follows: 3 Cu2+ + 2 Al(s) )* 3 Cu(s) + 2 Al3+ If a wastewater initially containing 300 mg/L Cu2+ and no dissolved aluminum is dosed with 100 mg/L of Al scraps and the above reaction proceeds until equilibrium is reached, what will the final concentrations of Cu2+ , Al3+ , Cu(s), and Al(s) be? 18. The following three reactions all describe the conversion of gas-phase hydrogen and oxygen into H2 O, with the water being liquid in two cases and a gas in the third. The equilibrium constant for Reaction 1 at 25 C is K = 1041.56 . The partial pressures of H2 (g) and O2 (g) in the earth’s atmosphere are 5⇥10 7 atm and 0.21 atm, respectively. Reaction 1: Reaction 2:
H2 (g) + 0.5 O2 (g) )* H2 O(l) 2 H2 (g) + O2 (g) )* 2 H2 O(l)
Reaction 3:
H2 (g) + 0.5 O2 (g) )* H2 O(g)
(a) Is Reaction 1 thermodynamically favorable under normal atmospheric conditions? (b) Under a given set of conditions, is the magnitude of DGr for Reaction 2 larger, smaller, or the same as DGr for Reaction 1? Explain your reasoning. 19. The following reaction for conversion of butyrate ions to acetate ions plays a central role in the anaerobic degradation of organic wastes. The standard Gibbs energy of formation of butyrate is 352.6 kJ/mol. C4 H7 O2 + 2 H2 O ! 2 C2 H3 O2 + 2 H2 (g) + H+ 210
(a) Determine the molar Gibbs energy of the reaction in a solution at pH 7 in which the acetate and butyrate concentrations are both 0.01 M, if PH2 (g) is 5 ⇥ 10 6 atm. Assume that the solutes behave ideally.
(b) Calculate the partial pressure of H2 (g), in atm, that would be required for the reaction to be in equilibrium for the solution conditions described in part (a).
20. Consider two containers (A and B) containing Ca2+ ions dissolved in 1.0 L and 2.0 L of water, respectively. The initial Ca2+ concentrations are 0.1 M in container A and 0.01 M in container B. The system is shown schematically in the figure below. Assuming that both solutions behave ideally, compute GCa2+ in each container as a function of the mass of Ca2+ that diffuses from A to B, and use the results to determine the distribution of the Ca2+ at equilibrium. (Note: in theory, water would also diffuse from one side of the system to the other, since its chemical potential is slightly different on the two sides of the system. However, we can ignore this small transfer and assume that the volumes of water in the two containers are constant.)
21. Most bacteria are able to oxidize simple organic compounds, converting the carbon in the compounds to carbonic acid (H2 CO3 ) or bicarbonate ions (HCO–3 ) and capturing the energy released by the reactions to carry out metabolic functions like food capture, waste elimination, and synthesis of new biomass. The chemical transformation of the carbon from its original form to species of the carbonate group is typically the same regardless of the other reactants that participate in the reaction or the overall solution composition. However, these latter factors place major constraints on the types of bacteria that can mediate the reaction and the amount of energy that can be acquired from the reaction. For example, most bacteria that can eat the organic matter by mediating a reaction between the organics and sulfate cannot survive if even a small amount of dissolved oxygen is present. Also, although some bacteria can mediate reactions of organic matter with either oxygen or nitrate, the amount of energy that they can obtain differs depending on which reactant is used. Consider three solutions containing methanol (CH3 OH) as the major organic constituent, along with the inorganic constituents listed below. Determine 211
the maximum amount of energy that bacteria could acquire by oxidizing the methanol based on each of the reactions shown. (Keep in mind that different bacteria would proliferate depending on the solution composition; no single bacterial species could mediate all three reactions. In fact, as noted above, bacteria that mediate the reaction with sulfate could not even survive in Solution 1.) GN2 (aq) = 18.23 kJ/mol. Solution 1
Solution 2
Solution 3
7.3
7.3
7.3
0.004
0.004
0.004
O2 (mg/L)
3.0
0
0
NO–3
15
15
0
N2 (mg/L)
8
8
8
SO2– 4 –
85
85
85
0
0
1
350
350
350
pH HCO–3
(equiv/L) N (mg/L)
(mg/L)
HS (mg/L as S) TOC (mg/L, assumed to be CH3 OH)
Reaction 1: CH3 OH + 1.5 O2 (aq) ! HCO3 + H2 O + H+
Reaction 2: CH3 OH + 1.2 NO3 + 0.2 H+ ! HCO3 + 0.6 N2 (aq) + 1.6 H2 O Reaction 3: CH3 OH + 0.75 SO24
! HCO3 + 0.75 HS + H2 O + 0.25 H+
22. The following reaction for dissolution of limestone by reaction with carbon dioxide and water is geochemically important. CaCO3 (calcite) + H2 CO3 )* Ca2+ + 2 HCO3 (a) Calculate the standard Gibbs energy change and the enthalpy change for the reaction at 25 C and 1 atm total pressure. (b) Calculate the equilibrium constant at 25 C and 1 atm total pressure. (c) Will an increase in temperature favor a shift toward products or reactants for this reaction? 23. Equilibrium constants are shown below for three reactions at 25 C. Use standard enthalpy data to prepare a plot showing log K as a function of temperature for the reactions over the range 5 C < T < 35 C. Show all three curves on a single plot. For each reaction, is dissociation favored or disfavored by an increase in temperature? HCO3 )* CO23 + H+ 212
log K = 10.33
+ * NH+ 4 ) NH3 (aq) + H H2 S(aq) )* HS + H+
log K = 9.24 log K = 7.02
24. Sketch plausible curves of G, H and T S vs. reaction coordinate for each of the following scenarios. (Note: The information provided establishes certain relationships that must be satisfied with respect to DGr , DH r , and T DSr . The patterns of G, H, and T S in the transition from reactants to products are less welldefined, but nevertheless must meet certain constraints. Try to draw curves that are consistent with all of these considerations.) (a) A reaction that is highly exothermic, approaches equilibrium very slowly, and can reach equilibrium by conversion of a small amount of reactants to products. Assume that the change in entropy favors the forward reaction, but the overall driving force for the reaction is dominated by the enthalpy change. (b) A reaction that is slightly endothermic, approaches equilibrium rapidly, and requires conversion of a large amount of reactants to products to reach equilibrium. 25. A solution at pH 8.8 and with an ionic strength is 0.003 M contains 3.9 ⇥ 10 5 M Ca2+ . The solution is in equilibrium with both the mineral calcite [CaCO3 (s)] and the CO2 (g) in the atmosphere. NaCl is then added to increase the ionic strength to 0.2 M while the pH is held constant. The calcite dissolution reaction is: CaCO3 (s) )* Ca2+ + CO23
K = 10
8.30
(Note: Thermodynamic data taken from different sources are often not entirely consistent. The value of K given here is based on the standard molar Gibbs energy values in Table 4.1. In later chapters, the equilibrium constant for the same reaction is given as 10 8.48 and is taken from a database associated with software for solving chemical equilibrium problems that is introduced in Chapter 7.) (a) What is the reaction quotient, Q, for the CaCO3 (s) dissolution reaction immediately after the salt addition, before any reaction has occurred? (b) Eventually, the solution re-equilibrates with the solid and gas phases, at which point the activity of CO2– 3 is the same as it was before the salt addition. (The basis for this result is explained in Chapter 9.) On a copy of the following graph, sketch the profiles of GCa2+ , GCaCO3 (s) and DGr for the calcite dissolution reaction as a function of time, defining t = 0 as the time of salt addition. The sketch should go from a few minutes before the salt is added until 15 minutes after the salt addition, at which time the 213
system is assumed to have reached equilibrium with both non-aqueous phases. Locations where G equals 0, GCa2+ , and GCaCO3 (s) are shown on the y-axis for reference.
214
5
ACIDS AND BASES, PART 1:
ACID/BASE SPECIATION AND EXACT SOLUTIONS TO ACID/BASE PROBLEMS
215
Contents
5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . THE DISSOCIATION OF WATER; Kw . . . . . . . . . . . . . . . THE STRUCTURE OF ACIDS AND BASES . . . . . . . . . . . . STRONG AND WEAK ACIDS, K a , AND CONJUGATE ACID/BASE PAIRS . . . . . . . . . . . . . . . . . . . . . . . . . . A BRIEF DIVERSION: THE STRUCTURE OF SOME IMPORTANT ORGANIC ACIDS AND BASES . . . . . . . . . . . COMBINING ACIDITY REACTIONS AND THE DEFINITION OF BASICITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . ACID/BASE SPECIATION AS A FUNCTION OF pH; DIAGRAMS OF LOG C VERSUS pH . . . . . . . . . . . . . . . . . . . . . . . LOG C-pH DIAGRAMS FOR MONOPROTIC ACIDS . . . . . . . EFFECTS OF NONIDEAL SOLUTE BEHAVIOR ON ACID/BASE SPECIATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LOG C-pH DIAGRAMS FOR MULTIPROTIC ACIDS . . . . . . . DETERMINING SPECIES AND RELEVANT EQUATIONS FOR SOLVING EQUILIBRIUM PROBLEMS . . . . . . . . . . . . . . NUMERICAL APPROACHES FOR SOLVING ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . THE pH OF SOLUTIONS CONTAINING ONLY WATER AND STRONG ACIDS OR BASES . . . . . . . . . . . . . . . . . . . . THE pH OF IDEAL SOLUTIONS CONTAINING WEAK ACIDS AND BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A SIMPLE SPREADSHEET ANALYSIS FOR DETERMINING ACID/BASE SPECIATION . . . . . . . . . . . . . . . . . . . . . . THE pH AND SPECIATION OF SYSTEMS CONTAINING MULTIPROTIC WEAK ACIDS AND BASES . . . . . . . . . . . ACID/BASE EQUILIBRIA IN NONIDEAL SOLUTIONS . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
217 219 221 222 229 233 238 243 247 249 256 266 266 269 274 279 280 283
5.19 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
5.1
INTRODUCTION
This chapter introduces the chemistry of acids and bases, groups of species that play a critical role in water quality. Acids and bases influence water quality both directly by controlling solution pH and indirectly by, for example, controlling the dissolution and precipitation of solids, altering the solubility of gases, catalyzing many other reactions, and affecting the interactions of chemicals with organisms. The control that solution pH exerts over the relative concentrations of hypochlorous acid (HOCl) and hypochlorite ion (OCl– ) in a solution has been alluded to in preceding chapters; because HOCl is a much more effective disinfectant than OCl– , the U.S. Environmental Protection Agency (EPA) has established different disinfectant dosing requirements for waters depending on their pH. Then, in Chapter 3 (Figure 3.3a), we saw that the rate of oxidation of ferrous iron [iron with a +2 charge, commonly represented as Fe(II)] to ferric iron [iron with a +3 charge, Fe(III)] by dissolved oxygen (Reaction 5.1) increases by approximately a factor of 100 for every unit increase in pH in the range 5 < pH < 8. 1 1 Fe (II) + O2 + H+ )* Fe (III) + H2 O 4 2
(5.1)
In addition to helping control the rate of the Fe(II)-to-Fe(III) conversion, the pH affects the ultimate solubility of the iron once the system equilibrates. For instance, at pH 7.0, the equilibrium solubility of ferrous iron is almost six orders of magnitude greater than that of ferric iron (Figure 5.1a). Solution pH also plays a central role in determining the affinity of many dissolved species for particles with which they come into contact. Figure 5.1b shows that the binding of dissolved natural organic matter to the surface of aluminum oxide particles increases dramatically with decreasing pH. A similarly strong dependence on pH characterizes the binding of many metals to oxide particles, except that in this case the trend is reversed: binding increases with increasing pH. To remove these species from solution in water and wastewater treatment plants, oxide particles are sometimes added to or generated in the raw water, and the solution pH is adjusted to the range where the contaminants bind strongly to the particles. The particles and associated contaminants are then removed by settling and/or filtration. Aside from being enormously important in their own right, acid/base reactions provide an appropriate starting point for studying the equilibrium speciation of a much larger class of chemical reactions, for three reasons. First, almost all acid/base reactions are extremely fast, so an assumption that equilibrium is attained (and that 217
Figure 5.1 (a) The effect of pH on total concentration of dissolved ferric [Fe(III)] and ferrous [Fe(II)] iron in equilibrium with the solids Fe(OH)3 (am) and Fe(OH)2 (am), respectively. (b) Effect of pH on binding of natural organic matter (4.7 mg DOC/L) onto g–Al2 O3 . After Davis (1982).1
the computed speciation is therefore the actual speciation in the system) is usually justified. Second, acid/base reactions are probably the simplest examples of a large class of reversible reactions in which two molecules collide and a portion of one of the molecules is transferred to the other one. In the case of acids and bases, the exchangeable unit is an H+ ion. These reactions are prototypical of many other types of reactions in which the attachment/detachment is more complicated. Finally, acids and bases are commonly encountered in everyday life. For instance, the widely used 1 Davis,
J.A. (1982) “Adsorption of natural dissolved organic matter at the oxide/water interface.” Geochim. Cosmochim. Acta 48, 679-691.
218
industrial chemical muriatic acid is the same material referred to in chemical contexts as hydrochloric acid (HCl); hypochlorite ion (OCl– ) is the active ingredient in bleach; acetylsalicylic acid (CH3 COO C6 H4 OH) is aspirin; ascorbic acid (C6 H8 O6 ) is vitamin C; sodium bicarbonate and sodium carbonate are baking soda and baking powder, respectively; and carbonic acid (H2 CO3 ) and phosphoric acid (H3 PO4 ) are important ingredients of many beverages. This chapter begins with some definitions that identify the key characteristics of acids and bases, continues with a description of how the strength of acids is quantified, and concludes with several examples showing how acid/base speciation can be evaluated in some simple systems. The discussion and many of the examples are presented in the context of answering the question: If a known amount of such-and-such acid or base is added to pure water, what will the solution composition be at equilibrium? In practice, the situation is rarely that simple. Solutions of interest usually contain a multitude of chemicals, and when an acid or base is added, the changes in water quality go well beyond those that are addressed in this chapter. Nevertheless, the acid/base reactions explored here are often of central importance, and studying the behavior of very simple systems can provide insights that serve us well in trying to understand the more complex systems we might encounter in natural or engineered environments.
5.2
THE DISSOCIATION OF WATER; Kw
In any aqueous solution, even the purest distilled water, some water molecules split apart in hydrolysis or dissociation reactions to form H+ and OH– ions. Like all equilibrium reactions, hydrolysis is reversible, so the reaction can be written as H2 O )* H+ + OH
(5.2)
The equilibration of water molecules with H+ and OH– ions is sufficiently rapid that the reaction is assumed to always be in equilibrium. The equilibrium constant for Reaction (5.2) is given a special symbol, Kw . Since the activity of H2 O is very nearly 1.0 in almost all solutions of interest, the equilibrium constant expression can be written as {H+ } OH Kw ⌘ ⇡ H+ OH (5.3) H2 O The value of Kw is 10 14.0 at 25 C and increases with increasing temperature, varying by approximately a factor of 30 over the range of temperatures encountered in natural aquatic systems and by a factor of ⇠ 500 between 0 and 100 C (Figure 5.2). Because {H+ } and {OH– } are related through Kw , any increase in {H+ } is accompanied by a decrease in {OH– }, and vice versa. A solution in which {H+ }>{OH– } is called acidic, and one in which {H+ }<{OH– } is called alkaline or basic. 219
Figure 5.2 The value of Kw from 0 to 100 C.
Although Reaction (5.2) is written as though H+ and OH– existed as distinct entities in solution, H+ ions are actually extremely unstable as free dissolved species, and almost every isolated H+ ion combines rapidly with one or more water molecules to form ions such as H3 O+ , H5 O+2 , and H7 O+3 (Figure 5.3). Thus, in reality, hydrolysis is not just the splitting of a water molecule; it is the transfer of an H+ ion from one water molecule to another in a reaction such as the following: H2 O + H2 O )* OH + H3 O+
(5.4)
Figure 5.3 A hydrated H+ ion (H3 O+ ), hydrogen-bonded to two other water molecules. (Image created by Xiaoshong Li.)
Under most circumstances, there is no need to distinguish among the various (H2 O)x H+ species. Although we will continue to use the symbol H+ , it should be understood that this symbol is a shorthand notation representing all species of the form H2x+1 O+x , of which H3 O+ is the most common. Collectively these ions are 220
called hydronium ions. Somewhat incorrectly, hydronium ions are often referred to as protons, as though they were indeed isolated H+ ions (an H+ ion has one proton and no neutrons in its nucleus, and no electrons orbiting it, so it is indeed simply a proton). Correspondingly, reactions in which an H+ ion binds to another molecule are commonly called protonation reactions.
5.3
THE STRUCTURE OF ACIDS AND BASES
In addition to the dissociation of H2 O molecules, H+ and OH– ions can be provided to solution by dissociation of chemicals that contain only one of these ions. For example, addition of hydrochloric acid (HCl) or acetic acid (CH3 COOH) to a solution can increase the H3 O+ concentration by the following acid dissociation reactions: HCl + H2 O )* H3 O+ + Cl CH3 COOH + H2 O )* H3 O+ + CH3 COO
K >> 1
(5.5)
K = 1.74 ⇥ 10
5
(5.6)
Molecules that act as acids via these types of reactions (transfer of an H+ ion to a water molecule) are referred to as Bronsted acids, and those that can combine with an H+ ion, such as Cl– and CH3 COO– in the reverse of Reactions (5.5) and (5.6), are referred to as Bronsted bases. Hydronium ions can also be generated by species that promote hydrolysis of water, even if those species themselves do not contain any H+ . For example, copper ions can react with water molecules as follows: Cu2+ + H2 O )* CuOH+ + H+
K = 10
8.0
(5.7)
In this reaction, the copper ion can be viewed as binding to a water molecule and inducing it to dissociate, so that the water molecule releases an H+ ion (which can combine with another water molecule to form H3 O+ ). The OH– that was part of the bound water molecule remains attached to the Cu2+ ion, so the net result is that copper causes the concentration (and activity) of H3 O+ ions in solution to increase, even though the copper itself has no H+ associated with it when it is initially added to solution. Acids that release H+ by such a mechanism are called Lewis acids, and the corresponding product of the reaction [CuOH+ in Reaction (5.7)], which can combine with H+ in the reverse reaction, is called a Lewis base.2 Based on the above discussion, acids (especially Bronsted acids) are sometimes referred to as proton donors (they “donate” protons, or H+ ions, to water molecules 2 The formal definition of a Lewis acid is a molecule that has the ability to accept a free electron pair. In the case of water binding to Cu2+ , the copper ion is an acid because it can accept an electron pair from the oxygen of the water molecule.
221
to form H3 O+ ) and bases as proton acceptors (they acquire protons from the donor molecules, i.e., the acids). Schematic representations of Reactions (5.4), (5.5), and (5.7) are shown in Figure 5.4.
Figure 5.4 Schematic representation of the donation of H+ from three acids — H2 O, HCl, and Cu2+ — to water, forming a hydronium ion. (Images created by Xiaosong Li.)
5.4
STRONG AND WEAK ACIDS, K a , AND CONJUGATE ACID/BASE PAIRS
The tendency of different acids to release an H+ ion to solution (or, equivalently, to donate an H+ ion to a water molecule) spans a wide spectrum. Logically enough, the strength of an acid is defined by its tendency to participate in such a reaction. For instance, as indicated by Reactions (5.5) through (5.7), hydrochloric acid (HCl), acetic acid (CH3 COOH) and cupric ion (Cu2+ ) can all cause the concentration of H+ ions in solution to increase (which is why all three are acids). When equal concentrations of these three acids are added to solution (either to separate solutions or all together), 222
the relative activities (and concentrations) of the various species at equilibrium are as follows:
Acids: Bases:
HCl < CH3 COOH < Cu2+ Cl > CH3 COO > CuOH
(5.8a) +
(5.8b)
From this sequence, we infer that hydrochloric acid is a stronger acid (induces release of more H+ ) than acetic acid, which is in turn stronger than cupric ion. By following this reasoning to its logical extreme, an infinitely strong Bronsted acid would completely dissociate, and an infinitely strong Lewis acid would hydrolyze one water molecule for every molecule of acid added, thereby leading to the maximum possible increase in the H+ concentration. Although the above comparison is useful, it is qualitative and not a particularly convenient way to discuss the relative strengths of acids. To address this issue, a shorthand has been developed by which the relative strengths of different acids can be assessed rapidly and quantitatively. Specifically, the acid dissociation constant or acidity constant (Ka ) is defined as the equilibrium constant for a reaction in which the acid is a reactant and one hydronium ion (i.e., one proton) is released as a product. Thus, for a generic acid HA, we can write HA + H2 O )* H3 O+ + A Ka ⌘
O+
{A } H3 = {HA} H2 O
(5.9)
{H+ } {A {HA}
}
(5.10)
Extremely acidic solutions can be generated in nature in locations where sulfide-containing minerals are exposed to air in limited amounts of water. This photograph is of a solution from the Richmond mine in Iron Mountain, CA, where pH values as low as 3.6 were recorded. (Charles N. Alpers and D. Kirk Nordstrom, USGS.)
223
Based on this definition, the equilibrium constants for Reactions (5.5) through (5.7) would all be called acidity constants. In addition, Equation (5.9) makes clear that the larger Ka is, the greater the acid’s tendency to dissociate, i.e., the stronger the acid. Acidity constants for several environmentally important acids are listed in Table 5.1.3 Note that water can be considered an acid with A– = OH– , as might be more apparent if Reaction (5.4) were written as H OH + H2 O )* H3 O+ + OH
(5.11)
The acidity constant for this acid dissociation reaction is Ka = {H+ }{OH }/{H2 O}, which is, of course, Kw . Some acids can undergo multiple acid dissociation reactions and thereby donate more than one proton to water molecules. For these acids, acidity constants are subscripted 1, 2. . . , starting with the most protonated species. Thus, for phosphoric acid (H3 PO4 ), {H+ }{H2 PO4 } H3 PO4 )* H+ + H2 PO–4 Ka1 = = 10 2.15 {H3 PO4 } H2 PO–4
)* H+ + HPO2– 4
3– + HPO2– 4 )* H + PO4
{H+ }{HPO24 } Ka2 = = 10 {H2 PO4 } Ka3 =
{H+ }{PO34 } = 10 {HPO24 }
7.20
12.38
In this acid/base group, the three species H3 PO4 , H2 PO–4 and HPO2– 4 can all act as acids, since each can release a proton. The Ka values indicate that H3 PO4 is a stronger acid than H2 PO–4 , which in turn is a stronger acid than HPO2– 4 . 4 Since all chemical reactions are reversible, the above reactions can also proceed from right to left, consuming H+ instead of releasing it. Consider, for instance, a solution prepared by adding Na3 PO4 to water. (Na3 PO4 is trisodium phosphate, which is sold commercially as the cleaning agent commonly referred to as TSP.) When the + Na3 PO4 first dissolves, it dissociates into Na+ and PO3– 4 ions. The Na ions are immediately surrounded by water molecules and are unlikely to react further. 3 Throughout this text,
the equilibrium constants given are consistent with the database in the chemical equilibrium computer program Visual MINTEQ, v.3.1 beta. This choice has been made to facilitate correspondence of computations carried out manually with the output from that program. Equilibrium constants for reactions of interest that are not in the program’s database have been selected from a variety of other sources. 4 As noted in Chapter 3, reactions are often referred to as irreversible if the reverse reaction is very slow compared to the forward reaction. Nevertheless, formally, all chemical reactions are reversible to some extent.
224
Table 5.1 Chemical formulas and acidity constants at 25 C of some important acids(a)
Name Hydrochloric acid Sulfuric acid Hydronium ion Chromic acid Oxalic acid Sulfurous acid Ferric ion Phosphoric acid Arsenic acid Salicylic acid Citric acid Hydrofluoric acid Benzoic acid Pentachlorophenol acid Acetic acid Propionic acid Carbonic acid Hydrogen sulfide Cupric ion Hypochlorous acid Hypobromous acid Zinc ion Arsenous acid Hydrocyanic acid Boric acid Ammonium ion Ferrous ion Silicic acid Phenol Cadmium ion Calcium ion
Formula HCl H2 SO4 H3 O+ H2 CrO4 (COOH)2 H2 SO3 Fe3+ H3 PO4 H3 AsO4 C6 H4 OHCOOH C3 H4 OH(COOH)3 HF C6 H5 COOH C6 Cl5 OH CH3 COOH CH3 CH2 COOH H2 CO3 H2 S Cu2+ HOCl HOBr Zn2+ H3 AsO3 HCN H3 BO3 NH+4 Fe2+ H4 SiO4 C6 H5 OH Cd2+ Ca2+
pKa1 <0 <0 0.00 0.20 1.250 1.85 2.02 2.148 2.20 2.972 3.128 3.18 4.20 4.7 4.757 4.874 6.352 7.02 7.497 7.53 8.63 8.997 9.17 9.21 9.236 9.244 9.397 9.84 9.98 10.097 12.697
pKa2 1.99 14.00 6.51 4.266 7.19 3.73 7.198 7.01 13.7 4.761
10.329 17.4 8.73
pKa3
pKa4
9.25 7.70 12.375 11.80 6.396
10.41
13.09
7.897 14.10
11.497 12.797
11.097 13.20
10.497
10.197
13.006 13.988
(a) Remember that, for logarithmic values, the number preceding the decimal point is an exponent, and the number of significant figures in the Ka values is only the number of digits to the right of the decimal point. The values in the table reflect the accuracy with which the particular Ka is known, but for example calculations, we will generally assume that Ka is known to two significant figures.
225
– + The rest of the solution then contains PO3– 4 ions, along with H2 O, H , and OH , – but no HPO2– 4 or H2 PO4 . Such a solution is out of equilibrium with respect to all three acid dissociation reactions shown above. To approach equilibrium, some PO3– 4 2– + combines with H+ ions to form HPO2– 4 , some HPO4 ions combines with H to form H2 PO–4 , etc. The process is shown schematically in Figure 5.5.
Figure 5.5 Schematic showing the stepwise protonation of phosphate ion, consuming dissolved H+ . 2– – The attachment of H+ to PO3– 4 , HPO4 , and H2 PO4 reduces the activity of dissolved H+ , so the product {H+ }{OH– } decreases and becomes transiently less than Kw . This process, in turn, causes a disequilibrium in the H2 O dissociation reaction, which is alleviated by the splitting of additional water molecules to increase the concentrations of H+ and OH– . The latter process continues until the product {H+ }{OH} is once again equal to Kw . The entire network of reactions proceeds until all the reactions are in equilibrium simultaneously. In the scenario described, when the final equilibrium state is achieved, the concentrations of H+ and OH– will not be equal; {OH– } will be greater than {H+ }, because equal amounts of H+ and OH– were added to solution by the water dissociation reaction, but some of the added H+ was then consumed by reactions with phosphate species, while no OH– was consumed. We will shortly develop the equations necessary to characterize the final equilibrium condition in such a solution quantitatively. For now, it is only important to note that the net effect of the addition of PO3– 4 is the removal of H+ from solution to form various Hx PO34 x species. Because PO3– 4 ions remove H+ from solution (or, put slightly differently, they accept H+ ions), we conclude that PO3– 4 acts as a base in this situation. 2– The species PO3– 4 and HPO4 are interconverted simply by the exchange of a pro2– ton. When HPO4 releases a proton to become a PO3– 4 ion, it is acting as an acid. The 3– PO4 ion, on the other hand, can combine with a proton to re-form HPO2– 4 , thereby acting as a base. In fact, the dissociation of any acid always generates a corresponding (potential) base, i.e., a species that, under different conditions, could combine with an H+ ion to regenerate the acid. Acid and base species that are related in this way are said to form a conjugate acid/base pair. Thus, HPO2– 4 is the conjugate base of the acid H2 PO–4 , and H2 PO–4 is the conjugate acid of the base HPO2– 4 . Note that 2– 3– HPO4 is also the conjugate acid of the base PO4 (Figure 5.6).
226
Figure 5.6 Conjugate acid/base relationships among phosphate species.
By conservation of charge, an acid always has a charge that is greater by 1 than its conjugate base. This is true regardless of whether the acid actually contains the “detachable” H+ (like HCl, CH3 COOH, and HOCl) or whether it releases an H+ ion by hydrolyzing a water molecule and combining with the OH– (like Cu2+ ). The way in which the charge on the molecule becomes redistributed when the proton is released plays a major role in determining the relative stability of the acid and conjugate base, and hence the value of Ka , as follows. The tendency of an atom to attract electrons is called its electronegativity. The relative electronegativity of most common elements has been characterized and quantified, so the tendency of electrons to associate primarily with one or another of the atoms in a given compound can be predicted. For instance, the partial negative charge that resides near the oxygen atom and partial positive charges that reside near the hydrogen atoms in a water molecule reflect the fact that oxygen is more electronegative than hydrogen. Colloquially, we might say that highly electronegative elements are more ‘comfortable’ carrying negative charge than less electronegative ones are. Applying this concept to solutions is a bit risky because interactions with water might change the electronic distribution in molecules, but in general it provides a good perspective from which to explore the relative acidity of various molecules. Some important acids (e.g., HCl and H2 S) have a very simple structure, consisting of one or two H+ ions attached to a single electronegative atom. The acidity of these species arises because when the H+ ion(s) is (are) released, the element that is left behind is strongly electronegative and hence comfortable carrying the negative charge. However, the vast majority of environmentally significant acids are more complex molecules. In most cases, these acids contain one or more oxygen atoms, and it is to these oxygen atoms that the acidic H+ ions are attached. Many inorganic acids (HNO3 , H2 SO4 , HClO4 , HClO, H2 CO3 ) fall into this category, as well as the most environmentally significant group of organic acids (carboxylic acids). If the acid is neutral or anionic, the conjugate base is anionic, and the negative charge resides primarily on the oxygen atoms. 227
Although they are useful in a myriad of biochemical and industrial applications, acids can also be destructive, as when they attack and dissolve pipe materials. This photograph shows a concrete sewer pipe that was severely damaged by “crown corrosion,” which occurs when hydrogen sulfide gas [H2 S(g)] is released from the sewage and dissolves into water droplets that have condensed on the top of the sewer pipe. Oxygen also dissolves into these droplets and reacts with the H2 S to form sulfuric acid (H2 SO4 ), which attacks the pipe structure. The ribbing occurs because the concrete degrades much more rapidly than the steel rebar. (Michael K. Stenstrom.)
One molecule that is a good example of such a system and that has special significance of its own is carbonic acid, H2 CO3 . This diprotic acid (containing two releasable H+ ions) can lose either or both of its H+ ions to form bicarbonate or carbonate ion (HCO–3 or CO2– 3 ), respectively. All three of these species are planar; i.e., the carbon atom and the three oxygen atoms lie in a plane, with the oxygen atoms roughly at the corners of an equilateral triangle surrounding the carbon (Figure 5.7). Although simplistic stick-and-ball representations usually show the negative charge on the molecules residing entirely on one oxygen atom in HCO–3 and on two oxygen atoms in CO2– 3 , the negative charge is actually always distributed among all three oxygen atoms. In HCO–3 , most of the negative charge is split between the two oxygen atoms that are not bound to a hydrogen ion. However, some of the negative charge resides on the oxygen atom to which the H+ is attached, increasing the strength of the O H bond. The dilution of the charge over multiple atoms stabilizes the molecule, 228
while the increased strength of the O H bond in HCO–3 causes that ion to be a weaker acid than H2 CO3 .
Figure 5.7 Schematic representation of the release of protons from carbonic acid to form bicarbonate and carbonate ions. In HCO–3 , most of the negative charge resides on the oxygen ions that are not bound to H+ (shown in the figure on the left side of the ion), but some charge resides on the third oxygen as well. In CO2– 3 , the charges are uniformly distributed across all three oxygen ions. (Images created by Xiaosong Li.)
Conceptually, much of the above discussion applies to bases as well, except that strong basicity is associated with electropositivity. The prototypical inorganic base is ammonia, with chemical formula NH3 . This molecule has a structure similar to that of water, except that a nitrogen atom, rather than an oxygen atom, is at the center of the tetrahedron, and three corners of the tetrahedron are occupied by hydrogen atoms instead of only two. The reaction in which ammonia acquires a fourth H+ ion to form NH+4 (thereby acting as a base, because it removes H+ from solution) (Figure 5.8) is very similar to the reaction by which water is protonated to H3 O+ . The NH+4 ion is referred to as ammonium ion (note the similarity with hydronium, H3 O+ ).
5.5
A BRIEF DIVERSION: THE STRUCTURE OF SOME IMPORTANT ORGANIC ACIDS AND BASES
Aside from being a critical species in all environmental systems, carbonic acid can be considered a prototype of a major class of organic acids known as carboxylic acids. These acids have a structure similar to H2 CO3 , except that the bond to one of the OH groups in H2 CO3 is replaced by a bond to another carbon atom, which 229
Figure 5.8 Structure of molecular ammonia, NH3 , and its protonation reaction. (Images created by Xiaosong Li.)
might be part of a large organic structure. Conventionally, any such structure (large or small) is represented by the letter R. Also, the HCO2 group is usually written as COOH, so the whole molecule is represented generically by the formula R COOH. For instance, one of the simplest carboxylic acids can be thought of as being formed by replacing one OH group in H2 CO3 by a methyl (CH3 ) group, to form acetic acid, with chemical formula CH3 COOH. Two structural representations of acetic acid are shown in Figure 5.9.
Figure 5.9 Structural representation of acetic acid. (Image on right created by Xiaosong Li.)
The bulk of the organic matter that enters natural waters is derived either from the microbial decay of plants in the local watershed, or by metabolic activity of organisms in the water itself. Many of these organic molecules are used by other microorganisms for growth and energy, but a small fraction are relatively resistant to further degradation and remain in the water. Thus, regardless of its source, the organic matter that we are likely to detect in natural waters consists primarily of decay-resistant molecules, for the simple reason that the more degradable molecules have already been consumed. Chemically, some of the earliest steps in the decay process break apart large organic molecules in the source material by an oxidation step that leaves a carboxyl group at the point where the new molecule had previously been joined to the original 230
structure. As a result, many of the organic molecules found in natural waters (collectively referred to as natural organic matter, or NOM) contain one or more carboxyl groups. The molecules also usually contain phenolic groups (aromatic rings with OH groups attached) that are slightly acidic (the proton can dissociate from the OH group, converting the phenol functional group to a phenolate group). Most NOM molecules are therefore polyprotic acids, capable of participating in several acid dissociation reactions. Current research suggests that many of these molecules contain 20 to 30 carbon atoms, at least a few of which are part of carboxylic and phenolic groups (Figure 5.10). Although the acidity of these groups depends on the structure of the entire molecule, most carboxylic groups have Ka values of 10 3 to 10 6 , while the phenolic groups are weaker acids, with Ka values in the range from 10 9 to 10 13 .
Figure 5.10 A conceptual model for a typical molecule of water-soluble natural organic matter. After Thurman, E.M. Organic Geochemistry of Natural Waters. Nijhoff, Boston (1985).
Just as H2 CO3 is an inorganic analog for the most common type of organic acid, NH3 is such an analog for the most common type of organic base. That is, most organic bases consist of an organic segment attached at some point to a nitrogen atom, which (in the simplest case) is attached to two H+ ions. Such a molecule, which can be represented generically as R NH2 , is shown in Figure 5.11a. In some cases, one or two additional organic segments are attached to the N, replacing the H atoms at other corners of the tetrahedron. Molecules that are attached to one, two, and three organic segments are referred to as primary, secondary, and tertiary amines, respectively; a secondary amine is shown in Figure 5.11b. 231
Natural organic matter can play either a beneficial or harmful role in many water treatment processes. This scanning electron micrographs shows NOM accumulated on the raw water side of a membrane. The membrane is intended to reject particulate contaminants while letting clean water pass through. The accumulated NOM dramatically increases the resistance to permeation of water through the membrane. [Reprinted from Journal AWWA, Vol.88, No.12 (December 1996). By permission. Copyright c 1996, American Water Works Association.]
Like an inorganic ammonia molecule, organic amines can bind to a proton via the unshared electron pair that is present in the neutral form of the molecules. Attachment of a proton in this way removes H+ from solution, so the amine is acting as a base; the conjugate acid of the secondary amine shown in Figure 5.11b is represented in part (c) of the figure. The acidity of most primary amines is fairly close to that of ammonium ion (Ka = 10 9.24 ), but the value varies depending on the structure of the rest of the molecule; the acidity of secondary and tertiary amines is often, but not always, of a similar magnitude. Amino acids are important biomolecules that contain both an organic nitrogen base and a carboxylic acid group. The structure of one such molecule, glycine, is shown in Figure 5.12, with the amine group in its acidic form and the carboxyl group in its basic form. Additional information about these relatively simple molecules, which are the building blocks of proteins and enzymes, can be found in any organic chemistry or biochemistry text. 232
Figure 5.11 Schematics of a (a) primary amine, (b) secondary amine, and (c) protonated secondary amine. The molecules in parts (a) and (b) are neutral, with an unshared electron pair occupying one orbital. These molecules can act as bases by accepting a proton, which bonds to the central N via the unshared electrons. The ion formed by reaction of the molecule in part (b) with a proton is shown in part (c). In that ion, a charge of approximately 0.5 esu resides on each H.
Figure 5.12 Structure of the amino acid glycine.
5.6
COMBINING ACIDITY REACTIONS AND THE DEFINITION OF BASICITY
When an acidity reaction is reversed, it is converted to a reaction in which an H+ ion is consumed. Adding that reaction to the reaction for the dissociation of water generates a new reaction that is analogous to an acidity reaction, but describes the release of one OH– ion rather than an H+ ion as follows:
+
Reverse of acid dissociation reaction
H+ + A– )* HA
(5.12)
Ka 1
Water dissociation
H2 O )* H+ + OH–
(5.13)
Kw
Basicity reaction
A– + H2 O )* HA + OH–
(5.14)
Kb = Ka 1 Kw
233
Reaction (5.14) represents a base being converted to its conjugate acid by the release of one OH– ion, just as Reaction (5.9) represents an acid being converted to its conjugate base by the release of one H+ ion. Reaction (5.14) is called a basicity reaction, and its equilibrium constant is designated Kb . Just as the magnitude of Ka indicates the strength of an acid, the magnitude of Kb indicates the strength of a base. As shown by the equilibrium constant for Reaction (5.14), Ka and Kb for a conjugate acid/base pair are related by Ka Kb = Kw
(5.15a)
pKa + pKb = pKw
(5.15b)
As shown by either form of Equation (5.15), the larger Ka is for an acid, the smaller Kb is for its conjugate base; i.e., strong acids always have weak conjugate bases, and vice versa. This conclusion is logical — since a strong acid is one that has a strong tendency to release H+ , it makes sense that the conjugate base of that acid would have only a slight tendency to acquire an H+ or, equivalently, to release an OH– . Example 5.1 The acidity constant for the second deprotonation reaction of phosphoric acid at 25 C is Ka2 = 10 7.198 . What is the corresponding basicity constant, and to what reaction does it apply? Solution The value of the basicity constant is Kw divided by the acidity constant, so Kb = 10 14.0 /10 7.198 = 10 6.802 . (Alternatively, we could note that pKa is 7.198, so pKb must be 14.0 7.198, or 6.802.) The acidity constant describes a reaction in which H2 PO–4 is the acid and HPO2– 4 is the conjugate base. The corresponding basicity reaction is for conversion of the base to the conjugate acid, accompanied by release of one hydroxyl ion; i.e., it is for the reaction HPO24 + H2 O )* H2 PO4 + OH
Example 5.2 When zinc perchlorate, Zn(ClO4 )2 , is added to water, the compound readily dissociates, and some of the Zn2+ hydrolyzes to form species with the generic formula ZnOH2x x . No dissolved species are known in which a Zn2+ is bound to one or more ClO–4 ions, and ClO–4 ions have negligible tendency to combine with H+ . Equilibrium constants for Zn-induced hydrolysis reactions are shown below. The symbols used to represent the reactions are conventional notation for various types of equilibrium 234
constants and are defined in greater detail later in the text; for now, all that is important is that the values shown are the equilibrium constants for the corresponding reactions. Zn2+ + H2 O )* ZnOH+ + H+
⇤K 1
Zn2+ + 2 OH– )* Zn(OH)2
b2 = 10+11.11
Zn(OH)2 +OH– )* Zn(OH)–3
K3 = 10+2.5
+ Zn2+ + 4 H2 O )* Zn(OH)2– 4 +4H
⇤b 2
= 10
= 10
9.00
41.19
(a) Using the above equilibria and Kw , determine the equilibrium constant (commonly designated b4 ) for the following reaction: Zn2+ + 4 OH )* Zn(OH)24
b4 =?
(b) At what pH or in what pH range do the following relationships hold? (i)
{Zn2+ } = {Zn(OH)+ }
(ii) {Zn(OH)2 } > {Zn(OH)24 } Solution (a) The desired equilibrium constant can be derived by combining the fourth reaction shown in the problem statement with four times the reverse of the water dissociation reaction: + Zn2+ + 4 H2 O )* Zn(OH)2– 4 +4H
⇤b 2
4 H+ + 4 OH– )* 4 H2 O
Kw 4 = 10+56.00
Zn2+ + 4 OH– )* Zn(OH)2– 4
b4 = 10+14.81
= 10
41.19
Note that since the dissociation of four water molecules is equivalent to the sum of four “normal” water dissociation reactions, the equilibrium constant for the reaction is Kw4 , and for the reverse reaction, K = Kw 4 . (b) (i) The relative activities of two species can be determined based on the equilibrium constant that relates them (in this case, the one designated ⇤ K1 ). Manipulating that constant, we find ⇤K {ZnOH+ } 10 9.00 1 = = {H+ } {H+ } {Zn2+ }
235
Based on this equation, {ZnOH+ } = {Zn2+ } when {H+ } = 10 pH = 9.00.
9.00 ,
i.e., at
(ii) We can follow a procedure analogous to that used in (b)(i), except that in this case we need to derive the equilibrium constant of interest. Combining the second reaction in the problem statement with the result of part (a), we obtain Zn2+ + 2 OH– )* Zn(OH)2
b2 = 10+11.11
– 2+ Zn(OH)2– 4 )* Zn + 4 OH
b4 1 = 10
– Zn(OH)2– 4 )* Zn(OH)2 + 2 OH
K = 10
10
3.70
{Zn(OH)2 } OH n o = Zn(OH)24
14.81
3.70
2
Manipulating this equilibrium constant expression to generate an expression for the relative activities of Zn(OH)2 and Zn(OH)2– 4 gives {Zn(OH)2 }
{Zn(OH)24 }
=
10
3.70
{OH }
2
The question asks us to identify conditions under which {Zn(OH)2 }> {Zn(OH)2– 4 }. For this inequality to apply, each side of the preceding equation must be > 1, so: OH
2
OH
< 10
< 10
3.70 1.85
Writing {OH– } in terms of {H+ }, and noting that pH is defined as we find Kw < 10 {H+ } log
log (1/{H+ }),
1.85
1 10 1.85 10 < log = log + {H } Kw 10
1.85 14.0
pH < 12.15
We therefore conclude that {Zn(OH)2 }>{Zn(OH)2– 4 } when pH < 12.15.
236
Example 5.3 Carbonic acid (H2 CO3 ) forms by hydration of dissolved carbon dioxide according to the reaction CO2 (aq) + H2 O )* H2 CO3 (aq) Although H2 CO3 and CO2 (aq) are different chemical species, for many applications the distinction between them is unimportant, and they are often lumped together and treated as a single type of molecule. In such cases, the sum ⇥ of the ⇤ concentrations ⇤ or activities of H2 CO3 and CO2 (aq) is often represented as H2 CO3 or {H2 CO3 ⇤ }, respectively, i.e.:5 H2 CO3 ⇤ ⌘ CO2 (aq) + H2 CO3 and
⇥
⇤ H2 CO3 ⇤ ⌘ [CO2 (aq)] + [H2 CO3 ]
The first dissociation constant for carbonic acid, (Ka1,H2 CO3 ), is approximately 10 3.50 , and the corresponding dissociation constant for H2 CO3 ⇤ is 10 6.35 , i.e., H2 CO3 ⇤ )* H+ + HCO3
Ka1,H2 CO3 ⇤ = 10
6.35
Compute the equilibrium constant for hydration of CO2 (aq) and comment on the relative concentrations of CO2 (aq) and H2 CO3 in an equilibrium solution. Assume activity coefficients are all 1.0. Solution Recognizing that the activity of water is essentially 1.0, the equilibrium constant that we seek (the first reaction shown in the problem statement) can be written as KCO2 hyd. =
H2 CO3 H2 CO3 = CO2 (aq) H2 O CO2 (aq)
The given acidity constants indicate that {H+ } HCO3 H2 CO3
= 10
3.50
{H+ } HCO3 H2 CO3 ⇤
= 10
6.35
5 This situation is analogous to that for H
represented collectively as
H+ .
+ 2x + 1 Ox
species, which are commonly lumped together and
237
The ratio of {H2 CO3 ⇤ } to {H2 CO3 } at equilibrium is given by the ratio of these K values, i.e., H2 CO3 ⇤ 10 3.50 = = 10+2.85 H2 CO3 10 6.35 Expanding the numerator in this expression and carrying out some algebra, we find H2 CO3 + CO2 (aq) CO2 (aq) = 1+ = 102.85 = 708 H2 CO3 H2 CO3 CO2 (aq) = 707 H2 CO3 KCO2 hyd. =
H2 CO3 1 = = 1.41 ⇥ 10 CO2 (aq) 707
3
This result indicates that the vast majority of the dissolved, undissociated carbon dioxide in an equilibrium solution is present as CO2 (aq) and not as H2 CO3 . In most texts and the technical literature, the term H2 CO3 is used to represent not just true H2 CO3 , but rather H2 CO3 ⇤ . This convention is also followed here unless explicitly stated otherwise. The preceding examples reinforce the point that chemical activities and concentrations can be added, multiplied, or manipulated in any other mathematically valid manner, and the result will be equally valid. In previous examples, we multiplied equilibrium constants in order to obtain a new constant for an overall reaction, and in Example 5.3 we added the activities of two distinct species to characterize a pair of similar species as a single one. In each case, the validity of the result is ensured simply by the validity of the mathematical operations, without any reference to the chemical implications of the calculations.
5.7 5.7.1
ACID/BASE SPECIATION AS A FUNCTION OF pH; DIAGRAMS OF LOG C VERSUS pH More Chemical Shorthand: a Notation
The speciation of acids and bases is frequently described by a shorthand known as a (alpha) notation. In this shorthand, the sum of the concentrations of all the species in solution that contain a chemical group A is defined as T OT A.6 For instance, if 6 The total concentration of a substance is represented in different ways by different authors. The choice here follows that of Morel and Hering [Morel, F.M.M., and Hering, J.G. Principles and Applications of Aquatic Chemistry. Wiley, New York, NY (1993)], which is consistent with the way that the term is usually represented in software for calculating chemical equilibrium.
238
we are interested in the speciation of the phosphate group in a solution, ‘A’ is PO4 , 3– and T OT PO4 is the sum of the concentrations of H3 PO4 , H2 PO–4 , HPO2– 4 , and PO4 . Returning to the generic case, a0 is defined as the fraction of T OT A that is in the most protonated form, and ai is defined as the fraction of the T OT A that is in a form that has lost i protons. Thus, for phosphoric acid, the values of a for the species 3– H3 PO4 , H2 PO–4 , HPO2– 4 and PO4 are defined as a0 =
[H3 PO4 ] T OT PO4
a1 =
[H2 PO4 ] T OT PO4
a2 =
[HPO24 ] T OT PO4
a3 =
[PO34 ] T OT PO4
By the definition of a, Â ai must always equal 1.0. As is shown next, if all the all i
species in an acid/base group behave ideally (i.e., if they all have activity coefficients of 1.0), their a values depend only on the solution pH. Thus, if the most protonated form of an acid/base group that behaves ideally is Hn A, the product ai (T OT A) expresses the concentration of the species Hn i Ai as the product of one term that depends on solution pH and is independent of T OT A, and another that depends on T OT A and is independent of solution pH.
5.7.2
Acid/Base Speciation as a Function of pH
To explore the relationships between ai and pH, we next prepare a chart showing, for a generalized monoprotic acid HA, the relative concentrations of HA and A– as a function of pH and pKa in an ideal solution. We begin with a slight manipulation of the Ka expression, as follows: {A } Ka = {HA} {H+ }
(5.16)
Taking logarithms of both sides gives log
{A } = log H+ + log Ka = pH {HA}
pKa
(5.17)
Because we are assuming that the solution is ideal, we can equate the activities of HA and A– with their molar concentrations. We could do the same for H+ . However, pH is defined as the negative logarithm of the H+ activity, not the H+ concentration, and it is a commonly measured parameter in both laboratory and field studies. By contrast, dissolved species other than H+ are much more commonly quantified in terms of their concentrations, not their activities. Therefore, to develop relationships that are applicable to the data that are most likely to be available, it makes sense to write the equations in terms of the concentrations of HA and A– , but the activity of H+ . We therefore rewrite Equations (5.16) and (5.17) as 239
[A ] Ka = [HA] {H+ }
⇥ ⇤ A log = pH [HA]
(5.18)
pKa
(5.19)
Equations (5.18) and (5.19) are important results that we will use extensively. For instance, Equation ⇥ ⇤ (5.19) indicates that every time the solution pH increases by 1, the ⇥ ⇤ A value of log also increases by 1, so the ratio A /[HA] increases by a factor [HA] of 10. This result applies to any two ideal species that form a conjugate acid/base pair, regardless of the value of Ka , T OT A, or the concentration of other species in the system. Adding an expression with a value of 1.0 to each side of Equation (5.18) and noting that the sum [A ] + [HA] is T OT A, we can write {H+ } [A ] [HA] Ka + = + [HA] [HA] {H+ } {H+ }
(5.20)
T OT A Ka + {H+ } = {H+ } [HA]
(5.21)
Inverting both sides of Equation (5.21) yields an expression for a0 as a function of {H+ } and Ka . The derivation for a1 is parallel to that for a0 . The resulting expressions are a0 =
{H+ } 1 = + Ka + {H } (Ka /{H+ } ) + 1
(5.22)
a1 =
Ka 1 = Ka + {H+ } 1 + ({H+ }/Ka )
(5.23)
The results of numerical calculations using Equations (5.22) and (5.23) are shown in Table 5.2 and Figure 5.13. Under conditions where the solution pH is one or more units less than pKa , corresponding to {H+ } values that are larger than Ka by a factor of about 10 or more, a0 is nearly 1.0 and a1 is nearly 0; that is, almost all of T OT A is present as the species HA. Conversely, at pH values significantly greater than pKa (i.e., under conditions where {H+ } is much smaller than Ka ), a0 ⇡ 0 and a1 ⇡ 1.0, so almost all of T OT A is present as A– . Only when the pH is between approximately pKa 1 and pKa + 1 are both species present in significant concentrations relative to T OT A. 240
241
10
0
1
2
3
4
5
6
7
8
1
101
102
103
104
105
106
107
108
1
3
3
10
1
4
4
10
10
5
5
10
2
6
6
10
7
7
10
2
8
8
⇤ Note:
Ka {H+ } = pH pKa
log
10
Ka {H+ }
8
7
6
5
4
3
3
1
✓
1.00
1.00
1.00
1.00
1.00
1.00
0.99
0.91
0.500
0.091
9.9 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
3
3
4
5
6
7
8
◆
1.0 ⇥ 10
{H+ } Ka = a1
1+
1
70
50
30
10
8
6
5
3
1
0.90
0.70
0.50
0.30
0.10
0.08
0.06
0.03
Ka {H+ }
1.85
1.70
1.48
1.00
0.90
0.78
0.70
0.48
0.00
0.046
0.15
0.30
0.52
1.00
1.09
1.22
1.52
Ka {H+ } = pH pKa
log
1
✓
Ka < 100. {H+ }
0.014
0.020
0.032
0.091
0.11
0.14
0.17
0.25
0.50
0.53
0.59
0.67
0.77
0.91
0.93
0.94
0.97
✓ ◆ Ka 1+ + {H } = a0
The final four columns are an expansion of the results in the region 0.01 <
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
9.9 ⇥ 10
0.091
0.500
0.91
0.99
1.00
1.00
1.00
1.00
1.00
1.00
✓ ◆ Ka 1+ + {H } = a0
◆
0.99
0.98
0.97
0.91
0.89
0.86
0.83
0.75
0.50
0.47
0.41
0.33
0.23
0.091
0.074
0.057
0.029
{H+ } Ka = a1 1+
Table 5.2 Values of a0 and a1 of a monoprotic acid as a function of the ratio of the acidity constant to the H+ activity⇤ 1
Figure 5.13 Changes in the distribution of the protonated (a0 ) and deprotonated (a1 ) forms of a monoprotic acid as a function of the difference between solution pH and the acid’s pKa . (a) Arithmetic representation. (b) logarithmic representation.
The point of intersection of the a0 and a1 curves indicates the condition where HA and A– are present at equal concentrations. Based on Equation (5.18) or (5.19), this equality is satisfied at {H+ } = Ka , or pH = pKa . Under those conditions, for a monoprotic acid, each species accounts for one-half of T OT A, so a0 = a1 = 0.5
At pH = pKa :
log a0 = log a1 = 0.3
(5.24)
Thus, the intersection of the curves representing a0 and a1 must be at pH = pKa , and in a system in which HA and A– are the only two possible forms of A, it must be at a0 = a1 = 0.5, log a0 = log a1 = 0.3. 242
5.8
LOG C-pH DIAGRAMS FOR MONOPROTIC ACIDS
Figure 5.13 can be converted to a plot of log a versus pH for a specific acid/base conjugate pair by inserting the appropriate value for pKa and plotting the corresponding values on the abscissa. For instance, log a versus pH for the acetic acid/acetate system (HAc/Ac– , pKa = 4.76) is plotted in Figure 5.14. Note that the intersection of the a0 and a1 curves has shifted from pH pKa = 0 in Figure 5.13 to pH = pKa = 4.76 in Figure 5.14, but that nothing else has changed. If, instead of the HAc/Ac– pair, we wanted to represent speciation of the hypochlorous acid/hypochlorite pair (HOCl/OCl– , pKa = 7.53), we would simply shift the curves in Figure 5.14 to the right, so that they intersected at pH = 7.53.
Figure 5.14 (a) a versus pH and (b) log a versus pH plots for the acetic acid/acetate pair. The graph is essentially identical to that in Figure 5.13, except that the x axis has been converted to a pH scale, with the value of 4.76 inserted for pKa .
243
The concentrations of HA and A– in a particular system can be found by multiplying the corresponding a values by T OT A; similarly, values of log [HA] and log [A ] can be found by adding log T OT A to log a, i.e., [HA] = (T OT A) a0 = (T OT A)
{H+ } Ka + {H+ }
log [HA] = log (T OT A) + log a0 ⇥ ⇤ Ka A = (T OT A)a1 = (T OT A) Ka + {H+ } ⇥ ⇤ log A = log (T OT A) + log a1
(5.25a) (5.25b) (5.25c) (5.25d)
Equations (5.25b) and (5.25d) indicate that a plot of log a versus pH can be converted to one of log c versus pH for a particular system simply by adding log (T OT A) to the values on the ordinate. A plot showing log [HA] and log [A ] at three different values of T OT A for a hypothetical acid with pKa of 5.50 is shown in Figure 5.15.
Figure 5.15 Speciation of an acid HA with acidity constant pKa = 5.5, in ideal solutions with three different concentrations of T OT A.
It is important to recognize that the curves in Figures 5.14 and 5.15 are simply plots of Equations (5.25a-d) applied to particular acid/base pairs in systems with specified values of T OT A. The plots are useful because they allow us to see at a glance trends that occur over many orders of magnitude of change in H+ , but ultimately they contain no information that is different from what is in the equations. Because the concentrations of all the species of interest can be plotted as a function of H+ (for a given acid/base conjugate pair, at a given total concentration), H+ is sometimes referred to as a master variable in such systems. Graphs such as Figure 5.15 are generically referred to as log (concentration)-pH or log C-pH graphs. 244
Because the only equations used to prepare the preceding graphs are the acidity constant and the definition of T OT A, the information in the graphs is very generic. That is, by choosing appropriate values for T OT A and Ka when carrying out the calculations, the same spreadsheet can be used to determine the speciation of any monoprotic acid present at any concentration in a solution. Plots of the speciation of acetic acid, ammonia, and hypochlorous acid as a function of pH are all shown together in Figures 5.16. Consistent with the preceding discussion, the curves for all three acid/base groups are essentially identical, but they are displaced from one another vertically, reflecting the different concentrations of the acids in system, and horizontally, reflecting the different strengths of the acids. In particular, the stronger the acid, the farther to the left the lines for HA and A– shift.
Figure 5.16 Distribution of acetate, hypochlorite, and ammonia species as a function of pH for a system containing 10 2.5 M T OT Ac, 10 3.7 M T OT OCl, and 10 3.1 M T OT NH3 . The pKa values for these acids are 4.76, 7.53, and 9.24, respectively.
Note that the slopes of the lines representing HA and A– are closely related: they change in tandem, so that when the slope of the [HA] line is 0, that of the A– line is 1, and when the former changes to 1, the latter changes to 0. This observation can be generalized to a broad range of systems by manipulating Equation (5.19) as follows: [A ] = pH pKa [HA] ⇥ ⇤ log A = log [HA] + pH log
(5.19) pKa
(5.26)
Differentiating both sides of Equation (5.26) with respect to pH yields d log [A ] d log [HA] = +1 d pH d pH 245
(5.27)
Equation (5.27) indicates that the slope of a curve of log [A ] versus pH is always greater by 1 than the slope of a curve of log [HA] versus pH. Thus, when one of these slopes changes, the other must change as well. This result, as most of the others derived above, is simply a consequence of the mathematical form of the Ka expression and hence applies to all conjugate acid/base pairs under all conditions. We will see shortly that lines representing [HA] and [A– ] can have slopes larger or smaller than those shown thus far; however, in all cases, the difference in their slopes must be 1.0, in accord with Equation (5.27). The graphical display of the dependence of speciation on pH provides additional insight into the definition of a strong acid. In particular, note that HOCl is almost completely dissociated at pH 10. Therefore, in a pH 10 solution, it behaves as a strong acid, releasing essentially all its protons. On the other hand, at pH 5, it is almost completely in the HOCl form; adding HOCl to a solution at pH 5 would hardly increase the dissolved H+ concentration at all, because virtually none of the added HOCl would dissociate. Thus, HOCl might behave as a strong acid in one solution and as a weak acid in another. Based on Table 5.2 and Figure 5.13, acids dissociate almost completely if pH> pKa + 1. This fact, in conjunction with the patterns observed in the logC-pH diagram, make it clear why individuals working in different subfields of water chemistry might have quite different ideas about what constitutes a strong acid. Since solutions with pH< 3 are rare in the natural world, any acid with pKa lower than about 2 is a relatively strong acid in that context; i.e., any acid with pKa < 2 will be almost completely deprotonated in any natural water system. On the other hand, solutions at pH near 0 are used fairly commonly in analytical aquatic chemistry. Correspondingly, an analytical chemist might consider an acid with pKa of 2.0 to be very weak, because only a small fraction of the acid is converted to its conjugate base in many solutions with which that chemist works. A similar comment applies to strong bases, which are defined as bases that are converted almost completely to their conjugate acids (by releasing OH– or combining with H+ ). Bases that have Kb values of 10 1 or 10 2 might be considered very strong in some contexts (specifically, if they are present in systems at pOH > 3, which is to say, pH < 11) and weak in others (e.g., at pOH < 1, corresponding to pH > 13). Graphs like those shown in Figures 5.14-5.16 are enormously valuable tools for understanding not only acid/base chemistry, but also its influence on numerous other reactions of interest. Indeed, it is fair to say that the development of these graphs transformed the study of aquatic chemistry. The major benefits of such graphs are that (1) they summarize in a concise format a great deal of speciation information, covering many orders of magnitude in the concentrations and activities of H+ and other solutes, and (2) they are essentially universal, applying to any monoprotic acid/base group. We will use such plots as a major tool for exploring water quality, the exchange of gases with solution, metal precipitation, and oxidation/reduction reactions throughout the remainder of the text. 246
5.9
EFFECTS OF NONIDEAL SOLUTE BEHAVIOR ON ACID/BASE SPECIATION
Much of the mathematics and discussion in the preceding section relies on the simplifying assumption that HA and A– behave as ideal solutes, so that their activities can be equated with their molar concentrations. That assumption is never strictly valid, and it is sometimes significantly in error. However, if the activity coefficients of those species are known, it is relatively easy to modify the equations to develop the appropriate logC-pH diagram. To do so, we expand and then simplify the ratio {A }/{HA} as follows: ⇥ ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ ⇤ A gA A / A gA A /1.0 mol/L gA A = = = {HA} gHA [HA]/[HA] gHA [HA]/1.0 mol/L gHA [HA]
(5.28)
Substituting this result into Equation (5.17). we obtain ⇥ ⇤ gA A log = pH gHA [HA] ⇥ ⇤ A log = pH [HA] where
pKa
pKa
pKa,eff = pKa + log
log
(5.29) gA = pH gHA
pKa,eff
gA gHA
(5.30)
(5.31)
In essence, Ka,eff is an effective or conditional equilibrium constant, relating the concentration of HA to that of A– in a solution by incorporating the activity coefficients of those species into the acidity constant. Because Equation (5.30) is identical to Equation (5.19) except with Ka,eff replacing Ka , the remainder of the derivation in the preceding section applies, yielding as final results: a0 = a1 =
{H+ } 1 = + Ka,eff + {H } (Ka,eff /{H+ } ) + 1 Ka,eff 1 = + + Ka,eff + {H } 1 + ({H }/Ka,eff )
(5.32) (5.33)
The implication is that the all the logC-pH plots we have drawn to this point still apply in solutions with nonideal solutes, but the crossover point where [HA] = [A ] is shifted from pH=pKa to pH=pKa,eff . 247
Furthermore, because the acid always has a charge that is one larger than its conjugate base, we can estimate the offset between pKa and pKa,eff based on the Davies equation as pKa,eff
gA pKa = log = log gA log g HA g HA ✓ ◆ 0.5 2 I = 0.51(zHA 1) 0.3I 1 + I 0.5 I 0.5 = 0.51 (2zHA 1) 1 + I 0.5
✓
I 0.5 0.51(zHA ) 1 + I 0.5 2
0.3I
◆
(5.34)
The magnitude of this term for a range of zHA values and ionic strengths is shown in Figure 5.17. This figure indicates that, for common environmental solutions, effective acidity constants can be expected to differ from tabulated values (which apply in ideal solutions) by 0.1 to 1.0 log unit.
Figure 5.17 The shift in the apparent acidity constant caused by nonideality. Activity coefficients have been estimated with the Davies equation.
Example 5.4 Sulfurous acid (H2 SO3 ) or its deprotonated forms (bisulfite [HSO3 ] or sulfite [SO23 ] ion) is commonly used to “quench” residual oxidizing agents before water is discharged to natural systems, to prevent reactions of those oxidants with chemicals or organisms in the receiving water. For instance, if wastewater is disinfected with hypochlorous acid, the HOCl and OCl– can be converted to the innocuous species chloride (Cl– ) before the water is discharged by adding sodium bisulfite (NaHSO3 ). The acidity constant for bisulfite is given in Table 5.1 as 10 7.19 . Compute the concentration ratio of HSO–3 to its conjugate base, SO2– 3 , at pH 7.0 in solutions with ionic 248
strengths of 0.002 and 0.1M, assuming the Davies equation applies. Solution In this case, the acid (HSO–3 ) has a charge of 1. Inserting that value and the ionic strength of 0.002 into Equation (5.34), we find: I 0.5 (2zHA 1) 1 + I 0.5 0.0020.5 = 7.19 + 0.51 ( 3) = 7.12 1.0020.5
pKa,eff = pKa + 0.51
An analogous calculation indicates that pKa,eff at an ionic strength of 0.1 is 6.73. The concentration ratios in the two solutions are therefore ⇥ ⇤ HSO3 {H+ } 10 7.0 ⇥ 2 ⇤ = I = 0.002 : = = 1.32 Ka,eff 10 7.12 SO3 I = 0.10 :
⇥
⇤ HSO3 {H+ } 10 ⇥ 2 ⇤ = = Ka,eff 10 SO3
7.0
6.73
= 0.54
The concentration ratio is shifted significantly toward SO2– 3 at the higher ionic strength.
5.10
LOG C-pH DIAGRAMS FOR MULTIPROTIC ACIDS
The relationships characterizing the concentrations and activities of the species comprising multiprotic acid/base systems can be derived by simple extensions of the analysis of monoprotic systems. Consider, for example, the speciation of carbonic acid, which undergoes the following two dissociation reactions: H2 CO3 )* HCO3 + H+
Ka1 = 10
6.35
HCO3 )* CO23 + H+
Ka2 = 10
10.33
From the acidity constants, we know the following relationships between each acid and its conjugate base, where the final equality applies if the solutes behave ideally: ⇥ ⇤ H2 CO3 H2 CO3 {H+ } ⇤ = = ⇥ (5.35) Ka1 HCO3 HCO3 HCO3 CO23
⇥ ⇤ HCO3 {H+ } = = ⇥ 2 ⇤ Ka2 CO3 249
(5.36)
The corresponding relationships between H2 CO3 and CO2– 3 can be obtained by taking the product of these two expressions: ⇥ ⇤ 2 H2 CO3 H2 CO3 {H+ } = = ⇥ 2 ⇤ (5.37) Ka1 Ka2 CO23 CO3
If the assumption of ideal solute behavior applies, these ratios can be used directly to derive an expression for a0 that depends only on the Ka ’s and {H+ } (or, equivalently, pH): ⇥ ⇤ ⇥ ⇤ H2 CO3 H2 CO3 ⇤ ⇥ ⇤ ⇥ ⇤ a0 = =⇥ T OT CO3 H2 CO3 + HCO3 + CO23 1 ⇤ ⇥ ⇤ ⇥ 2 ⇤ = ⇥ H2 CO3 HCO3 CO3 ⇥ ⇤+⇥ ⇤+⇥ ⇤ H2 CO3 H2 CO3 H2 CO3 =
1 Ka1 Ka1 Ka2 1+ + + {H } {H+ }2
(5.38)
Expressions for a1 and a2 can be derived by essentially identical manipulations, but with [HCO3 ] and [CO23 ], respectively, in the numerator of the first equation, yielding a1 =
1 {H+ } Ka1
a2 =
Ka2 +1+ + {H } 1
2 {H+ }
{H+ } + +1 Ka1 Ka2 Ka2
(5.39)
(5.40)
Results of calculations based on these equations are shown in Figure 5.18. For nonideal solutions, the same equations would apply, except that Ka,eff values would be substituted for the Ka values. The plots in Figure 5.18 bear many similarities to those for monoprotic systems. For example, in each system, the more highly protonated species are dominant at lower pH, and the less protonated ones at higher pH. The a value of the most protonated species (a0 ) decreases steadily with increasing pH, whereas that of the least protonated species (a1 for monoprotic acids, a2 for carbonic acid) increases. In the carbonate system, the dependence of a1 on pH is not monotonic, because bicarbonate ion (HCO–3 ) plays a double role as both the conjugate base of carbonic acid (H2 CO3 ) 250
Figure 5.18 Distribution of carbonate species as a function of pH. Carbonic acid is a diprotic acid with pKa1 = 6.35 and pKa2 = 10.33. The species distribution is represented as the fraction of T OT CO3 present in the most protonated (H2 CO3 , a0 ), partially 2– deprotonated (HCO2– 3 , a1 ), and fully deprotonated (CO3 , a2 ) forms.
and the conjugate acid of carbonate ion (CO2– 3 ). At pH up to about 8.3, the carbonate concentration is negligible (a2 is much smaller than a0 + a1 ), so bicarbonate behaves primarily as the conjugate base of carbonic acid, causing a1 to increase with increasing pH. However, at higher pH, the carbonic acid concentration is negligible (a0 is much smaller than a1 + a2 ), so bicarbonate behaves primarily as an acid, and a1 declines with increasing pH. Finally, for any of the conjugate acid/base pairs, the a values of the acid and base are equal when the pH equals the corresponding pKa , and at pH values that are more than about one unit away from pKa , the acid (if the pH is lower than pKa ) or the base 251
(if pH is higher than pKa ) is present at a much larger concentration than its conjugate partner. Like the plot for monoprotic acids, the log a-pH plot for carbonate species can be converted to a logC-pH plot by adjusting the vertical scale so that the upper bound (i.e., log a = 0) is reassigned a value of log T OT A (i.e., log T OT CO3 ). We saw earlier [Equation (5.27)] that the slope of the curve for an acid on a logC-pH diagram is always one less than that for its conjugate base. This result was derived solely from the Ka expression and is therefore applicable regardless of whether the acid is mono- or multi-protic. As a result, for a diprotic acid H2 A, the slope of the H2 A curve is always one less than that of HA– , and the slope of the HA– curve is always one less than that of A2– , meaning that the slope of the H2 A curve is always two less than that of A2– . Thus, for example, in Figure 5.18, the slope of the H2 CO3 line is always two less that of CO2– 3 : at pH 5.0, the slopes are 0 and +2, respectively; at pH 8.0, they are 1 and +1, respectively; and at pH 12.0, they are 2 and 0, respectively. Generalizing this result, for any multiprotic acid, the slopes of the curves for two species Hn A and Hm A differ by m n when plotted on a logC-pH diagram. The relationships needed to compute the acid/base speciation of dissolved phosphate are collected in Table 5.3a and graphed for a system containing 10 2 M T OT PO4 in Figure 5.19. All the calculations needed to derive these equations follow logically from the previous example describing speciation of the carbonate group. Note that, in the list of equations governing the phosphate system, the relationships in Table 5.3a are derived strictly from equilibrium constants and therefore relate activities of the various species. On the other hand, the ai expressions are, by definition, relationships among concentrations. It is when these two groups of relationships are combined to generate the final expressions shown for ai in Table 5.3b that the effects of nonideal behavior become important. The equilibrium constants in these equations are therefore shown as effective constants. The significance of nonideal behavior on speciation in this system is illustrated for a system with ionic strength of 0.1 M in Figure 5.19. Generalizing the preceding results, we can write ai for any acid that undergoes n acid dissociation reactions as follows for a system that behaves ideally:
ai =
{H+ } n
Â
+
H
i
i
’ Ka j j=0 k
k
’ Ka j j=0
k=0
!
(5.41)
where Ka0 ⌘ 1.0. As with all previous equations, the effects of nonideality can be accounted for by replacing Ka j with Ka j,eff . 252
253 K03 = Ka1 Ka2 Ka3 K13 = Ka2 Ka3 K02 = Ka1 Ka2
+ H2 PO–4 )* PO3– 4 +2H
+ H3 PO4 )* HPO2– 4 +2H
5
6
Ka3
3– + HPO2– 4 )* PO4 + H
3
+ H3 PO4 )* PO3– 4 +3H
Ka2
+ H2 PO–4 )* HPO2– 4 +H
2
4
Ka1
H3 PO4 )* H2 PO–4 + H+
Equilibrium constant
1
Equilibrium reaction
log
log
log
log
log
log
= log K13 + 2 pH = log K02 + 2 pH
PO34 H2 PO4 HPO24 H3 PO4
PO34 = log K03 + 3 pH H3 PO4
HPO24
= log K3 + pH
= log K2 + pH
HPO24 H2 PO4 PO34
= log K1 + pH
H2 PO4 H3 PO4
Relationship among species’ activities
Table 5.3a Key relationships needed to determine phosphate speciation as a function of pH
254
8
7
⇥
1 Ka1,eff Ka1,eff Ka2,eff Ka1,eff Ka2,eff Ka3,eff 1+ + + {H+ } {H+ }2 {H+ }3
1 ⇥ ⇤ ⇥ ⇤ ⇥ 3 ⇤ H2 PO4 HPO24 PO4 ⇤+⇥ ⇤ +⇥ ⇤ 1+ ⇥ H3 PO4 H3 PO4 H3 PO4
Alternatively, a1 can be derived from a0 as follows:
⇤ H2 PO4 1 ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ 3 ⇤= ⇥ ⇤ ⇥ ⇤ ⇥ 3 ⇤ a1 = ⇥ 2 H3 PO4 + H2 PO4 + HPO4 + PO4 H3 PO4 HPO24 PO4 ⇥ ⇤ +1+ ⇥ ⇤+⇥ ⇤ H2 PO4 H2 PO4 H2 PO4 1 = + Ka2,eff Ka2,eff Ka3,eff {H } +1+ + {H+ } Ka1,eff {H+ }2
=
⇤ H3 PO4 ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ ⇤= a0 = ⇥ H3 PO4 + H2 PO4 + HPO24 + PO34
⇥
a Relationship
Table 5.3b Key relationships needed to determine phosphate speciation as a function of pH
255
10
9
a3 = a0 {H+ }3
Ka1,eff Ka2,eff Ka3,eff
1
Ka1,eff Ka2,eff
2 {H+ }
+
Ka3,eff {H+ } +1+ {H+ } Ka2,eff
1
=
Ka1,eff Ka2,eff Ka3,eff
3 {H+ }
+
{H+ } {H+ } + +1 Ka2,eff Ka3,eff Ka3,eff
2
⇤ ⇥ ⇤ HPO24 HPO24 K K a0 ⇤ = a0 a1,eff a2,eff a2 = = a0 ⇥ = 2 2 + + T OT PO4 H3 PO4 {H } {H } /(Ka1,eff Ka2,eff )
⇥
or, if one had already computed a0 :
⇥ ⇤ H2 PO4 K a0 ⇤ = a0 a1,eff a1 = a0 ⇥ = + + {H } {H }/Ka1,eff H3 PO4 ⇥ ⇤ HPO24 ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ ⇤= a2 = ⇥ H3 PO4 + H2 PO4 + HPO24 + PO34
a Relationship
Table 5.3b – continued from previous page
Figure 5.19 Distribution of phosphate species as a function of pH for a system containing 10 2 M T OT PO4 . Phosphoric acid is a triprotic acid with pKa1 = 2.16, pKa2 = 7.20, and pKa3 = 12.375. The broken lines show speciation in a solution with 0.1 M ionic strength.
5.11
DETERMINING SPECIES AND RELEVANT EQUATIONS FOR SOLVING EQUILIBRIUM PROBLEMS
Many chemical equilibrium problems can be posed in the form of questions such as: If known amounts of known substances are added to an aqueous solution, what species will the system contain, and what will the concentrations of those species be once the system has reached equilibrium? Or conversely: If we wish to convert a known initial system to one containing a targeted amount of a specific species (e.g., if we wish to change solution pH from 5.0 to 7.0), what chemicals, and how much of those chemicals, should we add? These questions have both qualitative and quantitative components. The qualitative aspects can be dealt with based on a general understanding of the properties of the chemicals of interest, and this understanding is developed through study and experience. The immediate goal here, however, is to explore the quantitative aspects, a task that, it turns out, can be largely accomplished using basic algebra. Specifically, in most cases, the problem reduces to one of identifying the unique solution to a set of simultaneous algebraic equations. In this section, we begin to develop the skills for writing and solving those equations. As a brief review, recall that a system of n independent equations in n unknowns can be solved uniquely if the equations are linear. If the equations are nonlinear, there may be more than one solution. Most of the systems we will be dealing with are characterized by at least some nonlinear equations; therefore, if we have n un256
knowns, n equations might not uniquely define the system. However, it will always be possible to choose among the mathematically correct answers by imposing the restriction that the answers involve physically meaningful quantities. For instance, the concentrations of all constituents must be real and positive. With these restrictions, n equations will always be sufficient and necessary to define a system with n unknowns, regardless of whether the equations are linear or not. Equilibrium constant expressions will be among the equations used to solve virtually any such problem, so it is appropriate to review the conditions under which those equations are applicable. First, of course, equilibrium constants are applicable only if the reaction of interest is either known or assumed to reach equilibrium. While this is an important restriction, it is not one with which we concern ourselves here — we will always be interested in computing the expected speciation in a system that has reached equilibrium, regardless of whether we believe that state actually exists in the system or we are simply investigating a hypothetical condition toward which the system moves. Second, we should consider only equilibrium reactions and expressions that are independent of one another. For example, assume that the two following reactions occur in a system: A + B )* C + D
K1 =
{C} {D} {A} {B}
B + C )* F
K2 =
{F} {B} {C}
In such a case, we can write a completely valid third reaction by addition of the above two, and a corresponding equilibrium constant, as A + 2 B )* D + F
K3 =
{D} {F}
{A} {B}2
= K1 K2
An analysis of this system could consider the first two reactions and their equilibrium constants, but the inclusion of the third reaction and its equilibrium constant would be redundant, because those expressions do not provide any new information about the system. Of course, the expressions describing the first and third reactions, or the second and third, could be used just as well as those describing the first and second. That is, any two of the three reactions or equations completely describe the equilibrium relationships among A, B, C, D, and F; inclusion of the third equation is neither necessary nor helpful. Finally, an equilibrium expression is applicable to a system if and only if all the constituents on each side of the reaction are present. In such a case, by the very na257
ture of the equilibrium concept, some of those constituents will combine to form the species on the other side of the reaction, and the reaction will proceed until the equilibrium constant is satisfied. This process might involve conversion of a very small fraction of the existing constituents, almost all of them, or any amount in between; the actual amount is determined by the magnitude of the equilibrium constant. However, regardless of how much conversion occurs, we know that if the reaction can be characterized by an equilibrium constant, some conversion will take place. Indeed, one could take the preceding statement as a definition of an equilibrium reaction: equilibrium implies some type of balance between reactants and products, and if no conversion occurred when one of the reactants or products was absent, that would imply that no such balance, and no equilibrium reaction, existed. The preceding discussion suggests an approach for determining which equilibrium constants apply to a given system. Specifically, we can write all the constituents that we know have been added to a solution, consider what reactions occur when they dissolve, and then try to imagine all possible species that can form from the added constituents, either by combination of constituents with one another or by dissociation of one of the components. If we have reason to believe that a particular species can form by any of those processes, we must assume that it will do so. What, then, is the criterion we should use to conclude that a species can (or cannot) form by some hypothetical reaction that we imagine? Somewhat paradoxically, at least for the purposes of this text, it is the availability of an equilibrium constant for the reaction. That is, we will use the existence of an equilibrium constant describing the tendency for a reaction to occur as the only acceptable evidence that the reaction does occur and that it reaches an equilibrium condition. Of course, our inability to find the equilibrium constant for a reaction does not prove that the reaction cannot occur; such a situation might only indicate that the reaction has never been studied fully, or that the equilibrium lies so far toward the products that the reaction is conventionally treated as being irreversible (i.e., the equilibrium condition is approximated as complete conversion of reactants to products). If the possibility that a particular reaction could occur were critical to the analysis and we had difficulty finding an equilibrium constant for it, we might first try deriving the equilibrium constant from tabulated thermodynamic values. (Recall that equilibrium constants can be calculated based on the standard molar Gibbs energy of reaction.) Alternatively, we could try to evaluate the equilibrium constant experimentally. However, for reactions considered in this text, we will assume that any equilibrium reaction will be associated with a known or derivable equilibrium constant; i.e., we will take the absence of an equilibrium constant for a hypothetical reaction as an indication that the reaction is not known to occur, that it proceeds to such a slight extent that it can be ignored, or that it proceeds so extensively that it can be ignored (because a negligible amount of the reactants remains at equilibrium). 258
Example 5.5 A solution is made by adding hydrogen cyanide, HCN, to water. The acidity constant for dissociation of this species is Ka = 10 9.21 . Which of the following reactions should be considered in analyzing the acid/base balance in the solution? HCN )* H+ + CN H2 O )* H+ + OH CN )* C4+ + N3 OH )* H+ + O2 Solution The species in the first two reactions are known to be present in the solution, and equilibrium constants for those reactions (Ka and Kw ) are available, so they should be considered. In the absence of a detailed understanding of the chemical nature and reactivity of CN– and OH– , the third and fourth reactions are entirely reasonable reactions to hypothesize. However, lacking any information about their tendency to occur (via values for the equilibrium constants), we assume that they are not important reactions to consider in the analysis.
Example 5.6 A solution is made by adding sodium cyanide, NaCN, to water. The value of Kb for the base CN– is 10 4.79 . Which of the following reactions should be considered in analyzing the acid/base balance in the solution? H2 O )* H+ + OH HCN )* H+ + CN NaCN )* Na+ + CN CN + H2 O )* OH + HCN Solution Given that water is present and that Kw is known, it is clear that the first reaction needs to be considered. In addition, knowing that sodium is extremely stable in water and therefore has a tendency to “fall off” any compound to which it is attached when the compound dissolves in an aqueous solution, we can assume that NaCN dissociates at least partially in water to release Na+ and CN– ions. Therefore, since H+ is present from dissociation of water, we see that the constituents on the product side of the second reaction are all present. An equilibrium constant for this reaction is known (Ka ), so some HCN is bound to form from the available H+ and CN– . As a result, the second reaction must be considered as well. 259
We know that the third reaction proceeds when NaCN enters the solution, but we have no equilibrium constant characterizing the reaction. We therefore assume that it proceeds essentially to completion, i.e., until no NaCN remains, and we do not consider the reaction as a relevant one describing the final equilibrium solution.7 A similar assumption is made regarding Na+ , K+ , Cl– , NO–3 and other salt ions throughout the text, unless otherwise indicated. The fourth reaction is the basicity reaction for conversion of the base CN– to its conjugate acid, HCN. This reaction certainly occurs in the system, and its equilibrium constant is known. However, the basicity reaction provides no independent information about the system. The reaction and the corresponding equilibrium constant can be derived by combining the acidity reaction and the dissociation reaction of water (Kb = Kw /Ka ). One could therefore use any two of the three relevant reactions to completely define the system. That is, it is acceptable to use Kb as one of the constants describing the system quantitatively, in conjunction with either Kw or Ka . Similarly, it is acceptable to use Kw and Ka to describe the system quantitatively. However, it is not acceptable, or at least not useful, to try to consider all three equilibrium constants in the analysis of the system.
Assume we wish to analyze the equilibrium speciation in a system made by adding known quantities of various chemicals to water. The equilibrium activity or concentration of each species in the system represents one unknown. To solve for these unknowns, we need to write the same number of equations as there are unknowns; i.e., if there are five species at equilibrium, we need five equations. The following algorithm describes a process by which we can determine (1) how many species will be present at equilibrium and (2) how to write the requisite equations to solve for each concentration. Once these two things are done, the problem becomes simply an algebraic one. I Identify species expected to be present at equilibrium A The species H2 O, H+ , and OH– are always present in any aqueous solution. However, in all solutions of interest to us, {H2 O} ⇡ 1, so {H2 O} is not an unknown. Call H+ and OH– Type a species. B For every acid (or base) to be considered, there is a conjugate base (or acid). If one of these is present in the solution, the other must be there also. If an acid or base can undergo multiple dissociation reactions, all the 7 Another possible assumption to explain the absence of an equilibrium constant for the third reaction
would be that the NaCN does not dissociate at all. In the absence of any chemical knowledge, this assumption would appear to be just as likely as the assumption that dissolution is complete. Even with minimal chemistry background, however, the information in the preceding sections suggests that the assumption of complete dissociation is a better one.
260
species resulting from these reactions will be present. In general, an acid that undergoes n 1 acid dissociation reactions requires consideration of n species. For example, H3 PO4 , is a triprotic acid; therefore the PO4 “group” comprises four species (n = 4: H3 PO4 , H2 PO–4 , HPO2– 4 , and PO3– ), which are related to one another via three (n 1) reactions. Call 4 these Type b species. C If one of the ionic species added to the system is an extremely weak acid or base, it will be present at equilibrium as a salt ion, but the conjugate species will not be present in a significant concentration. For instance, Cl– , NO–3 , Na+ , and K+ do not combine measurably with H+ or OH– , so species such as HCl, HNO3 , NaOH, and KOH can generally be ignored in the equilibrated solution. Call salt ions that are extremely weak acids or bases Type c species. D Having listed all the species resulting from dissociation reactions, and all species generated via acid/base reactions, we now look for species which might form from combinations of Type b and Type c species. If there is reason to suspect that a new species can be formed from those that are listed, it should be included. For the purposes of this discussion, if a species can be formed from some combination of previously listed species, an equilibrium constant for the reaction will be provided. Call species formed in this way Type d species. II List equilibrium constants for reactions among the species Organized by species type, these equilibrium constants are as follows. Type a: Type b:
One equation (Kw ) For each acid-base group, n 1 acid dissociation constants, where n is the number of species associated with that group, i.e., HAc: One equation (n = 2: HAc, Ac– ) H2 CO3 : Two equations (n = 3: H2 CO3 , HCO–3 , CO2– 3 ) 3– H3 PO4 : Three equations (n = 4: H3 PO4 , H2 PO–3 , HPO2– 4 , PO4 )
Type c: Type d:
No equilibrium constants for these species. One equilibrium constant for each species in this category.
III List mass balances Define an ion type as any ion or molecule in its most dissociated form. For in2+ – + stance, CO2– 3 , Na , Cu , and NH3 (aq) are ion types, but HCO3 , CuCO3 (aq) and NH+4 are not. Although OH– could be considered to be an ion type according to this definition, the H+ /H2 O/OH– group is present in all solutions at a total concentration that is so much larger than that of any other ion type that 261
it is convenient to treat this group differently, so we will temporarily exclude it from the definition. The number of ion types in a system is equal to the number of weak acid/base systems (groups of Type b species) plus the number of Type c species. A mass balance can be written for every ion type in a system. A mass balance is simply a mathematical expression of the principle of conservation of mass, stating that if something is present in a system initially or is subsequently added to the system, it must be present, in some form, at equilibrium. When dealing with solution chemistry, it is convenient to define the aqueous phase as the system, in which case processes such as dissolution of a chemical from the gas phase or precipitation of a solid out of the solution would be considered an addition to or a removal from the system, respectively. If the system is defined in this way, then for each ion type, a mass balance can be written with the following form:
 (ci niA )initial or added = T OT A =  (ci niA )at equilibrium i
(5.42)
i
where A is an ion type, ci is the concentration of a species i which contains ion type A, and ni, A is the number of A groups in each molecule of i. If we know how the system was prepared, the individual ci and ni terms in the first summation in Equation (5.42) are known, so T OT A can be calculated. The mass balance equates this value with the second summation in the equation, in which the ci terms (the concentrations of the species at equilibrium) are the unknowns for which we wish to solve. IV Write the charge balance This equation, which reflects the requirement that all solutions are electrically neutral overall, can be written as  ci zi = 0, where ci is the concentration of i
species i in solution and zi is its ionic charge, including the sign. The summation must include all charged species in solution.8 Note that since the units of ci are moles of i per liter and those of zi are moles of charge per mole of i, ci zi has units of moles of charge per liter; i.e., the equation is a balance on the concentration of charge in the solution and not on the concentrations of individual chemical species. It is conventional to place all the cations on one side of the equation and all the anions on the other, in which case the charge balance can be expressed as (5.43) Â (ci |zi |) = Â (ci |zi |) cations
anions
a mass balance on H+ or OH– can be written and used instead of the charge balance. This approach is described in detail in Chapter 6. 8 Alternatively,
262
V Write expressions for the ionic strength of the equilibrium solution and the activity coefficients of all species in it These expressions are needed because the equilibrium constants identified in Step II express relationships among chemical activities, whereas the mass balances and charge balance in Steps III and IV express relationships among concentrations. Although most of the examples presented in the text are simplified by assuming that the solutes behave ideally, that assumption is not usually valid in real systems of interest. An example in which nonideal solute behavior is taken into account is presented later in the chapter. Inclusion of activity coefficients in the calculations is routine in computer programs written specifically to solve chemical equilibrium problems. Example 5.7 List the equilibrium species and write the relevant equations necessary to solve for the equilibrium speciation in solutions prepared with the following inputs. Assume ideal behavior of all solutes. (a) 10
4M
Na2 CO3 +10
3M
HCl+10
(b) H2 O + 10 2 MCdCl2 + 3 ⇥ 10 the following reactions:
3M
NaHCO3 +H2 O
3 MCd(Ac) 2
(cadmium acetate), considering
K1 : Ka for HAc (HAc )* Ac– + H+ ) K2 : K for the reaction Cd2+ + OH– )* CdOH+ K3 : K for the reaction CdOH+ + OH– )* Cd(OH)2 K4 : K for the reaction Cd2+ + Cl– )* CdCl+ Solution (a) 10
4M
Na2 CO3 +10
3M
HCl+10
3M
NaHCO3 +H2 O
Step I: List the species according to type. Type a: H+ , OH– Type b: The compounds added comprise only one weak acid/base system – (carbonate). Three species are associated with this system: CO2– 3 , HCO3 , and H2 CO3 . Type c: Na+ , Cl– Type d: We need to look at various combinations of Type b and Type c species to see if they can combine to form a Type d species. If a Type d species can form, we must have an equilibrium constant to describe the reaction forming it. For this example, there are no Type d species to consider. 263
Step II: Write the relevant equilibrium constant expressions. Although these expressions relate activities to one another, they are written here in terms of molar concentrations, taking advantage of the assumption of ideal behavior of the solutes. Type a species:
Kw = {H+ }{OH } = [H+ ][OH ]
Type b species: H2 CO3 )* HCO3 + H+
Ka1 =
{HCO3 }{H+ } [HCO3 ][H+ ] = {H2 CO3 } [H2 CO3 ]
HCO3 )* CO23 + H+
Ka2 =
{CO23 }{H+ } [CO23 ][H+ ] = {HCO3 } [HCO3 ]
Step III: Write a mass balance (MB) on each ion type other than H+ or OH– . These include MBs on CO3 , Na, and Cl. Ion Type
 (ci niA )initial or added i
4 )(1) + (10 3 )(1)
= Â (ci niA )at equilibrium i
= [CO23 ](1) + [HCO3 ](1) + [H2 CO3 ](1)
CO3
(10
Cl
(10 3 )(1)
Na
(10 4 )(2) + (10 3 )(1) = [Na+ ](1)
= [Cl ](1)
Step IV: Write the charge balance. [Na+ ](1) + [H+ ](1) = [Cl ](1) + [OH ](1) + [HCO3 ](1) + [CO23 ](2) The coefficient 2 following the [CO2– 3 ] term reflects the fact that each mole of CO2– accounts for two moles of negative charge. 3 Step V: Because we are assuming ideal behavior of all solutes, this step can be skipped.
Steps I through IV identify seven species whose equilibrium concentrations are unknown and seven equations (three equilibrium equations, three mass balances, and one charge balance) that relate these concentrations to one another. Simultaneous solution of these seven equations yields the concentration of all species at equilibrium.
264
(b) H2 O + 10
2M
CdCl2 + 3 ⇥ 10
3M
Cd(Ac)2
Step I: List species Type a: H+ , OH– Type b: Ac– , HAc, Cd2+ , CdOH+ , Cd(OH)2 Type c: Cl– Type d: CdCl+ Step II: Write the equilibrium equations Type a: Kw Type b: K1 , K2 , K3 from the list at the beginning of the problem Type c: None Type d: K4 from the list Step III: Write mass balances on the following ion types: Ac, Cd, Cl Ion Type
 (ci niA )initial or added
=
 (ci niA )at equilibrium
Ac
(3 ⇥ 10 3 )(2)
=
[HAc](1)+[Ac– ](1)
Cd
(10 2 )(1) + (3 ⇥ 10 3 )(1)
=
Cl
(10 2 )(2)
=
i
i
[Cd2+ ](1) + [CdOH+ ](1) + [Cd(OH)2 ](1) + [CdCl+ ](1) [Cl– ](1)+[CdCl+ ](1)
Note that the Type d species CdCl+ must be included in the mass balances for both Cd and Cl. Step IV: Write the charge balance [H+ ](1) + [Cd2+ ](2) + [CdOH+ ](1) + [CdCl+ ](1) = [OH ](1) + [Ac ](1) + [Cl ](1) Step V: Again, this step can be skipped The problem has now been reformulated as a combination of nine species (unknowns) and nine equations (five equilibrium equations, three mass balances, and the charge balance) and hence can be solved.
265
5.12
NUMERICAL APPROACHES FOR SOLVING ACID/BASE PROBLEMS
Having defined some of the nomenclature and established the basic principles of chemical equilibrium, we can now solve an impressively large array of important problems. In the following section, a few acid/base problems are solved analytically. In some cases, simplifying assumptions are introduced, and the reasoning behind them is discussed. In subsequent chapters, approaches are presented for solving the same types of problems graphically or numerically. The solutions to the example problems involve the use of unsophisticated and, at times, fairly tedious algorithms for solving a set of simultaneous equations. Once some fairly simple techniques are mastered, the solution approaches presented in subsequent chapters can be implemented to solve the problems more quickly and easily, especially for systems containing several acids and bases. In particular, we will see in Chapter 6 that diagrams of log a versus pH represent concise and highly informative summaries of the information embedded in several of the equations, so that use of the diagrams eliminates the need to do much of the algebra. Nevertheless, there is value in developing a thorough understanding of the manual, analytical, solution algorithm, for two reasons. First, it leads to a deeper understanding of how and why the alternative, faster approaches work. Indeed, the reason that those techniques are more efficient is that they combine several steps of the manual method into one, or they use more sophisticated mathematics; however, these very factors also make the techniques less transparent to the user. Second, some valuable insights can be gained from analysis of very simple systems, and those insights are most easily acquired by a step-by-step analysis, rather than by just considering the final result. The discussion accompanying each of the examples is intended to help provide some of those insights. In the following examples, all solutes are assumed to behave ideally, so both the concentration and the activity of solute i are represented as (i). Keep in mind, though, that equilibrium constant expressions are relationships among activities, whereas mass and charge balances are relationships among concentrations, and that the equilibrium constant expressions can be converted to relationships among concentrations by using values of Keff in place of K. For acidity constants, these two values are related by Equation (5.31); for other equilibrium constants, similar relationships between Keff and K exist, but they must be determined on a case-by-case basis.
5.13
THE pH OF SOLUTIONS CONTAINING ONLY WATER AND STRONG ACIDS OR BASES
If a solution is prepared that contains only water and a strong acid or base, determining the pH and speciation at equilibrium is fairly trivial, as shown in the following examples. 266
Example 5.8 Determine the pH of pure water. Solution
Species present: Relevant reaction:
Equilibrium relationship: Mass balances: Charge balance:
H+ , OH– (Type a) H2 O )* H+ + OH– Kw = (H+ )(OH ) = 10
14.0
N/A (H+ ) = (OH ) 2
Substitution of the charge balance into the Kw expression yields Kw = (H+ ) , which can be solved directly to give (H+ ) = 10 7.0 , i.e., pH= 7.0. According to the charge balance, then, (OH– ) is also 10 7.0 , and the problem is solved. Example 5.9 Determine the pH of a solution made by adding HCl or NaOH to water, each at a concentration of 10 2 mol/L. Solution Strong acids and bases completely dissociate in water, so that we need not consider the presence of undissociated species at equilibrium. Therefore, for the solution made by adding HCl, the system can be summarized as follows: Species present: Relevant reaction: Equilibrium relationship: Mass balance: Charge balance:
H+ , OH– (Type a) Cl– (Type c) H2 O )* H+ + OH– Kw = (H+ )(OH ) = 10 T OT Cl = 10
2
14.0
= (Cl )
(H+ ) = (OH ) + (Cl )
Since the solution has been made by adding an acid to water, we can assume tentatively that (H+ ) OH , so the charge balance simplifies to (H+ ) ⇡ Cl . The mass balance on Cl indicates that (Cl– )= 0.01, so this simplified charge balance indicates that (H+ ) = 10 2.0 . Substituting this value of (H+ ) into the Kw expression, we find OH = 10 12.0 . The computed concentrations confirm that (H+ ) OH , so the assumption was correct. The solution pH is therefore log 10 2.0 , or 2.0. 267
The analysis of the system in which NaOH is added to water is almost the same, except that in this case the inert salt ion is a cation (Na+ ), and we make the initial assumption that (OH ) (H+ ). The system summary is as follows: Species present: Relevant reaction: Equilibrium relationship: Mass balance: Charge balance:
H+ , OH– (Type a) Na+ (Type c) H2 O )* H+ + OH– Kw = (H+ )(OH ) = 10 T OT Na = 10
2
14.0
= (Na+ )
(H+ ) + (Na+ ) = (OH )
The calculations are left as an exercise. The assumption that (OH ) (H+ ) is confirmed by the calculations, and the computed equilibrium concentrations are (H+ ) = 10 12.0 , (OH ) = 10 2.0 , (Na+ ) = 10 2.0 . The pH of the solution is 12.0. Although it is not in the least surprising that addition of a strong acid (10 2 M HCl) lowers the solution pH, consideration of the details of the reactions leading to this result can be helpful. Prior to the acid addition, the activities of H+ and OH– are both 10 7.0 . When the acid is added, almost all of it dissociates, increasing the H+ concentration (and activity) to approximately (10 2 + 10 7 ) M. Assuming that this process occurs almost instantaneously, the OH– activity at that instant is still 10 7 , so the product (H+ )(OH ) is approximately 10 9 . Since this value is considerably larger than Kw , chemical reactions proceed that reduce the H+ and OH– concentrations, moving the H+ /OH /H2 O system toward equilibrium. Specifically, H+ and OH– ions combine to form H2 O. When water is formed, equal concentrations of H+ and OH– ions are consumed. Thus, if 10 7 mol/L of H+ and OH– combined to form 10 7 mol/L additional H2 O in this system, no OH– would remain in solution, and the new value of the product (H+ )(OH ) would be zero, indicating that the reaction had “overshot” equilibrium. Thus, the amount of new H2 O that forms must be less than 10 7 mol/L. Because, at most, 10 7 M H+ can combine with the available OH– , the loss of H+ is bound to be negligible compared to the total amount present (⇠10 2 M), so the final concentration of H+ will remain close to 10 2 M. To satisfy Kw , then, the OH– concentration must be approximately 10 12 M. Since the initial concentration of OH– was 10 7 M, the vast majority (99.999%) of the OH– originally present must combine with H+ in this process. This conclusion is, of course, exactly the same one we came to above, using a purely mathematical approach to solve the problem.
268
5.14
THE pH OF IDEAL SOLUTIONS CONTAINING WEAK ACIDS AND BASES
If, instead of a strong acid, we add a weak acid to solution, a sequence of events similar to that described above ensues. However, since the acid dissociates incompletely in this case, the increase in the H+ concentration and the corresponding decrease in the OH– concentration are less than in the example with HCl addition. Consider, for example, the speciation in a solution prepared by adding 10 3 M propionic acid to pure water. Propionic acid is a three-carbon carboxylic acid (H3 C CH2 COOH) which we will abbreviate as HPr. It is an intermediate product in the biological degradation of many organic compounds and is particularly important in anaerobic processes. When it dissolves, the carboxylic acid portion of the molecule can release its proton in a reaction for which Ka is 10 4.87 . A summary of the information needed to solve for the equilibrium speciation of the solution is as follows. Species present: H+ and OH– (Type a) HPr and Pr– (Type b) H2 O )* H+ + OH– HPr )* H+ + Pr–
Relevant reactions: Equilibrium relationships:
Kw = (H+ ) OH Ka =
Mass balance:
(H+ ) Pr (HPr)
T OT Pr = 10
Charge balance:
3.0
= 10 = 10
14.0 4.87
= (HPr) + (Pr )
(H+ ) = (OH ) + (Pr )
The equilibrium speciation can be determined by substituting expressions that depend only on (H+ ) for the terms on the right side of the charge balance. OH Pr
=
Kw (H+ )
= a1,HPr (T OT Pr) =
Ka (T OT Pr) (H+ ) + Ka
Thus, the charge balance can be rewritten as H+ =
Kw Ka + + (T OT Pr) + (H ) (H ) + Ka
Multiplying through by [(H+ ) + Ka ] (H+ ) gives H+
3
+ Ka H+
2
[Kw + (T OT Pr) Ka ] H+ 269
Kw Ka = 0
(5.44)
Propionic acid is a key chemical intermediate in the anaerobic degradation of organic matter. If propionic acid accumulates in the solution and is not neutralized or degraded, the treatment process fails. This photograph shows an egg-shaped anaerobic digester at a sewage treatment plant in Boston. (Michael K. Stenstrom.)
This expression is a cubic equation in one unknown, (H+ ), and can be solved by trial and error or by software embedded in most spreadsheet applications. The only positive, real value of (H+ ) that satisfies the equation is (H+ ) = 10 3.96 so pH = 3.96. The concentrations of the other solutes can then be found from Kw , the charge balance, and the mass balance, as follows: (OH ) =
Kw 10 = (H+ ) 10
(Pr ) = (H+ ) (HPr) = T OT Pr
14.0 3.96
10.04
= 10
(OH ) = 10 (Pr ) = 10
3.96 3
10
3.96
= 10
3.05
Recall that, in the analysis of a solution prepared by adding only HCl to water, the assumption (H+ ) (OH ) simplified the algebra needed to solve the problem. The assumption was checked once the problem was solved, and its validity was confirmed. Making the same assumption in this example might simplify the algebra similarly. To see if that is the case, we can re-solve the problem making that assumption, and see how far off the result is from the exact result obtained without any simplifying assumptions. If the assumption is made that (H+ ) (OH ), the charge balance becomes + (H ) ⇡ (Pr ). Substituting this approximate equality into the expression for Ka yields
270
Ka =
(H+ ) Pr (HPr)
= H+
2
=
(H+ ) Pr T OT Pr Pr
+ Ka H+
2
⇡
(H+ ) T OT Pr (H+ )
(T OT Pr)Ka = 0
(5.45)
As before, the problem can be represented as a polynomial equation in one unknown, but now the polynomial is only second-order (quadratic), whereas before it was third-order (cubic). The solution to the quadratic is (H+ ) = 1.1 ⇥ 10 4 ; pH= 3.96. That is, making the assumption caused essentially no error in the final result. Since (OH ) = Kw /(H+ ) = 10 10.04 , it is clear that the assumption (H+ ) (OH ) was justified. There is one further assumption we might have made. Knowing that HPr is a weak acid (not completely dissociated in typical solutions), we might try the assumption that (HPr) (Pr ) in addition to (H+ ) (OH ) . Then the mass balance on propionate would simplify to (HPr) ⇡ T OT Pr, and the expression for Ka would be Ka = H+
2 (H+ ) Pr (H+ ) = (HPr) T OT Pr q = 10 3 Ka = 10 3.93
(5.46)
Using the Kw expression in conjunction with this result, we find (OH ) = 10 10.07 , and from the charge balance, (Pr ) = 10 3.93 . The pH determined using these assumptions is in error by 0.03 unit. The general conclusion to be drawn from this exercise is that the more simplifying assumptions that are made, the easier it is to solve the problem, and the less exact the result becomes. Depending on how important it is to determine the pH and solution composition exactly, a given assumption might or might not be acceptable. An important observation from the above analysis is that both the exact result [Equation (5.44)] and the results obtained when the various simplifying assumptions are made [Equations (5.45) and (5.46)] can be generalized to solutions containing any weak acid at any concentration, simply by inserting the appropriate values of Ka and T OT A, as demonstrated in the following example. Example 5.10 Determine the pH of a solution of 10 3 M HOCl, using the exact equation developed previously and the equations applicable when the assumptions are made that (H+ ) (OH ) and that (HOCl) (OCl ). Are the assumptions valid in this case?
271
Solution The pH values we seek can be computed by substituting 10 7.60 for Ka and 10 3 for T OT OCl in Equations (5.44) to (5.46). The results are that pH is 5.30 in all cases, so both assumptions are good. Next, consider the composition of a solution containing the same total amount of Pr as in the preceding analysis (10 3 mol/L), but in which the Pr is added as the salt sodium propionate, NaPr. In this case, when the NaPr dissolves, the sodium ions are released and undergo no further reaction. The Pr– released can combine with H+ to form HPr. This reaction removes H+ from solution, so the solution becomes alkaline; i.e., the pH goes up. Following the same algorithm as previously, the system can be summarized as follows: Species present:
H+ and OH– (Type a) HPr and Pr– (Type b) Na+ (Type c)
Without any knowledge of the tendency for Na+ and Pr– to interact, we cannot know whether Type d species such as NaPr(aq) or NaPr–2 might be present at equilibrium. If we suspected they might be important species, we would have to search for equilibrium constants describing their formation from the ions; in this case, having no such information, we assume that they are not significant. The reactions and equations characterizing the system can therefore be summarized as follows: Relevant reactions: Equilibrium relationships:
H2 O )* H+ + OH– HPr )* H+ + Pr– Kw = (H+ ) OH Ka =
Mass balances: Charge balance:
(H+ ) Pr (HPr)
= 10 = 10
14.0 4.87
T OT Pr = 10 3.0 = (HPr) + Pr T OT Na = 10 3.0 = Na+ (H+ ) + Na+ = OH
+ Pr
Note that the reactions, equilibrium constants, and mass balance on Pr are all the same as in the case where HPr was the only constituent added to the solution. The system of equations is marginally more complex than in the previous example because (Na+ ) appears in the charge balance, but the mass balance on Na is so trivial that we can immediately substitute a known value for (Na+ ). As before, we can substitute into the charge balance to derive an equation that contains (H+ ) as the only unknown: 272
H+ + Na+ =
Kw + a1 (T OT Pr) (H+ )
(5.47a)
H+ + Na+ =
Kw Ka + + (T OT Pr) + (H ) (H ) + Ka
(5.47b)
Multiplying Equation (5.47b) through by (H+ ) and then (H+ ) + Ka , we obtain the following cubic equation, in which H+ is the only unknown: ⇤ ⇥ ⇤ 3 ⇥ 2 H+ + Ka + Na+ H+ Kw + T OT Pr Na+ Ka H+ Kw Ka = 0 (5.48) + 3 The result for T OT Pr = Na = 10 is pH= 7.93. Substituting this value into the other expressions characterizing the system, we find OH = (HPr) = 8.7 ⇥ 10 7 , and Pr = 9.99 ⇥ 10 4 . As in the preceding examples, simplifying assumptions could be used to reduce the order of the polynomial, if desired. For instance, since NaPr is a base, we expect the pH at equilibrium to be greater than 7.0, and hence we expect that (H+ ) will be much less than (Na+ ) (which we know is 10 3 ). Assuming that (H+ ) is negligible in the charge balance leads to the following quadratic expression, which also yields pH= 7.93: ⇥ ⇤ 2 Na+ H+ Kw + T OT Pr Na+ Ka H+ Kw Ka = 0 (5.49)
Note that the equations used to solve this problem, i.e., the mass balance on Pr and the Ka and Kw expressions, apply to any system containing water and propionate species, and that the charge balance applies to any system that contains H+ , OH– , Pr– , and Na+ as the only ions. That is, this set of equations is not restricted to systems where (Na+ ) = T OT Pr, but rather applies to solutions made by any combination of HPr, NaPr, and NaOH. Thus, as long as the proper values are used for (Na+ ) and T OT Pr, Equation (5.48) applies. Example 5.11 What is the pH of a solution prepared by adding 6 ⇥ 10 4 M HPr, 2 ⇥ 10 4 M NaPr, and 3 ⇥ 10 4 M NaOH to water? What is the speciation of the Pr at equilibrium? Solution The activity of Na+ at equilibrium will include contributions from the NaPr and the NaOH, both of which are assumed to dissociate completely when they dissolve. Similarly, T OT Pr will include contributions from both the HPr and NaPr additions. The equilibrium pH can therefore be determined by inserting values of 5 ⇥ 10 4 for (Na+ ) and 8 ⇥ 10 4 for T OT Pr into Equation (5.48). The equation is satisfied when 273
H+ = 7.8⇥10 6 , corresponding to pH= 5.11. The activities of the propionate species can be determined by computing a0 and a1 at this pH, and multiplying the respective values by T OT Pr, yielding (HPr) = 2.92 ⇥ 10 4 and (Pr ) = 5.08 ⇥ 10 4 . The equilibrium pH’s of solutions of 10 3 M HOCl, 10 3 M HPr, and 10 3 M NaPr have been determined in the preceding examples, and that of a solution of 10 3 M NaOCl can be computed by carrying out similar calculations, which are left as an exercise. The results of these calculations are summarized below. Solution 10 3 M HPr 10 3 M NaPr 10 3 M HOCl 10 3 M NaOCl
pH 3.90 7.93 5.30 9.30
Comparing the results for addition of HPr and HOCl, we note that both species would be considered weak acids because they dissociate only partially in water. Since an acid can be defined as a substance that donates protons to solution, the greater the tendency to donate H+ , the stronger the acid. Thus, propionic acid, which lowers the pH to 3.90 when added to pure water at a dose of 10 3 M, is a stronger acid than hypochlorous acid, which lowers the pH only to 5.30 when added at the same dose. Conversely, propionate and hypochlorite ions are bases, because they remove H+ from solution. Adding 10 3 M NaPr or NaOCl to water raises the pH to 7.93 or 9.30, respectively. Thus, NaOCl must be a stronger base than NaPr, consistent with the previous discussion which indicated that the stronger the acid (HPr > HOCl), the weaker the conjugate base (NaPr < NaOCl).
5.15
A SIMPLE SPREADSHEET ANALYSIS FOR DETERMINING ACID/BASE SPECIATION
The approach presented in the preceding section for determining the equilibrium acid/base speciation in a solution is straightforward and reasonably simple for systems in which only one monoprotic acid/base group is present. However, the order of the polynomial that must be solved increases with each additional acid/base group that is present, and also if multiprotic acid/base groups (e.g., H2 CO3 or H3 PO4 ) are present. In such cases, an alternative numerical approach can be used that avoids much of the algebra by solving the charge balance equation directly, after substituting expressions into it that characterize the ionic concentrations in terms of a values. For instance, as noted previously, the charge balance for a generic solution containing H+ , OH– , Pr– , and Na+ as the only ions is
274
(H+ ) + (Na+ ) = (OH ) + (Pr ) Kw = + + (T OT Pr)a1,Pr (H )
(5.50)
By preparing a spreadsheet with one column for pH (the master variable) and a separate column for each term in Equation (5.50), we can rapidly identify the pH where the equation is satisfied, along with the concentration (and activity) of each of the species. The relevant values can be computed based on the following equations, all developed earlier in the chapter: (H+ ) = 10
pH
(HPr) = a0 (T OT Pr)
(OH ) =
Kw (H+ )
(Na+ ) = T OT Na
(Pr ) = a1 (T OT Pr)
where a0 = (H+ )/[(H+ ) + Ka ] and a1 = Ka /[(H+ ) + Ka ]. Note that these calculations are essentially identical to those used earlier to develop the logC-pH diagram. The spreadsheet for a solution of 10 3 M NaPr is shown in Table 5.4. The table includes a column for (HPr), because the concentration of this species is of interest, even though it is not needed to solve the charge balance equation. The final column in the spreadsheet shows the value of the expression (H+ ) + Na+ OH Pr . If the charge balance is satisfied, this expression equals zero, so the magnitude of the expression at a given pH is an indicator of the computed charge imbalance at that pH. The values of (H+ ), (OH– ), (Na+ ), (Pr– ), and (HPr) in any row of the spreadsheet satisfy the mass balances on Na and Pr and the Ka and Kw equations for the given pH. However, the values in most rows are not consistent with the charge balance equation. This equation must be satisfied at equilibrium. Therefore, the equilibrium solution composition can be identified by scanning the final column to identify the pH where the charge imbalance equals zero. The key rows in the spreadsheet are shown in boldface. The top portion of the spreadsheet has increments of 1.0 pH unit and indicates that the electroneutrality condition is satisfied between pH 7.0 and 8.0; the lower portion, with increments of 0.1 pH unit, narrows the range to between 7.9 and 8.0 (the interpolated value is 7.96, consistent with the result obtained in the analytical solution). The spreadsheet approach offers the possibility of solving for the solution pH directly once the charge balance has been written. It has the substantial advantage over the algebraic approach that, for each additional acid/base group in the system, we need only add more columns to the spreadsheet, without making any changes to those that have already been programmed. Furthermore, the similarity of the expressions for a for different acid/base groups means that once we have developed the 275
Table 5.4 Speciation and net calculated charge as a function of pH in an ideal solution of 10 3 M NaPr pH
(H+ )
(OH )
(Na+ )
(Pr )
(HPr)
Net Charge
4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0
1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E
04 05 06 07 08 09 10 11
1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E
10 09 08 07 06 05 04 03
1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E
03 03 03 03 03 03 03 03
1.19E 5.74E 9.31E 9.93E 9.99E 1.00E 1.00E 1.00E
04 04 04 04 04 03 03 03
8.81E 4.26E 6.90E 7.36E 7.41E 7.41E 7.41E 7.41E
04 04 05 06 07 08 09 10
9.81E 04 4.36E 04 7.00E 05 7.36E 06 2.49E 07 9.92E 06 1.00E 04 1.00E 03
7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 8.0
1.00E 7.94E 6.31E 5.01E 3.98E 3.16E 2.51E 2.00E 1.58E 1.26E 1.00E
07 08 08 08 08 08 08 08 08 08 08
1.00E 1.26E 1.58E 2.00E 2.51E 3.16E 3.98E 5.01E 6.31E 7.94E 1.00E
07 07 07 07 07 07 07 07 07 07 06
1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E 1.00E
03 03 03 03 03 03 03 03 03 03 03
9.93E 9.94E 9.95E 9.96E 9.97E 9.98E 9.98E 9.99E 9.99E 9.99E 9.99E
04 04 04 04 04 04 04 04 04 04 04
7.36E 5.85E 4.66E 3.70E 2.94E 2.34E 1.86E 1.48E 1.17E 9.32E 7.41E
06 06 06 06 06 06 06 06 06 07 07
7.36E 06 5.81E 06 4.56E 06 3.55E 06 2.73E 06 2.05E 06 1.49E 06 9.96E 07 5.58E 07 1.51E 07 2.49E 07
expression for any given acid/base group, the expressions for other groups can be written by minor modifications to that template. A closely related approach that allows the equilibrium pH to be determined even more quickly is to program the charge balance expression into a cell in the spreadsheet and use embedded software (such as the Solver R routine embedded in Excel R )9 to find the value of (H+ ) that causes the equation to be satisfied. We can also use this approach to address an important and slightly different type of acid/base problem: determining the amount of acid or base that must be added, or must have been added, to systems where the equilibrium pH is known. To explore this issue, we consider an example solution whose pH has been lowered to 4.4 by addition of HPr to pure water, and we ask ourselves: How much HPr was needed? Although the variables whose values are known in this situation differ from those in the systems we investigated previously (the final pH is known, but T OT Pr is not), the equations used to solve the problem (mass balance, charge balance, and equilibrium constants) are identical. 9 Solver
and Excel are registered trademarks of Microsoft Corp. (Redmond WA).
276
H+ and OH– (Type a) HPr and Pr– (Type b)
Species present:
H2 O )* H+ + OH– HPr )* H+ + Pr–
Relevant reactions: Equilibrium relationships:
Kw = (H+ ) OH Ka,HPr =
Charge balance:
= 10
(H+ ) Pr (HPr)
(H+ ) = OH
14.0
= 10
4.87
+ Pr
Since the equilibrium value of (H+ ) is known, (OH– ) can be computed directly from Kw , allowing us to then compute (Pr– ) from the charge balance: From Kw :
OH 4.4
10 From the CB:
=
Pr
Kw = 10 (H+ )
9.6
= 10
9.6
+ Pr
= 10
4.4
= 4.0 ⇥ 10
5
Finally, (HPr) can be determined from Ka , and T OT Pr can be computed from the mass balance: (HPr) =
(H+ ) Pr Ka
=
(10
4.4
)(10
10
4.87
T OT Pr = HPr + Pr = 1.2 ⇥ 10
4
4.4
)
= 10
+ 4.0 ⇥ 10
5
3.93
= 1.2 ⇥ 10
= 1.6 ⇥ 10
The amount of HPr that must be added is T OT Pr (1.6 ⇥ 10 proximately 25% dissociates.
4
4
4 M),
of which ap-
Example 5.12 A solution is made by adding 10 3 M HOCl and 10 3 M NaPr to water. Then NaOH is added until the solution pH is 7.0. How much NaOH must be added? Solution Following the same procedure as in the preceding example, we write the system summary. Species present:
H+ and OH– (Type a) HPr and Pr– ; HOCl and OCl– (Type b) Na+ (Type c) 277
Relevant reactions:
H2 O )* H+ + OH– HPr )* H+ + Pr– HOCl )* H+ + OCl–
Equilibrium relationships:
Kw = (H+ ) OH Ka,HPr =
(H+ ) Pr (HPr)
Ka,HOCl = Mass balances:
= 10
= 10
(H+ ) OCl (HOCl)
T OT Pr = 10
3.0
T OT OCl = 10
14.0 4.87
= 10
7.53
= (HPr) + (Pr )
3.0
= (HOCl) + (OCl )
+
T OT Na =? = (Na ) Charge balance:
(H+ ) + Na+ = OH
Other known information:
pH = 7.0, i.e., H+ = 10
+ Pr
+ OCl
7.0
In this problem, T OT Na is unknown, since we do not know how much NaOH has to be added to adjust the pH to 7.0. Therefore, we have three equilibrium expressions, two mass balances, and a charge balance to work with (six independent equations). There are seven species in the system, but one of them is H+ , for which the final concentration is specified. Thus, the number of equations (6) matches the number of unknowns, and the problem can be solved. Substituting into the charge balance, we obtain H+ + Na+ =
Kw + a1,HPr (T OT Pr) + a1,HOCl (T OT OCl) (H+ )
The pH is known, so the a1 value of each weak acid is also known, and (Na+ ) is the only unknown. In this case, at pH 7.0, the a values are a1,HOCl =
Ka,HOCl 10 7.53 = (H+ ) + Ka,HOCl 10 7.0 + 10
a1,HPr =
Ka,HPr 10 4.87 = (H+ ) + Ka,HPr 10 7.0 + 10
7.53
4.87
= 0.228
= 0.993
Since a1,HOCl is much less than a1,HPr a smaller fraction of the HOCl (⇠23%) than the HPr (⇠ 99%) dissociates. This result is consistent with our expectations, since HOCl is a weaker acid than HPr. Substituting the a values and the values for T OT Pr, T OT OCl, Kw , and (H+ ) into the charge balance, we can solve for (Na+ ), which equals 1.221 ⇥ 10 3 M. Because 278
Na+ is the only form in which Na exists in the equilibrium solution, we can equate this value with T OT Na. In addition, we can write the mass balance on Na in terms of input species as (NaOH)in = T OT Na
(NaPr)in = 1.22194 ⇥ 10
We conclude that 2.22 ⇥ 10
5.16
4M
3
1.0 ⇥ 10
3
= 2.22 ⇥ 10
4
NaOH was added.
THE pH AND SPECIATION OF SYSTEMS CONTAINING MULTIPROTIC WEAK ACIDS AND BASES
The analysis of solutions containing multiprotic acid/base groups is a simple and direct extension of the analysis for monoprotic systems, as is shown in the following example. Example 5.13 A solution is made by adding 2 ⇥ 10 rium pH of the solution?
3M
NaHCO3 to water. What is the equilib-
Solution This problem is very similar to prior examples in which we found the pH of solutions to which HPr or NaPr had been added. The only difference is that in the current problem, we need to consider both acid dissociation reactions that relate the three carbonate species, and we need to consider all three carbonate species in the carbonate mass balance. It should be apparent from the previous few examples that we could write the charge balance equation with (H+ ) as the only unknown and could solve for the equilibrium pH without going through the preliminary steps of listing the species, reactions, equilibrium relationships, etc. Nevertheless, for the sake of completeness, we will carry out those steps. Species present:
H+ and OH– (Type a) H2 CO3 , HCO–3 , and CO2– 3 (Type b) Na+ (Type c)
Relevant reactions:
H2 O )* H+ + OH– H2 CO3 )* H+ + HCO–3 HCO–3 )* H+ + CO2– 3 279
Equilibrium relationships:
Kw = (H+ ) OH
Mass balances:
= 10
Ka1 =
(H+ ) HCO3 H2 CO3
Ka2 =
(H+ ) CO23 HCO3
14.0
= 10 = 10
T OT CO3 = 2 ⇥ 10
6.35
10.33
3.0
= (H2 CO3 ) + (HCO3 ) + (CO23 )
T OT Na = 2 ⇥ 10
3.0
= (Na+ )
(H+ ) + Na+ = OH
Charge balance:
+ HCO3 + 2 CO23
Substituting the known value of (Na+ ) and the expression for (OH– ) into the charge balance, we find H+ + 2 ⇥ 10
3
=
Kw + a1 (T OT CO3 ) + 2a2 (T OT CO3 ) (H+ )
As in the preceding examples with monoprotic acid/base groups, all the terms in the charge balance can be written as expressions that contain only known values and (H+ ). These expressions can be programmed into a spreadsheet, and the pH at which the charge balance is satisfied can then be identified. In this case, the equation is satisfied at (H+ ) = 10 8.28 , i.e., pH= 8.28. At that pH, almost all the carbonate (1.96 ⇥ 10 3 M) is present as HCO–3 , and the concentrations of H2 CO3 and CO2– 3 are 5 5 2.06 ⇥ 10 M and 1.87 ⇥ 10 M, respectively.
5.17
ACID/BASE EQUILIBRIA IN NONIDEAL SOLUTIONS
To this point, we have solved for the solution speciation only in systems with ideal solutes, i.e., solutes whose activity coefficients are 1.0, so that their activities can be equated with the numerical value of their molar concentrations. In most situations of interest, this simplification will not apply, so nonideal behavior is considered in a final example that closes the chapter. Example 5.14 Hydrofluoric acid is a weak acid (pKa = 3.17) that is used extensively in industry because F– attaches strongly to Fe3+ and Al3+ ions. As a result, it is an excellent reagent for removing surface scales from objects made of iron or aluminum before they are painted or other surface finishing operations are carried out. Compare the 280
pH and the concentrations of HF and F– in a solution prepared by adding 10 3 M HF to pure water, with that of a solution prepared the same way, except using water containing 0.1 M NaCl. Assume that the ionic strength is low enough in the NaCl-free solution that the solution behaves ideally, and use the Davies equation to estimate activity coefficients in the solution containing the NaCl. Keep in mind that pH is defined as log{H+ }, not log[H+ ]. Solution The mass balance on T OT F is identical in the two solutions: T OT F = 10
3
= [HF] + [F ]
The charge balances are also effectively identical because, although Na+ and Cl contribute to the total charge in the solution to which NaCl was added, these contributions exactly cancel each other. Therefore, in each case, the charge balance can be written as –
[H+ ](1) + [Na+ ](1) = [OH ](1) + [F ](1) + [Cl ](1) Equilibrium constants always relate species’ activities, so the relevant equilibrium constant expressions for this problem are Kw = {H+ }{OH } = 10 Ka =
{H+ }{F } = 10 {HF}
14.0
3.17
Based on the convention that the standard-state concentration of all solutes is 1.0 mol/L, we can rewrite the equilibrium constants in terms of species’ concentrations by expressing the activity of each solute as the product of its molar concentration and its activity coefficient, i.e., Kw = gH+ [H+ ]gOH [OH ] = 10 Ka =
gH+ [H+ ]gF [F ] = 10 gHF [HF]
14.0
3.17
Because all the activity coefficients are assumed to be 1.0 in the low ionic strength solution, that solution can be treated as having only four unknowns (the concentrations of the four species). We can solve for these unknowns using the mass balance, charge balance, and two equilibrium expressions. The solution containing NaCl, on the other hand, has the same four unknowns, along with four unknown activity coefficients. Therefore, four additional equations are needed; these additional equations relate the activity of each species to its concentration. 281
Because the ionic strength in the solution containing NaCl is known, the activity coefficients of the ions of interest (H+ , OH– , and F– ) can be computed without knowing the ultimate speciation of T OT F. Using the Davies equation, g for all three ions is estimated to be 0.78. The value of gHF is assumed to be 1.0, since HF is uncharged. Thus, the four additional equations that characterize the solution with NaCl are {H+ } = 0.78[H+ ]
{OH } = 0.78[OH ]
{F } = 0.78[F ]
{HF} = 1.0[HF]
Substitution of these expressions into the Kw and Ka expressions yields effective equilibrium constants (Keff ) that describe equilibrium relationships among species concentrations, rather than activities: Kw,eff ⌘ [H+ ][OH ] =
{H+ } OH gH+ gOH
=
10
14.0
(0.78)
2
13.78
= 10
{H+ } {F } {H+ } {F } + gH+ gF [H ][F ] 10 3.17 {HF} Ka,eff ⌘ = = g +g = = 10 H F {HF} [HF] (0.78)2 gHF gHF 1.0
2.95
The constants are conditional in the sense that they apply only under the conditions specified (0.1 M ionic strength), whereas the activity-based equilibrium constants apply at any ionic strength. The considerations described above show that once activity coefficients are taken into account, the system with NaCl reduces to four simultaneous equations, just like the system with no NaCl. The only difference is that the K values in the low ionic strength solution are replaced by Keff values in the high ionic strength solution. The mass balance and charge balance equations are identical in the two solutions. The results obtained by solving these equations are summarized below. [NaCl] = 0 Species +
H
OH– HF F– pH log [H+ ]
Concentration 5.49 ⇥ 10
4
1.82 ⇥ 10
11
5.49 ⇥ 10
4
4.51 ⇥ 10
4
[NaCl] = 0.1 M
Activity 5.49 ⇥ 10
Concentration 4
1.82 ⇥ 10
11
5.49 ⇥ 10
4
4.51 ⇥ 10
3.26
3.26
4
6.39 ⇥ 10
4
2.62 ⇥ 10
11
6.39 ⇥ 10
4
3.61 ⇥ 10
4
Activity 4.98 ⇥ 10
2.04 ⇥ 10
11
4.98 ⇥ 10
4
3.61 ⇥ 10
3.30
3.19
282
4
4
The solution compositions in the two cases are quite similar. In the solution with the NaCl, the concentrations of H+ and OH– are both larger than in the less salty solution, reflecting the fact that the product of the activities of these two species must be 10 14 in both solutions, but that the activity coefficients are lower in the salty water. This fact alone, however, has no effect on the solution pH; i.e., in water containing 0.1 M NaCl and no other solutes, the pH would be 7.0, just as it is in salt-free water. The water with NaCl has slightly higher H+ and F– concentrations than the saltfree solution, because the salt decreases the activity coefficients of these two species. Since the activity coefficient of HF is 1.0 in both solutions, the decreases in gH+ and gF force more HF to dissociate to satisfy the Ka expression, and this extra dissociation releases more H+ and F– to solution. Correspondingly, the concentration of HF that remains undissociated declines. Despite the increase in the concentrations of H+ and F– , the activity of each of these species declines, because their activity coefficients decline. The resulting, fractional decline in the product {H+ }{F } is exactly equal to the fractional decline in {HF}, so the ratio {H+ }{F }/{HF} is the same as in the solution with no NaCl, and equal to Ka . Because {H+ } is lower in the salty solution, pH is higher, as indicated. The effect of the salt would be greater if more 3– highly charged species were involved (for example, CO2– 3 or PO4 ).
5.18
SUMMARY
Acids and bases are, respectively, substances that increase and decrease the activity of H+ in solution. Many acids are species that have one or more “detachable” H+ ions, and these ions are the source of the compound’s acidity. Other acids cause H+ to enter solution by facilitating the splitting of water (hydrolysis) and then combining with the OH– ion that is released. The strength of an acid, i.e., its tendency to cause the H+ activity of a solution to increase, is quantified by its acid dissociation constant, Ka . Multiprotic acids have several Ka values, one for each acidity reaction they undergo. Each acid has a conjugate base; the stronger the acid, the weaker the conjugate base. The strength of the base is quantified by a basicity constant Kb , and the product Ka Kb equals the dissociation constant for water Kw . The fractional distribution of a compound among its acid and base forms depends on solution pH and is independent of the total amount of the compound in the system. If the solution pH happens to equal the pKa of an acid/base conjugate pair, then the acidic (protonated) and basic (deprotonated) forms of the compound are present at equal activities in an equilibrium solution. At pH more acidic than pKa (i.e., at pH
The determination of the pH in an equilibrium solution that has been prepared with known inputs, or the determination of solution speciation in a solution of known pH, involves writing and solving a set of simultaneous equations that include mass balances, equilibrium constants, and a charge balance. Chemical information is embedded in the values of the various equilibrium constants, but the solution of the equations is a purely mathematical exercise, requiring no specialized chemical knowledge.
5.19
PROBLEMS
In solving these problems, use K a data from Table 5.1 and assume ideal solution behavior unless the question states otherwise or the ionic strength is given. If you prepare a log C-pH diagram, label the axes and the individual curves, and include a caption. The range of values on the axes will depend on the details of the problem, but they should typically be 8 to 14 pH units on the abscissa and 6 to 10 log units on the ordinate. 1. How much HOCl must be added to pure water to make a solution of pH 4.3? pH 6.5? Do not assume (H+ ) (OH ) at pH 6.5. 2. Identify the strongest acid and the strongest base in the following reaction, using only the K value given. H2 S + OCl )* HOCl + HS
K = 10+0.51
3. The bicarbonate ion concentration in a solution at pH 7.50 is 10 3.2 M. Calculate the concentrations of all carbonate species and the total dissolved inorganic carbon (DIC) concentration in mol/L and g/L C. 4. A diprotic acid has pKa1 = 2.8 and pKa2 = 8.4. The standard Gibbs energy of formation of the most deprotonated species in this group (A2– ) is 85.2 kJ/mol. What is GH A ? 2
5. The transfer of a proton from butyric acid (stomach acid) to acetate can be described by the following reaction: CH3 CH2 CH2 COOH + CH3 COO )* CH3 CH2 CH2 COO + CH3 COOH Butyric acid + Acetate )* Butyrate + Acetic acid (a) The equilibrium constant for this reaction is K = 0.87, and Ka for acetic acid is 1.74 ⇥ 10 5 . Using those values and Kw , compute the acid dissociation constant for butyric acid. 284
(b) Write the reaction and determine the value of Kb (the basicity constant) for butyrate. (c) A solution is made by adding some acetic acid and some sodium butyrate to water. List all the species in the equilibrium solution that can act as acids, and rank them from most acidic to least acidic. Do the same for all species that can act as bases. 6. Dissolved copper ion (Cu2+ ) can bond to several different molecules in aqueous solution, including some weak bases. Following are some of the reactions in which Cu2+ participates. Cu2+ + Cl– )* CuCl+ 2+
log K = 0.30
–
Cu + 2 Cl )* CuCl2
log K = 0.26
Cu2+ + CO2– 3 )* CuCO3
log K = 6.77
Cu2+ + OH– )* CuOH+
log K = 6.50
Cu2+ + NH3 )* Cu(NH3 )2+
log K = 4.02
Cu2+ + 2 NH3 )* Cu(NH3 )2+ 2
log K = 7.40
(a) What is the equilibrium constant for the following reaction? 2+ + * Cu(NH3 )2+ + NH+ 4 ) Cu(NH3 )2 + H
(b) List the species that would be present at equilibrium if 10 3 M Cu(HCO3 )2 and 10 3.3 M NaCl were added to water, and write the equations required to solve for the equilibrium concentrations of all these species. Include numerical values in the equations, if they are known. (c) What is Kb for the base CuOH+ ? 7. Write the charge balance for the following solutions and list the equations that would have to be solved to determine the equilibrium solution composition. (a) 10
3M
benzoic acid
(b) 10
3M
Na2 S
(c) 10
3M
(NH4 )2 CO3
(d) 10
3M
(NH4 )2 CO3 plus 5 ⇥ 10
3M
(NH4 )3 PO4 plus 10
3M
CaCO3
8. Using algebraic techniques and/or a spreadsheet approach, determine the equilibrium pH of each solution in Problem 7. 9. Prepare a log C-pH diagram for each solution in Problem 7. Use a full sheet of paper for each plot, and show the ranges 1 pH 13 and 10 logC 1. 285
10. When PbCl2 is added to water, it dissociates into Pb2+ and Cl– ions, and the Pb2+ ions then participate in the following reactions: Pb2+ + OH )* PbOH+ Pb2+ + 2 OH )* Pb(OH)2
log K = 6.40 log K = 10.91
(a) What are pKa1 and pKa2 for Pb2+ ? (b) How much PbCl2 must be added to water to yield a solution of pH 5.1? 11. Determine the values of Kb1 , Kb2 , and Kb3 for sodium arsenate, Na3 AsO4 , and write the reactions to which these constants apply. Also determine K for the reaction in which arsenic acid (H3 AsO4 ) completely dissociates to form arsenate ion (AsO3– 4 ). 12. Na2 CO3 is added to pure water until the pH is 9.45. (a) What is T OT CO3 in the solution? (b) How much HCl would have to be added to the solution to lower the pH to 7.0? (c) How much H2 CO3 would have to be added to the solution to lower the pH to 7.0? 13. Cyanide ion, CN– , is important in metal plating industries because it can keep metals dissolved under conditions where they would otherwise form solids (precipitate) and settle out of solution. The details of how this occurs are presented in Chapter 11. It is important to maintain pH>10.5 in these solutions to minimize release of toxic hydrogen cyanide gas. If a solution is prepared by dissolving 10 2 M NaCN in water, will the pH be in the region where the solution is safe? What is the HCN concentration in the solution? 14. A solution containing 0.04 M NaCl and 5 ⇥ 10 4 M of a base NaX has a pH of 8.9. Estimate Ka of the acid HA, taking nonideal solute behavior into account. Hint: There are many approaches for solving the set of equations that characterize this problem. One approach is to guess the ionic strength of the solution, solve the remaining equations based on that guess, and determine a new value for the ionic strength based on the computed composition. Then, revise the original guess to equal the computed ionic strength and repeat the process until the guessed and computed values converge. 15. A solution contains 10 4 M T OT OCl, 90% of which is present as HOCl. The solution also contains 7 ⇥ 10 4 M H2 CO3 . What is T OT CO3 in the solution?
286
16. Uncharged ammonia [NH3 (aq)] is toxic to fish at a concentration of 0.01 mg/L NH3 N. What is the maximum concentration (mg/L) of total ammonia nitrogen (the sum of the N concentrations in NH3 and NH+4 ) that can be in solution without causing toxicity at pH 7.0? At pH 9.0? 17. You wish to add enough NaOCl (sodium hypochlorite) to a 150-m3 (⇠400,000gal) swimming pool to provide a dose of 5.0 mg/L T OT OCl as Cl2 . (a) How much NaOCl (kg) should you add? The EW of NaOCl is based on the reaction: NaOCl + 2 H+ + 2 e– )* Na+ + Cl– + H2 O. (b) The pH in the pool after adding the NaOCl is 8.67. To improve disinfection, you want at least 90% of the T OT OCl to be in the form of HOCl. Assuming that HOCl/OCl– is the only weak acid/base group in the solution, what volume (L) of 10 N HCl must be added to the stock solution to achieve the goal? 18. Several simple acids and bases are common constituents of household items. Among these are acetic acid (vinegar), bicarbonate of soda (NaHCO3 ), ammonia, ascorbic acid (vitamin C [C6 H8 O6 ], a diprotic acid with pKa1 = 4.10 and pKa2 = 12.35), sodium hypochlorite (bleach), and trisodium phosphate (in many cleansers). (a) A six-year-old playing chemist finds some of these items in a cupboard and decides to mix some vinegar, vitamin C, bleach, and cleanser, hoping to make something exciting happen (like an explosion, or at the very least some serious fizzing). Alas, nothing dramatic occurs, and she decides to devote her energy to theory rather than experimentation. Hence, she decides to list: (i) all the chemical species that she thinks were in the mixture; (ii) the species that could act as acids and those that could act as bases, listing each from weakest acid to strongest (including the acid/base reactions of H2 O); and (iii) the equations necessary to solve for the solution pH. Being rather precocious, she does this perfectly. Reproduce the lists that she prepared. (b) Having opened the bottle of ammonia, she finds the odor offensive and decides to eliminate it. The odor, she knows, is from evaporation of NH3 , because ammonium ions (NH+4 ) cannot enter the gas phase in significant concentrations. Having already used most of the other ‘reagents’ available to her, she has only bicarbonate of soda, vinegar, and some cleanser containing trisodium phosphate remaining. Which should she add? Explain briefly. 287
19. What is the pH of a solution made by adding 10 2 M sodium benzoate (C6 H5 COONa) to pure water, taking into account nonideal solute behavior? Benzoate is the conjugate base of benzoic acid, a common preservative added to food and beverages. It might be useful to see the Hint in Problem 14. 20. What is the composition of a solution made by adding 0.01 M Ca(OCl)2 to pure water, assuming ideal behavior? How much would the computed concentration of OCl– change if nonideal solute behavior were taken into account? It might be useful to see the Hint in Problem 14. 21. The concentration of ammonia in a gas phase [NH3 (g)] that is in equilibrium with a solution depends on the dissolved concentration of NH3 (aq) but is independent of the dissolved NH+4 concentration. (a) Calculate pH and NH3 (aq) in solutions made by adding 10 4 , 10 3 , 10 2 , 10 1 , and 1.0 M NH3 to pure water. Assume ideal solution behavior in all cases. Plot pH and the NH3 (aq) concentration vs. the logarithm of the dose of NH3 added. Why is the NH3 (aq) concentration not proportional to the amount of ammonia added (T OT NH3 )? (b) Carry out similar calculations as in part (a) for the same concentrations of T OT NH3 added, but in this case calculate the pH and concentration of NH+4 when NH4 Cl is added to the solution. In this case, the concentration of NH+4 is (almost) proportional to T OT NH3 added. Why? 22. Would the acidity constant, Ka , for the NH+ 4 /NH3 (aq) acid/base pair increase, decrease, or remain the same if the reference state environment were changed from infinite dilution in pure water to infinite dilution in major-ion seawater? Explain your reasoning. 23. An acid H2 A is stable in solution in both the fully protonated (H2 A) and the fully deprotonated (A2 ) forms, but the intermediate species (HA– ) is negligibly stable and is never present at a significant concentration. (a) A solution of 3 ⇥ 10 3 M H2 A has a pH of 3.7. Find the product Ka1 Ka2 , assuming ideal solute behavior. (b) What are the pH and composition of a solution prepared by adding 0.05 M CaCl2 , 3 ⇥ 10 4 M Na2 A and 3 ⇥ 10 4 M NaHA to water? (Even though HA– is unstable in water, it can be present in a dry salt that can be added to a solution.) Ignore the acid/base reactions of Ca2+ , and assume that Davies equation applies. When computing activity coefficients, note that T OT Ca and T OT Cl are much greater than T OT A. 24. The log C-pH diagram shown below shows the speciation of Cu2+ in a solution containing 3 ⇥ 10 4 M T OT Cu and with the Cl– concentration and ionic 288
strength of seawater (⇠0.5 M, and ⇠0.7 M, respectively). (Some polymeric species have been excluded from the calculations.) Activity coefficients were determined with the Davies equation. Using only information from the graph and Kw , determine the following equilibrium constants: (a) Kb for the dissolved base Cu(OH)2 . (b) Keq for the reaction: Cu2+ + 2 Cl– )* CuCl2 .
25. As indicated in Table 5.1, citric acid (which we can abbreviate as H3 Cit) is a triprotic carboxylic acid. (a) A solution is made by adding lemon juice to water until the pH is 2.20. Assuming all the acidity is from dissociation of citric acid, find the total concentration of citrate species (i.e., T OT Cit) and the concentration of HCit2– . (b) The solution in part (a) is diluted 1 : 10 and partly neutralized by addition of sodium bicarbonate (NaHCO3 ). Designating the total concentration of carbonate species added as T OT CO3 , write all the equations necessary to compute the new pH. You need not solve the equations. (c) If the final pH is 6.0, what is the ratio of (HCO–3 ) to (H2 CO3 ) in the solution? What is the ratio of (CO2– 3 ) to (H2 CO3 )? Would these ratios change if the pH were still 6.0, but T OT Cit were doubled? 26. A brine with an ionic strength of 0.1 M is at pH 8.8 and contains 2.2 mg/L T OT Zn. (a) Determine the Ka,eff values for formation of Zn(OH)2–x species for 1 x x 4. 289
(b) What are the concentrations and activities of Zn2+ and Zn(OH)o2 in the solution? 27. One hundred micromoles (10 4 moles) of solid calcium phosphate (Ca3 (PO4 )2 (s)) is added to 1.0 L of pure water. When the solution reaches equilibrium, the pH is 9.34. (a) Explain why the pH increased above that of the pure water (7.0). (b) Write the equations that would have to be solved to determine the equilibrium pH if the system had not been investigated experimentally. Assume that the solution behaves ideally. (c) When the system reaches equilibrium, the concentrations of total dissolved Ca (T OT Ca) and total dissolved phosphate (T OT PO4 ) are 3.34 ⇥ 10 5 M and 2.23 ⇥ 10 5 M, respectively. If an additional 1.0 L of pure water is added to the suspension, and the system then re-equilibrates, will the pH increase, decrease, or remain the same as before the water was added? Explain briefly.
290
6
ACIDS AND BASES, PART 2: USE OF Log C-pH DIAGRAMS AND THE TOTH EQUATION
291
Contents 6.1
GRAPHICAL SOLUTIONS FOR SETS OF SIMULTANEOUS EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 USING LOG C-pH DIAGRAMS TO SOLVE WEAK ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 THE MASS BALANCE ON H (THE TOTH EQUATION AND PROTON CONDITION) . . . . . . . . . . . . . . . . . . . . . . . 6.4 DEFINING THE COMPOSITION OF THE SYSTEM: COMPONENTS, SPECIES, AND THE SYSTEM TABLEAU . . . 6.5 EXPRESSING INPUT AND EQUILIBRIUM CONCENTRATIONS IN TERMS OF COMPONENTS; COMPONENT MASS BALANCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 IDENTIFYING DOMINANT SPECIES . . . . . . . . . . . . . . . 6.7 COMPARISON OF APPROACHES FOR SOLVING ACID/BASE PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 APPENDIX 6A: ARITHMETIC OPERATIONS AND LOG-LOG DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
292 293 308 310
316 327 341 342 343 345
GRAPHICAL SOLUTIONS FOR SETS OF SIMULTANEOUS EQUATIONS
In Chapter 5, the key equations that characterize acid/base equilibria were presented and solved analytically, in some cases after applying various simplifying assumptions. Then, a spreadsheet was developed that allowed us to identify the approximate equilibrium condition simply by scanning a column that characterizes how closely the charge balance is satisfied at various pH values. Log C-pH diagrams were also introduced as a convenient vehicle for visualizing acid/base speciation over a wide 292
range of conditions, and the point was made that these diagrams convey essentially the same information as is in the spreadsheet. In this chapter, we combine those ideas to demonstrate how the equilibrium pH (and indeed, the speciation of the whole solution) can be determined using log C-pH diagrams in conjunction with either the charge balance or a new equation (the ‘T OT H equation’ or ‘proton condition’) that can be viewed as a mass balance on H+ . We also introduce a concise format for conveying all the equilibrium and mass balance data associated with such problems. Before computers were widely available, this approach was used almost universally to solve complex acid/base problems, using hand-drawn log C-pH diagrams. For many types of problems, the graphical approach can be carried out very rapidly and can yield very good approximations for the equilibrium speciation, even using rough back-of-the-envelope sketches. However, in systems where both acid/base and other types of reactions are important, this approach becomes impractical, and software packages written specifically for solving chemical equilibrium problems offer the only reasonable means of solving the problems. This type of software is introduced in Chapter 7 and is used to analyze problems later in the text. Of course, simple problems as well as complex ones can be solved using the software that is introduced in Chapter 7. However, solving the problems manually in conjunction with log C-pH diagrams can help us develop an intuitive feel for the key relationships that control solution behavior in acid/base systems, and that approach is therefore demonstrated here before introducing the computerized approach in Chapter 7.
6.2
USING LOG C-pH DIAGRAMS TO SOLVE WEAK ACID/BASE PROBLEMS
In Chapter 5, we used both algebraic and spreadsheet analyses to determine the equilibrium pH of a solution of 10 3 M HPr (pKa = 4.87). Consider now how we might solve the same problem with the help of a log C-pH graph. The graphical solution must, of course, satisfy the same set of equations as a numerical solution. In the current case, these equations are the relevant equilibrium constant and mass balance expressions, and the charge balance.1 We start by showing the speciation of HPr and Pr– on a log C-pH diagram, using the techniques developed in Chapter 5. The dependence of log(H+ ) and log(OH ) on pH is trivial, and we plot those values on the same log C-pH graph. The result is a single plot, shown in Figure 6.1, that tells us what the equilibrium values of (H+ ), (OH– ), (HPr), and (Pr– ) would be at any pH in a system with the given T OT Pr. 1 As has been noted in previous chapters, equilibrium constants relate activities, while mass and charge balances relate concentrations. Here, and in the remainder of the text, we assume all solutes behave ideally unless otherwise indicated, and we continue to use parentheses to represent both the activity and the molar concentration when the assumption applies. Parentheses are also used to represent the activity of water, but the concentration of water is represented by square brackets.
293
Figure 6.1 A log C-pH diagram for a solution containing 10
3M
T OT Pr.
To expand on a point that was made in Chapter 5, this log C-pH diagram displays exactly the same information, no more and no less, as is contained in the Ka and Kw expressions and in the mass balance on Pr. That is, just as a straight line on a graph is completely equivalent in information to an equation of the form y = mx + b, the group of lines representing (H+ ), (OH– ), (HPr), and (Pr– ) on the log C-pH diagram has equivalent information content to the set of Ka , Kw , and mass balance equations that characterize the system. (The same can be said for the columns in a spreadsheet showing the concentrations of these species as a function of pH.) Although the values of (H+ ), (OH– ), (HPr), and (Pr– ) shown in Figure 6.1 at any pH satisfy the Ka , Kw , and T OT Pr equations, at most pH’s those values do not satisfy the charge balance. In fact, the charge balance is satisfied at only one pH, so only at that pH are all the equations characterizing the system satisfied simultaneously. We can find that pH by plotting curves representing the two sides of the charge balance on the log C-pH diagram. The charge balance equation is (H+ ) = (Pr ) + (OH )
(6.1)
If we draw curves on the diagram representing the left-hand side [LHS = (H+ )] and right-hand side [RHS = (OH ) + (Pr )] of this equation, the point where those curves intersect will be the pH where the charge balance is satisfied; that is, they will intersect at an x value corresponding to the equilibrium pH. The curve for the LHS is trivial to draw: it simply overlays the (H+ ) line. The curve for the RHS is a bit trickier, since we want to represent a linear summation on a logarithmic scale; i.e., we want to plot a curve representing log[(OH ) + (Pr )]. To plot this curve exactly, we would need to carry out some simple, but nevertheless tedious, calculations. (Recall from basic math that log[(OH ) + (Pr )] does not 294
equal log(OH ) + log(Pr )!) It turns out, however, that a very close approximation of log(RHS) can be plotted without any such calculations. To understand this approximation, consider the data in Table 6.1, showing the results of the relevant calculations at a few pH values. Table 6.1 Values of log[(OH ) + (Pr )] at various pH values for a system containing 10 3.0 M T OT Pr
log(OH– )
pH 4.0 6.0 8.0 10.0 11.0 12.0 14.0
log(Pr– ) 3.92 3.03 3.00 3.00 3.00 3.00 3.00
10.00 8.00 6.00 4.00 3.00 2.00 0.00
log[(OH– )+(Pr– )] 3.92 3.03 3.00 2.96 2.70 1.96 0.00
The values in the final column of the table correspond to log(RHS); i.e., they are the values we want to plot. Comparison of these values with log(OH ) and log(Pr ) shows that, over most of the pH scale, log(RHS) is almost exactly equal to either log(OH ) or log(Pr ), whichever is larger (the values shown in boldface type in the table). As a result, over most of the graph, a curve representing log(RHS) simply overlays the higher of the log(OH ) and log(Pr ) curves. The only region where this approximation does not hold is where OH ⇡ Pr , in this example near pH 11. When OH = Pr , log[(OH ) + (Pr )] equals log[2(OH )], which is the same as 0.3 + log(OH ). Thus, at the pH where that equality applies (i.e., at the intersection of the OH and Pr curves), log(RHS) is exactly 0.3 log unit larger than log(OH ) and log(Pr ). Considering all of the preceding information, we can plot RHS in regions where OH and Pr are significantly different from one another, and also at the point where they exactly equal each other; we can then sketch the curve near the intersection point to obtain a good approximation of RHS across the whole pH range. Both LHS and RHS are shown in bold in Figure 6.2a, and Figure 6.2b shows an expanded view of the region where the RHS curve shifts from being dominated by Pr to being dominated by OH . The intersection of the RHS and LHS lines is circled in Figure 6.2a and is approximately at pH = 3.95. At this pH, the concentrations of all the species in solution can be read from the graph as (H+ ) = 10 (OH ) = 10
3.95 10.05
295
(HPr) ⇠ = 10 ⇠ 10 (Pr ) =
3.03 3.95
Figure 6.2 (a) A reproduction of Figure 6.1, with additional lines (in bold) indicating the values of the left-hand side (LHS) and right-hand side (RHS) of the charge balance equation for a solution of 10 3 M HPr. (b) An expanded view of the pH region from pH 9.5 to 12.5, where OH is close to Pr .
The accuracy of the result can be checked by evaluating the charge balance (CB) and the Ka expression using these estimates. As shown below, both equations are indeed satisfied to a close approximation. Check CB:
10
? 3.95 =
10
Check Ka :
10
? 4.87 =
(10
3.95 + 10 10.05 3.95 )(10 3.95 )
10
3.03
= 10
4.87
If we needed a more accurate answer than the approximate one determined using the log C-pH diagram (which is unlikely), we could at least use the graph to note that 296
at the equilibrium condition (OH ) ⌧ (H+ ) and then solve the problem numerically as shown in Chapter 5. In such a case, the graph would provide a quick way to determine which assumptions are reasonable; more often, however, an answer within ±0.05 of the exact pH is obtainable and entirely satisfactory, so the graphical method can be used to solve the problem without carrying out any calculations at all. The preceding discussion demonstrates (not surprisingly) that the spreadsheet, graphical, and analytical solutions all yield essentially the same information. However, the graphical approach makes it much easier to visualize the behavior of the various species as a function of pH than does either of the other two approaches, because in the analytical and numerical solutions, all the information about the system is embedded in equations or lists of numbers. This advantage of the graphical approach is marginal in the very simple example system analyzed above, but it becomes dramatic when one is analyzing more complex systems, as shown in the remainder of the chapter. The simplicity and benefits of the graphical analysis derive primarily from the fundamental similarity of the equations characterizing acid/base speciation in all systems. That is, the equations for a0 and a1 are essentially identical for any monoprotic acid/base system, differing only in the value of Ka , and the equations describing (H+ ) and (OH ) as a function of pH are the same in all systems. As a consequence, the graphs “look” very similar for any monoprotic acid or base system: the lines for (H+ ) and (OH ) always have slope 1 and +1, respectively, and always intersect at (7.0, 7.0), and the curves for (HA) and (A ) have consistent, characteristic shapes, regardless of which acid/base pair is under consideration. Increasing the total concentration of the acid/base group (i.e., T OT A) moves both the HA and the A curves “up” the graph, and increasing pKa moves both to the right (higher pH). Regardless of the total concentration and pKa value, however, the shapes of these curves and their relationship to one another never change – they always intersect at pH = pKa and at a value on the ordinate that is 0.3 log unit below T OT A. Furthermore, it is very common for each side of the charge balance equation to be dominated by a single term at any given pH, as was the case in the preceding example. In such cases, the LHS or RHS curve is just a linked series of sections, each overlaying the species that makes the largest contribution to the LHS or RHS summation in that pH region. Occasionally, two or more terms contribute significantly to one side of the charge balance over a fairly wide pH range. A simple algorithm for drawing the LHS and RHS lines in such situations is described in Appendix 6A. An additional advantage of the graphical approach is demonstrated in the following example. Example 6.1 Use a graphical analysis to determine the equilibrium concentrations of all species in a solution of 10 3 M NaPr. 297
Solution As in the analysis of the 10 3 M HPr solution, we start by drawing the log CpH diagram that characterizes the system. When we identify the equations that are needed to draw the (H+ ), (OH– ), (HPr), and (Pr– ) curves on the graph, however, we reach an extremely useful conclusion: Ka , Kw , and the mass balance on Pr are identical to those in the previous problem, and those are the only equations that are used to draw the four curves. As a result, all the lines in Figure 6.1 are applicable to this problem as well, and we can use the same graph to solve the problem! Since we want to represent the concentrations of all dissolved species on the graph, we need to add a line for (Na+ ), but this is trivial, because (Na+ ) = 10 3 , independent of pH. We also need to include Na+ in the charge balance, which becomes (Na+ ) + (H+ ) = (Pr ) + (OH ) As before, we can draw lines on the graph to represent the LHS and RHS of the charge balance, and again, in each case, the LHS or RHS curve simply overlays the higher of the two curves in the corresponding summation. Unfortunately, though, these curves do not readily identify a single pH where the charge balance is satisfied. Rather, they indicate that the equation is approximately satisfied over the entire range 6 pH 10 (Figure 6.3a). Thus, the graph is not sufficiently precise to allow us to solve the problem using the charge balance directly. Note that the graph is still a perfectly valid representation of the system, and the RHS and LHS lines actually do intersect at a single point which characterizes the equilibrium condition. The problem we are encountering is strictly a practical one, related to the fact that we cannot draw lines thin enough to pick out the exact point of intersection. The difficulty in identifying the pH at which RHS = LHS in this problem can be circumvented in at least two ways. One approach would be to return to using a spreadsheet to solve the problem, taking advantage of the fact that the numerical processing in the spreadsheet is much more precise than can be displayed graphically. Although that approach is effective, it would still be nice to take advantage of the features in the graphical approach for solving the problem, and this can be accomplished by combining the charge balance with two of the mass balances. Specifically, if we equate the mass balances for Na and Pr (since T OT Na = T OT Pr = 10 3 ) and then substitute the resulting expression for (Na+ ) in the charge balance, we obtain CB: Equating the two MBs: Substituting into the CB: Modified CB:
(Na+ ) + (H+ ) = (Pr ) + (OH ) (Na+ ) = (Pr ) + (HPr) (Pr ) + (HPr) + (H+ ) = (Pr ) + (OH ) (HPr) + (H+ ) = (OH ) 298
Figure 6.3 (a) A log C-pH diagram for a solution of 10 3 M NaPr, with the RHS and LHS of the charge balance shown in bold. (b) The same diagram as in part (a), but with the RHS and LHS based on an equation developed by combining the charge balance with the mass balance on Pr.
By plotting the LHS and RHS of this modified charge balance equation on the graph (Figure 6.3b), it is easy to identify their point of intersection. This intersection point indicates that the equilibrium pH is 7.95, consistent with the numerical solution in Chapter 5 (where we found the exact pH to be 7.94). Reading values from the graph at pH 7.95, the concentrations of the various species at equilibrium are (Pr ) = 10
3.00
(H+ ) = 10
(HPr) = 10
6.05
(OH ) = 10 299
7.95 6.05
(Na+ ) = 10
3.00
Checking the value of Ka as a way of confirming the result, we see that the computed concentrations do indeed meet the criterion for equilibrium: Ka = 10
4.90 ?
=
7.95 )(10 3.00 )
(10
10
6.05
= 10
4.90
The preceding example points out that although the graphical analysis might provide the best big-picture view of the relationships of interest, it suffers from the drawback that, at least in some cases, we need to combine the charge balance with one or more mass balances to solve the problem. Two obvious questions arise from this outcome: (1) How can we know which equation(s) to combine with the charge balance to obtain the best equation to use with the graph? and (2) Can we write the desired equation directly, without going through the algebra of combining multiple equations? The answer to the first question is fairly straightforward. For the NaPr system, each side of the charge balance consists of the sum of two numbers: (Na+ ) + (H+ ) = (Pr ) + (OH ) The problem we encounter when trying to solve this equation directly is that, over a fairly wide region near where it is satisfied, one of the numbers on each side [(Na+ ) on the left, (Pr– ) on the right] is much larger than the other. As a result, even though the equation is satisfied exactly at only one pH, it is satisfied approximately over a wide range where (Na+ ) ⇡ (Pr ). To “home in” on the pH where the equation is satisfied exactly, we need to eliminate the large number from each side of the equation; this need was the basis for using the mass balances on Na and Pr to convert the charge balance into a more useful equation for the graphical analysis. Generalizing this result, whenever a charge balance appears to be satisfied over a wide pH range, we need to combine the charge balance with mass balances that allow us to eliminate the largest number on each side of the equation. Additional examples employing this approach and extending the analysis to systems containing mixtures of acids and bases or multiprotic acids are provided next; we defer answering the second question until after those examples. Example 6.2 Find the equilibrium concentrations of H+ , HOCl, OH– , and OCl– when the following chemicals are added to pure water. Recall that pKa,HOCl = 7.53. (a) 10
3M
KOCl
(b) 10
4M
KOCl + 9 ⇥ 10
4M
HOCl 300
Solution (a) The charge balance for a solution of 10 CB:
3M
KOCl is
(K+ ) + (H+ ) = (OCl ) + (OH )
Plotting the two sides of the charge balance directly on a log C-pH diagram for a system with 10 3 M T OT K and 10 3 M T OT OCl (Figure 6.4a), we run into the same problem as in the NaPr system: the unique intersection point of the RHS and LHS cannot be discerned from the graph. Noting that (K+ ) and (OCl– ) are the largest terms on the left and right sides of the charge balance, respectively, we can equate the mass balances on K and OCl and subtract the resulting equation from the charge balance to obtain the desired equation: CB:
(K+ ) + (H+ ) = (OCl ) + (OH )
MB:
(K+ ) = (OCl ) + (HOCl) (H+ ) + (HOCl) = (OH )
Plotting LHS and RHS of the combined CB/MB equation (Figure 6.4b), we find their intersection at pH = 9.26
pOH = 4.74 ?
Ka =
log(HOCl) = 4.74
(10
9.26 )(10 3.01 )
10
4.74
= 10
log(OCl ) = 3.01 7.53
(b) The charge balance for a solution of 10 4 MKOCl+9⇥10 4 M HOCl is shown below, and the two sides of the equation are plotted in Figure 6.5. CB:
(K+ ) + (H+ ) = (OCl ) + (OH )
Although this charge balance has the same form as the one in part (a), it differs in that the K+ concentration is only 10 4 M. As a result, the intersection point of the LHS and RHS is easily identified on the graph without any modification of the equation. The equilibrium condition is such that (K+ ) ⇡ (OCl ), at which point the composition of the solution is pH = 6.58
log(OCl ) = 4.00
log(OH ) = 7.42
log(HOCl) = 3.05 301
log(K+ ) = 4.00
Figure 6.4 LogC-pH diagram for a solution containing 10 3 M T OT OCl. The LHS and RHS of (a) the CB, and (b) a combined CB/MB are shown for a solution made by adding all the OCl as KOCl.
Example 6.3 Prepare log C-pH graphs for the following solutions, and determine the speciation in each system. pKa for HAc is 4.76, and that for NH+4 is 9.24. (a) 10
3M
HAc + 5 ⇥ 10
4M
(b) 10
3M
NaAc + 2.5 ⇥ 10
NH4 Cl
4M
NH3 + 2.5 ⇥ 10
4M
NH4 Cl
Solution Because both solutions contain 10 3 M T OT Ac and 5 ⇥ 10 4 M T OT NH3 , a single graph can be used to determine the equilibrium composition of both of them 302
Figure 6.5 LogC-pH diagram for a system containing 10 3 M T OT OCl. The RHS and LHS of the charge balance are shown in bold for a system comprised of 10 4 M KOCl plus 9 ⇥ 10 4 M HOCl.
(Figure 6.6). However, in both cases, the charge balance must be combined with a mass balance to determine the equilibrium pH. The various manipulations of the equations and the final results are summarized below.
Figure 6.6 LogC-pH diagram for a system containing 10 T OT NH4
303
3M
T OT Ac and 5 ⇥ 10
4M
(H+ ) + (NH+ 4 ) = (OH ) + (Ac ) + (Cl )
(a) +
(Cl ) = (NH+ 4 ) + (NH3 ) (H+ ) = (OH ) + (Ac ) + (NH3 )
pH ⇡ 3.91, where (H+ ) ⇡ (Ac ) + + (NH+ 4 ) + (Na ) + (H ) = (Ac ) + (OH ) + (Cl )
(b)
(Ac ) + (HAc) = (Na+ )
+
+ (NH+ 4 ) + (H ) + (HAc) = (OH ) + (Cl ) + (NH+ 4 ) + (H ) + (HAc) = (OH ) + 2.5 ⇥ 10
4
pH ⇡ 9.19 (⇡ pKa,NH+ ), where (NH3 ) = (NH+ 4 ) = 2.5 ⇥ 10
4
4
Note that, in Example 6.3b, the (HAc) term contributes negligibly to the LHS of the combined CB/MB at the point where the equation is satisfied, and (Ac– ) does not appear in the equation at all. As a result, the acetate species have essentially no effect on the result; i.e., the pH is the same in this solution as it would be if we had added the ammonia species but no Ac at all. The log C-pH diagram makes it clear why this is so. At the equilibrium pH, almost all the Ac in the system is present as acetate ion, Ac– . Since that is also the form in which all the Ac was added to the solution, we conclude that acetate enters this solution as Ac– and then undergoes no further reaction. In particular, virtually none of the Ac– ions combine with H+ to form HAc. In essence, then, in this system, NaAc acts not like a base but rather like an inert salt. This result applies because the NH3 increases the pH to a value where the Ac– has no significant tendency to become protonated. Example 6.4 (a) What are the concentrations of all species in solutions of (i) 10 (ii) 10 2 M NaHCO3 ; (iii) 10 2 M H2 CO3 ?
2M
Na2 CO3 ;
(b) A solution has three times as much H2 CO3 as CO2– 3 . What is its pH? Solution (a) As in the previous examples, the chemical relationships in this system can be expressed by a combination of mass balances, equilibrium constants, and a charge balance. The mass balance on T OT CO3 and the equilibrium equations 304
are the same for all three solutions, so they can be characterized by a single log C-pH graph, which is shown in Figure 6.7.
Figure 6.7 log C-pH diagram for a system containing 10
(i)
CB :
2M
T OT CO3
(Na+ ) + (H+ ) = (OH ) + (HCO3 ) + 2(CO23 )
This charge balance differs from the ones analyzed in the previous examples because, under conditions where the last term on the right dominates that side of the equation, RHS ⇡ 2(CO23 ). Thus, under these conditions, the curve representing RHS will not overlap the (CO2– 3 ) curve, but rather 2– will coincide with a curve representing 2(CO3 ). To determine where such a curve lies on the graph, we can rewrite the approximation as follows: ⇥ ⇤ log(RHS) ⇡ log 2(CO23 ) = log 2 + log(CO23 ) = 0.3 + log(CO23 )
This result indicates that, when RHS is dominated by the term 2(CO23 ), the curve for log(RHS) is 0.3 log unit above the log(CO23 ) curve, as shown in Figure 6.7. The two sides of the charge balance are shown in Figure 6.8, labeled as RHS-(i) and LHS-(i), respectively. It is not too difficult to estimate the equilibrium pH by using the charge balance in this case, but that pH can be identified more accurately if the largest terms in the charge balance are eliminated. Based on the intersection of the RHS-(i) and LHS-(i) curves, the largest terms in the equation at the equilibrium pH are (Na+ ) and 2(CO2– 3 ) respectively. Eliminating these terms from the equation is 305
easily done, since T OT Na = 2 T OT CO3 . Subtracting that equality from the CB equation, we obtain (Na+ ) + (H+ ) = (OH ) + (HCO3 ) + 2(CO23 ) (Na+ ) = 2(H2 CO3 ) + 2(HCO3 ) + 2(CO23 ) 2(H2 CO3 ) + (HCO3 ) + (H+ ) = (OH )
Figure 6.8 LogC-pH diagram for Example 6.4. The bold lines show the two sides of the unmodified charge balance equation for part (i). The solutions for part (i) using the modified charge balance and for parts (ii) and (iii) are also shown.
(H+ ) = 10
11.13
(H2 CO3 ) = 10 (ii)
CB :
7.64
(OH ) = 10 (HCO3 ) = 10
2.87 2.86
(CO23 ) = 10
2.06
(Na+ ) + (H+ ) = (OH ) + (HCO3 ) + 2(CO23 )
Once again, identifying the pH where the RHS and LHS are equal is difficult, and we can overcome the problem by substituting to eliminate the largest terms. In this case, the mass balances are related by T OT Na = T OT CO3 . Combining this expression with the CB yields (Na+ ) + (H+ ) = (OH ) + (HCO3 ) + 2(CO23 ) (Na+ ) = (H2 CO3 ) + (HCO3 ) + (CO23 ) (H2 CO3 ) + (H+ ) = (OH ) + (CO23 ) 306
The resulting equation is satisfied at (H2 CO3 ) ⇡ (CO23 ), pH 8.34. This point is circled and labeled (ii) in Figure 6.8. The composition of the solution is (H+ ) = 10
8.34
(H2 CO3 ) = 10 (iii)
CB :
(OH ) = 10
4.00
(HCO3 ) = 10
5.66 2.01
(CO23 ) = 10
4.00
(H+ ) = (OH ) + (HCO3 ) + 2(CO23 )
The point where this CB is satisfied can be identified directly on the graph. The result is that the equilibrium pH is 4.18, where (H+ ) ⇡ (HCO3 ). (H+ ) = 10
4.18
(H2 CO3 ) = 10
(OH ) = 10
2.00
(HCO3 ) = 10
9.82 4.17
(CO23 ) = 10
10.32
(b) In this problem, we know the relative concentrations of two species and want to determine the solution pH. One approach for doing this is to generate a reaction relating the two species of interest and H+ by adding the reactions for dissociation of H2 CO3 and HCO–3 . Recalling that when we add reactions, we must multiply equilibrium constants, we find H2 CO3 )* HCO–3 + H+
Ka1
+ HCO–3 )* CO2– 3 +H
Ka2
+ H2 CO3 )* CO2– 3 +2H
(CO23 )(H+ )2 = 10 (H2 CO3 )
Ka02 = Ka1 Ka2 = 10
16.68
16.68
Substituting the given information that (H2 CO3 )/(CO23 ) = 3, we can compute the solution pH: (H+ )2 = 3 ⇥ 10 (H+ ) = 10
16.68
8.10
= 10
16.20
pH = 8.10
To summarize, the preceding discussion and examples demonstrate how a log CpH diagram can be used to determine the pH of a system with known inputs. The 307
graph contains information about the relevant acidity constants (indicated by the intersection points of conjugate acid/base pairs), the dissociation of water (indicated by the positions of the H+ and OH– lines), and the total amount of each weak acid/base group in the system (indicated by the largest value that log C approaches for the individual acid and base species). Once the graph is drawn, all we need to do to solve for the equilibrium speciation of the system is to write the charge balance equation and find the unique pH where that equation is satisfied. In some cases, we can determine the equilibrium pH directly from the graph and charge balance; however, in other cases, it is necessary to derive a new equation by combining the charge balance with a mass balance before the graphical analysis can be used. While knowing how to convert the charge balance into a more useful form is helpful, it would be even better if we could write the final form of the equation directly, without going through the charge balance at all. An approach for accomplishing that goal is presented next.
6.3
THE MASS BALANCE ON H (THE TOTH EQUATION AND PROTON CONDITION)
Recall that, when writing the mass balances that characterize acid/base systems, we chose not to evaluate the mass balance on H. The reason for this decision is most easily understood by contrasting the mass balance on H with that for other substances in the system. Consider a hypothetical system containing 10 6 M total arsenate (75 µg AsO4 /L). The arsenate acid/base group is triprotic, forming arsenic acid (H3 AsO4 ) when it is fully protonated; the general chemistry of this group is very similar to that of the phosphate group. However, whereas phosphate is a major component of nucleic acids and adenosine triphosphate (ATP, the key chemical used to store energy in most cells), arsenate is a poison at relatively low levels. It has been used as an herbicide in the past, but its use in such applications and in industrial processing is now severely limited.2 The acidity constants of arsenic acid are close to those of phosphoric acid, so the forms most likely to be found in natural waters are the mono- and divalent ions 3– H2 AsO–4 and HAsO2– 4 , with arsenic acid and arsenate ion (AsO4 ) present at much lower concentrations. Nevertheless, in the mass balance, we consider the contribu2 Most current environmental concerns about arsenic in the developed world are related to its presence in drinking water and to the cleanup of a few severely contaminated but localized sites of soil pollution. As of 2013, the maximum contaminant level (MCL) for arsenic in the United States was 10 µg/L, and the maximum contaminant level goal (MCLG) was zero. In a few locations in the world, water that has been used as a potable source has arsenic concentrations that far exceed safe levels, leading to disastrous epidemiological consequences. In those cases, the source of the contamination has usually been natural, arsenic-bearing rocks in the aquifer.
308
Increased awareness of the risks associated with arsenic ingestion have caused several governments to lower the allowable level of arsenic in drinking water. In Bangladesh, natural concentrations of arsenic in groundwater have had a devastating e↵ect on some communities. The photograph on the left shows some of the symptoms of arsenic poisoning, and that on the right shows sheep being dipped in an arsenic-based solution to kill external parasites. (Left: http://bdhalchall.blogspot.com/2010/03/arsenic-pollution-in-bangladesh.html; Right: G.R. Roberts/G.R. “Dick” Roberts Photo Library.)
tions of all four species to T OT AsO4 . Thus, in the example solution, the mass balance on arsenic is T OT AsO4 = 10
6.0
= (H3 AsO4 ) + (H2 AsO4 ) + (HAsO24 ) + (AsO34 )
The baseline arsenic concentration that we use when we do this accounting is zero. That is, the total amount of arsenic present is quantified by comparison with a solution containing zero arsenic. This choice is so logical and obvious that it is normally taken for granted. However, there are times when it is more useful to quantify the amount of some substance in solution by comparison with a baseline other than zero. One easy way to understand this concept is to consider a nonchemical example, such as banking. Banks provide monthly statements that tell people how much money they have in their accounts, and these statements are based on a balance of zero as the baseline. However, let’s say you have an account that provides free checking if you maintain a balance of at least $500. If you want to make sure that you always maintain the minimum balance in your checkbook, you might keep track of your available funds, meaning the balance above $500. In this case, you might record a balance of $650 on the bank’s statement as a balance of only $150 in your checkbook. Furthermore, if your real balance did dip below the minimum (say, to $350), your records would show that your balance had become negative ( $150). This scenario is perfectly rational (at least I think so; I used to use it!), and it is as accurate and reliable as a more conventional accounting practice. The only critical requirement is that you know what the accounting rules are in order to interpret the numerical data correctly. In the absence of a good reason to define a nonzero baseline for a chemical mass balance, we generally do not do so, and as noted above, we generally do not even 309
take note of the fact that we are choosing “zero substance” as the baseline condition. When writing a mass balance on H, however, we do have a good reason for using a nonzero baseline. Specifically, the concentration of H atoms associated with H2 O molecules is enormous (2 ⇥ 55.6 M) compared to the concentration present in all other species in solution. This fact makes it very inconvenient to try to keep track of all the H in the system. Rather, it is much easier to do the accounting by defining a large absolute concentration of H (approximately equal to the H concentration present as part of H2 O molecules, but also including a few other terms) as the baseline, and to keep track of H concentrations relative to that baseline. The mass balance on H using a nonzero baseline has traditionally been written in two forms. One form is exactly analogous to the mass balances on other components of solution, except for the nonzero baseline. In this equation, the total concentration of H relative to the baseline is designated as T OT H, and the equation is called the TOTH equation. Because of its similarity to the other mass balances that characterize the system, the T OT H equation is particularly well-suited for inputting the data into computer software for solving chemical equilibrium problems. In the other, only slightly different form of the equation, terms are shifted so that the LHS and RHS of the mass balance each involves only additions and no subtractions. This version of the equation is called the proton condition (PC); the proton condition is better suited than the T OT H equation for solving problems graphically using log C-pH diagrams. In the next section, an algorithm is presented for simultaneously developing all the mass balance equations applicable to a system, including the T OT H equation. The conversion of the T OT H equation to the PC is then demonstrated, and the PC is used to solve several problems graphically; the T OT H equation is used to solve similar problems in Chapter 7, where chemical equilibrium software is introduced, and in subsequent chapters.
6.4
DEFINING THE COMPOSITION OF THE SYSTEM: COMPONENTS, SPECIES, AND THE SYSTEM TABLEAU
We begin by assigning special meanings to two terms that have been used in a more general way to this point. Specifically, we define a system’s species as the group of chemicals expected to be present in the system at equilibrium, and the components for the system as any group of chemicals that satisfies the following two criteria: (1) combinations of the components can be used to generate all the system’s species, and (2) none of the components can be generated by a reaction involving only other components. Note that, based on these criteria, components only have meaning when defined as an entire group. Those familiar with the terminology of linear algebra will recognize these restrictions as establishing that the components ‘span the space’ 310
of species. Formation of a group of four species from three components is shown schematically in Figure 6.9.
Figure 6.9 Schematic showing the formation of four species from various combinations of three components. The numbers next to the arrows indicate the stoichiometric coefficients for forming each species.
As indicated in the figure, it is allowable to either add or subtract components from one another to form species. For instance, Spec 3 is formed by subtracting (i.e., removing) one unit of Comp C from one unit of Comp B. An example of such a reaction would be the formation of the species OH– by removing one H+ ion from a water molecule. The figure also indicates that Spec 1 consists of one unit of Comp A and nothing else. Thus, Spec 1 and Comp A are chemically identical. However, the operations that we will carry out on components are different from those on species, so we will distinguish between them, even if in some cases a component is the same chemical entity as a species. To make the process of defining species and components more explicit, consider a system made by adding 2 ⇥ 10 3 M HAc to water. The chemicals expected to be present at equilibrium (i.e., the system’s species) include H2 O, H+ , Ac– , OH– , and HAc. Say we chose H2 O, H+ , and Ac– as components. Three of the species (H2 O, H+ , and Ac– ) are chemically identical to components, so obviously those species could be formed from the components directly. The remaining two species can be generated by appropriate combinations of components: OH– can be made by “subtracting” an H+ from an H2 O molecule, and HAc can be made by adding an H+ to an Ac– . When written as chemical reactions, these ways of generating OH– and HAc bear a close and obvious relationship to reactions for the dissociation of water (the Kw reaction) and the acetic acid dissociation reaction, respectively: 311
Reaction forming species as a combination of components H2 O
H+ )* OH–
H+ + Ac– )* HAc
Reaction written conventionally H2 O )* H+ + OH–
(6.2)
HAc )* H+ + Ac–
(6.3)
The (conceptual) ability to form all the species by various combinations of the components satisfies the first criterion for using H2 O, H+ , and Ac– as a set of components for the system. Since it is not possible to generate any of the components by a reaction involving only other components (for instance, it would not be possible to form H+ by any combination of H2 O and Ac– without simultaneously generating or removing other chemical species), the second criterion is satisfied as well, and the proposed set of components is acceptable. Next, the stoichiometry for forming each species from the components is formalized. The simplest way to do this is to fill in the stoichiometric coefficients for the components in reactions like the following for forming each species: nA, j Comp A + nB, j Comp B + ... + nN, j Comp N )* 1Spec j
(6.4)
where ni, j is the stoichiometric coefficient for component i in the reaction forming species j. The same set of components is used to form all the species in the system. Thus, for example, for a solution of HAc in water, the reactions would be: 1 H2 O +
0 H+ + 0 Ac– )* 1 H2 O
(6.5)
0 H2 O +
1 H+ + 0 Ac– )* 1 H+
(6.6)
0 H2 O +
0 H+ + 1 Ac– )* 1 Ac–
(6.7)
1 H2 O +
1 H+ + 0 Ac– )* 1 OH–
(6.8)
0 H2 O +
1 H+ + 1 Ac– )* 1 HAc
(6.9)
Although the choice of H2 O, H+ , and Ac– as components is acceptable for this system, this set is not unique; several other choices are equally acceptable. For instance, the group H2 O, H+ , and HAc meets the criteria for an acceptable component set, as does the group H2 O, HAc, and Ac– . On the other hand, it would not be allowable to choose H+ , HAc, and Ac– as a component set, both because it is possible to generate Ac– by a reaction involving only the other components (by HAc dissociation) and because it is impossible to generate H2 O or OH– solely by a combination of these three components. These examples reinforce the idea that no individual chemical entity is inherently acceptable or unacceptable as a component; components can only be acceptable or unacceptable as complete sets, and any system can be represented by many different sets. 312
It is even possible to choose chemicals that are not present in the system or that are not realistic molecules as components. For instance, if we thought it would be useful for some reason, we could specify ways to make all the species in the example system using H2 O, Ac– and H4 Ac3+ as components, even though H4 Ac3+ has never been detected in solution. The stoichiometry for forming the five species from this set of components would be 1 H2 O +
0
Ac– +
0
H4 Ac3+ )* 1 H2 O
0 H2 O +
1
Ac– +
0
H4 Ac3+ )* 1 Ac–
1 H2 O +
0.25 Ac– +
0.25 H4 Ac3+ )* 1 OH–
0 H2 O +
0.25 Ac– +
0.25 H4 Ac3+ )* 1 H+
0 H2 O +
0.75 Ac– +
0.25 H4 Ac3+ )* 1 HAc
Thus, if we include the possibility of using unrealistic components, an infinite number of acceptable component sets can be used to describe any system. However, once a component set is chosen, there is only one way to write the reaction forming each species from those components; i.e., the stoichiometric coefficients for forming each species are uniquely defined once the components have been chosen. Each reaction forming a species from the components can be assigned a unique equilibrium constant. In the baseline system, the equilibrium constants for forming H2 O, H+ , and Ac– , i.e., for Reactions (6.5) through (6.7), are trivial: they must be 1.0, as is obvious if the reactions are rewritten leaving out the components that have coefficients of zero. These reactions are repeated below, with the components shown in bold to distinguish them from the corresponding species. 1 H2 O )* 1 H2 O
KH2 O = 1.0
(6.10)
1 H+
)* 1 H+
KH+ = 1.0
(6.11)
1 Ac– )* 1 Ac
KAc = 1.0
(6.12)
Keep in mind that, in terms of our formal algorithm, these reactions are not just saying that one water molecule forms one water molecule [Reaction (6.10)] or that one acetate ion forms one acetate ion [Reaction (6.12)]. Rather, Reaction (6.7) [or its abbreviated version, Reaction (6.12)] indicates that the species Ac– is formed by combining one unit of the component Ac– plus zero units of the component H2 O plus zero units of the component H+ , and then applying an equilibrium constant of 1.0 to that reaction. The equilibrium constant for Reaction (6.8) can be derived formally as follows: 313
H2 O )* H+ +OH–
Kw
(H+ )* H+ ) 1 H2 O
K = 1.0
1 H+ )* OH
K(6.8) = Kw K = 10
14.00
Similarly, the equilibrium constant for Reaction (6.9) is 1 Ac– + 1 H+ )* HAc
K(6.9) = Ka 1 = 10+4.76
(6.13)
The five equilibrium constant equations relating the species to the components are summarized in both algebraic and logarithmic form in Table 6.2. The same information is also presented in a more concise matrix format in Table 6.3, with each row in the matrix containing the stoichiometric coefficients and log K value for forming a particular species from the components. This type of matrix is widely referred to as a tableau. Table 6.2a Formal algebraic expressions for relating species activities to component activities for an aqueous solution of HAc
Reaction (6.5) (6.6) (6.7) (6.8) (6.9)
Algebraic equations to compute species activity at equilibrium (H2 O) = K(6.5) (H2 O) = K(6.5) (H2 O)1 (H+ )0 (Ac– )0 (H+ ) =
K(6.6) (H+ )
= K(6.6) (H2 O)0 (H+ )1 (Ac– )0
K(6.7) (Ac– ) = K(6.7) (H2 O)0 (H+ )0 (Ac– )1 (H O) (OH ) = K(6.8) 2+ = K(6.8) (H2 O)1 (H+ ) 1 (Ac– )0 (H ) (Ac ) =
(HAc) = K(6.9) (H+ )(Ac– ) = K(6.9) (H2 O)0 (H+ )1 (Ac– )1 Table 6.2b Logarithmic forms of the equations in Table 6.2a
Reaction
Logarithmic equation
(6.5)
log(H2 O) = log K(6.5) + 1 log(H2 O) + 0 log(H+ ) + 0 log(Ac– )
(6.6)
log(H+ ) = log K(6.6) + 0 log(H2 O) + 1 log(H+ ) + 0 log(Ac– )
(6.7)
log(Ac ) = log K(6.7) + 0 log(H2 O) + 0 log(H+ ) + 1 log(Ac– )
(6.8)
log(OH ) = log K(6.8) + 1 log(H2 O)
(6.9)
log(HAc) = log K(6.9) + 0 log(H2 O) + 1 log(H+ ) + 1 log(Ac– )
314
1 log(H+ ) + 0 log(Ac– )
Table 6.3 Tableau showing the key features of the equilibrium expressions in Table 6.2
Component: Species H2 O H+ Ac– OH– HAc ⇤ For
Stoichiometric Coefficient⇤ H2 O H+ Ac– 1 0 0 1 0
0 1 0 1 1
0 0 1 0 1
log K 0.00 0.00 0.00 14.00 4.76
forming the species in a given row from the components.
Example 6.5 Prepare a tableau like that shown in Table 6.3 for the same system (addition of HAc to water), but using H2 O, OH– , and HAc as components. Solution The reactions and equilibrium constants for forming the five species from the new set of components are shown below, and the information is summarized in the following matrix. Convince yourself that the equilibrium constants are correct for the reactions shown. 1 H2 O + 0 OH– + 0 HAc )* 0 H2 O + 1 OH– + 0 HAc )*
1 H2 O
K = 1.00
1 OH
–
K = 1.00
0 H2 O + 0 OH + 1 HAc )* 1 H2 O + 1 OH– + 1 HAc )* 1 H2 O + 1 OH– + 0 HAc )*
1 HAc
K = 1.00
–
–
1 Ac
K = 109.24
1 H+
K = 10
Stoichiometric Coefficient H2 O
OH–
HAc
log K
H2 O
1
0
0
0.00
OH
–
0
1
0
0.00
HAc
0
0
1
0.00
–
Ac
1
1
1
9.24
H+
1
1
0
14.00
Component: Species
315
14.00
Example 6.6 Prepare the tableau for a solution of 10 3 M NaAc, using H2 O, Na+ , H+ , and Ac– as components. Note that, in this case, some sodium-containing chemical must be added as a component, in order to be able to “make” the species Na+ . Sodium ion is the simplest choice for such a component. Solution The matrix is shown below. Confirmation of the values shown for the equilibrium constants is left as an exercise. Stoichiometric Coefficient H2 O
H+
Na+
Ac–
log K
H2 O
1
0
0
0
0.00
Na
0
0
1
0
0.00
H+
0
1
0
0
0.00
–
0
0
0
1
0.00
–
1
1
0
0
14.00
HAc
0
1
0
1
4.76
Component: Species +
Ac
OH
6.5
EXPRESSING INPUT AND EQUILIBRIUM CONCENTRATIONS IN TERMS OF COMPONENTS; COMPONENT MASS BALANCES
Because all the species in a system can be represented as combinations of the components, we can imagine a process in which the species in the equilibrium solution are all decomposed into a mixture of components and nothing else. Using the same terminology as in Equation (6.4), if the system contains Meq species, the total concentration of component i in the mixture of decomposed species will be Meq
T OTieq = ni,1 c1 + ni,2 c2 + ... + ni,M j cMeq =
 ni, j c j
(6.14)
j=1
where ni, j is the stoichiometric coefficient for component i in the reaction forming species j, and c j is the concentration of species j in the solution at equilibrium. In words, each term in the summation on the right side of Equation (6.14) can be viewed as the concentration of component i “embedded” in species j in the equilibrium solution, so the summation is the concentration of i embedded in all species in the whole 316
solution. The coefficients ni, j are listed in the column for component i in the tableau so, for example, the three applications of Equation (6.14) to the data in Table 6.3 are: T OT H2 Oeq = 1[H2 O]eq + 1(OH )eq
(6.15)
T OT Heq = 1(H+ )eq
(6.16)
1(OH )eq + 1(HAc)eq
T OT Aceq = 1(HAc)eq + 1(Ac )eq
(6.17)
where the subscript emphasizes that the concentrations are those in the equilibrium solution. Note that we can represent the concentrations of the solutes as being equal to their activities, but for H2 O, we need to distinguish between those two terms and use the concentration when calculating T OT H2 Oeq . Equation (6.14) is based on the (conceptual) decomposition of the equilibrium species into a mixture of the components. Since these species were all either input into the system directly or generated via reactions among the input chemicals, it must be the case that the input chemicals can also be decomposed into the components. We can therefore write a T OTi expression based on the inputs by an equation analogous to (6.14), as follows: T OTiin = ni,1 c1 + ni,2 c2 + ... + ni,Min cMin =
Min
 ni,k ck
(6.18)
k=1
where ni,k is the stoichiometric coefficient for component i in the reaction forming the input chemical k, ck is the dose of k added (moles of k input per liter of solution), and Min is the number of input chemicals used to prepare the system.3 Applying this equation to the example system, which was prepared using two inputs (Min = 2: H2 O and HAc), we find: T OT H2 Oin = 1[H2 O]in + 0[HAc]in = 55.6 T OT Hin = 0[H2 O]in + 1[HAc]in = 2 ⇥ 10
T OT Acin = 0[H2 O]in + 1[HAc]in = 2 ⇥ 10
(6.19) 3
(6.20) 3
(6.21)
We can incorporate this information into the tableau by adding a row to the bottom for each input chemical. When we do that, and also add a column for the input and equilibrium concentrations, the tableau is expanded as shown in Table 6.4. 3 As
used here, ‘inputs’ include all chemicals that have entered the solution prior to the time when it reaches equilibrium. In some cases, our interest is in a solution where all the inputs are known; in others, we might know the composition of the solution at some initial point without knowing exactly what chemicals were added to prepare that solution, and also know the inputs that entered subsequently. In such cases, we can consider the ultimate solution to be a combination of the initial solution and the later inputs, and we can calculate T OTiin based on that conceptualization. Example 6.8 demonstrates this idea.
317
Table 6.4 Expanded tableau showing inputs and equilibrium species for the example system
Component:
Stoichiometric Coefficient H2 O H+ Ac–
Species H2 O H+ Ac– OH– HAc
1 0 0 1 0
0 1 0 1 1
0 0 1 0 1
Inputs H2 O HAc
1 0
0 1
0 1
log K
Conc’n
0.00 0.00 0.00 14.00 4.76
[H2 O]eq [H+ ]eq [Ac ]eq [OH ]eq [HAc]eq 55.6 2 ⇥ 10
3
Finally, we can equate the total concentration of each component in all the equilibrium species [Equations (6.15)-(6.17)] with that in all the inputs [Equations (6.196.21)] to obtain T OTieq = T OTiin
(6.22)
i =H2 O:
[H2 O]eq + (OH )eq = 55.6
(6.23)
i =H:
(H+ )eq
i =Ac:
(HAc)eq + (Ac )eq = 2 ⇥ 10
(OH )eq + (HAc)eq = 2 ⇥ 10 3
3
(6.24) (6.25)
Equations (6.23) through (6.25) can be considered mass balances on the three components. However, of the three equations, only Equation (6.25) looks like a conventional mass balance. For example, the mass balance on H+ [Equation (6.24)] does not include any of the H+ associated with water molecules, and it counts the concentration of OH– as a negative contribution to T OT H, even though OH– ions do contain H+ . Similarly, the mass balance on H2 O [Equation (6.23)] includes OH– , even though OH– molecules do not contain a full molecule of H2 O. These initially counterintuitive results arise because the mass balances shown are written not in terms of conventional chemical groups, but rather in terms of the molecules we have chosen as components. Thus, the species HAc is included in the mass balance on Ac– because, according to our algorithm, one unit of the component Ac– is required to ‘make’ the species HAc. Similarly, H3 O+ and OH– are included in the mass balance on H+ with coefficients of 1 and 1, respectively, because 1 and 1 ‘units’ of H+ are used when these species are made from the components. 318
In terms of the discussion that introduced this section, the component choices we have made can be viewed as shifting the baseline for the mass balances on those components. For example, the baseline for H+ is defined to include all the components other than H+ , meaning that those components are treated as though they contain zero H+ , regardless of their actual molecular composition. The baselines for the other components (H2 O and Ac– ) are defined in analogous ways. Like in the banking example given earlier, this way of accounting is perfectly valid and, as shown next, it can have significant advantages over more traditional methods. Because of the centrality of H2 O to all systems we will explore, we will always choose it as a component, and so a mass balance on H2 O will always be one of the equations that must be solved to determine the equilibrium speciation. However, this mass balance is invariably like Equation (6.23), in that it includes a summation of [H2 O] plus one or more other species that are present at orders of magnitude lower concentrations. Thus, the mass balance can always be approximated as [H2 O] = 55.6, so there is no need to “solve” for the equilibrium concentration of H2 O. Similarly, unless otherwise indicated, we will always assume that the activity of H2 O is 1.0. Therefore, in the subsequent discussion, we will include H2 O in the list of components, but will not explicitly list it as an input chemical or write the accompanying mass balance. Example 6.7 (a) Prepare two tableaus for a solution made by adding 10 2 M each of HAc and Na2 CO3 to water, using the two different sets of components listed below. Component set #1:
H2 O, H+ , HAc, Na+ , CO2– 3
Component set #2:
H2 O, H+ , Ac– , Na+ , HCO–3
(b) Convert the resulting T OT H equations to proton conditions (PCs), and use the PCs in conjunction with a log C-pH diagram to solve for the equilibrium pH. What are the dominant species of each acid/base group in the equilibrated solution? Solution (a) The two tableaus are shown below. (The final column in the top portion of the tableau is unnecessary and has therefore been left blank.) The species that are present at equilibrium are, of course, identical in the two tableaus, but the way in which they are generated is not; as a result, the equilibrium constants for generating some of the species differ in the two tableaus. Study the tableaus carefully to make sure you understand how the stoichiometric coefficients and equilibrium constant for each species and each input chemical were derived. 319
Tableau using component set #1 Stoichiometric Coefficient H2 O
H+
HAc
Na+
CO2– 3
log K
1
0
0
0
0
0.00
0
1
0
0
0
0.00
HAc
0
0
1
0
0
0.00
+
Na
0
0
0
1
0
0.00
CO2– 3 –
0
0
0
0
1
0.00
OH
1
1
0
0
0
14.00
Ac–
0
1
1
0
0
4.76
HCO–3
0
1
0
0
1
10.33
H2 CO3
0
2
0
0
1
16.68
HAc
0
0
1
0
0
Na2 CO3
0
0
0
2
1
Component:
Conc’n
Species H2 O +
H
Inputs 1.0 ⇥ 10
1.0 ⇥ 10
2 2
Tableau using component set #2 Stoichiometric Coefficient H2 O
H+
Ac–
Na+
HCO–3
log K
1
0
0
0
0
0.00
0
1
0
0
0
0.00
HAc
0
1
1
0
0
4.76
+
0
0
0
1
0
0.00
CO2– 3
0
1
0
0
1
10.33
–
1
1
0
0
0
14.00
–
Ac
0
0
1
0
0
0.00
HCO–3
0
0
0
0
1
0.00
H2 CO3
0
1
0
0
1
6.35
HAc
0
1
1
0
0
Na2 CO3
0
1
0
2
1
Component:
Conc’n
Species H2 O +
H
Na
OH
Inputs
320
1.0 ⇥ 10
1.0 ⇥ 10
2 2
(b) The T OT H equations for the two component sets are written by equating T OT Heq [from Equation (6.14)] with T OT Hin [from Equation (6.18)]. The results are as follows: T OT H equation for component set #1: H+
OH
Ac
+ HCO3 + 2 H2 CO3 = 0 1.0 ⇥ 10
2
+ 0 1.0 ⇥ 10
2
=0
2
1 1.0 ⇥ 10
2
=0
T OT H equation for component set #2: H+
OH
+ (HAc)
CO23
+ H2 CO3 = 1 1.0 ⇥ 10
The T OT H equations can be converted to PCs by moving terms around so that both sides of the equation contain only summations, i.e., Proton condition for component set #1: H+ + HCO3 + 2 H2 CO3 = OH
+ Ac
Proton condition for component set #2: H+ + (HAc) + H2 CO3 = OH
+ CO23
A log C-pH diagram for the system is shown in Figure 6.10, with the LHS and RHS of each PC highlighted. The equilibrium pH is not easily identified using the PC from the first component set. If we had investigated only that set of components, we would have had to combine the T OT H equation and/or PC with some other equation (another mass balance or the charge balance) to obtain an equation that was easily evaluated in conjunction with the graph. On the other hand, the PC derived from the second component set can be used directly with the graph to identify the equilibrium pH, which is 8.34. At this pH, almost all the acetate in the system is present at Ac– , and almost all the carbonate is present as HCO–3 . Note that, in the solution of this problem, the log K values listed in the tableau were not used explicitly. The reason for this is that the same information was already embedded in the log C-pH diagram. This situation will apply to all the problems we analyze in the remainder of the chapter. Nevertheless, we will continue to populate the log K columns of the tableaus, because that information becomes useful (and necessary) when we explore computer-based approaches for solving these problems in Chapter 7. 321
Figure 6.10 LogC-pH diagram for the example system. The solid bold lines are the LHS and RHS of the PC based on component set #1, and the broken bold lines are for the PC based on component set #2. The oval and circle indicate the regions where LHS⇡RHS for the two equations.
The preceding example demonstrates that, even though there are many acceptable choices for the component set for a given system, not all choices are equally useful when applied in conjunction with a log C-pH diagram to find the equilibrium solution composition – some choices lead to a PC whose LHS and RHS nearly overlap in the region where the equation is satisfied, making it difficult to identify the equilibrium pH. This is, of course, the same problem we encountered earlier when we tried using the charge balance in conjunction with a log C-pH diagram. In all these cases, the problematic situation arises when the largest term on each side of the equation is insensitive to pH in the region where the equation is satisfied. Correspondingly, the problem can be overcome by subtracting from both sides a value equal to the largest term on one side. The question then becomes: How can we choose components wisely, so that we derive equations that can be used directly to find the equilibrium pH, without going through the step of modifying the PC? To answer this question, we take advantage of two observations. First, the line representing a species on a log C-pH diagram is nearly horizontal (i.e., insensitive to pH) only in the pH region where that species is the dominant one in its acid/base group. For example, for the carbonate group, the dominant species is H2 CO3 at pH < 6.35, HCO–3 between pH 6.35 and 10.33, and CO2– 3 at pH > 10.33. In each of these regions, the line for the dominant species has a slope near zero, while the lines for the other (nondominant) species have slopes close to ±1 or ±2. Thus, if we can choose components in a way that assures that the T OT H equation (and therefore the PC) does not include the dominant species of any group in the pH region where the 322
equation is satisfied, then we will have assured that the lines representing the RHS and LHS of the PC will not be nearly horizontal in that region, and hence will not (nearly) overlap. Second, according to our algorithm, the baseline for the mass balance on H+ includes all the components other than H+ . Therefore, if a species has the same chemical formula as a component, the coefficient for that species in the H+ column of the tableau will be 0, and the T OT H equation will not include the concentration of the species. For example, reviewing the two T OT H equations and PCs in Example 6.7, we see that when HAc and CO2– 3 were chosen as components (set #1), neither (HAc) 2– nor (CO3 ) appeared in the resulting T OT H equation or PC. The same applies to set #2, where Ac– and HCO–3 do not appear in the T OT H equation or the PC because they were chosen as components. Combining these observations, we conclude that the way to develop a PC that can be used directly to find the equilibrium pH is to choose the species that are dominant in the equilibrium solution as components. The following example shows how we can take advantage of this conclusion. Example 6.8 A wastewater at pH 7.0 contains 250 mg/L total organic carbon (TOC) and 1.5 ⇥ 10 3 M total inorganic carbon (TIC, which is typically taken to be T OT CO3 ). The TOC is distributed among a variety of molecular forms. However, since the waste is thought to be mostly carbohydrates and is highly biodegradable, you have decided to treat all the TOC as if it were the simple sugar glucose (C6 H12 O6 ). Write the T OT H equation and determine the equilibrium pH of the solution after microorganisms consume essentially all the TOC, respiring one-half of it to carbonic acid by the reaction C6 H12 O6 +6 O2 )* 6 H2 CO3 and converting the other half to new biomass. Assume that neither glucose nor the biomass participates in any acid/base reactions. Use H2 O, H+ , and HCO–3 as components, at least in the initial analysis. Solution At pH 7.0, the alpha values for the carbonate system [computed using Equations (5.38) through (5.40)] are a0 = 0.18, a1 = 0.82, and a2 = 10 3.33 . The concentrations of H2 CO3 , HCO–3 , and CO2– 3 in the initial solution can be computed as ai T OT CO3 and are 2.70 ⇥ 10 4 M, 1.23 ⇥ 10 3 M, and 7.0 ⇥ 10 7 M, respectively. The statement that the solution contains 250 mg/L TOC means that 250 mg C/L is present as part of organic molecules. If one-half of this TOC is respired, 125 mg C/L becomes part of H2 CO3 molecules. From a chemical perspective, the appearance of H2 CO3 in solution as the result of a reaction is no different from addition of the same chemical from outside, so we can treat this conversion as an input. On a molar basis, this input is
✓
mg TOC 125 L
◆✓
1 mol C 12,000 mg C
◆✓
1 mol H2 CO3 mol TOC 323
◆
= 1.04 ⇥ 10
2 mol
H2 CO3 L
Therefore, after the biological conversion, the solution can be thought of as a combination of the initial solution containing the carbonate concentrations in that solution and a subsequent addition of 1.04 ⇥ 10 2 M H2 CO3 . The tableau for the system, using the specified components, is shown below. Note that the carbonate species in the initial solution are included in the lower part of the tableau along with the H2 CO3 that was added by respiration, since all of those species are considered ‘inputs’ that were added to pure water to generate the solution. The T OT H equation and PC are shown below the tableau. Stoichiometric Coefficient H2 O
H+
HCO–3
log K
H2 O
1
0
0
0.00
H
0
1
0
0.00
HCO–3 –
0
0
1
0.00
OH
1
1
0
14.00
H2 CO3
0
1
1
6.35
0
1
1
10.33
0
1
1
0
0
1
† CO2– 3
0
1
1
H2 CO3 ‡
0
1
1
H2 O
1
0
0
Component:
Conc’n
Species +
CO2– 3
Inputs H2 CO3 † HCO–3 †
† From
TIC present initially.
‡ From
respiration of TOC.
T OT Hin = 2.7 ⇥ 10
+ 1.04 ⇥ 10
4 2
(1) + 1.23 ⇥ 10
(1) = 1.07 ⇥ 10
2.7 ⇥ 10
4
7.0 ⇥ 10
7
1.23 ⇥ 10 1.04 ⇥ 10
3
2
55.56
3 2
(0) + 7.0 ⇥ 10
T OT H equation:
(H+ ) + H2 CO3
OH
CO23
PC:
(H+ ) + H2 CO3 = OH
+ CO23
7
( 1)
= 1.07 ⇥ 10 + 1.07 ⇥ 10
2 2
The LHS and RHS of the PC are shown by the solid bold lines in the log C-pH diagram in Figure 6.11. The intersection is in the range between pH 4.5 and 5.5, but is difficult to specify more precisely. As in previous examples, we could try to determine expressions that we could subtract from each side of the PC to obtain a more 324
useful form of that equation. However, the preceding discussion in the text suggests that we can resolve the problem immediately by rewriting the T OT H equation using the dominant species at equilibrium as components. Even though we cannot determine the equilibrium pH from the graph, we can see that, at equilibrium, the dominant carbonate species will be H2 CO3 . We therefore revise the tableau, choosing H2 CO3 as a component. This change alters the stoichiometric coefficients for formation of all the carbonate species in both the inputs and the equilibrium solution.4 The new tableau, T OT H equation, and PC are shown below. Plotting the LHS and RHS of this PC in Figure 6.11 (broken bold lines) allows us to easily identify the equilibrium pH of 5.4.
Stoichiometric Coefficient H2 O
H+
H2 CO3
log K
1
0
0
0.00
H+
0
1
0
0.00
HCO–3 –
0
1
1
6.35
OH
1
1
0
14.00
H2 CO3
0
0
1
0.00
0
2
1
16.68
0
0
1
0
1
1
CO2– 3
0
2
1
H2 CO3
0
0
1
Component:
Conc’n
Species H2 O
CO2– 3
Inputs H2 CO3 HCO–3
T OT Hin = 2.7 ⇥ 10
+ 1.04 ⇥ 10
4 2
T OT H equation: (H+ ) PC:
(0) + 1.23 ⇥ 10
3
4 The change in components also alters the log
7.0 ⇥ 10
7
1.04 ⇥ 10 3
( 1) + 7.0 ⇥ 10
7
3
2
( 2)
3
HCO3
(H+ ) + 1.23 ⇥ 10
4
1.23 ⇥ 10
(0) = 1.23 ⇥ 10 OH
2.7 ⇥ 10
= OH
2 CO23
= 1.23 ⇥ 10
3
+ HCO3 + 2 CO23
K values for forming the carbonate species. However, because the correct equilibrium relationships are already embedded in the diagram, the only information we need from the tableau is the set of stoichiometric coefficients for the mass balance on H+ , i.e., the coefficients that appear in the T OT H equation. Therefore, although the log K values in the tableau change when we choose new components, we need not compute the new values. Nevertheless, the correct values are shown in the tableau here for completeness.
325
Figure 6.11 LogC-pH diagram for the example system.
In addition to illustrating how choosing the dominant species as a component can facilitate the identification of the equilibrium pH, the preceding example illustrates two useful points about the relationship between the T OT H equation and the charge balance. First, note that the problem statement specified the pH and TIC of the original solution, but it did not provide any information about other solutes. We know that at least one additional cation must have been present to balance the negative charge of the carbonate species, and the solution might have contained many additional cations and anions, but the problem statement provides no way to assess which ones or in what proportions. In such a situation, we can write a generic charge balance that acknowledges our limited information about the ionic composition of the solution. For this example, that charge balance would be H+
init
+
Â
all other cations i
ci |zi | = OH
init
+ HCO3
init
+ 2 CO23
init
+
Â
cj zj
all other anions j
(6.26) where ci and c j are the concentrations of the cations and anions, respectively, that are in the solution but are not represented explicitly in the rest of the equation, and |zi | and |z j | are the absolute values of the charges on those ions.5 The charge associated with the known inputs can be computed based on the information in the bottom portion of the tableau. For instance, the net charge associated with the inputs in both tableaus in Example 6.8 is 1.23 ⇥ 10 3 M, indicating that 5 The
summations
Â
all other cations i
ci |zi | and
Â
all other cations j
c j z j are commonly represented as CB and CA , respec-
tively, based loosely on the assumption that all unspecified cations and anions are monovalent salt ions that entered the solution as part of a strong base (B) like NaOH or a strong acid (A) like HCl. Equation (6.26) is more general and also clarifies that the summation is on charge, not mass.
326
the unknown constituents of the initial solution must carry a net positive charge of +1.23 ⇥ 10 3 M. The net charge associated with the unknown inputs is preserved as the solution equilibrates, so, at equilibrium, the unknown constituents will still carry a charge of +1.23 ⇥ 10 3 M. The known constituents in the equilibrated solution must therefore carry a charge of 1.23 ⇥ 10 3 M; that is, the charge on the known constituents is the same before and after the equilibration process. Although we have been referring to the charge-related equation that is used in the solution algorithm as a charge balance, it is more correctly described as a statement of charge conservation, establishing that the charge on the known input chemicals equals that on the known equilibrium species. The preceding discussion establishes that this equation is valid regardless of whether the total computed charge is balanced (yielding zero net charge) or not. Furthermore, we have seen that if we satisfy the charge conservation equation, we guarantee that the full charge balance equation will be satisfied as well, even though we do not know all the terms that would be included in that equation. Thus, we can use the charge conservation equation considering only known species and be assured of obtaining a correct result for the equilibrium speciation of the solution. A related observation is that, because all charges associated with the inputs are preserved when the inputs are expressed as an equivalent mixture of components, the net charge is reflected entirely in the T OT H equation if all the components other than H+ are neutral species. For instance, when component set #2 was used to analyze the solution in Example 6.8, the components other than H+ were the neutral compounds H2 O and H2 CO3 . As a result, all the charge associated with the inputs ( 1.23 ⇥ 10 3 M) appeared as the numerical constant in the T OT H equation.
6.6
IDENTIFYING DOMINANT SPECIES
We have now established that, to solve problems using log C-pH diagrams, (1) it is useful to develop a T OT H or PC equation that does not contain the concentrations of species that are dominant in the equilibrium solution, and (2) we can accomplish that by choosing components that are identical to the dominant species. We have also seen that, by choosing an arbitrary set of components, we can quickly identify the dominant species. The final issue we address is how we can identify the dominant species initially, without going through the process of developing a preliminary tableau and then revising it. In this section, a procedure is described for making good guesses about which species will be dominant at equilibrium, so that we can choose components that will allow us to solve acid/base problems graphically using the T OT H equation or PC directly. In addition to assisting in the choice of a good set of components for use with the graphical analysis, the procedure provides insight into the behavior of acid/base systems when those systems are titrated with other acids or bases. Such titrations, which are central to many water treatment processes and are frequently used in chemical analysis protocols, are discussed in detail in Chapter 8. 327
The basis for the procedure is the idea that equilibrium solutions have no “memory”. That is, once a solution has reached equilibrium, it is impossible to tell which chemicals were added, or the order in which they were added, to prepare the solution; all that one can say for sure is that all the species in the equilibrium system were added in some form. To be more explicit, if a solution contains 3 ⇥ 10 3 M HOCl, 1 ⇥ 10 3 M OCl– , and 1 ⇥ 10 3 M Na+ at equilibrium, we can be certain that 4 ⇥ 10 3 M T OT OCl was added to the solution, but not whether that T OT OCl was added entirely as HOCl (perhaps followed by some NaOH), as a mixture of HOCl and NaOCl (followed by a smaller amount of NaOH), or as a mixture of only HOCl and NaOCl. A corollary of the idea that equilibrium solutions have no memory is that we are free to imagine that the solution was concocted in any way we like and, as long as we account for all the species that are present in the actual solution, our imaginary concoction will behave identically to the real system. In fact, in the procedure about to be described, we will imagine that solutions can be prepared in a totally unrealistic way: by adding everything but the H+ ions first, then adding the H+ later. In reality, it is impossible to add individual ions to solutions; anytime we add an anion, we must add an equal and opposite amount of cationic charge. Nevertheless, this imaginary procedure proves useful for predicting the dominant species in the system, and because solutions have no memory, we can analyze systems as though they were prepared in this way without any risk of drawing incorrect conclusions. For the purposes of this procedure, the key attribute of acid/base systems that we will rely on is that, except at pH values quite near pKa , each conjugate acid/base pair is present almost entirely as the acid or almost entirely as the base. For instance, in a solution at pH 6 containing some OCl species (pKa = 7.53) and some Ac species (pKa = 4.76), almost all the OCl is protonated [i.e., T OT OCl ⇡ (HOCl)], whereas almost all the Ac is deprotonated [T OT Ac ⇡ (Ac )]. Recall that bases are sometimes called proton acceptors; the greater the tendency to accept a proton, the greater the strength of the base. Thus, the higher degree of protonation of the OCl species reflects the fact that OCl– is a stronger base than Ac– . Generalizing the behavior in the example system to other combinations of acid/ base groups, we make the approximation that the H+ ions added to a system combine sequentially with the various bases in the system, starting with the strongest base and working toward the weakest, and that weaker bases do not begin to protonate until stronger ones have become fully protonated. That is, although in reality all bases in a solution are at least slightly protonated at any pH, and no bases are 100% protonated, we treat them as though each base does go from 0% to 100% protonated before the next weaker base becomes protonated at all. Now consider what happens in the hypothetical procedure in which we add all 328
other chemicals to the solution before adding the H+ . In such a case, a concentration T OTi of each acid/base group is present initially, all as the most deprotonated species of the group. We then imagine adding the H+ in small increments. The first such increment will combine (in our idealized scenario) exclusively with the strongest base in the system. After we have added enough H+ to convert all of that base to its conjugate acid, protonation of the next strongest base begins, and the process repeats itself. Successively weaker bases become protonated until we have added all the H+ available. At that point, we can review the speciation of each acid/base conjugate pair and determine which species in each group is dominant. Consider, for example, a system made by adding 10 3 M HAc, 2 ⇥ 10 3 M NaAc, 3 ⇥ 10 4 M NaOCl, and 5 ⇥ 10 4 M NaOH to water. We could imagine that all the salt ions and bases were added first, before any H+ . At that point, the solution would contain 2.8 ⇥ 10 3 M Na+ , 3.0 ⇥ 10 3 M Ac– , 3.0 ⇥ 10 4 M OCl– , and 5 ⇥ 10 4 M OH– , and we would have 10 3 M H+ still “waiting” to be added. Listing the bases in the system from strongest to weakest (highest to lowest pKa ), we can fill in the first three columns in Table 6.5. Table 6.5 Hypothetical sequential protonation of bases in the example system.
Fraction Protonated
TOTi
Base/Acid
pKa
OH– /H2 O
14.00
5 ⇥ 10
4
OCl– /HOCl
7.60
3 ⇥ 10
4
Ac– /HAc
4.76
3 ⇥ 10
3
0% 100% 0% 100% 0% 6.7%
H+ Remaining 1.0 ⇥ 10
3
5.0 ⇥ 10
4
5.0 ⇥ 10
4
2.0 ⇥ 10
4
2.0 ⇥ 10
4
0
The 10 3 mol/L of available H+ is expected to combine first with the OH– , converting it to its conjugate acid, H2 O. This conversion requires 5 ⇥ 10 4 mol/L H+ , so once it is complete, we still have 5 ⇥ 10 4 mol/L H+ to add. Conversion of the OCl– to HOCl consumes 3 ⇥ 10 4 mol/L of these H+ ions, leaving 2 ⇥ 10 4 mol/L. These remaining protons combine with Ac– , but since T OT Ac is 3 ⇥ 10 3 M, the available H+ can convert only 6.7% of the Ac– to HAc. These steps are all summarized in the final two columns of the table. We therefore conclude that, at equilibrium, the dominant species of the weak acid/base groups will be HOCl and Ac– . 329
Example 6.9 Predict the acid/base species that will be dominant at equilibrium in a system prepared by combining 10 3 M HCl, 10 2 M NH3 , 10 2 M NaHCO3 , and 10 3 M butyric acid (HBut). Butyric acid is a four-carbon carboxylic acid, in the same chemical family as but larger than formic acid (one carbon), acetic acid (two carbons), and propionic acid (three carbons). It is the highly distinctive, foul-smelling constituent of stomach acid that can induce (or be an unpleasant consequence of) nausea. It also sometimes accumulates in anaerobic waste treatment systems that are not operating properly, to the dismay of all around. The pKa for HBut is 4.82, very close to that of its smaller analogs. Solution A table listing the acid/base conjugate pairs in the system in order of decreasing basicity along with the corresponding T OTi values is shown below. The two acid/base pairs of the carbonate system are listed independently, and water is included as a base that could be protonated to form H3 O+ . TOTi
Fraction Protonated
H+ Remaining
Base/Acid
pKa
– CO2– 3 /HCO3
10.33
10
2
NH3 /NH+4
9.24
10
2
HCO3 /H2 CO3
6.35
10
2
0%
0
But /HBut
4.82
10
3
0%
0
H2 O/H3 O+
0.00
55.6
0%
0
0% 100% 0% 20%
1.2 ⇥ 10
2
0.2 ⇥ 10
2
0.2 ⇥ 10
2
0
As before, we imagine that the solution initially contains only inert salt ions (10 3 M Cl– and 10 2 M Na+ ) and the bases that were added to the solution (10 2 M – + 2 NH3 , 10 2 M CO2– 3 , and 10 M But ). The concentration of H available to add to 2 this hypothetical solution is 1.2 ⇥ 10 M, consisting of the protons added with the NaHCO3 , HBut, and HCl. The strongest base in the solution is CO2– 3 , with a total concentration of 10 2 M. We assume that 10 2 M protons combine with the CO2– 3 to convert it all to HCO–3 , while all the other bases remain completely deprotonated. 330
This leaves 0.2 ⇥ 10 2 M H+ to combine with the other bases. The next strongest base is NH3 . It would take 10 2 M H+ to convert all the NH3 , to NH+4 , but only 0.2 ⇥ 10 2 M H+ is available. We therefore expect the available protons to convert approximately 20% of the NH3 to NH+4 . Correspondingly, approximately 80% of the NH3 and 100% of the HCO–3 and But– remain in the basic form, so the dominant species in the equilibrium solution are expected to be HCO–3 , NH3 , and But– . The whole process is summarized in the preceding table. Example 6.10 Which species would you expect to be dominant in solutions prepared by adding the following chemicals to water? (a) 5 ⇥ 10 (b) 10
4M
3M
2M
KH2 PO4 + 10
H2 CO3 + 10
4M
NaOH + 10
NH3 + 10
2M
2M
Na2 HPO4
HCl
Solution (a) The deprotonated forms of the acid/base groups in the solution are PO3– 4 and – OH , and the available proton concentration is [H+ ]avail = 2(5 ⇥ 10 3 ) + (1 ⇥ 10 2 ) = 2 ⇥ 10
2
The expected sequence of protonation reactions is summarized in the following table. The protons will bind first to the OH– (1 ⇥ 10 2 M) and will then 2– 2– convert two-thirds of the PO3– 4 to HPO4 , so we predict that HPO4 will be the dominant form of phosphate at equilibrium. Fraction Protonated
TOTi
H+ Remaining
Base/Acid
pKa
OH– /H2 O
14
10
PO34 /HPO24
12.38
1.5 ⇥ 10
2
HPO24 /H2 PO4
7.20
1.5 ⇥ 10
2
0%
0
H2 PO4 /H3 PO4
2.15
1.5 ⇥ 10
2
0%
0
0%
2
331
100% 0% 67%
2.0 ⇥ 10
2
1.0 ⇥ 10
2
1.0 ⇥ 10
2
0
(b) The available proton concentration in this solution is [H+ ]avail = 2(1 ⇥ 10 4 ) + 10
2
⇥ 10
2
The protons will bind to the bases in the system in the following sequence and quantities: 1 ⇥ 10 1 ⇥ 10 1 ⇥ 10
4M
– H+ to convert CO2– 3 to HCO3
4M
H+ to convert NH3 to NH+4
4M
H+ to convert HCO–3 to H2 CO3
As shown in the following table, after all these weak bases are converted to their fully protonated forms, the dominant species will be NH+4 and H2 CO3 , and 0.99 ⇥ 10 2 M H+ will still be available. These protons will combine with the next-weakest base in the system, H2 O, forming H3 O+ . The concentration of H2 O is 55.6 M, so clearly its capacity to acquire H+ will never be exhausted. We therefore conclude that the dominant species in this solution will be NH+4 , H2 CO3 , and (as always) H2 O. In addition, we can estimate that the final solution will contain 0.99 ⇥ 10 2 M H3 O+ , so the pH will be approximately 2.0. TOTi
Base/Acid
pKa
– CO2– 3 /HCO3
10.33
10
4
NH3 /NH+ 4
9.24
10
4
HCO3 /H2 CO3
6.35
10
4
H2 O/H3 O+
0.00
55.6
Fraction Protonated 0% 100% 0% 100% 0% 100% 0% 0.02%
H+ Remaining 1.02 ⇥ 10
2
1.01 ⇥ 10
2
1.01 ⇥ 10
2
1.00 ⇥ 10
2
1.00 ⇥ 10
2
0.99 ⇥ 10
2
0.99 ⇥ 10
2
0
A little thought leads to the conclusion that [H+ ]avail (i.e., the concentration of H+ available to protonate the bases in our hypothetical procedure) is identical to T OT Hin 332
for an analysis in which we choose water and the fully deprotonated bases as components. (As an exercise, confirm this statement for the initial solutions in Example 6.10.) Recognition of this fact is helpful in the analysis of acid/base titrations in Chapter 8. The method described above will almost always correctly identify the dominant species in a system. The only time it may fail is when there are several acid/base pairs in the system with nearly identical Ka values. In that case, when we try to evaluate the PC graphically, it will be evident which species are dominant, and we can revise the tableau and solve the problem. Having developed the basic procedures for drawing log C-pH diagrams, identifying the species expected to be dominant at equilibrium, and writing the T OT H equation or PC, we are in a position to determine the equilibrium composition of virtually any mixture of acids and bases. A couple examples of this process are provided next. In reading through the examples, it is worthwhile to mentally compare this approach with the numerical approaches presented in Chapter 5. To review, the steps involved in the graphical approach include: • Drawing the log C-pH diagram to understand how speciation varies with pH; • Assessing the strength of the bases in the system and comparing their concentrations with the concentration of available H+ to predict which species will be dominant at equilibrium; • Preparing the system tableau, using the species that are predicted to be dominant at equilibrium as components; • Writing the T OT H equation, and converting it to a PC; and • Adding curves to the graph for the LHS and RHS of the PC, and determining the equilibrium solution composition based on their intersection. Example 6.11 A model anaerobic waste solution is prepared as a mixture of 10 2 M NaHCO3 , 10 4 M Na2 S, and 4 ⇥ 10 3 M HAc. Upon dissolution, sodium sulfide (Na2 S) releases sulfide ion (S2– ), a strong base that can protonate to form bisulfide ion (HS– ) or hydrogen sulfide (H2 S). Hydrogen sulfide can enter the gas phase to generate a rotten egg smell that can be detected even at extremely low concentrations. Organic sulfides (R SH) also tend to have strong, unpleasant odors and are also generated in anaerobic environments, as well as being found in some industrial wastes (e.g., from pulping). The values of pKa1 and pKa2 for H2 S are 7.02 and 17.04, respectively. Determine the composition of the model waste solution at equilibrium.
333
Solution The species expected to be dominant at equilibrium can be determined by imagining the solution to be prepared by adding the salts and deprotonated bases (Na+ , 2– – + 2 CO2– 3 , S and Ac ) and then adding the available H , in this case equal to 10 M 3 from the NaHCO3 plus 4 ⇥ 10 M from HAc. Based on that value and the following summary table, the dominant species are likely to be H2 S, HCO–3 , and Ac– . Base/Acid 2–
pKa
–
Fraction Protonated
TOTi
S /HS
17.04
10
4
CO23 /HCO3
10.33
10
2
HS /H2 S
7.02
10
4
HCO3 /H2 CO3
6.35
10
2
Ac /HAc
4.76
4 ⇥ 10
H+ Remaining
0% 100% 0% 100% 0% 100% 0%
2
1.39 ⇥ 10
2
1.39 ⇥ 10
2
0.39 ⇥ 10
2
0.39 ⇥ 10
2
0.38 ⇥ 10
2
0.38 ⇥ 10
2
0
38% 3
1.40 ⇥ 10
0
0%
The system tableau using the expected dominant species as components is shown on the following page. The corresponding T OT H equation and PC are T OT H :
H+ (1) + OH +
PC:
CO23
( 1) + HS
( 1) + S2 4
( 2) + H2 CO3 (1)
( 1) + (HAc) (1) = (1 ⇥ 10 )( 2) + (4 ⇥ 10 3 )(1)
H+ + H2 CO3 + (HAc) = OH
+ HS
334
= 3.8 ⇥ 10
+ 2 S2
+ CO23
3
+ 3.8 ⇥ 10
3
335
1 0 0 0
2–
H2 CO3 0
HAc 0 0 0
NaHCO3 Na2 S HAc
Inputs
0
CO2– 3
S
HS
OH –
0
0
–
Ac
HCO–3 –
Na
0
0
+
H2 S
H
0
1
H2 O
+
H2 O
Species
Component:
1
2
0
1
1
1
2
1
1
0
0
0
0
1
0
H+
0
1
0
0
0
0
1
1
0
0
0
0
1
0
0
H2 S
0
2
1
0
0
0
0
0
0
0
0
1
0
0
0
Na+
0
0
1
0
1
1
0
0
0
0
1
0
0
0
0
HCO–3
Stoichiometric Coefficient
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
Ac–
4.76
10.33
6.35
24.06
7.02
14.00
0.00
0.00
0.00
0.00
0.00
0.00
log K
4.0 ⇥ 10
1.0 ⇥ 10
1.0 ⇥ 10
Conc’n
3
4
2
The log C-pH diagram is shown in Figure 6.12, indicating that the equilibrium pH is 6.57, at which the dominant acid/base species are HCO–3 , Ac– , and H2 S, as we anticipated. The composition of the equilibrium solution is (H+ ) = 10
6.57
(H2 CO3 ) = 10
(OH ) = 10 2.42
7.43
(HCO3 ) = 10
2.20
(HAc) = 10
4.21
(Ac ) = 10
2.41
(H2 S) = 10
4.13
(HS ) = 10
4.58
Figure 6.12 LogC-pH diagram for a system consisting of 10 and 10 4 M Na2 S.
(CO23 ) = 10 (S2 ) = 10
2M
5.96
15.10
NaHCO3 , 4⇥10
3M
HAc,
Example 6.12 Prepare system tableaus using the dominant species as components for each of the following solutions, and write the corresponding PCs. Then, construct a log CpH diagram for a system containing 0.10 M T OT PO4 and determine the equilibrium pH and the concentration of all species in the solutions. (a) 0.10 M NaH2 PO4 (b) 0.10 M Na2 HPO4 (c) 0.10 M Na3 PO4 336
Solution Because all three solutions have the same T OT PO4 , they can all be characterized by a single log C-pH diagram (Figure 6.13). However, we anticipate that the T OT H equations and PC’s will be different for each system, causing them to have different compositions at equilibrium.
Figure 6.13 LogC-pH diagram for a solution containing 10
2M
T OT PO4 .
Using the approach outlined in the preceding section, we surmise that H2 PO–4 , 3– HPO2– 4 , and PO4 are likely to be the dominant PO4 -containing species in parts (a), (b), and (c), respectively. (Carry out the calculation to convince yourself of this result.) System tableaus using the anticipated dominant species as components, the corresponding T OT H equation and PC, and the log C-pH diagram with curves added showing the right and left sides of the PC are provided on the following pages. Note that, in parts (b) and (c), the LHS of the PC contains terms that are 2 or 3 times the concentrations of particular species; i.e., the PC contains terms like 3(H3 PO4 ) and 2(H2 PO–4 ). As before, we incorporate these coefficients in the graphical analysis based on the following relationships: ⇥ ⇤ log 3 H3 PO4 = log 3 + log H3 PO4 = 0.48 + log H3 PO4 ⇥ ⇤ log 2 H2 PO4 = log 2 + log H2 PO4 = 0.30 + log H2 PO4
Thus, curves representing 3(H3 PO4 ) and 2(H2 PO4 ) are, respectively, 0.48 and 0.30 log unit above the (H3 PO4 ) and (H2 PO4 ) curves.
337
(a) 0.10 M NaH2 PO4 Stoichiometric Coefficient H2 O
H+
Na+
H2 PO–4
log K
H2 O
1
0
0
0
0.00
H
+
0
1
0
0
0.00
Na+
0
0
1
0
0.00
H2 PO–4 –
0
0
0
1
0.00
OH
1
1
0
0
14.00
H3 PO4
0
1
0
1
2.15
0
1
0
1
7.20
0
2
0
1
19.57
0
0
1
1
Component:
Conc’n
Species
HPO2– 4 3– PO4 Inputs
NaH2 PO4 T OT H :
H+ (1) + OH
( 1) + H3 PO4 (1) + HPO24
PC:
0.10
(H+ ) + H3 PO4 = OH
338
+ HPO24
( 1) + PO34 + 2 PO34
( 2) = 0
(b) 0.10 M Na2 HPO4 Stoichiometric Coefficient H2 O
H+
Na+
HPO2– 4
log K
H2 O
1
0
0
0
0.00
H
0
1
0
0
0.00
Na+
0
0
1
0
0.00
H2 PO–4 –
0
1
0
1
7.20
OH
1
1
0
0
14.00
H3 PO4
0
2
0
1
9.35
0
0
0
1
0.00
0
1
0
1
12.38
0
0
2
1
Component:
Conc’n
Species +
HPO2– 4 3– PO4 Inputs
Na2 HPO4 T OT H :
H+ (1) + OH
0.10
( 1) + H3 PO4 (2)
+ H2 PO4 (1) + PO34
PC:
(H+ ) + 2 H3 PO4 + H2 PO4 = OH
339
+ PO34
( 1) = 0
(c) 0.10 M Na3 PO4 Stoichiometric Coefficient H2 O
H+
Na+
PO3– 4
log K
H2 O
1
0
0
0
0.00
+
H
0
1
0
0
0.00
+
Na
0
0
1
0
0.00
H2 PO–4 –
0
2
0
1
19.57
OH
1
1
0
0
14.00
H3 PO4
0
3
0
1
21.72
0
1
0
1
12.38
0
0
0
1
0.00
0
0
3
1
Component:
Conc’n
Species
HPO2– 4 3– PO4 Inputs
Na3 PO4 T OT H :
H+ (1) + OH
0.10
( 1) + H3 PO4 (3)
+ H2 PO4 (2) + HPO24
PC:
(H+ ) + 3 H3 PO4 + 2 H2 PO4 + HPO24
340
= OH
(1) = 0
The equilibrium pH and concentration of all species are summarized in the following table. pH
6.7
log(H3 PO4 )
log(H2 PO–4 )
log(HPO2– 4 )
log(PO3– 4 )
(a)
4.69
3.54
1.00
3.51
11.20
(b)
9.74
11.14
3.54
1.00
3.64
(c)
12.58
17.23
6.80
1.42
1.21
COMPARISON OF APPROACHES FOR SOLVING ACID/BASE PROBLEMS
To conclude the chapter, we compare the various approaches that have been presented thus far for solving acid/base speciation problems. We saw in Chapter 5 that the problem of determining the equilibrium solution composition for a system with known inputs, or of determining the required input of a particular chemical to attain the desired composition (e.g., to reach a specified pH), could be formulated as one of evaluating a number of unknowns by solving an equal number of independent simultaneous equations. We also identified the unknowns as the concentrations of all the dissolved species at equilibrium, and showed that the mass balances on different ‘ion types’, equilibrium constants for formation of various species, and the charge balance (or, more generally, a statement of charge conservation) provided the required number of equations. We then solved those equations, either analytically or numerically, for several example problems. We also saw that, if the solutions were not ideal, the same approach could be used, but ‘effective’ equilibrium constants that accounted for the nonideality had to be used instead of conventional equilibrium constants. LogC-pH diagrams were also developed in Chapter 5, and we saw that those diagrams provide a concise visual representation of speciation over a wide pH range in systems where the total concentrations of acid/base groups and salt ions are known. It was emphasized that the diagrams contain all the same information as is used in the analytical or numerical solution of the problems, except for the charge balance. Therefore, the equilibrium speciation could be determined by identifying the unique pH where the charge balance was satisfied. The tableau format introduced in this chapter is, in one sense, just another way of organizing and presenting the input information for these types of problems. One could even argue that it adds complexity to the problem by requiring us to choose a group of components whose concentrations need to be determined; i.e., the equilibrium concentrations of the components are additional unknowns. Thus, formally, 341
the tableau formulates a problem with N components and Meq equilibrium species as having N + Meq unknowns. However, this complexity is more than balanced by the simple, consistent, and concise way in which the required equations are organized. In essence, each of the Meq rows in the upper portion of the tableau provides the information for an equation relating the concentration of one of the species to the concentrations of the components, and each of the N columns provides the information for a mass balance on a component. Thus, the tableau implicitly summarizes all N + Meq equations needed to solve the problem. The log C-pH diagram contains all the information that is in the tableau, except for the mass balance on H+ . Thus, we can determine the equilibrium condition by scanning the diagram to find the unique pH where the T OT H equation is satisfied. When we carry out this scanning exercise manually, it is convenient to convert the T OT H equation to a PC. Looking back, it should be evident that the modified CBs developed early in the chapter were, in fact, the PCs that are obtained when the dominant species are chosen as components. The most important difference between the equations included implicitly in the tableau and those used previously to solve these kinds of problems is that, in the tableau format, the charge conservation equation is replaced with a mass balance on H+ (i.e., the T OT H equation). The advantages of using the T OT H equation are that (1) by choosing the components wisely, we can be assured that the PC will allow us to identify the equilibrium pH directly, without additional steps, and (2) because the formulation of the T OT H equation is essentially identical to the formulation of the mass balances on the other components, it is more easily incorporated into an input algorithm for computer solutions to these kinds of problems; such solutions are introduced in Chapter 7. Thus, the T OT H equation has certain practical advantages over the charge balance (or charge conservation) equation for this particular application. However, we also saw that the charge balance is implicitly satisfied when the T OT H equation is solved, so we ultimately reach the same conclusion about the equilibrium speciation regardless of which equation is used.
6.8
SUMMARY
This chapter demonstrates the development and use of graphical approaches for analyzing acid/base systems. The graphs that are used in these analyses highlight certain universal features of such systems. In particular, they emphasize the fact that the acid form of an acid/base conjugate pair is present at a larger activity than the basic form at any solution pH less than pKa , and the reverse is true at pH’s greater than pKa . At a pH equal to the pKa , the acid and base are present at equal activities. If the assumption of ideal behavior of the solutes applies, then these same statements apply to the species’ concentrations. The species concentrations shown at any pH on a log C-pH graph satisfy the 342
mass balances on the acid/base groups that are present, the Ka expressions, and the Kw expression. Under the circumstances, only one additional, independent equation need be satisfied to establish the equilibrium composition of the system. The most common choices for this equation are the charge balance, the T OT H equation, and the proton condition, any of which can be used, and all of which can be derived from one another and the mass balances. The pH at which the additional equation is satisfied is sometimes difficult to identify directly on the log C-pH graph, but the difficulty can be overcome by writing a new equation that eliminates the largest value from each side of the equation. The resulting equation is, in effect, a mass balance on protons using a nonzero baseline concentration. This equation can be written directly by developing the T OT H equation using dominant species as components, and then converting that equation into a PC by moving terms so that each side contains only summations and not differences. A simple algorithm can be used to predict which acid/base species will be dominant once equilibrium is attained. This algorithm treats the system as though it could be prepared by first adding all the bases in their fully deprotonated forms, while withholding the H+ ions (i.e., the available protons). The protons are then added, and an approximation is made that they bind first to the strongest base in the system and then to progressively weaker ones, until all the protons that were present in the real system have been taken into account. The basis of this algorithm is that solutions have no memory, so that any approach that generates the same total inputs of species yields the same predicted equilibrium composition, regardless of how or in what order the chemicals were added. Once the dominant species have been identified, they can be chosen as components when developing the system tableau. In such a case, the T OT H and PC equations will not contain terms for the dominant species and will therefore be easy to use in conjunction with the log C-pH diagram to determine the equilibrium solution composition.
6.9
APPENDIX 6A: ARITHMETIC OPERATIONS AND LOG-LOG DIAGRAMS
One of the keys to using either a charge balance, TOTH equation, or proton condition in conjunction with a log C-pH diagram is recognizing that each side of the equation is often dominated by a single term, so that the summation can be approximated as that term alone. Although that situation is common, there are times when two or more terms must be considered over a fairly wide pH range. Most often, this complication arises when two or more lines on the diagram are parallel and fairly close to one another. In such a case, it might not be an acceptable approximation to ignore the contribution of the species represented by lower line to the summation. The complication that arises when we try to draw a line representing the sum of 343
the other two is that the values on the graph are logarithmic, while the summation is arithmetic. To compute the sum correctly, we need to take the anti logarithm of the log C values for the two lines, add these values, and then take the logarithm of this sum and plot it. The equation of the line representing the summation can be derived as follows. Let CU be the concentration represented by the upper line, and CL be the concentration represented by the lower line. Define x as the vertical distance between the two lines on a logarithmic scale, as shown in Figure 6.14. log CU = log CL + x
(6.27)
Figure 6.14 A logarithmic plot of the values of two individual terms (CL and CU) and the value of their sum.
Taking the antilogarithm of each side of the Equation (6.27), we obtain 10log CU = 10log CL+x log CL
10
= 10
(6.28)
log CU x
CL = CU 10
(6.29)
x
CU + CL = CU 1 + 10
(6.30) x
log (CU + CL) = log(CU) + log 1 + 10
(6.31) x
(6.32)
Thus, a line representing the logarithm of the sum of the two concentrations is displaced vertically from the upper line by an amount log (1 + 10x ). If x = 0, then the two lines are on top of each other, and the vertical displacement is log (1 + 1), or 0.3 units. This is the maximum displacement possible. If the lower line is one log 344
unit below the upper one, the shift is log (1.1), or 0.04 log units. This is about the smallest displacement we would logically be concerned about. Thus, if two lines are separated by less than one log unit, we would probably want to consider that the sum of the corresponding concentrations is somewhat larger than the value represented by the upper line alone. On the other hand, if the lower line is more than one log unit below the upper one, we can probably ignore it without affecting our conclusions substantially.
6.10
PROBLEMS
In solving these problems, use K a data from Table 5.1 and assume ideal solution behavior unless the question states otherwise or the ionic strength is given. If you prepare a log C-pH diagram, label the axes and the individual curves, and include a caption. The range of values on the axes will depend on the details of the problem, but they should typically be 8 to 14 pH units on the abscissa and 6 to 10 log units on the ordinate. 1. Using a graphical approach, solve for the pH of the following solutions.
(a) 10
3M
HAc
(e) 10
3M
HCN
(b) 10
3M
HOCl
(f) 10
3M
Na2 CO3
(c) 10
3M
NaCN
(g) 10
3M
NaHCO3
(d) 10
3M
Na2 Oxalate
(h) 10
3M
(NH4 )2 CO3
2. Prepare a log C-pH diagram for a solution containing 10 4 M chromic acid (H2 CrO4 ). Indicate the values of the slopes of the lines on the diagram. What is the predominant species of chromium in this solution, and what is its concentration, at pH 6? at pH 8? 3. What are the concentrations of all species in a solution of 10 10 2 M NaHCO3 ? 10 2 M H2 CO3 ?
2M
Na2 CO3 ?
4. Identify the dominant species and prepare a tableau using those species as components for the solutions described below. Prepare a log C-pH diagram and determine the equilibrium pH and the concentrations of all species in each system. (a) 10
3M
NaHCO3 +10
2M
HCl +10
2M
NaOH
(b) 10
3M
NaHCO3 +10
3M
HCl +10
2M
NaOH
(c) 10
3M
NaHCO3 +10
2M
HCl +10
3M
NaOH
345
5. Write the T OT H equation for each of the following systems, using the dominant species as components. (a) 0.02 M HCl (b) 10
3M
NaCN
(c) 10
3M
HOCl
(d) 10
4M
NaHCO3 +10
(e) 10
3M
NaCN +10
4M
3M
NH4 Cl
HOCl
6. Write the T OT H equation for the following systems. (a) 10 3 M Na3 PO4 +10 nent.
4M
Na2 HPO4 + H2 O, using PO3– 4 as a compo-
(b) Same system as (a), using HPO2– 4 as a component. (c) 10 3 M NH4 Cl +5 ⇥ 10 3 M HNO3 +2 ⇥ 10 dominant species as components.
3M
NaOH +H2 O, using
7. The effluent from an anaerobic digester is simulated as a mixture of 5⇥10 4 M each of NaH2 PO4 and Na2 HPO4 , and 6 ⇥ 10 4 M each of NH3 , H2 CO3 , and NaHCO3 . (a) Identify the species that you think will be dominant in each acid/base group in the equilibrated solution. (b) Write the T OT H equation for the system, using H2 O, H+ , H2 PO–4 , NH3 , HCO–3 , and Na+ as components. (c) Estimate the equilibrium pH, assuming ideal solute behavior. (You can sketch a log C-pH diagram if you’d like, but you need not do so to answer the question.) 8. The organic chemical ethylene diamine tetra-acetate (EDTA4– ) is used extensively in industrial processes to prevent metals from precipitating as solids. The pKai values for H4 EDTA are 2.11, 3.12, 6.27, and 10.95 for i equal to 1 through 4, respectively. (a) At what pH or over what pH range is the activity of HEDTA3– greater than that of H4 EDTA? (b) Write the expression and compute the value of Kb for the reaction in which H3 EDTA– serves as the conjugate acid. (c) Prepare a log C-pH diagram for a solution of 3 ⇥ 10 3 M T OT EDTA. What is the slope of the curve for H4 EDTA at pH = pKa3 ?
346
(d) A solution of 1 ⇥ 10 3 M Na2 H2 EDTA and 2 ⇥ 10 3 M Na3 HEDTA is prepared. Develop the tableau for the solution using the EDTA species that you expect to be dominant as a component. Write the T OT H equation, and determine the speciation of EDTA at equilibrium. 9. A solution is made by adding 2 ⇥ 10 4 moles of oxalic acid (which we will represent as H2 Ox), 5 ⇥ 10 4 moles Na2 HPO4 , and 5 ⇥ 10 4 moles CuCl2 to 1.0 L of pure water. (a) Prepare a log C-pH diagram showing the speciation of T OT Ox, T OT PO4 , and T OT Cu in the system. (b) H3 PO4 is added to the original solution until the pH is 5.5. Designating the dose of H3 PO4 added (in mol/L) as x, write out the T OT H equation for the solution, using the species that are expected to be dominant in the new solution as the components. 10. A wastewater from an electroplating process is at pH 2.5, due primarily to the presence of nitric acid, HNO3 , with pKa = 1.5. The solution contains 6 ⇥ 10 4 M T OT Cu, almost all of which is present as Cu2+ . The wastewater is to be neutralized using a solution that was originally prepared as 0.2 M NaOH. However, the NaOH solution has been exposed to the atmosphere for some time, and CO2 has dissolved into it, so that T OT CO3 is now 0.02 M. (a) Prepare a log C-pH diagram for a solution containing 0.02 M T OT CO3 and use it to determine the pH of the NaOH solution after the CO2 dissolves. (b) Add curves to the log C-pH diagram for Cu2+ and Cu(OH)2x x species for x 4. Write a T OT H equation for the waste solution after 30 mL of the alkaline, neutralizing solution has been added per liter of wastewater. Because the volume of titrant added is only 3% of the volume of wastewater solution, dilution of the wastewater can be ignored. What is the pH of the mixture? 11. A log C-pH diagram for a solution containing 10 is provided below.
3M
hydrogen sulfide (H2 S)
(a) Label the lines with the species they represent and assign correct values to the axes. (b) Draw lines for H+ and OH– on the diagram. (c) What is the pH of a solution of 10
4M
NaHS plus 9 ⇥ 10
4M
H2 S?
(d) Would a solution of 0.5 ⇥ 10 4 M Na2 S plus 9.5 ⇥ 10 4 M H2 S be more acidic, more alkaline, or the same as the solution in part (c)? Explain your answer in one or two sentences or equations. 347
12. Consider an acid H2 A with pKa1 = 5.5 and pKa2 = 9.5, in a solution with 10 2 M T OT A. (a) At what pH or in what pH range (if any) will the following conditions be met? (i) The concentration of A2– increases by approximately a factor of 100 for every increase of one pH unit? (ii) The concentration of HA– decreases by approximately a factor of 10 for every increase of one pH unit? (iii) The concentration of H2 A increases by approximately a factor of 10 for every increase of one pH unit? (iv) The ratio (H2 A)/(A2 ) decreases by approximately a factor of 100 for every increase of one pH unit? (v) The value of p(HA– ) is approximately 2.3? (vi) The value of p(H2 A) is approximately 4.0? (b) What is the pH of a solution of 0.005 M H2 A + 0.005 M NaHA? (c) Write the T OT H equation and determine the pH of a solution of 5 ⇥ 10 3 M Na2 A +5 ⇥ 10 3 M H2 A.
(d) Will a solution of 10
2M
NaHA be acidic, neutral, or alkaline?
(e) Write the equilibrium equation and the form and the value of Kb for HA– , i.e., HA– acting as a base. 13. The following figure is a log C-pH diagram for oxalic acid. (a) What is T OT Ox? What are pKa1 and pKa2 ? (b) Over what range of pH values is HOx– the predominant oxalate species? (c) Is H2 Ox a relatively strong or weak acid? 348
(d) If 10
2M
NaHOx were added to water, what would the solution pH be?
14. A log C-pH diagram for an acid/base system containing 10 2.7 M T OT A is shown below. The fully protonated form of the acid is H4 A. However, the curves are shown for only three species, because one of the potential Hn A4–n species is so unstable that its concentration in this system is always less than 10 10 M. The ‘s’ values indicate the approximate slopes of the curves in the given region, and the pH values indicate the intersection points of the various curves. (a) What is the equilibrium constant for the reaction H3 A )* HA + 2 H+ ? (b) What is the value of Kb for a reaction in which HA is the acid?
15. Nickel can combine sequentially with three OH– ions to form ‘nickel hydroxo complexes’ via the following reactions. 349
Ni2+ + OH– )* NiOH–
log K = 4.60
Ni2+ + 2 OH– )* Ni(OH)2
log K = 8.96
Ni2+ + 3 OH– )* Ni(OH)–3
log K = 11.98
(a) Combine these reactions with Kw so that they have the form of Ka ’s. Determine which species is dominant at any given pH in the range from 0 to 14. (b) Prepare a log C-pH diagram for a solution containing 10 2 M T OT Ni in the pH range 6 to 13 and for log C values from 2 to 8. (Hint: the diagram might be difficult to draw at first, because the acidity of various nickel species is unusual in a way that will be apparent from the Ka values. Despite the unusual features, the diagram must reflect the basic equations that characterize any acid/base system: the mass balances and equilibrium constants. On the other hand, there is nothing fundamental about the “look” of graphs that you have become accustomed to. If you are having difficulty drawing the graph, use a spreadsheet to compute the a values of the different species over a range of pH values, and plot the results. Once you have drawn the graph, you will be able to see how and, hopefully, why the appearance of this graph is slightly different from that of most log C-pH diagrams you have seen previously.) (c) What is the pH of a solution of 10
2M
Ni(NO3 )2 ?
16. Chromic acid (H2 CrO4 ) is a diprotic acid which has important uses in industrial processes and in preparation and analysis of environmental samples. Under certain conditions it can “dimerize” according to the following reaction: 2 HCrO )* Cr2 O27 + H2 O
log K = 1.54
The product shown is called dichromate ion and is the deprotonated form of a strong diprotic acid. The acidity constants for chromate are pKa1 = 0.20 and pKa2 = 6.51, and those for dichromate are pKa1 < 0; pKa2 = 0.07. Find the pH and composition of the following two solutions: (a) 0.2 M Na2 CrO4 + H2 O (b) 0.2 M NaHCrO4 + H2 O (Hint: Because of the dimerization reaction, the total molar concentration of dichromate species in a system where most of the Cr is dimerized is different from the total molar concentration of chromate in the system when most of the Cr is present as the monomer. Write out the mass balance and equilibrium 350
equations as the first step in your analysis. Then either solve those equations or develop a log C-pH diagram by substituting into the mass balance on Cr until it contains H+ and one Cr containing species as the only variables. You can then solve that equation at various pH values and use the results to determine the concentrations of the other Cr containing species, so that the complete diagram can be drawn. Note that only one Cr-containing species can be chosen arbitrarily as a component for the T OT H equation.) 17. Citric acid is triprotic (H3 Cit), with acidity constants pKa1 = 3.13, pKa2 = 4.76, and pKa3 = 6.40. (a) Write out the expressions for a0 , a1 , a2 , and a3 as a function of the Ka values and (H+ ). (b) Identify the region where each citrate species is dominant. (c) Determine which terms in the denominators for the a values are significant at pH 7.5 (terms contributing less than 5% to the summation can be considered negligible). For a solution with 10 1 M total citrate, write out an approximate equation of the form: [log C = constant + n pH] for each citrate species at pH 7.5, approximating the dominant species’ concentration as T OT Cit. (d) Draw a log C-pH diagram for a solution with 10 pH range 1.0 < pH < 8.0.
1M
T OT Cit over the
(e) Find the pH of the following solutions: (i) 10 1 M H3 Cit + water (ii) 10 1 M Na2 HCit + water (iii) 0.05 M Na3 Cit + 0.05 MNaH2 Cit + water 18. A wastewater is simulated as an ideal solution containing 10 and 10 3 M NH4 Cl.
2.7
M NaHCO3
(a) Prepare a log C-pH diagram for the system. (b) Predict the dominant species in the equilibrium solution, and prepare a system tableau using those species as components. (c) Write the T OT H equation and use it to determine the pH of the solution. (d) The utility hopes to remove nitrogen from the solution by driving out NH3 gas. To prepare for that step, 148 mg/L Ca(OH)2 (hydrated lime) will be added to raise the pH and convert NH+4 to NH3 (aq). Calculate the pH after lime addition, but before any NH3 exits the solution.
351
19. Acid drainage from coal mines can make the receiving waters uninhabitable by fish. To remediate one such situation, a consultant has recommended dissolving limestone, CaCO3 , in the drainage water to neutralize the acid. The pH of the untreated water is 2.40 due to the following overall reaction: FeS(s) + 2.25 O2 (aq) + 2.5 H2 O ! Fe(OH)3 (s) + SO24 + 2 H+ + (Note: This reaction is shown only to illustrate how the SO2– 4 and H enter the water. The following questions focus just on the concentrations of these ions in solution; you need not consider the other chemicals in the reaction when answering the questions.)
(a) Determine the concentration of T OT SO4 in the drainage water using manual calculations. Use the Davies equation to model activity coefficients. Is HSO–4 a strong acid in this solution? (Hint: write out the mathematical expressions that relate the concentrations of H+ , HSO–4 , and SO2– 4 to known values and/or each other. Some of these relationships will include activity coefficients. Then, make a guess for the ionic strength of the solution, compute the activity coefficients, and solve for the species concentrations. Finally, test whether the ionic strength based on those concentrations is consistent with your guess. If not, iterate on the guess until it converges with the computed value.) (b) Prepare the system tableau for the solution after it has been neutralized to pH 7.0 by addition of CaCO3 , and write the corresponding T OT H equation. Use the expected dominant species as components. (c) What is T OT Ca in the stream after the pH has been raised to 7.0? As in part (a), use the Davies equation to estimate activity coefficients.
352
7
SOFTWARE FOR SOLVING CHEMICAL EQUILIBRIUM PROBLEMS
353
Contents 7.1 7.2 7.3 7.4 7.5 7.6 7.7
7.1
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . THE COMPUTER ALGORITHM FOR SOLVING A SIMPLE ACID/BASE PROBLEM . . . . . . . . . . . . . . . . . . . . . . . SYSTEMS IN WHICH THE EQUILIBRIUM ACTIVITIES OF ONE OR MORE SPECIES ARE KNOWN . . . . . . . . . . . . . . . . USING CHEMICAL EQUILIBRIUM SOFTWARE TO DEVELOP LOG C pH DIAGRAMS . . . . . . . . . . . . . . . . . . . . . . MODIFYING THE COMPONENT AND SPECIES DATABASES IN VISUAL MINTEQ . . . . . . . . . . . . . . . . . . . . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
354 355 363 369 373 375 375
INTRODUCTION
In the preceding chapters, numerical and graphical approaches for determining the equilibrium composition of solutions containing acid/base groups have been presented. The types of problems that have been analyzed can be divided into two broad categories. In one, the concentrations of all chemicals used to prepare the system are known, and the final equilibrium concentrations are sought. In the other, at least one of the final conditions is known (e.g., pH), and we wish to infer something about the chemical inputs needed to get to that point (e.g., how much acid or base must be added to adjust the solution to the known final pH?). The approach for solving either type of problem involves identifying the unknowns in the system, writing a set of independent equations that relate the unknowns to one another, and solving the equations simultaneously. In all cases, the relevant equations include mass balances, chemical equilibrium expressions, and a charge balance, proton condition, or T OT H equation. The consistent way in which the equations characterizing these systems are formulated makes them ideally suited for a computerized solution algorithm, and such 354
algorithms have been available since at least the 1970s. Although the applications of the software have expanded greatly over the years and the interfaces available to run it have become more user-friendly, the underlying solution approach has hardly changed at all. Basically, the user inputs information about the types of chemicals in the system and the total mass input or equilibrium activity of any chemicals for which that information is known. Information about the relevant equilibrium expressions is then either input manually or retrieved from a database that the program can access. Finally, the program initiates a numerical routine that is carried out iteratively until all the equations are solved to within an acceptable tolerance. A number of programs with various user interfaces are in common use today. The software that is used as the default program for chemical equilibrium calculations in this text is Visual Minteq, which is freeware available for downloading from the web.1 This program is a widely used member of the MINEQL family of programs, which includes MINEQL, MINTEQ, and MINEQL+ R , among others. The central organizational element of this family of software is the tableau that was introduced in Chapter 6. While this chapter does not delve into the details of how the software manipulates the inputs to solve for the equilibrium solution composition, it does provide an outline of how the programs work, so that you, the user, can better understand both the potential applications and the limitations of the software. To help readers become familiar with the program capabilities, some program-specific instructions are provided and some input and output screens are shown, despite the certainty that some of these details will evolve more rapidly than the text can be updated. Readers are strongly encouraged to explore the program on their own and to utilize the online tutorials and other help features embedded in the program to get more up-to-date information.2
7.2
THE COMPUTER ALGORITHM FOR SOLVING A SIMPLE ACID/BASE PROBLEM
As is the case when we solve a chemical equilibrium problem manually, an essential step when we use a chemical equilibrium software package is to identify all the species expected to be present in the equilibrium system. Recall that these species can be ‘made’ using various groups of components. If the problem is to be solved manually, we can choose any set of components that meet the criteria described in 1 The
examples and screenshots from Visual Minteq shown in this text were run using version 3.1. The latest version of the software can be downloaded from : http://vminteq.lwr.kth.se. 2 In each of the remaining chapters of this text, one or more sections are devoted to a discussion of how Visual Minteq can be used to solve the types of problems covered in that chapter. Readers are encouraged to visit the Visual Minteq Help page to familiarize themselves with updates to the program before reading the corresponding sections.
355
Chapter 6 (specifically, that components can be combined to form all species in the equilibrated system, but that no component can be formed strictly by combining other components). In addition, we saw in Chapter 6 that it is advantageous to choose the dominant species as components if the problem is to be solved in conjunction with a log C pH graph. Because numerical processing can be carried out with far greater precision than a graphical analysis, there is little advantage to be gained by choosing dominant species as components when the problem is solved numerically, and most software packages have a preselected set of components that are used regardless of which species are expected to be dominant. Consider, for example, the simple system containing 2 ⇥ 10 3 M HAc that was evaluated in Section 6.5. The tableau that we developed for this system using H2 O, H+ , and Ac– as components was shown in Table 6.4 and is reprinted here as Table 7.1. For convenience, we assume that the solution is ideal, so the activities of the solutes equal their molar concentrations, and the activity of H2 O is 1.0. Table 7.1 Tableau for 2 ⇥ 10
Component:
3M
HAc, using H2 O, H+ , and Ac– as components
Stoichiometric Coefficient H2 O H+ Ac–
Species H2 O H+ Ac– OH– HAc
1 0 0 1 0
0 1 0 1 1
0 0 1 0 1
Inputs H2 O HAc
1 0
0 1
0 1
log K
Conc’n
0.00 0.00 0.00 14.00 4.76
[H2 O] (H+ ) (Ac– ) (OH– ) (HAc) 55.6 2 ⇥ 10
3
The equilibrium activity of any species can be computed based on the equilibrium constant for forming that species from the components and the components’ activities. All the information needed to carry out that calculation for a given species is shown in a single row in the upper portion of the tableau. For example, the coefficients in the HAc row (i.e., the bottom row of the upper part of the tableau) indicate that the activity of the species HAc can be computed as log(HAc) = 0 log(H2 O) + 1 log(H+ ) + 1 log(Ac– ) + 4.76
(7.1)
Thus, if we made guesses for the equilibrium activities of the three components, the corresponding equilibrium activity of HAc could be calculated. For instance, if 356
our guesses were (H2 O) = 1.0, (H+ ) = 10 activity of HAc would be: log(HAc) = 0 ⇥ log 1.0 + 1 ⇥ log 10
8.0
8.0 ,
and (Ac– ) = 10
+ 1 ⇥ log 10
3.0
3.0 ,
the computed
+ 4.76 = 6.24
(7.2)
Similar calculations can be carried out for the other four species, and the results can be incorporated into the final column in an updated version of the tableau (Table 7.2). Table 7.2 Summary of the computed system composition for the specified initial guesses of the components’ activities
Component:
Stoichiometric Coefficient H2 O H+ Ac– 0.0⇤ 8.00⇤ 3.00⇤
Species H2 O H+ Ac– OH– HAc
1 0 0 1 0
0 1 0 1 1
0 0 1 0 1
Inputs H2 O HAc
1 0
0 1
0 1
log K
Conc’n†
0.00 0.00 0.00 14.00 4.76
55.6 10 8.00 10 3.00 10 6.00 10 6.24 55.6 2 ⇥ 10
3
⇤
Initial guess of log ai , where i is the corresponding component
†
Concentrations shown in the upper part of the tableau are computed based on the current guesses
of ai , assuming all solutes behave ideally. These guesses are updated in each iteration until the mass balances are satisfied, at which point they are the final, equilibrium values. Water is treated specially by the programs, so that an activity of 1.0 is always associated with a concentration of 55.6 M.
Similarly, all the information needed to write the mass balance equations that characterize the system is shown in the columns of the tableau. For instance, the coefficients in the Ac– column imply that the mass balance on that component is ?
T OT Aceq = T OT Acin Meq
j=1
Min
?
 ni, j c j =
 ni,k ck
k=1
357
(7.3a) (7.3b)
0 [H2 O]eq + 0 [H+ ]eq + 1 [Ac ]eq + 0 [OH ]eq + 1 [HAc]eq ?
= 0 [H2 O]in + 1 [HAc]in
(7.3c)
where the subscripts i, j, and k refer, respectively, to the components, the species in the equilibrium solution, and the input chemicals used to prepare the system. For the given guesses of the component activities, the computed value of T OT Aceq is only about one-half of T OT Acin (0 ⇥ 55.6) + (0 ⇥ 10
8.0
) + (1 ⇥ 10
3.0
) + (0⇥10
6.0
) + (1 ⇥ 10
6.24
?
)=
(0 ⇥ 55.6) + (1 ⇥ (2 ⇥ 10 3 ))
1.0006 ⇥ 10
3
6= 2 ⇥ 10
3
An analogous calculation shows that the mass balance on H+ (i.e., the T OT H equation) is also not satisfied for the given component activities; the mass balance on H2 O is approximately satisfied, but only because the concentration of H2 O (55.6 M) overwhelms the other terms in the summation for both the input and equilibrium solution. These results indicate that the guessed values of the components’ activities are incorrect, so the computed species’ concentrations do not represent the final equilibrium composition of the solution. If we wished to continue using this algorithm to solve manually for the equilibrium solution composition, we would next attempt to make better guesses for the component activities, using the errors in the mass balances to guide our choices. This is essentially what the software does, using conventional, goal-seeking approaches to simultaneously improve the guesses for the activities of all the components. The steps in the overall process, including both those carried out by the user and by the software, can be summarized as follows: 1. Based on the known chemical inputs, identify an acceptable set of components. If the software has a preselected set of available components, choose from that list if possible. (All programs include the option for the user to add new components if necessary.) 2. By representing the input chemicals in terms of the components, determine the total input of each component to the system; i.e., for each component X, determine T OT Xin . 3. Make an initial guess for the equilibrium activity of each component. Often, the software will make such guesses by default if the user does not input specific guesses. 4. Compute the activities of the species, based on the guessed activities of the components. 358
5. Represent the computed concentrations of the species as an equivalent mixture of components. 6. Determine whether the total amount of each component embedded in the species (T OT Xeq , per Step 5) equals that in the input (T OT Xin , per Step 2). If so, the computed solution composition is the equilibrium composition, and the problem has been solved. If not, use numerical techniques to improve the guesses for the component activities, and repeat Steps 2-6 until the guesses cause all the mass balance equations to be satisfied to within an acceptably small error. Note that, if the equilibrium activity (or concentration) of a species is known in advance, the initial and revised guesses for the activities of the components are constrained to assure that the computed activity of that species equals the known value. For example, if we want to specify that the equilibrium activity of the species H2 CO3 is 10 5.0 , we make sure that the guesses for the components’ activities are always consistent with this desired outcome. Typically, the constraint that (H2 O) = 1.0 is included in the software by default. The solution procedure is characterized schematically in Figure 7.1. Once the components are selected, most programs (including Visual Minteq) query an embedded database to identify all species that might form and the corresponding equilibrium constants for forming those species from the components. Therefore, the user need not provide that input unless a species is to be considered that is not already in the database, or the user has updated information for, e.g., a log K value. The preceding algorithm is applicable if the activity coefficients of all components and species are 1.0. If nonideal solution behavior is to be considered, the ionic strength can be computed at the end of each iteration, activity coefficients can be computed according to some preselected model (e.g., the Davies equation), and an appropriate adjustment can be made to account for the nonidealities. By default, Visual Minteq uses the Davies equation to compute activity coefficients for ionic solutes. However, options are provided to choose other commonly used equations for this calculation on the “Various default settings” page, which is accessed via the “Parameters” pulldown menu on the home page. The program uses Equation (2.8) with kneut = 0.10 to compute activity coefficients of neutral species. Example 7.1 Use Visual Minteq to find the equilibrium solution composition of 2 ⇥ 10 3 M HAc. What is the activity coefficient for Ac– in the solution, according to the Davies equation? Solution The tableau for this problem has already been developed and is provided in Table 7.1. To solve for the solution composition using Visual Minteq, install the pro359
Figure 7.1 Flowchart of calculations used by many computer software packages to compute equilibrium speciation in a system containing only soluble species, and assuming ideal solution behavior.
gram on your computer and click the “Show organic components” box on the opening screen. Then pull down the “Component name” list and select acetate ion (shown in the list as Acetate-1). The input concentration of T OT Acetate-1 can be specified using various units, as shown in the “Concentration unit” box. Leaving the default choice of molal units, type 0.002 in the “Total concentration” box. (Recall that molal units correspond to moles per kg of water which, for our purposes, is essentially identical to molar units [moles per liter of solution].) Click the “Add to list” button to accept 0.002 M Acetate-1 as an input to the system. Leave the other options at their default values, including • Have the program calculate the equilibrium pH (as opposed to specifying a known final pH) 360
• Have the program calculate the final ionic strength (as opposed to specifying a known final value) and use the Davies equation to calculate activity coefficients. (Different choices for the equation used to calculate activity coefficients can be made from the Parameters/Various Default Settings pull-down menu.) • T = 25 C
Repeat the above process to add 0.002 M H+ as an input.3 Thus, the real input of 0.002 M HAc is being input as an equivalent combination of components (0.002 M Acetate-1 plus 0.002 M H+ ). Since these are the only two inputs of components for this simple problem, click the “View/edit list” button to open a secondary screen that summarizes the inputs; errors in the inputs can be corrected here or back on the home screen. Assuming the inputs are what you intended, click on the “Species tableau” button to see the tableau. The tableau, shown below as Figure 7.2, contains similar information to that in Table 7.1, but differs in a few ways. First, the first column lists only species that are chemically different from the components. This choice is simply for compactness; all the components are implicitly included as species, with each species made up of one unit of that component and nothing else. Second, a column has been added for the component H2 O, which is assumed to be present in all cases. Finally, a column shows the standard enthalpy of reaction (DH r , shown in the tableau as delta Hr ) for the reaction forming each species from the components. Recall from Chapter 4 that this value can be used to adjust the equilibrium constant to account for temperature, if the system temperature is different from the standard value (25 C).
Figure 7.2 Visual Minteq tableau for a solution of 2 ⇥ 10
3M
HAc
At this point, it is useful to save the input file, using the File pull-down menu on the home screen. (For such a simple problem, saving is not particularly important, but for problems with many inputs, saving the file speeds the process of making adjustments to the inputs later, if desired.) Then, the file can be run by clicking the “Run MINTEQ” button. Once the calculations converge, a screen appears H+ is present in all aqueous solutions, it is included as a component by default and will appear in the component list even if it is not selected manually. However, one does have to input the value of 0.002 M for T OT Hin , and that is most easily done by selecting H+ as a component in the same way that Acetate-1 is selected. 3 Because
361
showing a summary of the computed solution composition, including the concentration and activity of each species, the solution pH (which, remember, is log aH+ , not log cH+ ), and various other pieces of information (Figure 7.3). For this example, the equilibrium pH is 3.749, and the concentrations of HAc and Ac– are 1.819 ⇥ 10 3 M and 1.812 ⇥ 10 4 M, respectively. The final ionic strength is only 1.81 ⇥ 10 4 , because most of the acetate in the solution is present as the neutral species HAc. The “View species distribution” button opens a screen that shows the distribution of T OT Acetate-1 on a percentage basis; the other buttons provide information that is not useful for this problem, but will become useful in future problems.
Figure 7.3 Visual Minteq output for a solution with 2 ⇥ 10
3M
HAc as the only input
Because the standard-state concentration of Ac– is 1.0 M, and we are assuming that molar and molal concentrations are equivalent, the activity coefficient of Ac– can be computed as gAc =
aAc cAc /cAc
=
1.784x10 4 = 0.985 1.812x10 4 M/1.0 M
Recall that, by making appropriate assumptions about the “missing” species, the equilibrium composition of a solution can be determined even if not all the inputs are known, and if the known inputs do not satisfy the criterion of zero net charge. We saw in Chapter 6 that, in such a case, we could solve for the equilibrium composition by assuming that the unknown inputs are inert salts (like Na+ or Cl– ) and satisfying the charge conservation equation (i.e., the criterion that the charge imbalance is the same after equilibrium is reached as it was based on the known inputs). In Visual 362
Minteq, this approach is implemented by default, in which case the pH is said to be “Calculated from mass balance” (i.e., the mass balance on H+ , expressed by the T OT H equation). Note that this approach leads to a slight internal inconsistency, in that inert salts are assumed to be present to neutralize the charge, but those salts are not considered in the calculation of ionic strength or activity coefficients. An alternative option for such a situation is to assume that the charge imbalance is attributable to unknown inputs of H+ and/or OH– , rather than unknown inputs of salt ions. Consider, for example, a solution that is known to contain 10 3 M T OT Ac and 1.5 ⇥ 10 3 M T OT Na, but that other inputs are unknown. If the charge imbalance were attributable to inert salts, Visual Minteq would be run using the default option, and the equilibrium solution composition would be reported to have the same charge imbalance as the inputs (a net charge of +5.0 ⇥ 10 4 M). However, if the charge imbalance were attributed to H+ and OH– , the input would be modified to include 5.0 ⇥ 10 4 M T OT H. The computed, equilibrium composition would therefore have a substantially higher pH than if the default option were used. The option to balance the charge using H+ /OH– is implemented by specifying that pH is “Calculated from mass & charge balance” on the main menu. If the charge balance is satisfied by the specified inputs (i.e., if the net charge from the inputs is zero), then the program output is identical regardless of which option for calculating pH is utilized. To assist the user in determining whether the inputs are charge-balanced, an option is provided to have the program report the total positive and negative charge in solution; this option is activated by clicking the “Show initial charge balance” entry in the Parameters pull-down menu.
7.3
SYSTEMS IN WHICH THE EQUILIBRIUM ACTIVITIES OF ONE OR MORE SPECIES ARE KNOWN
If the final concentrations of one or more species are known in advance, we can use Visual Minteq to solve for the equilibrium speciation using almost the same procedure as described in the preceding section, with minor variations. For ideal solutions with all inputs known, a system that can be characterized by N components and Meq species has N + Meq unknowns (the concentration of each component and each species), and we can find the values of all those unknowns by solving the corresponding N mass balances and Meq equilibrium constant expressions. If the solution is not ideal, an additional N + Meq + 1 unknowns (the activities of the components and species, and the solution ionic strength) appear, along with N + M equations for the activity coefficients and one equation for computing the ionic strength from the solution composition. 363
The elimination of one of the unknowns in such a problem is unavoidably associated with the appearance of a different one. For instance, if the final pH is specified, the activity of H+ in the equilibrium solution is no longer an unknown, but the input concentration of some acid or base becomes a new unknown. Consider, for example, a solution of 1.6 ⇥ 10 3 M HAc plus 0.4 ⇥ 10 3 M NaAc that is then adjusted to pH 7.5 by adding NaOH. Using H+ , Ac– , and Na+ (and, implicitly, H2 O) as components, we could prepare the system tableau shown in Table 7.3. A few changes have been made to make the tableau more concise (the labels have been removed, the designations for the species concentrations at equilibrium have been deleted, and the final two columns have been merged), but all the essential information for setting up the problem remains. Table 7.3 Tableau for 1.6 ⇥ 10 3 M HAc plus 0.4 ⇥ 10 and Na+ as components
3M
NaAc at pH 7.5, using H+ , Ac– ,
H+
Ac–
Na+
H+ Ac– Na+ OH– HAc
1 0 0 1 1
0 1 0 0 1
0 0 1 0 0
0.00 0.00 0.00 14.00 4.75
HAc NaAc NaOH
1 0 1
1 1 0
0 1 1
1.6 ⇥ 10 0.4 ⇥ 10 x
3 3
The tableau has the same kinds of entries as previous tableaus we have considered, but the input concentration of NaOH is listed as x, since we do not know how much NaOH was needed to prepare the solution. The same types of equations that have been used to solve previous problems – in this case, three mass balances on the components and four equilibrium expressions for computing the activities of the species from those of the components – are still available to solve for the equilibrium composition. The only difference from previous problems is that, in this problem, the amount of NaOH added (x) replaces the equilibrium activity of the species H+ as an unknown. Thus, we still have seven unknowns and seven equations (or, if the solution is not ideal, 15 unknowns and 15 equations), all but one of which are identical to those applicable in a system in which the inputs are all known. The problem can therefore be solved. As has been noted, virtually all chemical equilibrium programs include a feature to specify that the equilibrium activity of H2 O is always 1.0. In addition, Visual Minteq allows the user to specify the activity of any other species in the system. 364
Two approaches are available in the program for accomplishing this goal. One is applicable if the desired activity is achieved by adding that species to solution (e.g., if the desired activity of H+ is reached by adding H+ to the solution), and that approach is presented next. If the desired species activity is achieved by adding some other input species (e.g., if a certain H+ activity is reached by adding HAc to the solution), a different modeling approach must be used; that approach is presented in Chapter 8. Say, for example, that we have prepared a solution of 10 4 M NaAc, and we want to know how much HAc must be added to it to obtain a solution in which the equilibrium activity of HAc is 0.001 M. We can model such a process in Visual Minteq by selecting Na+ , Acetate-1 and H+ as components, and specifying appropriate concentrations of those components to account for the known inputs. In this case, the known inputs are 10 4 M each of Na+ and Acetate-1, and zero H+ . We then specify the known activity of HAc by clicking the button for a “Fixed activity” input on the program’s home page. This action changes the name of the pulldown menu on the left from “Component name” to “Species name”. When we look at the list of names in that menu, we see the species (H Acetate, Na Acetate, OH– , and Na OH) that can be made from the components that have already been specified. By selecting H Acetate and inputting a fixed activity of 0.001, we indicate that the program should add HAc (i.e., the components H+ and Acetate-1, in a 1:1 ratio) until the equilibrium activity of the species H Acetate is 0.001. The key part of the screen at this point is shown in Figure 7.4.
Figure 7.4 Input screen for specifying a fixed activity of 0.001 for the species HAc
Clicking the “View/edit list” button takes us to the screen showing the list of components, from which we can click other buttons to display the species tableau or the “List of fixed species”, where H Acetate appears with the desired fixed activity of 0.001. Returning to the home page, we can run the program; the equilibrium solution composition appears on the main output page (Figure 7.5a) and shows that the final activity of H Acetate is indeed 0.001. By clicking the “Equilibrated mass distribution” button on this page, we can see the total concentrations of all the components (i.e., T OT Xin or T OT Xeq ) in the system (Figure 7.5b). These total concentrations account not only for the NaAc that was present initially, but also for the HAc that was added in the simulation in order to bring the HAc activity to the desired value. The concentration of HAc added can 365
then be calculated offline based on the mass balance on either Acetate-1 or H+ , as follows: [HAc]added = [Acetate-1]eq
[Acetate-1]in,init = 1.934 ⇥ 10 = 9.34 ⇥ 10
[HAc]added = [H+ ]eq
[H+ ]in,init = 9.34 ⇥ 10
5
4 5
1.0 ⇥ 10
0 = 9.34 ⇥ 10
4
5
Figure 7.5 Output screens for a system containing 10 4 M NaAc, to which HAc is added until {HAc} = 0.001. (a) Overall summary of solution composition; (b) Equilibrated mass distribution, showing the total concentrations of the components at equilibrium.
In the preceding example, the species whose activity we wanted to specify (H Acetate) was not also a component. The activity of a species that is a component can be specified following similar procedures, but in this case the value of the fixed activity is input when the component is first selected. Since the program will adjust the total input concentration of the component to force the species to have the desired activity at equilibrium, the component’s initial input concentration is irrelevant and cannot be specified. Thus, for instance, if we wished to determine how much strong acid (e.g., HCl) would be required to adjust a solution of 10 4 M HAc to pH 4.0, we would input 0.0001 M Acetate-1 and select H+ as a second component, but we would click the “Fixed activity” button when inputting H+ and type 1e 4 into the box. No other input is needed. We then run the program, generating the output shown in Figure 7.6. (Note that, if this approach is used, the Cl– that is added with the 366
H+ is ignored. If we wish, we can compute the input of Cl– offline after the program has run. Even if we do that, though, the contribution of Cl– to the ionic strength is not considered within the program’s calculations, so the ion activity coefficients and the final composition computed by the program are slightly in error. In the rare case where this error is considered unacceptable, an alternative approach that simulates a titration of the solution with HCl can be used. This approach is described in Chapter 8.)
Figure 7.6 Output screens for a system containing 10 3 M HAc which is then adjusted to pH 7.0 by addition of strong base. (a) Overall summary of solution composition; (b) Equilibrated mass distribution.
Once again, we can compute the amount of reagent added by writing the mass balance, this time on H+ : [HCl]added = [H+ ]eq
[H+ ]in,init = 9.51 ⇥ 10
4
1.00 ⇥ 10
4
= 8.51 ⇥ 10
4
Of the 9.507 ⇥ 10 4 M T OT H in the equilibrium solution, 1.013 ⇥ 10 4 M is present as free H+ (i.e., H3 O+ ), generating an H+ activity of 10 4.0 . The rest of the T OT H (8.494 ⇥ 10 4 M) is bound with acetate in HAc molecules. As expected, since the pH of 4.0 is lower than pKa for HAc (4.74), the protonated species is present at a larger activity (8.494 ⇥ 10 4 ) than deprotonated Ac– (1.486 ⇥ 10 4 ). By far, the species whose activity we wish to fix most often is H+ (when we wish to explore conditions at a particular pH). Therefore, a short-cut is provided for those cases – one can use the pull-down menu for pH on the initial screen to select ‘Fixed at. . . ’ and then input the desired pH. Confirm for yourself that inputting 0.001 M Acetate-1 as the only specified component and using the pH menu to fix the pH at 4.0 gives the same results as shown in Figure 7.6. 367
Example 7.2 When a gas phase containing CO2 is bubbled through a solution, some CO2 dissolves and combines with H2 O to form H2 CO3 . Some of the H2 CO3 then dissociates, + generating HCO–3 and CO2– 3 and releasing H . If such a gas is bubbled into a solution of 10 4 M NaOH, determine how much CO2 must dissolve per liter of water to cause the activity of H2 CO3 to reach 10 5.0 . What is the pH of the solution? Solution We can use Visual Minteq to analyze the system, choosing H+ , Na+ , and CO2– 3 as components with input concentrations of 10 4 , 10 4 , and 0 M, respectively. (Note that the addition of the species OH– is represented as a negative input of H+ , based on the stoichiometry: 1 H2 O 1 H+ = 1 OH– . Formally, we should also add 10 4 M H2 O as an input required to form the OH– . However, since T OT H2 O is 55.6 M and the activity of H2 O is fixed at 1.0, the input of H2 O has no effect on the calculations and can be ignored.) We also specify that the species H2 CO3 is present at a fixed activity of 10 5 . The output from the calculation is shown at the top of the following page, indicating that the equilibrium pH is 7.345. T OT CO3 and T OT H are reported on the “Equilibrated mass distribution” page as 1.097 ⇥ 10 4 M and 1.194 ⇥ 10 4 M, respectively. Since dissolution of CO2 (g) is the only way that CO3 enters the solution, the final value of T OT CO3 (1.097 ⇥ 10 4 M) can be interpreted as the amount of CO2 that dissolves per liter of water. The final value of T OT H is larger than the initial input of H+ because the program represents the added CO2 as an equivalent addition of components, based on the reaction 2 H+ + 1 CO2– 3
1 H2 O )* 1 CO2
The calculation for the final T OT H is therefore ⇥ ⇤ T OT H = H+ init + 2 [CO2 (g)]dissolving = 10 4 + 2(1.097 ⇥ 10 4 ) = 1.194 ⇥ 10
7.4
4
USING CHEMICAL EQUILIBRIUM SOFTWARE TO DEVELOP LOG C pH DIAGRAMS
In addition to providing rapid and accurate numerical solutions to the simultaneous equations that characterize equilibrium problems, chemical equilibrium software offers the option to analyze a series of systems that are similar, but differ with respect to one or a few specific parameters. For example, the software can analyze a system with a given set of T OT X values for all components other than H+ , and with a sequence of specified values for the activity of the species H+ . By spanning a range of 368
H+ activities from, say, 10 2 to 10 12 , the program thus determines all the information needed to prepare a log C pH diagram over the pH range from 2 to 12. This information can then be exported to a spreadsheet program to prepare the diagram. As has been noted, if we are just interested in determining the composition of a solution with specified inputs, the software can provide the desired information without drawing the diagram. However, the diagram can still be helpful as a tool for visualizing the trends in composition over a range of pH values and for understanding how and why the system responds to various perturbations. To have Visual Minteq generate the data for a log C pH diagram, we provide the usual input regarding the components. As mentioned previously, H+ can be selected as a component, but even if it is not selected, the program will insert it. Once the components and the corresponding T OT X values are input, we select the “Multiproblem/Sweep” option at the top of the main menu and click on the “Sweep: one parameter is varied” box on the page that appears. We then choose pH as the sweep component and fill in the boxes with the initial value of pH, the increment between successive values, and the number of values to be tested. For example, if we wanted to prepare a log C pH diagram over the range 2.0 < pH < 12.0 by evaluating the solution composition at pH’s of 2.00, 2.25, 2.50. . .12.00, we would choose 2.0 as the starting value and 0.25 as the increment, and instruct the program to run 41 problems. We next select the output we want the program to report. By default, the program provides output for the solution composition for each pH investigated. In addition, we have the option to see how individual parameters change as the pH changes from one run to the next. For instance, if the system of interest contained some T OT PO4 369
and T OT CO3 , we might request information for log ci as a function of pH for all the acid/base species in those groups, as well as for H+ and OH– . The appearance of the “Multi-problem menu” page after those requests have been input is shown in Figure 7.7.
Figure 7.7 The “Multi-problem/Sweep” page of Visual Minteq after the range of parameters and pH values to be modeled have been specified.
Once the desired output is specified, we can return to the main menu and run the program. For an input of 7 ⇥ 10 3 M T OT CO3 and 2 ⇥ 10 3 M T OT PO4 , the output page that first appears (reproduced in Figure 7.8) shows the solution composition for the first run in the sequence (for this example, at a fixed pH of 2.0). By using the pulldown menu, we can choose to display the composition for any of the other runs. Alternatively, we can view the trends in the parameter(s) specified previously by clicking the “Selected sweep results” button (Figure 7.9). From that page, we can export the results to Excel c (which Visual Minteq opens automatically) and plot the log C-pH diagram (Figure 7.10). 370
Figure 7.8 Initial Visual Minteq output screen for analysis of a solution containing 7 ⇥ 10 3 M T OT CO3 and 2 ⇥ 10 3 M T OT PO4 over a range of pH values from 2 to 12.
Figure 7.9 Selected sweep results for the example problem, showing log C values of various solutes as a function of pH. (Additional columns for Log Concentration of other species are not shown in this view but are accessible by scrolling to the right on the screen.)
371
Figure 7.10 Log C pH diagram based on Visual Minteq output for a solution containing 7 ⇥ 10 3 M T OT CO3 and 2 ⇥ 10 3 M T OT PO4 . Data points are the Visual Minteq output; smooth curves inserted in Excel.
Although the curves in Figure 7.10 closely approximate those that we might draw manually for the system of interest, close inspection of the figure or the underlying data indicates some differences. For example, the lines for H+ and OH– intersect when the concentration of each species is 10 6.99 , not 10 7.00 , and the concentrations of H2 PO–4 and HPO2– 4 equal one another at pH 7.095, whereas the pKa relating these two species is 7.198. These are small differences, but still more than can be accounted for by numerical imprecision. The source of the discrepancies is that Visual Minteq accounts for activity coefficients, whereas when drawing the log C pH diagrams manually, we assumed ideal solute behavior. We could cause all the intersection points to correspond to the pKa values by plotting species’ activities rather than concentrations on the ordinate, but in that case the sum of the activities for all the species in an acid/base group would not equal T OT X (which is, by definition, a concentration-based summation). The other alternative is to instruct Visual Minteq to assume all species behave ideally by specifying that the ionic strength is zero. While the consideration of activity coefficients in the development of log C pH diagrams arguably leads to more accurate representations of speciation, the effects are often small and generally do not outweigh the convenience of assuming that the log C curves for conjugate acid/base pairs intersect at the corresponding pKa . For this reason, subsequent example calculations with Visual Minteq will be carried out with the ionic strength assigned a fixed value of zero (i.e., for ideal solute behavior) unless stated otherwise. 372
7.5
MODIFYING THE COMPONENT AND SPECIES DATABASES IN VISUAL MINTEQ
Although the Visual Minteq databases include the components and species that are of interest in most scenarios, we occasionally need to add a new entry in one of those categories or change the parameter values for a component or species that is already in the database. Changes to the species database are much more common than those to the component database, so the option to make the former changes is included in the main body of the software; changes to the component database are possible, but must be implemented outside of the software. The following algorithm describes the steps needed to add the dissolved species CuBr2 (aq) to the species database; the description is reprinted from the Visual Minteq Help file with only minor modifications. This algorithm can be used as a template for addition of any other species. 1. Select Cu2+ and Br– as components from the main menu. 2. Select “Database management/Aqueous species” from the main menu. 3. Click “Add new species”. A number of new text and command boxes appear on the screen. 4. In the “Species name” box, enter the name of the new species [CuBr2 (aq)]. 5. Under “Component no. 1”, enter the stoichiometric coefficient and name of the first component (1 and Cu+2, respectively) that is needed to form the new species. 6. Click “Next component” and repeat the process for the second component (2 and Br , respectively). Continue as needed until all the components needed to form the new species have been entered. After entering the data for the final component, rather than clicking “Next component”, click “Suggest ID number and mol. wt.”. This step causes the information that has been entered to be transferred to the table higher up on the screen. 7. Fill in the remaining fields (log K, dHr, Charge and References). 8. When the table looks satisfactory, click “Save changes to database”. The new species has now been added to the database. Properties of existing species can be modified by selecting “Database management/Aqueous species” from the main menu, selecting the species to be modified in the “Species name” box, changing the parameter values as needed, and clicking “Save changes to database”. 373
As noted previously, modifications to the component database are rarely needed and cannot be made within the program; rather, this database can be edited only by opening the core files within a text editor. Two such files exist – one (comp 2008.vdb) that includes all the components represented in the form that they actually exist in solution, and a second (compel 2009.vdb) that includes a subset of these same components, represented in some alternative form. For instance, sulfate is represented in comp 2008.vdb as “SO4–2” with a molecular weight of 96.0616, but in compel 2009.vdb, it is represented as “S (SO4)” with a molecular weight of 32.064. Both SO4–2 and S (SO4) appear in the list of components on the main page of the program, and either can be selected as a component. When one of these components is selected, the program recognizes that the chemical of interest is SO2– 4 ; the only difference between the two choices is that, if S (SO4) is selected, the program interprets the input concentration to be SO2– 4 as S. Thus, if S (SO4) is chosen as the component, an input concentration of 32.064 mg/L is interpreted as 1 mmol/L SO2– 4 , whereas the same concentration would be specified as 96.06 mg/L if the component were SO4–2. Both options are provided because the user might want to input the data in either form, but they are in different databases to prevent the user from including both (and thereby double-counting the input); an error message appears if one does try to include both components in a single problem. The compel 2009.vdb database also includes some components that have the same molecular weight as a component in the comp 2008.vdb database, but with a different spelling of the component name. Thus, hypochlorite ion is included in comp 2008.vdb as OCl–1 and in compel 2009.vdb as ClO–1. These two options are provided because both ways of writing the chemical formula for hypochlorite ion are used in the water chemistry literature. Once again, placing them in separate databases prevents accidental input of the same component twice. To add a new component to either database, open the appropriate database with a text editing program and add a line containing the six required parameter values, separated by commas. The first entry is an identification number for the component, and the second is the component’s name enclosed in quotation marks. The next four entries are the component’s charge, two size parameters for use in the Debye-H¨uckel equation, and the atomic or molecular weight of the component. The first DebyeH¨uckel size parameter is the nominal size, in angstroms, of the hydrated molecule; this is the size parameter that was discussed in Chapter 2 and presented in Table 2.2b. The second size parameter makes a minor adjustment to the computed activity coefficient for an ion and can be ignored (i.e., set to zero) for any new component that a user of Visual Minteq is likely to add. Once these entries are made and the revised database is saved, the new component will be available for use in the program.
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7.6
SUMMARY
Software specifically designed to solve for the equilibrium speciation of chemical systems has been available for a few decades and is used widely to simulate chemical systems. The inputs to the software are essentially the same as those needed to solve for speciation manually; however, much of the data regarding plausible reactions and the associated equilibrium constants are available in the program’s databases, so that the effort required to run the programs (and the likelihood of errors in data input) is reduced. Many of the available programs, including Visual Minteq, organize the data in a concise tableau like those developed in Chapter 6, in which the system information is organized around the choice of components. Although numerous choices can be made for sets of components, many software packages have preselected components that are used by default. The system tableau includes information regarding the chemical inputs to the system, the species expected to be present once equilibrium is attained, and the reactions that can occur in the system, along with their equilibrium constants. Once that information is input, the programs solve the mass (mole) balances and equilibrium expressions (mass action laws), subject to any other specified constraints (such as the known activity of a species in the equilibrium solution). By carrying out repetitive calculations for systems with gradually varying inputs, the software can quickly and efficiently generate information needed to develop log C pH diagrams.
7.7
PROBLEMS
1. The easiest, and perhaps only, way to familiarize yourself with the capabilities and limitations of Visual Minteq or similar software is to use it. Numerous acid/base example problems have been solved using manual or spreadsheet approaches in Chapters 5 an 6, and many other problems without solutions are provided at the ends of those chapters. Solve or re-solve some of those problems using Visual Minteq, practicing all techniques introduced in this chapter (e.g., drawing log C pH diagrams, exploring the effects of ionic strength on speciation, and defining new components and species that are not in the Visual Minteq data base). In each case, compare the results of the calculations using the different methods and try to understand the basis of any differences. Your instructor may specify some problems for you to hand in, but it will behoove you to go beyond those assignments to become as proficient as possible at using the program. In presenting the results of your simulations, you should provide an explana375
tion of how the problem was set up and solved with the program, give the results in a well formatted and titled table, and include appropriate axis labels and a caption for all graphs. Program output can be appended to your solution, but it is not acceptable to turn in only raw computer output. Note that Visual Minteq reports results for all species considered, even if they are present at absurdly low concentrations. When presenting your results, show only data that is relevant to the question. Although some exceptions exist, this guideline means that the lowest value of log C on your plots should typically be in the range of 8 to 12.
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8
TITRATIONS AND BUFFERS
377
Contents
8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15
8.1
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . REACTIONS OCCURRING DURING A TITRATION: QUALITATIVE CONSIDERATIONS . . . . . . . . . . . . . . . . QUANTITATIVE INTERPRETATION OF TITRATION DATA . . USING VISUAL MINTEQ TO SIMULATE TITRATIONS . . . . . TITRATIONS WITH WEAK ACIDS OR BASES . . . . . . . . . . THE EFFECT OF THE ACIDITY OF WATER ON TITRATION CURVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TITRATION OF SOLUTIONS CONTAINING UNKNOWN ACIDS AND BASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TITRATION EQUIVALENCE POINTS . . . . . . . . . . . . . . . ALKALINITY AND ACIDITY . . . . . . . . . . . . . . . . . . . ACID/BASE TITRATIONS OF NATURAL ORGANIC MATTER . BUFFERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . APPENDIX 8A: DETERMINING ALKALINITY FACTORS AND INSERTING THEM INTO THE VISUAL MINTEQ DATABASE . APPENDIX 8B: DERIVATION OF THE BUFFER INTENSITY OF MULTIPROTIC ACIDS . . . . . . . . . . . . . . . . . . . . . . . PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
378 379 384 393 402 405 406 409 413 438 442 453 455 456 457
INTRODUCTION
In previous chapters, our focus has been on determining the pH and speciation in solutions with known inputs, or the amount of acid or base that must be added to adjust a solution to a known final pH. These calculations are useful because, at times, 378
we do have a fairly good idea of the major acid/base species in a solution, and also because such systems provide a convenient vehicle for introducing some general principles about acid/base chemistry. However, both natural waters and process waters typically contain complex mixtures of acids and bases and cannot be analyzed completely; that is, it is not possible to identify and quantify every acid and base present. In this chapter, we use the ideas developed in previous chapters to estimate the concentration and strength of acid/base groups in solutions with unknown composition. A key tool for addressing this issue is acid/base titrations. In an acid/base titration, known quantities of an acid or base are added to a solution, and the resulting change in the solution pH is determined. Based on this response, we can make reasonable inferences about the (unknown) acids and bases present in the original solution. Titrations can also help us predict the pH change that will result from the mixing of two waters (e.g., a waste stream and a river) or from the addition of a process chemical to a water undergoing treatment (e.g., addition of HOCl or NaOCl to disinfect the water, addition of Al2 (SO4 )3 to coagulate particles and remove certain dissolved contaminants, or injection of air to add carbonate species or strip out ammonia). All these processes involve the mixing of various acids and bases, and an understanding of titrations will help us predict the characteristics of the mixtures. Also, many natural geochemical cycles are controlled by processes that can be viewed as titrations of basic rocks by acids that might be introduced into the ecosystem via volcanic activity, metabolic processes, or pollution. Since life developed under conditions controlled by these processes, and since many human activities can alter their rates, it serves us well to understand them, so that we can evaluate how large an effect our activities have. The types of questions we will address include: • Why does pH stay relatively steady and then change precipitously in some titrations? • Why can two solutions with the same initial pH respond very differently to acid or base addition? • How can we buffer the pH of a solution so that it remains in a narrow range when strong acids or bases enter the solution? • What is alkalinity?
8.2
REACTIONS OCCURRING DURING A TITRATION: QUALITATIVE CONSIDERATIONS
We begin by considering what reactions occur when a solution containing a mixture of weak acids and bases is titrated with a strong acid or base. For this exercise, it is useful to recall the approach we used to predict which acid/base species would be 379
dominant in a solution with known inputs. In that case, we imagined that the solution could be treated as though it had been prepared by first adding all the bases and then adding H+ ions incrementally, and we made the simplifying assumption that the H+ ions combine with the bases sequentially, converting all of one base to its conjugate acid before beginning to combine with the next weaker base. As a reminder of that analysis, consider the reactions that occur if we add strong acid to a pH 10 solution containing 10 4 M T OT Ac. A certain amount of acid will be required to lower the pH to, say, 6.76. Because pKa for HAc is 4.76, essentially all the acetate in the solution is present as Ac– at pH 10, and it is still in this form at pH 6.76, as indicated by the log C-pH diagram in Figure 8.1. That is, the acetate barely reacts at all with added protons over this pH range. As a result, its presence has no more effect on the solution than would the presence of an inert anion such as chloride. On the other hand, if we continue adding acid until the pH decreases to, say, 4.76 (= pKa ), one-half of the acetate is converted to HAc. During this process, the protonation of dissolved Ac– ions consumes many of the H+ ions added, and many more protons must be added to lower the pH than would be required if the acetate were not present, or if chloride were present instead of acetate.
Figure 8.1 Log C-pH diagram for a solution containing 10 4 M T OT Ac. The HAc/Ac speciation changes little in the pH regions far from pKa (labeled A), but it changes substantially in the region near pKa (labeled B).
If the acid addition continued, almost all the acetate ions would be converted to acetic acid molecules as the pH was lowered another two pH units, to 2.76. Once again, the number of H+ ions required to accomplish this pH change would be substantially greater than if the solution contained chloride instead of acetate. On the other hand, once the pH reached 2.76, the acetate would be distributed as 99% HAc and 1% Ac– (assuming ideal behavior). Acidifying this solution further would not be significantly different from acidifying a solution containing chloride instead of 380
acetate, because both the acetate and chloride would be virtually inert; that is, they would not undergo significant changes in speciation as H+ is added. Comparing the two systems from start to finish, we see that the total amount of H+ needed to titrate the system containing acetate from pH 10.0 to 2.76 would be approximately equal to the total amount needed to titrate the system without acetate, plus the amount of H+ needed to convert all the Ac– to HAc, i.e., T OT Ac. Furthermore, the effect of the acetate on the titration would be manifested almost entirely in the pH range near pKa . If a different weak base had been present, say, OCl– , exactly the same argument would apply, except that the “extra” acid needed would be required to change the pH from approximately 9.53 (i.e., pKa +2) to 5.53 (pKa 2); once the system reached pH 5.53, further additions of H+ would cause the pH to change about as much whether OCl was present or not. It should also be clear that these systems are completely reversible. That is, the amount of base needed to bring the pH back to 10.0 from 2.76 would be the same as the amount of acid needed to lower the pH initially, and the region where “extra” acid was needed to lower the pH in the systems with the weak acid/base groups (near the pKa ) would also be a region where “extra” base would be needed to raise the pH. The response of solution pH to additions of strong acid or base is analogous in many ways to the response of water temperature when thermal energy is added or removed. Consider, for example, the change in temperature when heat is extracted from pure water initially at 25 C (Figure 8.2). At first, the removal of thermal energy causes the temperature to decline steadily, at a rate determined by the heat capacity of liquid water. However, when the temperature reaches 0 C, further extraction of heat causes the water to freeze, while the temperature stays constant. Only after all the water has frozen does the temperature start dropping again, this time at a rate determined by the heat capacity of ice. In the latter (physical) system, thermal energy can be added or removed, and temperature is a measure of the instantaneous state of the system. Similarly, in the chemical system, protons can be added or removed, and the H+ activity (or the pH) is a measure of the instantaneous state of the system. Over most ranges of temperature (all temperatures other than 0 C, in the example), all the heat added or removed affects the temperature; however, in a limited range, the heat exchange mediates a transition in the form of the water, and none of it is available to alter the bulk temperature of the system. Similarly, in the chemical system, almost all the H+ that is added to or removed from the system under most conditions leads to a change in the concentration (and activity) of free H+ and therefore in the pH. However, in some ranges (near the pKa values of dissolved acid/base groups), added H+ is used mostly to mediate the conversion of A– to HA, and little remains as free H+ , so the pH change is minimal; if H+ is being removed, the same general trends apply, but the conversion is from HA to A– . In the physical system, a second transition exists at 100 C, and in the chemical system, transitions exist at the pKa value of every dissolved acid in the solution. Also, 381
Figure 8.2 Similarities between the response of water to gradual extraction of thermal energy (broken line) and titration of a solution with an acid (solid line). In each case, the response parameter (temperature or pH) changes rapidly over some regions and slowly over others. The ‘flat’ portions of the curve for a thermal titration occur where the water undergoes a change of state; those for the chemical titration occur where a solute undergoes a change in speciation.
in both systems, the amount of the “titrant” (energy or H+ ) needed to pass through the transition is proportional to the amount of material being converted (the mass of H2 O in the physical system, T OT A in the chemical system). The main difference between the physical and chemical processes is that water freezes and/or melts at a single, fixed temperature, whereas HA and A– are interconverted more gradually, over a small range of pH values, rather than entirely at a single pH. Example 8.1 A wastewater at pH 10.5 contains 4 ⇥ 10 3 M pentachlorophenol (a wood preservative sometimes referred to as PCP or simply penta, with formula C6 Cl5 OH and pKa = 4.7) and 2 ⇥ 10 3 M p chlorophenol (C6 H4 ClOH, pKa = 9.43). It is necessary to neutralize the water to pH8 to prepare the waste for further treatment. Which of these two acid/base groups exerts a greater demand for acid during the neutralization process? Solution At the initial pH of 10.5, both acid/base couples are mostly (> 90%) deprotonated. However, p chlorophenol (pKa = 9.43) is a much weaker acid than PCP (pKa = 4.7); correspondingly, deprotonated p chlorophenol (‘p chlorophenate’) is a much stronger base than deprotonated PCP. As the pH is lowered from 10 to 8, p chlorophenate becomes mostly protonated, consuming approximately 2 ⇥ 10 3 M 382
H+ , whereas the speciation of the PCP barely changes. Therefore, the p chlorophenol species will exert almost all the demand for acid in the neutralization process. Note that we can draw this conclusion without having any information about how the solution was originally prepared, i.e., whether the species of interest were added as acids or bases, and whether other strong acids or bases might also have been added. All we need to know is the initial and final pH, along with the pKa values. From the preceding discussion and our prior knowledge, we can draw several important qualitative conclusions related to acid/base titrations. First, of course, it is possible to change the pH of any system by addition of strong acid or base. Second, when a weak acid or base is present in the system, it affects the amount of strong acid or base needed to cause a given pH change only to the extent that the weak acid/base couple reacts with the H+ or OH– that is being added, by the following reactions: Titration with acid
A– + H+ )* HA
(8.1)
Titration with base
HA + OH– )* A– + H2 O
(8.2)
Because the speciation of a weak acid/base pair (HA/A ) shifts most dramatically at pH values near pKa , the presence of some A in solution has the greatest effect on a titration when the titration moves through the pH region near pKa,HA . Finally, if the titration goes from an initial pH at which T OT A is almost fully deprotonated to one at which it is almost fully protonated, the extra acid needed compared to a system with no A present is approximately T OT A, and the same amount of extra base
Streams contaminated by acids can be neutralized in various ways. (a) Lime being added to a river in Sweden. The solid dissolves and hydrolyzes to release OH– ions (CaO + H2 O )* Ca2+ + 2 OH– ), which consume some of the acidity of the water. (b) An acidic stream (on the right) containing high concentrations of dissolved Al and Fe mixing with near-neutral water from another stream. When the pH of the acidic stream increases, Al and Fe oxides precipitate. [(a) Martin Bond/Science Source/Photo Researchers, Inc.; (b) Chris Fuller, USGS.]
383
would be needed to reverse the process. Armed with this general understanding of what happens during a titration, we next consider the quantitative details of titration curves.
8.3 8.3.1
QUANTITATIVE INTERPRETATION OF TITRATION DATA Units for Quantifying Titration Data: Equivalents and mg/L as CaCO3
Additions of titrant are often expressed in terms of equivalents (equiv). As explained in Chapter 1, this term is used in a number of ways in chemistry, but in all cases the implication is that one equivalent of substance X either combines with or releases one equivalent of substance Y; i.e., equivalents express information about reaction stoichiometry. For acids and bases, equivalents are defined with reference to H+ and OH– , respectively. That is, one acid equivalent is defined as 1 mole of H+ , one base equivalent is defined as 1 mole of OH– , and one equivalent (1 equiv) of any other substance is the amount of that substance that can release or react with 1 mole of H+ or OH– . Thus, in any acid/base titration, the number of moles of H+ or OH– added per liter can be referred to as the number of equivalents added per liter.1 The equivalent weight (EW) of an acid or base is the weight (or, more often, mass) of the substance that can combine with or release 1 mole of H+ or OH– , respectively. For instance, 1 mole of a triprotic acid can potentially release 3 moles of H+ , so 1/3 mole of the acid is “equivalent” to 1 mole of H+ ; the equivalent weight of the acid is therefore 1/3 of its molecular weight. When carrying out this calculation, the number of acid equivalents attributed to a given species is typically based on the maximum number of H+ ions that the species can release, and the number of base equivalents is based on the maximum number with which it can combine. Thus, + for example, a concentration of x mol/L HPO2– 4 could potentially release x mol/L H 3– to form PO4 or combine with 2x mol/L H+ to form H3 PO4 . It would therefore be considered to correspond to x equiv/L of acid or 2x equiv/L of base. Care is required when reporting or interpreting equivalency, since the number of equivalents that actually react in a given situation is not always the number that could potentially react. For instance, in a solution at pH 9.0 that contains phosphate species, almost all the T OT PO4 is present as HPO2– 4 . If that solution were titrated to pH 4.5, almost all the T OT PO4 would be converted to H2 PO–4 . Thus, although + 3 HPO2– 4 has the capacity to react with two H ions (so that a concentration of 10 M 2– 2– 3 HPO4 might be referred to as 2 ⇥ 10 equiv/L of HPO4 ), each mole of HPO2– 4 1 Formally, one should always distinguish between an equivalent of acid and an equivalent of base. In practice, however, both quantities are just referred to as ‘equivalents’, and the meaning must be inferred from context.
384
would consume only one equivalent of H+ during the titration. A second common way to express titration data (especially when reporting alkalinity, which is discussed later in the chapter) is as milligrams per liter as CaCO3 . This type of representation was discussed in Chapter 1, and it would be worthwhile to review that section if you have forgotten the details. Briefly, expressing a concentration of acid or base in this way indicates that the acid or base is present in the given solution at the equivalent concentration (i.e., the same number of equivalents per liter) as the indicated concentration of CaCO3 . The concentration of CaCO3 is used as the basis for this type of reporting both for historical reasons and because of the convenient coincidence that its molecular weight is 100. Since each molecule of CaCO3 can potentially bind two protons (when the molecule dissociates, the Ca2+ does not react further, and the CO2– 3 can potentially be converted to H2 CO3 ), the equivalent weight of CaCO3 is 100/2, or 50. Thus, a dose of OH– added during a titration and reported as 50 mg/L as CaCO3 corresponds to 1.0 milliequivalent per liter (meq/L) of the base.
8.3.2
Titration of a Strong Base with a Strong Acid
Consider the conditions during the titration of a 10 3 M NaOH solution with strong acid (HCl). In the most general sense, assuming ideal solute behavior, we can identify (H+ ), (OH– ), and (Na+ ) as unknowns in the initial solution. We can solve for these unknowns using the relevant mass balances (on T OT Na and T OT H) and Kw . Since NaOH is a strong base, all the added NaOH dissociates, and the result is that the initial solution pH is 11.0. Once some HCl is added, the equilibrium concentration of Cl– becomes an additional unknown, and the mass balance on Cl is an additional independent equation characterizing the system. Designating the concentration of HCl added as x, the tableau, T OT H equation, and proton condition (which, in this case, is just the charge balance) at any point during the titration are as follows:2
H+
H2 O
H+
Na+
Cl–
0
1
0
0
0.00
–
1
1
0
0
14.00
+
0
0
1
0
0.00
–
Cl
0
0
0
1
0.00
NaOH
1
1
1
0
10
HCl
0
1
0
1
x
OH Na
2 Normally,
3
the titrant is added as a concentrated reagent, so that the volume of the solution being titrated changes negligibly as the titration proceeds. We make that assumption throughout the chapter, unless stated otherwise.
385
T OT H equation:
(H+ )
Proton condition:
10
(OH ) = 10
3 + (H+ ) =
3 +x
(OH ) + x
Substituting various values for x, we can solve for the pH after different additions of acid using any of the approaches described in the preceding chapters [i.e., writing (OH– ) as Kw /(H+ ) and solving analytically; programming the expression into a spreadsheet and solving numerically; using the PC in conjunction with a log C–pH diagram; or inputting the information into chemical equilibrium software, which then uses the T OT H equation to solve for the speciation]. The results of such calculations are shown in Figure 8.3.
Figure 8.3 Titration curve for a solution containing 10
3M
NaOH, titrated with a strong acid.
The titration curve indicates that a significant amount of HCl must be added to lower the pH from 11 to 10, but only a small amount of additional HCl is then required to cause the pH to decrease to 4. This outcome reflects the fact that titrating the solution from pH 11.0 to pH 10.0 [i.e., from (OH ) = 1.0 ⇥ 10 3 to 0.1 ⇥ 10 3 ] requires neutralization of 90% of the free OH– ions, which consumes 0.9 ⇥ 10 3 M H+ . Then, decreasing the pH to 4 requires 0.1 ⇥ 10 3 M H+ to neutralize almost all of the remaining free OH– ions, plus an additional 0.1 ⇥ 10 3 M H+ to build up the free H+ concentration from 10 10 to 10 4 M. Thus, the pH of the system declines from pH 10 to 4 with the addition of only approximately 22% as much acid as is required to decrease pH from 11 to 10 (0.2 ⇥ 10 3 M compared to 0.9 ⇥ 10 3 M).
8.3.3
Titration of a Weak Base with a Strong Acid
Now consider what would happen if we carried out a titration similar to that described above, but with a solution that initially contained 10 3 M NaOCl in addition 386
to the 10 3 M NaOH. Being a strong base, essentially all the NaOH would once again dissociate, releasing 10 3 M OH– . Dissociation of the NaOCl and protonation of some of the OCl– can only increase the OH– and pH even more, so we know that the pH of the initial solution will be 11. However, because OCl– protonates negligibly at such high pH’s, the NaOH would contribute as much OH– as in the prior example, while the NaOCl would contribute essentially none, so the initial pH would once again be close to 11.0. Given that (OCl– ) and (H+ ) are known (approximately), (HOCl) can be calculated as (HOCl) =
(H+ )(OCl ) (10 11.0 )(10 ⇡ Ka 10 7.53
3.0 )
= 10
6.47
As H+ is added to this solution, it protonates both OH– and OCl– in such a way that the two equilibrium constants (Kw and Ka ) are satisfied. That is, each time an increment of base is added, some H+ ions participate in each of the following reactions: OCl + H+ )* HOCl OH + H+ )* H2 O
(8.3) (8.4)
In the system considered earlier containing only NaOH, all the H+ added either neutralized OH– or remained in the solution as H+ ions. Because, in the system with NaOCl, some H+ is consumed in the protonation of OCl– , a given addition of HCl lowers the pH less than in the OCl-free solution. As a result, to reach a given final pH, we need to add more HCl to the solution containing OCl than to the OClfree solution. To quantify this effect, we prepare the following tableau and T OT H equation.
H+
H2 O
H+
Na+
Cl–
OCl–
0
1
0
0
0
0.00
–
1
1
0
0
0
14.00
+
0
0
1
0
0
0.00
–
OH Na Cl
0
0
0
1
0
0.00
–
OCl
0
0
0
0
1
0.00
HOCl
0
1
0
0
1
7.53
NaOH
1
1
1
0
0
10
3
NaOCl
0
0
1
0
1
10
3
HCl
0
1
0
1
0
x
T OT H equation:
(H+ )
(OH ) + (HOCl) = 10 387
3 +x
Table 8.1 T OT H and solution pH for various additions of HCl to a solution containing 10 3 M NaOH and 10 3 M NaOCl⇤
HCl added 0 2.0 ⇥ 10 4.0 ⇥ 10 6.0 ⇥ 10 8.0 ⇥ 10 1.0 ⇥ 10 1.2 ⇥ 10 1.4 ⇥ 10 1.6 ⇥ 10 1.8 ⇥ 10 2.0 ⇥ 10 2.2 ⇥ 10 2.4 ⇥ 10 2.6 ⇥ 10 2.8 ⇥ 10 3.0 ⇥ 10
⇤ T OT H
4 4 4 4 3 3 3 3 3 3 3 3 3 3 3
T OT H Equation 1.0 ⇥ 10 0.8 ⇥ 10 0.6 ⇥ 10 0.4 ⇥ 10
0.2 ⇥ 10 0.2 ⇥ 10 0.4 ⇥ 10 0.6 ⇥ 10 0.8 ⇥ 10 1.0 ⇥ 10 1.2 ⇥ 10 1.4 ⇥ 10 1.6 ⇥ 10 1.8 ⇥ 10 2.0 ⇥ 10
pH
3
= (H+ )
(OH ) + (HOCl)
11.00
3
= (H+ )
(OH ) + (HOCl)
10.90
3
= (H+ )
(OH ) + (HOCl)
10.78
3
= (H+ )
(OH ) + (HOCl)
10.60
3
= (H+ )
(OH ) + (HOCl)
10.30
0 = (H+ )
(OH ) + (HOCl)
9.26
3
= (H+ )
(OH ) + (HOCl)
8.13
3
= (H+ )
(OH ) + (HOCl)
7.70
3
= (H+ )
(OH ) + (HOCl)
7.35
3
= (H+ )
(OH ) + (HOCl)
6.93
3
= (H+ )
(OH ) + (HOCl)
5.27
3
= (H+ )
(OH ) + (HOCl)
3.70
3
= (H+ )
(OH ) + (HOCl)
3.40
3
= (H+ )
(OH ) + (HOCl)
3.22
3
= (H+ )
(OH ) + (HOCl)
3.10
3
= (H+ )
(OH ) + (HOCl)
3.00
+
–
+
–
based on use of H , Cl , Na , and OCl as components
Once again, the pH after various additions of HCl can be computed using any of the techniques introduced in prior chapters. At any point in the titration, the T OT H equation is identical to that shown above, with different values of x corresponding to different amounts of HCl added. The results are summarized in Table 8.1 and plotted as the ‘NaOH + NaOCl’ curve in Figure 8.4a. The titration curve developed previously for a solution containing only 10 3 M NaOH is also shown in the figure for reference. The titrations in the presence and absence of OCl are almost identical until slightly less than 10 3 M H+ has been added, corresponding to pH near 10; in both 388
Figure 8.4 (a) Titration curves for solutions containing 10 3 M NaOH, either alone or in addition to 10 3 M of a weaker base. (b) Speciation of the weak acid/base groups considered in part (a) of the figure.
systems, the pH decreases gradually down to that pH and then starts decreasing more steeply. However, whereas the steep decrease in solution pH continues to around pH 4.5 in the system with no OCl, it ceases at pH⇠9 (i.e., about 1.5 pH units above pKa ) in the system with OCl. At that point, the latter curve becomes quite flat again; i.e., a large amount of H+ must be added to generate even a slight decrease in pH. This flattening occurs because much of the H+ added to the solution is reacting with the OCl– to convert it to HOCl. The curve remains flat until enough H+ has been added to lower the pH to around 6.0, at which point the steep increase in pH begins again. The total H+ addition required to adjust the pH to 3.0 is 2.0 ⇥ 10 3 M in the system 389
without OCl and 3.0 ⇥ 10 3 M in the system with it. The difference, 1.0 ⇥ 10 3 M, is the amount of H+ that is consumed in the conversion of essentially 100% of the hypochlorite from OCl– to HOCl; i.e., it is T OT OCl. Two other titration curves are included in Figure 8.4a and are shown as broken lines. These curves are for titrations of systems initially containing either 10 3 M NaAc or 10 3 M NH3 in addition to 10 3 M NaOH. The shapes of these curves are similar to that of the NaOH + NaOCl curve, except that the flat regions of the curves appear at different pH’s, in each case near the pKa of the weak acid/base group that is present. Figure 8.4b shows the speciation of the weak acid/base couples as a function of the amount of acid added in these three systems. Whereas the titration curves show that the pH’s are quite different in the three systems after a given addition of base, the speciation of the acid/base couples is very similar. For instance, all the couples are split equally between their acid and base forms when 1.5 ⇥ 10 3 M H+ has been added, even though the pH’s at this point range from 4.7 to 9.2. The reason for this similarity becomes clear when we think about the fate of the H+ added during the titration. Initially, all the systems are at pH near 11.0, i.e., (OH ) = 10 3 . The strongest base in the system at this point is OH– , so any added H+ is likely to neutralize this base before reacting with weaker ones. The conversion of any of the weak bases to their conjugate acids is therefore negligible until about 10 3 M H+ has been added. Then, as the next 10 3 M H+ is added, it reacts almost entirely with the weak base (NH3 , OCl– , or Ac– ), converting the base to its conjugate acid. The pH range over which most of this protonation occurs is approximately pKa 1.5 to pKa + 1.5; i.e., it is different in the three systems. In each case, after most of the weak base has been protonated, the pH decrease accompanying further additions of acid is more or less as in the blank. Thus, the speciation of T OT A between the forms HA and A– is very similar in the three systems as a function of the amount of base added, even though the pH at which most of the conversion from HA to A– occurs is different. Generalizing the results in Figure 8.4a, we infer that titration curves are relatively flat (i.e., the solution pH changes very gradually) when the pH is near the pKa of any conjugate acid/base pair that is in solution. Furthermore, the explanation for this behavior suggests that, if more than one weak acid is present, the titration curve will have flat portions near the pKa of each of them. These flat portions may be easily distinguishable from one another, or they may be indistinct if the pKa ’s are close enough to one another that the flat portions overlap. For example, Figure 8.5a shows the titration curve for a solution containing 1.0 ⇥ 10 3 M NaOH, 0.4 ⇥ 10 3 M NaAc, 0.25 ⇥ 10 3 M NaOCl, and 0.35 ⇥ 10 3 M NH3 . The graph is relatively flat near the pKa ’s, and the width of each flat region is proportional to T OT A for the corresponding acid, but the flat portions run into one another and are not easily distinguished. Nevertheless, Figure 8.5b, which shows the speciation of the three weak acids during this titration, makes it clear that, as the pH declines, each base is virtually completely converted to its conjugate acid 390
before the next weaker base begins to acquire protons. That is, almost none of the ammonia becomes protonated until most of the OH– has been neutralized (H+ added ⇠1.0⇥10 3 M), almost none of the hypochlorite protonates until most of the ammonia has been converted to ammonium ion (H+ added ⇠1.35⇥10 3 M), and almost none of the acetate protonates until almost all the hypochlorite has been converted to hypochlorous acid (H+ added ⇠1.6⇥10 3 M).
Figure 8.5 (a) Titration curves for solutions containing either only a strong base or a mixture of weak bases in addition to the strong base. (b) Speciation in the system with several weak bases characterized in part (a). The dominant reaction involving the added H+ in each pH range is shown at the top of the figure.
391
Example 8.2 Strong acid is added to the solution described in Example 8.1 [containing 4 ⇥ 10 3 M PCP (pKa = 4.7) and 2 ⇥ 10 3 M p chlorophenate (C6 H4 ClOH, pKa = 9.43)], causing the pH to decrease from 10.5 to 4.0. Describe qualitatively and semiquantitatively the features of the titration curve. Solution The initial pH, 10.5, is 1.07 pH units above pKa of p chlorophenol, so only approximately 8% of this acid/base group is protonated (a0 = 0.08). When acid is first added, the change in pH will be gradual due to both the relatively large concentration of free OH– in the solution (3.16 ⇥ 10 4 M) and the consumption of H+ via protonation of the p chlorophenate. The gradual change continues until most of the p-chlorophenate has been converted to its conjugate acid, p-chlorophenol, at pH ⇠7.9 (1.5 pH units below pKa ). Virtually none of the PCP becomes protonated over this pH range. Therefore, to reach pH 7.9, the acid requirement includes approximately 3.16 ⇥ 10 4 M H+ to neutralize almost all of the OH– in the initial solution and 1.78⇥10 3 M H+ to protonate ⇠89% of the p chlorophenate (the difference between 8% protonated at pH 10.5 and 97% protonated at pH 8.4), for a total required addition of 2.10 ⇥ 10 3 M H+ . Thereafter, the pH will decrease rapidly until the PCP begins to protonate significantly, at pH ⇠ 5.7. At pH 4.0, a0 of PCP is 0.83. Therefore, to lower the pH from 7.9 to 4.0 requires an amount of acid approximately equal to the 3% of the p-chlorophenate remaining at pH 7.9, plus about 83% of T OT PCP (3.32 ⇥ 10 3 M). In addition, 10 4 M H+ is needed to supply the free H+ at pH 4.0. The total amount and the fate of the H+ added during the whole titration from pH 10.5 to 4.0 can be summarized as follows: H+ to neutralize OH– H+ to protonate p chlorophenate H+ to protonate PCP
1.84 ⇥ 10 3.32 ⇥ 10
Increase in free H+
0.10 ⇥ 10
Total H+ requirement
8.3.4
0.32 ⇥ 10
5.58 ⇥ 10
3 3 3 3 3
Titrations of Multiprotic Acids and Bases
Titration curves for systems containing a multiprotic acid are similar in many respects to those for systems with two or more independent monoprotic acid/base pairs. 392
As each pKa is passed, an additional amount of base equal to T OT A is required to convert H3 A to H2 A, or H2 A to HA, or HA to A. As is the case when chemically distinct acids are present, if the pKa ’s of the multiprotic acid are widely separated, each will lead to a distinct flat region of the titration curve, whereas if the pKa ’s are close to one another, the flat regions may merge into a single broad and indistinct zone. A titration curve for addition of a strong acid to a solution of 10 3 M NaOH plus 10 3 M Na2 CO3 is shown in Figure 8.6.
Figure 8.6 Titration curves for addition of strong acid to solutions of either 10 10 3 M NaOH plus 10 3 M Na2 CO3 .
8.4
3M
NaOH or
USING VISUAL MINTEQ TO SIMULATE TITRATIONS
The sequential addition of several doses of a strong acid or base to a solution corresponds to step-wise changes in various mass balances that characterize the solution. For example, if HCl is being added, each increment of acid dosed increases T OT H and T OT Cl by equal amounts, whereas if H2 SO4 is dosed, the increase in T OT H is twice the increase in T OT SO4 . If a base is added, each mole of OH– entering solution corresponds to addition of one mole of H2 O and removal of one mole of H+ , so, for example, NaOH addition can be represented as identical increases in T OT H2 O and T OT Na and an equal decrease in T OT H. In addition to the changes in the T OTi values noted above, with each dose of titrant added to the sample, the T OTi values of the components in the original sample solution change as well, because the addition of titrant increases the overall solution volume and therefore dilutes the components in the original solution. Frequently, though, the volume of titrant added is small compared to the sample volume, so that 393
the effect of dilution can be ignored, and the T OTi values for components that are not in the titrant solution can be assumed to remain constant. In such a case, for titration with a strong acid or base, the only parameters that change as the titration proceeds are the T OTi values of H2 O, H+ , and the salt ion that is part of the acid or base. When solving for the solution pH during a titration, the mass balance on H2 O can always be ignored, because the addition of the titrant never changes the total concentration of H2 O significantly. Furthermore, in many cases, the salt ion that is added with the titrant has a negligible effect on speciation in the solution, in which case the mass balance on that ion can be ignored as well. Such a situation would arise, for example, if the titrant were HCl or NaOH, and if the added Na+ or Cl– neither reacted with other solutes nor changed the ionic strength significantly. Thus, if the conditions of negligible volume change and negligible impact of the salt ion in the titrant are met, the only system parameter that changes significantly between sequential titration points with an acid or base is T OT H. Problems involving steadily changing T OTi values for one or several components are easy to program into chemical equilibrium software, so many software packages provide built-in options for simulating titrations. In Visual Minteq, slightly different approaches are used to input the relevant data depending on whether the titration can be simulated by a change in only one mass balance or whether multiple mass balances must be changed with each titrant addition. Both scenarios are described in this section.
8.4.1
Simulations Varying Only TOTH
To use Visual Minteq to simulate a titration in which T OT H is the only parameter that must be changed, we first input the components of the sample solution in the conventional way, and then go to the “Multi-problem/Sweep” page (the same one used to generate the data for log C–pH diagrams). There, we check the box labeled “Sweep: one parameter is varied”, click the button labeled “Total concentration, any component” and use the pulldown menu to select H+ as the component whose total concentration will be changed between runs. Next, we specify the range of T OT H values to be considered by inputting an initial value, the (fixed) increment to be added in sequential runs, and the number of runs to be carried out. For example, to simulate titration of a solution of 3 ⇥ 10 4 M NaHCO3 by addition of 0 to 5 ⇥ 10 4 M HCl in increments of 0.25 ⇥ 10 4 M, the initial value of T OT H would be 3 ⇥ 10 4 , the increment would be 0.25 ⇥ 10 4 , and 21 runs would be required (the initial run with no acid addition, and then 20 more runs with the specified incremental additions, leading to a final run in which T OT H is 8.0 ⇥ 10 4 M). For titration of the same solution with NaOH, the same inputs would be used, but the increment of T OT H between runs would be 0.25 ⇥ 10 4 (yielding a final T OT H value of 2.0 ⇥ 10 4 M). 394
The final step before running the program is to specify the parameters that we wish to follow as the titration proceeds. The parameter that is being varied during the titration (T OT H in this case, which appears in the list as the total concentration of dissolved H+ ) is reported by default. The program also provides the option to track various types of information (ci , log ci , ai , etc.) about any of the species that are expected to be present in the equilibrium solution. For this problem, the only parameter that we need to monitor is pH, so we add the “log activity of H+ ” to the selection list near the bottom of the page. The final form of the “Multi-problem” menu page for the titration of NaHCO3 with HCl over the desired range is shown in Figure 8.7.
Figure 8.7 The Multi-problem menu page in Visual Minteq for titration of 3.0 ⇥ 10 NaHCO3 with 20 additions of 0.25 ⇥ 10 4 M HCl each.
4M
When we carry out the simulation, the output page shows the equilibrium solution speciation from the first run (with no titrant added). The corresponding information for any other run can be viewed by using the pulldown tab at the top of the “Output” page. Alternatively, and of more interest to us in this simulation, is the summary of changes from run to run; this summary is accessed via the “Selected sweep results” button on the main “Output” page and is shown in Figure 8.8. The summary can be exported to Excel, where the data can be plotted to produce the graph shown in Figure 8.9. 395
Figure 8.8 Selected sweep results for titration described in Figure 8.7, assuming ideal solution behavior.
Figure 8.9 Simulated titration of 3 ⇥ 10 4 M NaHCO3 with HCl, assuming ideal solute behavior. Data points are output from Visual Minteq; smooth curve inserted in Excel.
396
8.4.2
Simulations Varying TOTi of More Than One Component
Although simulated titrations in which the total concentration of only one component changes are often sufficient to provide the information we are seeking, there are times when a more complete accounting of the changes in the solution composition during the titration is required. These situations include, for example: • titrations in which both the anionic and cationic portion of the titrant can react with constituents of the original solution (e.g., titrations with NH4 CO3 ) or in which the titrant addition causes a significant change in the activity coefficients of some species of interest; • titrations that are intended to simulate mixing of two complex solutions in different proportions (e.g., mixing of freshwater and seawater in various proportions in estuaries). • titrations in which the volume of titrating solution added is sufficient to significantly dilute the original solution, so that the component concentrations change during the titration in ways other than via chemical reactions. In some cases, if the titrant is an acid or base that can be made by a combination of H2 O, H+ , and just one other component (e.g., SO2– 4 if H2 SO4 is being added, or + Na if NaOH is being added), a simple approach can be used to account for both the H+ and the salt ion that are added with each dose of titrant. To use this approach, we follow the same steps as when we vary T OT H between titration steps, but instead we vary the total input concentration of the salt ion and select the “calculated from mass & charge balance” option on the home page as the method for determining the equilibrium pH. This option instructs to program to solve the mass balances on all components other than H+ in the usual way, but to determine T OT H based on the charge balance rather than on some user-specified input value. For example, say we wanted to simulate addition of 10 doses of titrant to a 10 3 M NaHCO3 solution, with each dose adding 10 4 mol of H2 SO4 per liter of original solution. We could input the NaHCO3 in the usual way and add SO2– 4 as an input component with zero concentration. We would then invoke the “calculated from mass & charge balance” option for pH and specify on the “Multi-problem” page that T OT SO4 should be increased in increments of 10 4 M for 11 runs. In the program’s first run, simulating the solution before any H2 SO4 was added, it would determine that T OT H had to be 10 3 M. Then, in each subsequent run, it would determine that T OT H would have to increase by 2 ⇥ 10 4 M with each increase of 10 4 M in T OT SO4 to maintain the charge balance. This approach is more realistic than one in which only T OT H is changed between runs, because it accounts for the effect of the salt ion on the solution ionic strength and also for any potential interactions between the salt ion and other constituents of the solution. However, it still does not account for dilution of the original sample by the titrant, and it is not applicable for titration 397
with compounds that cannot be represented as a combination of H2 O, H+ , and one other component (e.g., NH4 Cl). Titration scenarios that cannot be modeled using one of the preceding approaches can be simulated in Visual Minteq by providing a full description of the two solutions that are to be mixed, and specifying a range of volume ratios for the mixing process. To do this, we first follow the conventional approach to input the composition of the solution to be titrated, and then move to the “Multi-problem” menu and select “Titration/mixing with a titrant with known composition”. A button labeled “Go to Titration Manager” appears, which, when clicked, takes us to a screen where the volume of the solution to be titrated, the composition of the titrant solution, and the volume of the titrant solution to be added in each step can be input; the number of titration steps (i.e., mixing volume ratios) is specified on the “Multi-problem/sweep” page. The units for the volumes of the two solutions are arbitrary, but must be the same, so that their ratio corresponds to the actual ratio of the titrant volume added in each step to the original solution volume. The protocol for inputting the composition of the titrant solution, which can contain up to 10 components, is essentially the same as that used to input the composition of the original solution on the home page. Once all these inputs are provided, the program can be instructed to track the desired outputs via the “Multi-problem” menu and can be run as described previously for titrations with a single component. Consider, for example, a scenario where leachate from a municipal landfill is leaking into a nearby stream, and we wish to find the pH and the concentration of un-ionized ammonia (NH3 (aq), which is considerably more toxic to fish than NH+4 is) for mixing ratios from 0.5% to 5% leachate by volume. The compositions of the two solutions are as shown in Table 8.2. Note that a concentration is reported for ammonium (NH4 N), but not for ammonia (NH3 N). When this is done, the usual interpretation is that what was actually measured was T OT NH4 N, and that this value is being reported as NH4 N based on the assumption that NH3 is a small contributor to the total. The composition of the stream after it mixes with various amounts of leachate can be determined using Visual Minteq by inputting the known components of the stream as the solution to be titrated, and the known components of the leachate as the titrant solution. The components of the stream are entered in the usual way on the – program‘s home page, noting that the H+ and CO2– 3 associated with both HCO3 and H2 CO3 contribute to T OT H and T OT CO3 . Thus, T OT H equals [0.80 + 2(0.02)] M, or 0.84 M, and T OT CO3 equals (0.80 + 0.02) M, or 0.82 M. (The concentration of CO2– 3 could be computed and included in T OT CO3 , but its contribution is negligible at the pH’s of the two solutions.) The components of the leachate are entered in the same way, but on the “Titration Manager” page.3 3 Be sure to click “Save and Next” if another component is to be added, but “Save and Back to Multi/Sweep menu” after the final component has been input.
398
Table 8.2 Composition of leachate and stream water for mixing example
Conc’n in leachate⇤
Conc’n in stream
53.0 24.5 14.7 23.4 5.6 82.0 194.5 5.8
1.17 0.40 0.01 0.90 0.13 0.80 0.02 7.9
+
Na Ca2+ NH4 N Cl– SO2– 4 HCO–3 H2 CO3 pH ⇤
All concentrations in mmol/L, except pH (in pH units)
The “Titration Manager” page is also where information about the range of mixing ratios of interest is entered. For the current simulation, it is convenient to carry out the simulation in two stages – one exploring the range from 0 to 0.5% leachate in 0.1% increments, and a second exploring the range 0.5% to 5% in 0.5% increments. To carry out the first stage, we enter 100 as the “Volume of solution to be titrated” and 0.1 as the “Volume of titrant” (i.e., the volume to be added in each titration step). Then, on the “Multi-problem/sweep” menu, we specify that the titration should comprise six runs (representing the initial point plus five steps), and we choose the parameters in the equilibrium solution that we want the program to report as part of the “Selected sweep results”. In the current case, we are interested in the pH (so we specify log of the activity of H+ ) and the un-ionized ammonia concentration (so we specify concentration of NH3 ); to assess the fraction of the total ammonia that is in the form of NH3 , we also include the “Total dissolved NH+4 ” on this list. [The designation “Total dissolved” is available only for components (not species), and refers to the total concentration of the component in the solution, i.e., T OT NH4 ]. Images of the “Titration Manager” and “Multi-Problem” pages at this point are shown in Figures 8.10a and 8.10b, respectively. When we run the program, the output shown in Figure 8.10c is displayed on the “Selected sweep results”. This output can be exported to Excel R , where a column can be added to the spreadsheet showing either the leachate-to-stream volume ratio or the percentage of leachate in the mixture based on the following calculation: % leachate =
Vleach Vleach = Vleach +Vstream Vleach + 100
(8.5)
After repeating the same procedure for higher leachate percentages (10 steps, starting at Vleach of 1.0 and incrementing Vleach by 0.5 in each step, but this time 399
allowing titrant addition in the first step), the results of the two sets of runs can be combined in the spreadsheet and plotted as in Figure 8.11. The plot demonstrates that even at a volume ratio of only 0.1%, the leachate causes a significant decline in the solution pH, and the pH continues to decline, albeit less rapidly, as the leachate percentage increases. Despite the fact that T OT NH4 more than doubles when the first increment of leachate is added, the concentration of NH3 (aq) declines by approximately 75%, and its concentration remains lower than that in the original stream water even when the leachate fraction increases to 5%. The reason for this initially surprising result is that the lowering of the pH shifts the speciation of the NH+4 /NH3 (aq) couple toward the protonated species so much that it more than compensates for the increase in T OT NH4 . Thus, this example demonstrates both the complex and sometimes counterintuitive interactions that can control water chemistry when waters mix, and also the ease with which those interactions can be simulated and then interpreted with the help of Visual Minteq or similar programs.
Figure 8.10 Visual Minteq pages when simulating the initial, small doses of leachate solution into the stream. (a) Titration Manager.
400
Figure 8.10 (Continued) (b) Multi-problem/sweep page; (c) Selected sweep results.
8.4.3
Simulating Evaporation or Extraction of Pure Water from a Solution
Visual Minteq also provides a simple approach for one other titration-like process — the gradual removal of pure water from a solution, thereby concentrating the solutes in the solution left behind. Calculations using this feature of the program can simulate the changes in a solution when, for example, some of the water evaporates or is removed by permeation through a reverse osmosis membrane (minor amounts of some dissolved constituents might also come out of solution in both of these processes, but these changes are not accounted for in the program). To utilize this function, we navigate to the “Multi-problem/sweep” page and click on the “Simulate evaporation/ concentration” box. Then, when a value is typed into the “Number of problems” field, a column of fields appears in which we can input the factors by which the original T OT X values are to be multiplied for the various runs. In each run, all the original T OT X values are multiplied by the same value, 401
Figure 8.11 Solution pH and NH3 (aq) concentration for various mixtures of leachate and stream water.
e.g., a value of 2 corresponds to doubling of the concentrations, reflecting removal of one-half of the water compared to the original solution. After these concentration fields are filled in, the remaining steps in the simulation are identical to those for other titration options.
8.5
TITRATIONS WITH WEAK ACIDS OR BASES
Titrations with strong acids or bases are common analytical procedures, and addition of strong acids or bases to change the pH of process waters is also common. Permits for many industries require that the wastewaters they discharge be within one to three pH units of neutrality, and addition of a strong acid or base is often the most convenient and inexpensive way to meet that requirement. However, if a solution contains only small amounts of weak acids or bases in addition to some strong acids or bases, it might have a titration curve like that for the solution containing only HCl in Figure 8.3, in which case adjusting and stabilizing the pH in the desired range can be very difficult. This difficulty is exacerbated if the composition of the water changes irregularly, as is the case in many industries. The challenge of neutralizing a solution whose pH is highly sensitive to the addition of strong acid or base can be largely overcome by titrating with a weak acid or base instead. For instance, say we wished to neutralize a solution containing 10 3 M HCl to a pH between 6.0 and 9.0. If we did so with NaOH we would need to add at least 0.999 ⇥ 10 3 M NaOH, but no more than 1.010 ⇥ 10 3 M; the difficulty of maintaining control of such a system is obvious. 402
On the other hand, if we used Na2 CO3 to neutralize the acidity instead of NaOH, the behavior of the system would be quite different. The first doses of Na2 CO3 would consume two H+ ions per molecule of base added, as the carbonate was converted to carbonic acid: Na2 CO3 + 2 H+ )* H2 CO3 + 2 Na+
(pH less than about 5.5)
However, as more base was added and the pH increased, the carbonic acid that formed previously would begin to dissociate, neutralizing some of the CO2– 3 added. In effect, the titration curve would become flatter due to the presence of a weak acid in the system, the weak acid being the protonated form of the previously added titrant. With each addition of Na2 CO3 the acidity of the solution is partially neutralized, but the concentration of weak acids in solution (H2 CO3 and HCO–3 ) also increases. Curves characterizing the titration of 10 3 M HCl with NaOH and with Na2 CO3 are shown together in Figure 8.12, from which the advantages of titrating with carbonate are apparent. The range of Na2 CO3 additions yielding a final pH between 6 and 9 is 1.2 ⇥ 10 3 to 2.1 ⇥ 10 3 equiv/L (0.6 ⇥ 10 3 to 1.05 ⇥ 10 3 M). This range of acceptable reagent additions, in equiv/L, is more than 80 times as large as when NaOH is used to neutralize the acid. Clearly, at least in this example, the pH can be controlled much more easily and can be maintained at a more stable value by using carbonate instead of hydroxide as the neutralizing agent.
Figure 8.12 Titration of a solution containing 10 3 M HCl with a strong base (NaOH) or with Na2 CO3 . The values on the abscissa are in equivalents of base added per liter of original solution. The dose of NaOH in mmol/L is the same as the value in meq/L, but for Na2 CO3 , the dose in mmol/L is one-half of the value in meq/L (e.g., 1 mmol Na2 CO3 /L = 2 meq/L).
403
A bank of reverse osmosis (RO) membrane modules for desalination of seawater. Although RO is capable of removing almost all the solutes from a solution, the pressure required to force water through the membrane while leaving solutes behind is large, so the process can be energyintensive and costly. For this reason, RO treatment is typically preceded by ’pretreatment’ processes designed to remove as much material as possible from the feed and thereby reserve the RO system for the contaminants that are most difficult to remove. (YuJung Chang.)
Example 8.3 Compute the pH and speciation in a solution after 4 ⇥ 10 3 equiv/L of NaOH is added to 10 3 M H2 SO4 . Repeat the calculation for a system in which the same number of equivalents per liter of base is added as Na2 CO3 . Solution The initial solution contains 2 ⇥ 10 3 equiv/L of strong acid. Each mole of NaOH corresponds to one equivalent, so addition of 4 ⇥ 10 3 equiv/L of NaOH corresponds to addition of 4 ⇥ 10 3 mol/L. The first 2 ⇥ 10 3 mol/L of NaOH would exactly neutralize the acidity of the initial solution, and the second 2 ⇥ 10 3 mol/L would dissociate to Na+ and OH– , increasing the solution pH to ⇠11.3. On the other hand, each mole of Na2 CO3 corresponds to two equivalents of base, so addition of 4 ⇥ 10 3 equiv/L of Na2 CO3 corresponds to addition of 2 ⇥ 10 3 mol/L. The pH of the solution after addition of this amount of Na2 CO3 can be determined by the techniques developed in previous chapters. The fact that the Na2 CO3 is added in a titration after 404
the H2 SO4 is irrelevant to the calculations (because the solution has no memory); the problem is identical to one in which 2 ⇥ 10 3 M Na2 CO3 , and 10 3 M H2 SO4 are mixed initially. The final pH is computed to be 8.3. The pH is lower when Na2 CO3 is added than when an “equivalent” amount of NaOH is added. This is not surprising, since we know that CO2– 3 is a weaker base than OH– . In truth, though, the difference between the two systems is not that CO2– 3 is a weaker base than OH– , but rather that HCO–3 is a weaker base than OH– . That + is, essentially every CO2– 3 ion that is added to the system combines with an H , just – 2– as every OH that is added does. However, once all the CO3 has been converted to HCO–3 , the pH is 8.3, and the HCO–3 acts as a virtually inert ion, rather than a base: almost none of the HCO–3 combines with an H+ to form H2 CO3 . Thus, it is the weakness of HCO–3 as a base that causes the two systems to behave + differently. Even though each CO2– 3 ion has the potential to react with two H ions – 2– (and is equivalent to two OH ions in that respect), most of the CO3 ions react with only one H+ under the specified conditions. Put another way, rather than acting as a strong diprotic base, in this system the CO2– 3 acts as a strong monoprotic base. – The weakness of the new base that is generated when CO2– 3 is protonated (HCO3 ) causes the solution pH to increase much less than when an equivalent dose of NaOH is added.
8.6
THE EFFECT OF THE ACIDITY OF WATER ON TITRATION CURVES
All the titration curves shown thus far are relatively flat at pH< 4 and pH> 10, suggesting that some characteristic of the solutions overrides the effect of weak acids and bases and causes all the systems to behave similarly in these pH ranges. The key to understanding this behavior is the recognition that (1) pH is a logarithmic function, so that when the pH changes from 3.0 to 2.0, the change in H+ activity is 1000 times as great as when pH changes from 6.0 to 5.0, and (2) typical total concentrations of weak acid/base groups in environmental solutions are rarely greater than about 10 2 M. As we have seen, in the region extending from one pH unit above to one pH unit below pKa for a monoprotic acid, a0 changes from 0.09 to 0.91. Therefore, when the pH decreases over such a range, the acid/base group consumes an amount of H+ equal to 0.82 ⇥ T OT A. Correspondingly, even for an acid/base group at a high environmental concentration, the consumption of H+ by the group when the pH changes by two units is, at most, about 0.82 ⇥ 10 2 M, and that large a value would apply only if the the pKa of the acid was exactly in the middle of the range over which the pH changes. 405
However, when the pH of any solution is lowered from 4.0 to 2.0, the H3 O+ concentration increases by 0.99 ⇥ 10 2 M. Thus, over this two-unit change in pH, the amount of H+ needed just to increase the H3 O+ concentration is likely to exceed the amount needed to protonate virtually any weak acid/base group in the solution. As a result, regardless of which weak acid/base groups are present, and regardless of their total concentration, the shape of a titration curve for an environmental solution at pH less than approximately 4.0 is likely to be dominated by the effect of free H+ ions. It is for this reason that all titration curves tend to look alike under those conditions. A similar situation applies at relatively high pH (⇠10 or greater), where the increase in the OH– concentration needed to cause a given increase in pH tends to overwhelm the amount needed to convert any weak acids that are present to their conjugate bases. To generalize, then, the acid/base chemistry of most environmental solutions at low or high pH is dominated by the H+ /H2 O/OH– group and is therefore insensitive to the presence of weaker acids or bases. This point is further elaborated later in the chapter, in the section on pH buffering.
8.7
TITRATION OF SOLUTIONS CONTAINING UNKNOWN ACIDS AND BASES
The ideas developed in the preceding sections allow us to use titration data to infer information about the pKa ’s and quantities of acid/base groups in a solution of unknown composition. For instance, imagine that a solution at an initial pH of 6.8 is split and titrated with acid and base to pH 3.0 and 11.0, respectively, yielding the titration curve shown in Figure 8.13. As indicated in the figure, the requirements of strong acid and strong base to reach these pH endpoints are 3 and 2 meq/L, respectively, so 5 meq/L H+ would be required to titrate the sample from pH 11 to 3. The titration curve for a blank (pure water) is also shown, indicating that additions of 1 meq/L H+ and 1 meq/L OH– are required to reach pH 3 and 11, respectively. Thus, adjusting the sample between pH 3 and 11 requires 3 meq/L more H+ or OH– than a corresponding adjustment of the blank. Based on these results, we could infer that the sample contains a total of 3 meq/L of unidentified weak acids or bases that become protonated between pH 11 and 3. Furthermore, since the titration curve is flattest near pH 6.5, we could infer that the dominant weak acid in the system has a pKa near 6.5. (A more precise approach for estimating Ka values from titration curves is discussed shortly.) This type of information can be obtained without having any idea, in advance, of the identities of the acids in the solution. It is this aspect of titrations that makes them valuable analytical techniques.
406
Figure 8.13 Titration of a sample of unknown composition and of a blank (pure water) with strong acid or base. Addition of strong base is represented as a negative addition of H+ .
Example 8.4 Titration curves for a water sample and a blank solution are shown below. NaOH was added to the blank to adjust its pH to the same value as the sample before the titration was carried out. Several data points are shown on each curve, indicating the amount of acid added and the pH at intermediate stages of the titration; the numerical values associated with these data points are also provided. The analyst hypothesizes that the solution contains a diprotic acid with pKa ’s of 5.5 and 8.5. Do you agree with this interpretation? Why or why not?
407
pH
H+ added (meq/L) Sample Blank
10.10
0
0
8.50
3
0.123
7.10
6
0.126
5.50
11
0.129
3.75
16
0.31
2.35
20
4.62
Solution Titration of the sample from pH 10.10 to 3.75 required the addition of 16.0 meq/L + H , far more than for a blank (0.3 meq/L); on the other hand, the further titration of the sample from pH 3.75 to 2.35 required about as much H+ as a blank (4.3 meq/L vs. 4.0 meq/L). The “extra” H+ required to titrate the sample is attributable to the presence of weak acid/base groups that become protonated during the titration. Apparently, these groups are almost fully protonated once the pH has been lowered to 3.75, since little extra H+ is required to continue the titration to pH 2.35. Visual inspection of the titration curve suggests that the weak bases in the sample become protonated in two distinct steps: one centered near pH 8.5 that accounts for ⇠6 meq/L of extra H+ , and the other centered near pH 5.5 that accounts for ⇠10 meq/L H+ . Since protonation reactions proceed most extensively when the pH is near a pKa , it is reasonable to infer that the solution contains weak bases with pKa ’s near 8.5 and 5.5, as the analyst suggests. The concentration of each weak acid in the sample can be correlated with the concentration of extra H+ that must be added to titrate the sample compared to the blank. If the only weak acid/base group in the sample were diprotic (e.g., H2 A), then the extra H+ needed to convert A2– to HA– (near pH 8.5) would be identical to the amount needed to convert HA– to H2 A (near pH 5.5). Since the concentration of extra H+ required is different in the regions near the two pKa values, the analyst is incorrect in attributing the shape of the titration curve to the presence of a single diprotic acid. Rather, it appears that two different weak monoprotic acids might be present, one with pKa near 5.5 and at a total concentration of 0.010 M, and the other with pKa near 8.5, present at a concentration of 0.006 M. Alternatively, the solution might contain 0.006 M of a diprotic acid with pKa ’s near 5.5 and 8.5, but it would then also have to contain a monoprotic acid with pKa near 5.5, at a concentration of 0.004 M. In either case, the titration curve is not completely consistent with the analyst’s interpretation.
408
8.8 8.8.1
TITRATION EQUIVALENCE POINTS Definition of Equivalence Points
Consider three solutions prepared by adding 10 3 M HAc, HOCl, or NH4 Cl to water. Using the techniques developed in previous chapters, we can determine that the pH of these solutions would be 3.91, 5.27 and 6.12, respectively. We can also compute the expected change in pH as each solution is titrated with a strong base, yielding the curves shown in Figure 8.14.
Figure 8.14 Titration curves for addition of strong base to solutions containing 10 HOCl, or NH4 Cl.
3M
HAc,
When 10 3 mol/L NaOH has been added to these solutions, the titration is said to be at the equivalence point, because the amount of base added (expressed in equivalents) is the same as the amount of acid initially present. Multiprotic acids have one equivalence point for each proton that they can release. For example, a triprotic acid H3 A has three equivalence points, and if a solution of 10 3 M H3 A is titrated with base, the conditions when 10 3 , 2 ⇥ 10 3 , and 3 ⇥ 10 3 mol/L NaOH have been added are referred to as the first, second, and third equivalence points, respectively. For the three solutions of monovalent weak acids shown in Figure 8.14, the chemical additions at the equivalence point include 10 3 M HA (where A is Ac– , OCl– , or NH3 ) and 10 3 M NaOH (plus, in the solution with ammonia, 10 3 M Cl– that was added with the NH+4 ). However, the fact that the solution has no memory means that, if we imagine the chemicals to have been added in any other forms, the real and imaginary solutions will have the same equilibrium composition as long as the overall chemical additions are the same. Based on this idea, we imagine that each acid was first separated into its conjugate base and H+ ions, that the NaOH was separated 409
into Na+ and OH– ions, and that the H+ ions from the acid and the OH– ions from the base were combined before all the constituents were added to the water. The result is that the real solutions have the same compositions as ones made by adding 10 3 M each of Na+ , A– , and HOH (i.e., H2 O) to pure water (with the ammonia-containing solution also having 10 3 M Cl– ). That is, 10 3 M HAc + 10 3 M NaOH = 10 3 M Ac
+ 10 3 M Na+ + 10 3 M H2 O
10 3 M HOCl + 10 3 M NaOH = 10 3 M OCl + 10 3 M Na+ + 10 3 M H2 O 10 3 M NH4 Cl + 10 3 M NaOH = 10 3 M NH3 + 10 3 M Na+ + 10 3 M H2 O The 10 3 M H2 O, Na+ , and Cl– have no effect on the solution pH, so we conclude that, at the equivalence point, each solution is identical to one prepared by adding only 10 3 M of the conjugate base plus some inert salt ions to pure water. Correspondingly, for a 10 3 M solution of a hypothetical triprotic acid H3 A, solutions at the first, second, and third equivalence point would be identical to solutions made by adding 10 3 M NaH2 A, Na2 HA, and Na3 A, respectively, to pure water. The amount of acid or base added during a titration is sometimes quantified by comparison with T OT A. Specifically, the fraction titrated, f , is assigned a value of zero in a solution made by adding T OT A in its most protonated form to water. Then, at various points during a titration with base, f is defined by f⌘
Base added (equiv/L) T OT A (mol/L)
(8.6)
Thus, f equals 1.0 at the (only) equivalence point for a monoprotic acid, or at the first equivalence point for a multiprotic acid; f = 2.0 at the second equivalence point for a multiprotic acid; etc. When f = 0.5, the amount of OH– added is equal to one-half of T OT A. Following the same reasoning as above, a solution of HA that is titrated to f = 0.5 is identical to one in which, instead of adding all the T OT A as HA and then adding a concentration of NaOH equal to 0.5 T OT A, the T OT A was added one-half as HA and one-half as NaA (along with an amount of H2 O equal to 0.5 T OT A). Adding equal amounts of an acid and its conjugate base to pure water yields a solution with pH close to pKa , so we expect the pH to be very near pKa when f = 0.5. Review of Figure 8.14 shows that this expectation is met for the three acids shown. In principle, characterizing the amount of acid or base needed to reach a specified equivalence point could be quite informative. Unfortunately, though, such a characterization is impossible for most real solutions of interest, because the composition of the solution must be known to determine the pH of the equivalence point, and that composition is, typically, what we are trying to determine. Therefore, for environmental solutions, an assumption is commonly made that the carbonate acid/base 410
group is the dominant one in solution, and equivalence points are computed assuming the solution contains a typical environmental concentration of T OT CO3 and no other acid/base groups. The formalities of carrying out this characterization are described next.
8.8.2
Equivalence Points in the Carbonate System
In environmental aquatic systems, T OT CO3 is commonly in the range of 10 4 to 10 2 M. A log C–pH diagram showing the speciation of the carbonate group for three T OT CO3 values over this range is shown in Figure 8.15.
Figure 8.15 Log C–pH diagram for solutions containing 10 4 , 10 3 , and 10
2M
T OT CO3 .
A generic system tableau for solutions in which the carbonate group is the only weak acid/base group is shown in Table 8.3, where the solution is presumed to have been prepared by adding x mol/L H2 CO3 and y mol/L NaOH to pure water. The + T OT H equation for such a solution, using H+ , CO2– 3 , and Na as components, is H+
OH
+ HCO3 + 2 H2 CO3 = 2x
y
(8.7)
The value of y can be expressed as f x, so this equation can be written as H+
OH
+ HCO3 + 2 H2 CO3 = (2
f )x
(8.8)
Noting that x = T OT CO3 and f x = T OT Na, Equation (8.8) can be solved to determine the pH and solute concentrations for any combination of H2 CO3 and NaOH inputs. The results for the equilibrium pH in solutions with 10 4 , 10 3 , and 10 2 M T OT CO3 at f values of 0, 1, and 2 values are indicated in Figure 8.15, and those results are summarized along with a few others in Table 8.4. In addition, titration curves for the solutions are shown in Figure 8.16. The key result from this analysis is 411
that, for T OT CO3 values that are characteristic of environmental systems, the pH’s of the equivalence points fall in fairly narrow ranges (roughly, 4.2-5.2, 8.0-8.4, and 9.9-11.1 for f equal to 0, 1, and 2, respectively). This result serves as the basis for the definitions of alkalinity and acidity, as described next. Table 8.3 System tableau for solutions prepared by adding various amounts of H2 CO3 and NaOH to water
H2 O
H+
Na+
CO2– 3
H2 O H+ Na+ CO2– 3 OH– HCO–3 H2 CO3
1 0 0 0 1 0 0
0 1 0 0 1 1 2
0 0 1 0 0 0 0
0 0 0 1 0 1 1
0.00 0.00 0.00 0.00 14.00 10.33 16.68
H2 CO3 NaOH
0 1
2 1
0 1
1 0
x y
Figure 8.16 Titration curves for solutions of 10 4 , 10 3 , and 10 base as a titrant.
412
2M
H2 CO3 , using strong
Table 8.4 pH of solutions containing various concentrations of H2 CO3 , after titration with strong base to various values of f
T OT CO3 10 10 10
4.0 3.0 2.0
0.0
0.5
f 1.0
5.19 4.68 4.18
6.36 6.35 6.35
8.09 8.30 8.34
1.5
2.0
9.55 10.10 10.30
9.87 10.56 11.13
Example 8.5 On a summer morning, a pond whose acid/base behavior is dominated by the carbonate group is at 20 C and pH 8.10, and it contains 7.42 ⇥ 10 4 M T OT CO3 . (In Chapter 9, we will see that this T OT CO3 causes the water in the pond to be in equilibrium with CO2 (g) in the atmosphere.) Determine the pH at which f = 0 for the pond water. Solution The pH at which f = 0 is, by definition, the pH that the solution would have if the only input had been H2 CO3 , added at a dose of T OT CO3 . In that case, if CO2– 3 is chosen as a component, T OT H would be 2⇥T OT CO3 , or, for the scenario of interest here, 1.48 ⇥ 10 3 M. Knowing T OT H, T OT CO3 , and the relevant equilibrium constants, we can solve for the pH. Using either a log C-pH diagram in conjunction with the known T OT H, or Visual Minteq, we find that pH= 4.75. Given that T OT CO3 is between 10 4 and 10 3 M, this result is consistent with those shown in Figure 8.16 and Table 8.4.
8.9 8.9.1
ALKALINITY AND ACIDITY Conceptual Basis and Definition
Because many anthropogenic inputs to water bodies are acidic, the response of receiving waters to acidification is typically of greater interest than its response to bases. Table 8.4 shows that, for typical environmental solutions, acid addition to the point where f = 0 for the carbonate group yields pH’s in the range of 4.7 ± 0.5. Furthermore, as a general rule, the biodiversity of aquatic food webs declines markedly at pH’s lower than about 5.0. For these reasons, titrations to pH endpoints of 4.5 to 413
5.0 have been adopted as tools to measure of the capacity of solutions to absorb acid without major ecological consequences. The fact that the critical pH for significant ecological effects is usually near the pH corresponding to f = 0 for carbonate-dominated systems reflects the fact that, for millions of years, biological evolution has proceeded in aquatic systems whose pH has been substantially controlled by the carbonate acid/base group. As shown later in this chapter, the presence of this group helps maintain (buffer) the pH at values where f > 0. Therefore, most aquatic organisms have not developed mechanisms to protect themselves from pH values lower than those corresponding to f = 0, i.e., lower than approximately 4.5. For this reason, the alkalinity is sometimes referred to as the acid-neutralizing capacity (ANC) of the water, based on the assumption that once this capacity is used up, severe ecological effects become much more likely. The amount of strong acid needed to titrate a solution to a preselected pH near 4.5 is called the alkalinity (ALK). Alternatively, alkalinity is sometimes defined specifically in terms of the carbonate group, as the amount of strong acid needed to titrate the solution to the pH where f = 0 for the carbonate system.4 The former definition is used most often in practice in the US, both because it is more convenient to carry out the titration to the same pH endpoint in all cases than to compute a solution-specific endpoint for each sample, and because the resulting value is more meaningful for solutions containing several acid/base groups. Nevertheless, to partially account for the effect of T OT CO3 on the pH at which f = 0, the endpoint for alkalinity titrations is sometimes adjusted to reflect the expected solution composition. For instance, Standard Methods5 recommends endpoints between pH 4.3 and 4.9, with the higher values used for solutions with lower alkalinity. The European standard for analyzing alkalinity (ISO 9963-2) specifies that the solution be titrated to pH 5.4 while being bubbled with pure N2 (g) which, as explained in Chapter 9, removes essentially all the T OT CO3 from the solution. This approach accounts for essentially the same contribution to alkalinity from the carbonate group as does the conventional US approach (titrating to pH 4.5 without bubbling), but it yields smaller values for the contributions of other bases in the solution. Unless stated otherwise, titration to a fixed endpoint of pH 4.5 is used as the definition of ALK throughout this text. The alkalinity of a solution is sometimes subdivided into a few categories defined by titration to pH’s corresponding roughly to f values of 2.0, 1.0, and 0.0 for 4 When this definition is used, the pH at f = 0 typically must be determined after the titration has been carried out, since it depends on T OT CO3 , which is usually not known in advance. A numerical strategy known as the Gran titration technique can be used to determine the pH endpoint based on the titration curve. Descriptions of this technique can be found on numerous websites. 5 Standard Methods for the Evaluation of Water and Wastewater, 13th ed. (2012) is a compendium of standardized methods commonly used in water quality analysis. It is jointly authored by the American Public Health Association (APHA), the American Water Works Association (AWWA), and the Water Environment Federation (WEF), and is published by APHA (New York, NY).
414
the carbonate system. Alkalinities based on these endpoints are called the caustic alkalinity, the carbonate or phenolphthalein alkalinity, and the total alkalinity, respectively. If the term ‘alkalinity’ is used without a qualifier, it usually refers to the total alkalinity. The term ‘phenolphthalein alkalinity’ refers to the fact that the organic chemical phenolphthalein changes color at pH near 8.3 (i.e., near f = 1.0). Before pH meters and electrodes were as inexpensive as they are now, alkalinity was widely evaluated by adding this indicator to the solution being titrated and using the color change to signal the condition when f = 1.0. The experimental determination of alkalinity using the conventional endpoint of pH 4.5 is illustrated in Figure 8.17, which is a reproduction of Figure 8.13 with some additional notes. Based on the amount of acid required to titrate the sample to pH 4.5, the alkalinity of the sample is 2.05 meq/L.6
Figure 8.17 Titration curve from Figure 8.13, with some additional information related to the alkalinity.
A concept that is exactly analogous to alkalinity but that measures the baseneutralizing capacity of a solution is the acidity. This quantity represents the concentration of OH– ions that must be added to titrate the solution to pH near 10.5 (near f = 2.0 for a solution dominated by the carbonate group). If the initial pH is < 4.0, the OH– addition required to raise the pH to ⇠4.5 is identified as the mineral acidity. (Because such waters typically contain strong acids, an endpoint at pH 3.7 is recommended by Standard Methods.) Note that, by definition, alkalinity and acidity are concentration-based, not activity-based, quantities. That is, if one mmol/L of a weak base (e.g., HCO–3 ) is present 6 The
alkalinity is sometimes defined as the extra acid needed to titrate the solution to pH 4.5, compared to the amount needed to titrate a blank (a solution containing no weak acids or bases) over the same pH range. However, the correction for the blank is small if the initial pH is less than 10, so it is often ignored.
415
in the original solution, and essentially all of this base becomes protonated (e.g., to H2 CO3 ) when the solution is titrated to the alkalinity endpoint, then the base consumes one mmol/L of H+ in the process and hence contributes one meq/L of alkalinity to the original solution. This calculation applies regardless of whether the base behaves as an ideal or highly nonideal solute. Of course, nonideality can affect the equilibrium concentration of the base in the original and titrated solutions, but that is a separate matter; the point is that the alkalinity and acidity are calculated based on the change in species’ concentrations during the titration, not their activities.
8.9.2
Computing Alkalinity from Titration Data or Based on Solution Composition
When we titrate a solution with strong acid, T OT H increases by an amount equal to amount of acid added. Because, by definition, the alkalinity equals the amount of strong acid that must be added to reach a specific pH, the alkalinity can be equated with the increase in T OT H when a solution is titrated to the endpoint. One mole of H+ corresponds to one acid equivalent, so we can write ALK =
equiv H+ added during titration
L
=
mol H+ added during titration
L
= T OT Hafter
T OT Hbefore
(8.9)
where the subscripts refer to the solution before and after the alkalinity titration. Thus, the alkalinity of any solution equals the increase in T OT H when the solution is titrated to the alkalinity endpoint. Note that, when calculating this change in T OT H, the numerical values of T OT Hbefore and T OT Hafter individually depend on which chemicals we choose as components, but the difference T OT Hafter T OT Hbefore is independent of these choices.
8.9.3
Assessing the Contribution of Different Species to Alkalinity
We next explore the relative contributions of different solutes to alkalinity by considering a hypothetical domestic wastewater at pH 8.0, containing 5 ⇥ 10 3 M T OT CO3 , 1 ⇥ 10 4 M T OT PO4 , and 1 ⇥ 10 3 M T OT NH4 as its major weak acid/ base groups. We begin by preparing the upper portion of the system tableau for the solution before and after it has been titrated with HCl. This portion of the tableau is shown in Table 8.5 and is identical before and after the titration, except for the inclusion of Cl– as a component and a species after the titration; to remind us of this distinction, the column and row for Cl– are shown in italic. Because Cl– is the only new species added during the titration, and this species does not contribute to T OT H, the same generic expression can be used to calculate T OT H before and after the titration, although the numerical values of the individual terms in the equation do change when the titration is carried out. This generic expression is 416
T OT H = H+
OH
+ HCO3 + 2 H2 CO3 + HPO24
+ 2 H2 PO4 + 3 H3 PO4
NH3
(8.10)
As noted in the preceding section, the alkalinity of any solution equals the change in T OT H when the solution is titrated with strong acid to the alkalinity endpoint. The alkalinity of the original solution can therefore be computed as ALK = H+
final init
+ 3 H3 PO4
final + init
OH
final +2 init
HCO3
H2 PO4
final +2 init
final + init
H2 CO3
HPO24
final init
final init
NH3
final init
(8.11)
where the designation final init indicates that the value being computed equals the species concentration after the titration minus that in the original solution. The concentration of each species that is part of a weak acid/base group can be represented as a product of the form ai T OT A. Making that substitution and inserting values for the pH at the beginning and end of the titration, we can rewrite Equation (8.11) as Table 8.5 Upper (equilibrium species) portion of the tableau for a simulated domestic wastewater containing carbonate, phosphate, and ammonia species, titrated with HCl
H2 O H+ CO2– 3 PO3– 4 NH+4 Cl– OH– HCO–3 H2 CO3 HPO2– 4 H2 PO–4 H3 PO4 NH3
H2 O
H+
CO2– 3
PO3– 4
NH+4
Cl–
1 0 0 0 0 0 1 0 0 0 0 0 0
0 1 0 0 0 0 1 1 2 1 2 3 1
0 0 1 0 0 0 0 1 1 0 0 0 0
0 0 0 1 0 0 0 0 0 1 1 1 0
0 0 0 0 1 0 0 0 0 0 0 0 1
0 0 0 0 0 1 0 0 0 0 0 0 0
417
0.00 0.00 0.00 0.00 0.00 0.00 14.00 10.33 16.68 12.38 19.57 21.72 9.24
ALK = H+
pH4.5 pH8.0
OH
pH4.5 + T OT CO3 pH8.0
⇣ ⌘pH4.5 a1,CO3 + 2a0,CO3 pH8.0 ⌘pH4.5
⇣ + T OT PO4 3a0,PO4 + 2a1,PO4 + a2,PO4 ⇣ ⌘pH4.5 T OT NH3 a1,NH33 pH8.0
pH8.0
(8.12)
Although Equations (8.11) and (8.12) look cumbersome, we will simplify them shortly. They are written out in detail to highlight some important points about the alkalinity concept. First, they make clear that every weak acid or base in solution affects the alkalinity; when the pH is lowered to the alkalinity endpoint, each acid/base group becomes more protonated, and all the protons needed to carry out this process contribute to the alkalinity. Second, the equation makes clear that the contributions of different acid/base groups to the alkalinity are independent and additive. That is, if a solution at pH 8.0 contains phosphate species, the contribution of those species to the⌘alkalinity equals ⇣ the difference in the product T OT PO4 3a0,PO4 + 2a1,PO4 + a2,PO4 between pH 4.5 and pH 8.0, regardless of how much or what types of other acid/base groups are in solution. This makes the alkalinity a conservative quantity. That is, the contribution of each acid/base group to the alkalinity of a solution can be computed based only on its concentration and the initial pH, and the alkalinity of the whole solution can then be computed as the sum of the alkalinities contributed by the individual groups. Finally, the contribution of an acid/base group to the alkalinity is proportional to the total concentration of that group and to the change in its speciation when the solution is titrated to the alkalinity endpoint. For the example solution, the computed contributions of carbonate, phosphate, and ammonia species are 4.84, 8.66 ⇥ 10 2 , and 5.39 ⇥ 10 2 meq/L, respectively. When the contributions of H+ (3.26 ⇥ 10 2 meq/L) and OH– (1.01 ⇥ 10 3 meq/L) are included, the total alkalinity of the solution is determined to be 5.02 meq/L.7 For this solution, the contribution of the carbonate acid/base group to ALK is significant, because it is present at a substantial total concentration and because its speciation changes from almost all HCO–3 to almost all H2 CO3 when the solution is titrated from pH 8.0 to 4.5. Thus, essentially every CO3 -containing molecule in the original solution consumes one H+ ion during the titration. Most of the dissolved phosphate is present as HPO2– 4 in the original solution as – H2 PO4 in the titrated solution. Therefore, as is true for CO3 -containing molecules, one H+ ion is consumed for almost every PO4 -containing molecule initially present. 7 Note that, for the terms in Equation (8.11) or (8.12) that are preceded by a minus sign, the value at pH 4.5 is less than that at pH 8.0, so the net effect of each of those terms is a positive contribution to alkalinity.
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However, since T OT PO4 is only 2% as large as T OT CO3 , the contribution of phosphate species to ALK is much smaller than that of carbonate species. The total concentration of the NH3 group is less than that of the CO3 group by a factor of five in the original solution. However, the contribution of the NH3 group to ALK is much less than can be accounted for just by that factor of five. The reason is that almost all the T OT NH3 is present as NH+4 in the initial, pH 8.0 solution and also at the end of the titration (⇠ 95% initially, > 99% at pH 4.7). Because such a small a portion of T OT NH3 is present as NH3 initially, the NH+4 /NH3 group has a very small capacity to combine with H+ ions during the titration; this group therefore contributes very little to ALK (even less than the phosphate group). The alkalinity of the original solution and the contributions of individual species or groups of species to that value can be determined easily using Visual Minteq. To do so, we use the “Multi-problem/sweep” menu to carry out runs at two fixed pH’s — the initial pH and the alkalinity endpoint. For instance, in this case, we would specify pH 8.0 as the initial value and instruct the program to complete two runs, using a pH increment of 3.5 (so that the pH of the second run is 8.0 3.5, or 4.5). We then indicate that all the species that contribute to alkalinity should be reported in the “Selected sweep” results, and export those results to Excel. The alkalinity of the original solution can be determined by comparing the values of T OT H in the two simulations (as shown on the ‘Equilibrated mass distribution’ page), and contributions of different species to the alkalinity can be determined based on the change in their concentration when the titration is carried out [e.g., based on Equation (8.11) for this example]. Example 8.6 (a) Determine the pH and alkalinity (in meq/L) of a solution of 1.5 ⇥ 10 3 M Na2 CO3 plus 5 ⇥ 10 4 M NaHS. How would this alkalinity be expressed in mg/L as CaCO3 ? (b) Interpret the alkalinity qualitatively, in terms of how protons are consumed during the titration from the initial pH to pH 4.5. Assume that the titrant is HCl. Solution (a) A log C–pH diagram for the system is shown below, along with a system tableau. Sulfide ion (S2– ) is present at concentrations less than 10 8 M throughout the pH range shown and therefore does not appear in the graph. In the tableau, entries shown in normal text apply both before and after the titration, whereas those shown in italic are relevant only in the titrated solution. The amount of HCl added during the titration is unknown and is designated as x. 419
Because the number of moles per liter of HCl added to reach pH 4.5 is the alkalinity, x =ALK.
H2 O
H+
CO2– 3
HS–
Na+
Cl–
1
0
0
0
0
0
0.00
H+
0
1
0
0
0
0
0.00
CO2– 3 –
0
0
1
0
0
0
0.00
H2 O
HS
0
0
0
1
0
0
0.00
+
0
0
0
0
1
0
0.00
–
Cl
0
0
0
0
0
1
0.00
OH–
1
1
0
0
0
0
14.00
HCO–3
0
1
1
0
0
0
6.35
H2 CO3
0
2
1
0
0
0
16.68
H2 S
0
1
0
1
0
0
7.02
S
0
1
0
1
0
0
17.4
Na2 CO3
0
0
1
0
2
0
NaHS
0
0
0
1
1
0
1.5 ⇥ 10
HCl
0
1
0
0
0
1
Na
2–
420
5.0 ⇥ 10 x
3 4
The T OT H equation and PC for the initial solution (before the titration) are
T OT H equation:
(H+ )
(OH )+(HCO3 ) + 2(H2 CO3 ) + (H2 S)
(S2 )
= 0 [Na2 CO3 ]in + 0 [NaHS]in = 0 Proton condition:
(H+ ) + (HCO3 ) + 2(H2 CO3 ) + (H2 S) = (OH ) + (S2 )
The pH of the solution before the titration can be determined either by solving the PC in conjunction with the log C–pH diagram or by using Visual Minteq. To use the latter approach, we specify the input concentrations of components as T OT CO3 = 1.5 ⇥ 10 3 , T OT HS = 5.0 ⇥ 10 4 , T OT Na = 3.5 ⇥ 10 3 , and T OT H = 0, and also specify that the ionic strength is zero (to apply the assumption of ideal solute behavior). The output screen (Figure 8.18) indicates that the pH of the solution is 10.66, corresponding to the point where (HCO3 ) = (OH ) on the log C–pH diagram. As commonly occurs, species that are not normally considered when preparing log C–pH diagrams manually (e.g., NaCO–3 , NaHCO3 (aq), and NaOH(aq)) appear in the output table because they are present in the Visual Minteq database. These species contribute only slightly to the total concentrations of the components, so ignoring them has no significant effect on the computed pH or concentrations of other species.
Figure 8.18 Visual Minteq output for a solution of 1.5 ⇥ 10 NaHS.
421
3M
Na2 CO3 plus 5.0 ⇥ 10
4M
The alkalinity equals the change in T OT H when the solution is titrated with strong acid to the alkalinity endpoint. In this case, T OT H before the titration is zero, so ALK is simply T OT H after the titration, which can be computed as T OT Hafter = H+
OH
+ HCO3 + 2 H2 CO3 + (H2 S)
S2
where all the values are for the solution at pH 4.5. If we are carrying out the analysis manually, we can read the concentration values directly from the log C–pH diagram to find ALK = T OT Hafter = H+ = 10
4.5
10
= 3.51 ⇥ 10
3
9.5
+ 10
OH + HCO3 +2 H2 CO3 +(H2 S) 4.68
+ 2(10
2.83
) + 10
3.30
S2
0
equiv/L = 3.51 meq/L
Alternatively, we can use Visual Minteq to determine T OT H after the titration by running the program with the same inputs of T OT CO3 , T OT HS, and T OT Na as were used previously, but specifying that pH= 4.5. In this case, the program adds H+ as needed until the specified pH is reached. The value of T OT H at that point (and therefore ALK) is reported on the “Equilibrated mass distribution” output page as 3.51 ⇥ 10 3 M, consistent with the preceding calculation using values from the log C–pH diagram. To convert the alkalinity from units of equiv/L to mg/L as CaCO3 , recall that 1 meq ALK corresponds to 50 mg CaCO3 . Therefore, ✓ ◆ ⇣ mg CaCO3 meq ⌘ mg ALK = 3.51 50 = 176 as CaCO3 L meq L
. interpret these results qualitatively, we note that the initial solution was (b) To prepared with Na2 CO3 and NaHS. Since the Na+ ions are released when the molecules dissolve and are unreactive thereafter, the input species that can con– tribute to the alkalinity are CO2– 3 and HS . The alkalinity titration lowers the – pH to 4.5, converting the CO2– 3 and HS almost entirely to H2 CO3 and H2 S, respectively. Thus, we expect each mole of carbonate ion that was added to the solution to consume two moles of H+ and thereby contribute two equivalents of alkalinity, and each mole of bisulfide ion to contribute one equivalent of alkalinity. The alkalinity of the original solution can therefore be estimated as 2(Na2 CO3 )added + (NaHS)added . This sum equals 3.50 meq/L, which is indeed very close to the computed result; the difference is due primarily to the small error in the approximation that the ions are 100% protonated at the final pH and the failure to take into account the 10 4.5 M free H+ present at the end of the titration.
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Because carbonate species so often contribute substantially to the alkalinity, it is useful to derive an expression for the contribution of just the carbonate species to alkalinity as a function of pH and T OT CO3 . Equations (8.11) and (8.12) make it clear that this contribution can be expressed as ⇥ ⇤final ⇥ ⇤final ALKCO3 ⌘ HCO3 init + 2 H2 CO3 init = T OT CO3 (a1 + 2a0 )final init
(8.13a) (8.13b)
where ‘init’ and ‘final’ refer to the pH values in the initial solution and at the alkalinity endpoint, respectively. These expressions are based on the idea that the alkalinity equals the change in T OT H during the titration and on the choice of CO2– 3 as a component for the computation of T OT H. The value of alkalinity does not depend on which chemicals are chosen as components, so it is equally acceptable to compute ALKCO3 using H2 CO3 as a component instead of CO2– 3 . In that case, the coefficients – 2– on HCO3 and CO3 in the T OT H equation are 1 and 2, respectively, and the expression for ALK becomes ALKCO3 ⌘
⇥
HCO3
⇤final init
= T OT CO3 ( a1
⇥ ⇤final 2 CO23 init 2a2 )final init
= T OT CO3 (a1 + 2a2 )init final
(8.14a) (8.14b) (8.14c)
To a very good approximation, all of the HCO–3 and CO2– 3 in the initial solution become fully protonated during the titration, so the concentrations of those species are negligible in the titrated solution. Based on that approximation, we can express the alkalinity contribution of just the carbonate species as ALKCO3 ⇡ [HCO3 ]init + 2[CO23 ]init = (a1 + 2a2 )T OT CO3
(8.15a) (8.15b)
Correspondingly, T OT CO3 can be determined from ALKCO3 and the pH by T OT CO3 =
ALKCO3
a1 + 2a2
(8.16)
The usefulness of the preceding equations will be illustrated shortly.
8.9.4
The Contribution of Added Species to Alkalinity
The preceding discussion has focused on the contribution to alkalinity of the species that are already in a solution. The same logic leads to an interesting implication 423
regarding the alkalinity contributions of species that are added to a solution. Specifically, keeping in mind that the alkalinity contributed by any species equals the concentration of H+ with which that species combines when it is in a solution that is titrated to pH 4.5, we can write: pH 4.5
ALKadded i = T OT Hi |as added
(8.17)
where the right side of the equation represents the difference in T OT H associated with the added i when it is converted from the form in which it is added to the distribution of species that it forms at pH 4.5. For example, phosphate species are present almost entirely as H2 PO–4 at pH 4.5. Therefore, if Na3 PO4 is added to a solution, essentially every added Na3 PO4 molecule will combine with two H+ ions when the solution is titrated to pH 4.5. Correspondingly, the alkalinity of the solution will increase by 2 meq/L for each mmol/L of Na3 PO4 added, regardless of the pH of the original solution. This result applies even though the added chemical might become substantially protonated and thereby increase the pH when it is added to some solutions (for instance, if the initial solution pH was ⇠6), but protonate minimally and have almost no effect on the pH when added to others (e.g., a solution at a much higher initial pH). Table 8.6 illustrates this result for a few relatively simple solutions and also demonstrates the additivity of the alkalinity contributions from different species. For example, solution (b) in the table represents the result of adding 1.5 ⇥ 10 3 mol/L Na2 CO3 to solution (a) (pure water). The table indicates that this addition increases the alkalinity of the solution by 2.98 meq/L. A similar comparison for solutions (a) and (c) indicates that adding 5 ⇥ 10 4 mol/L NaOCl to pure water increases the alkalinity by 0.50 meq/L. Because ALK is conservative, the alkalinity of solution (g), which can be viewed as a combination of solutions (a), (b), and (c), is 0.03 + 2.98 + 0.50, or 3.51 meq/L. Example 8.7 How does addition of the following substances affect the alkalinity and pH of the solution? Assume that the initial solution contains only water and strong acid or base. (a) 10
4M
NH3 , if the initial pH is 10.0.
(b) 10
4M
NH3 , if the initial pH is 6.0.
(c) 10
4M
NH4 Cl, if the initial pH is 6.0.
Solution As noted in the preceding discussion, the increment in alkalinity associated with addition of a given chemical does not depend on the initial pH; it depends only on 424
Table 8.6 pH and alkalinity of various ideal solutions†
Solution (a) No added chemicals (pure water) (b) 1.5 ⇥ 10 3 M Na2 CO3 (c) 5 ⇥ 10 4 M NaOCl (d) 5 ⇥ 10 4 M NaAc (e) 2 ⇥ 10 3 M NaAc (f) 10 3 M HOCl (g) 1.5 ⇥ 10 3 M Na2 CO3 +5 ⇥ 10 4 M NaOCl (h) 1.5 ⇥ 10 3 M Na2 CO3 +10 3 M HOCl (i) 1.5 ⇥ 10 3 M Na2 CO3 +5 ⇥ 10 4 M NaAc †
pH
ALK, meq/L
7.00 10.67 9.11 7.73 8.03 5.27 10.67 9.92 10.67
0.03 3.01 0.53 0.35 1.32 0.03 3.51 2.51 3.33
Alkalinity defined by titration to pH 4.5.
the change in the chemical’s speciation between the form in which it is added and the end of the alkalinity titration. The effect on solution pH, on the other hand, depends on how the chemical interacts with protons when it is added to the solution. (a) At pH 10, most of the T OT NH3 in solution is present as the neutral species NH3 (aq). NH3 added to such a solution does not protonate very significantly, so it does not affect the solution pH very much. However, NH3 does become almost fully protonated when the solution is titrated to pH 4.5. Therefore, each NH3 molecule added to solution combines with one H+ by the time the alkalinity titration has been completed, regardless of the initial pH. As a result, the addition of 10 4 M NH3 contributes 10 4 equiv/L of alkalinity, but it has only a minimal effect on solution pH (calculations with Visual Minteq indicate that the pH would increase only to 10.05). (b) Because the contribution of a chemical to alkalinity does not depend on the composition of the solution to which it is added, addition of 10 4 M NH3 contributes 10 4 equiv/L of alkalinity to this solution, as it does in part (a). At pH 6.0, however, almost all the NH3 added to a solution becomes protonated, so the pH of the solution increases much more in this solution than in part (a); the computed equilibrium pH is 8.54. (c) Addition of 10 4 M NH4 Cl adds essentially no alkalinity to a solution, because the ions that are added (NH+4 , Cl– ) are in the same form at the end of the alkalinity titration as the form in which they were added. If the initial pH of the solution is 6.0, addition of NH4 Cl will also not induce a significant 425
pH change, since the added ions do not combine with or release protons to a significant extent at that pH; the equilibrium pH is 5.97.
8.9.5
The Implications of Alkalinity as a Conservative Quantity
Because alkalinity is conservative, if a given solution contains x meq of alkalinity and a substance is added that has an alkalinity of y meq, the alkalinity of the new solution is exactly x + y meq. Similarly, if two solutions with different alkalinities are mixed, the alkalinity of the mixture is the weighted average of the alkalinities of the original solutions (weighted according to their mass or volume). This property of alkalinity is demonstrated mathematically by Equations (8.11) and (8.12) and is also apparent from computation of the alkalinity of solutions containing various combinations of weak acids and bases. In contrast to the alkalinity, it is clear from all the examples in this and preceding chapters that pH is a highly nonlinear, nonconservative parameter. That is, one cannot predict the pH of a solution generated by mixing two other solutions based simply on the pH values of the original solutions; the complete acid/base composition of those couples must be known in order to predict the pH of the mixture. Students often find alkalinity to be a challenging concept, and the contrast between the conservative nature of alkalinity and the nonconservative nature of pH is often at the core of this confusion. This contrast leads to the counterintuitive result that, under certain circumstances, some chemicals can affect the alkalinity of a solution without changing the pH, others can change the pH without affecting the alkalinity, and some can even cause the alkalinity and pH to change in opposite directions. These possibilities emphasize the key conceptual difference between pH and alkalinity: pH describes the current acid/base state of a solution, while alkalinity describes how difficult it is to change that state to a specified, different state. The fact that alkalinity is conservative can be extremely useful for solving certain types of problems, especially those related to systems in which carbon dioxide is exchanged between a solution and a gas phase. Gas-liquid equilibrium is described in detail in Chapter 9, but it is worthwhile to explore the effect of CO2 exchange on solution pH and alkalinity at this point. When CO2 enters solution and becomes hydrated, it forms carbonic acid (H2 CO3 ), which can then deprotonate. As a result, dissolution of CO2 tends to lower solution pH. Similarly, removal of CO2 from solution by transfer into a gas phase causes the solution pH to increase. Now consider what happens if some CO2 (g) dissolves and the solution is then titrated to pH 4.5. At the end of the titration, virtually 100% of the carbonate in the system is present as H2 CO3 , which can be thought of as a CO2 molecule bonded to H2 O; i.e., essentially all the carbon that entered solution from the gas phase is in a form that has neither gained nor released H+ compared with 426
the form in which it originally entered the solution. As a result, and consistent with Equation (8.17), CO2 dissolution has no effect on the solution alkalinity; the same is true, of course, for CO2 transfer from solution into the gas phase.8
8.9.6
pH and Alkalinity in Carbonate-Dominated Systems
As has been noted, the alkalinity of many solutions of environmental interest is dominated by the carbonate acid/base group. In such cases, it is convenient to compute alkalinity considering that group as the only weak acid/base group in solution. If we compute T OT H using H+ and H2 CO3 as components, the expressions for T OT H and alkalinity are T OT H = (H+ ) ALK = T OT Hfinal = H+
(OH )
2(CO23 )
(HCO3 )
(8.18)
T OT Hinit
final init
(8.19a)
final init
OH
HCO3
final init
2 CO23
final init
(8.19b)
We can reverse the signs and the final/init designations on the terms on the right side, and then expand the resulting expression to obtain ALK =
H+
init + final
=
H+
init
OH
+ H+
HCO3
final
init + final
final
HCO3
+ OH
+ 2 CO23
init
init
init +2 final
OH
2 CO23
CO23
final
init final
+ HCO3
init
(8.20)
final
As was pointed out previously, at the titration endpoint, the vast majority of T OT CO3 is present as H2 CO3 , and the concentrations of HCO–3 and CO2– 3 are both much less than in the initial solution. Typically, the final concentrations of H+ and OH– are also much less than the initial concentration of one or both carbonate species. Thus, all the terms for final concentrations in Equation (8.20) can be ignored, and the alkalinity can be approximated based strictly on concentration values in the initial solution, as + 2 CO23
ALK =
H+
init
+ OH
init
+ HCO3
=
H+
init
+ OH
init
+ T OT CO3 (a1 + a2 )init
8 The
init
init
(8.21a) (8.21b)
effect on ALK of addition or removal of H2 CO3 is exactly zero if ALK is determined by titration to f = 0 for the carbonate system. In that case, the endpoint of the titration is slightly different for the solution before and after the H2 CO3 exchange. If ALK is defined by titration to a fixed pH endpoint, the effect of H2 CO3 addition or removal is almost always negligible, but it is not exactly zero. Specifically, addition of n mol of H2 CO3 to a solution decreases ALK by a1 n equivalents, where a1 is evaluated at the endpoint of the titration. For instance, if the titration endpoint is 4.5, each mole of H2 CO3 added to an ideal solution decreases ALK by 0.014 equiv.
427
If the pH of the initial solution is less than ⇠9, the equation can be simplified even more to ALK = HCO3 init (8.22) Unfortunately, Equation (8.22) can add to confusion about the effect of H2 CO3 on alkalinity. H2 CO3 addition usually leads to an increase in the concentration of HCO–3 , and the approximation suggests that such an increase should increase ALK. However, note that the generation of HCO–3 (or CO2– 3 ) when H2 CO3 dissociates is accompanied by an equivalent generation of H+ : H2 CO3 )* HCO3 + H+ )* CO23 + 2 H+ If we apply one of the more complete expressions for the alkalinity of a carbonatedominated solution [Equation (8.20) or (8.21)] to these reactions, we see that the increase in ALK caused by the increases in the HCO–3 and CO2– 3 concentrations when CO2 dissolves and dissociates is exactly counteracted by the increase in the H+ concentration, so the addition of CO2 has no net effect on ALK, as we saw earlier. Example 8.8 The alkalinity of a solution at pH 8.0 containing 5 ⇥ 10 3 M T OT CO3 , 10 4 M T OT PO4 , and 10 3 M T OT NH4 was computed earlier to be 5.02 meq/L. Compare that value with the alkalinity that would be determined using Equations (8.21) and (8.22). Solution The a values for the carbonate system at pH 8.0 are a0 = 0.0218, a1 = 0.974, and a2 = 0.0046. The alkalinity computed according to Equation (8.21b) or (8.22) is therefore (8.21b) :
ALK =
H+
pH=8.0 3
+ OH
pH=8.0
+ T OT CO3 (a1 + 2a2)pH=8.0
= 4.92 ⇥ 10 eq/L (8.22) :
ALK = T OT CO3 ⇥ a1,pH=8.0 = 4.87 ⇥ 10 3 eq/L
These two estimates of alkalinity differ by, at most, 3% from the value computed with the more complete equation that considers all contributions. The approximations in Equations (8.21) and (8.22) therefore seem acceptable in this case. Equations (8.21) and (8.22) are convenient approximations, but their validity and implications have sometimes been overemphasized; indeed, in many introductory treatments of alkalinity, Equation (8.21) is given as the definition of alkalinity, which is problematic for two reasons. First, and most obviously, that claim is simply not 428
true — the alkalinity always includes contributions from every acid/base group in a solution. Even if the contributions of noncarbonate species are small, they should be acknowledged in the defining equation. Second, and just as importantly, Equation (8.21) gives the impression that the alkalinity is a measure of what species are present in a solution, whereas it is more correctly thought of as an indicator of how much H+ must be added to induce a specified change in the solution speciation. These two ideas are related, of course, but they are not the same thing. As shown in the following example, some acid/base groups make a contribution to alkalinity that is considerably less than their total concentration. This result is very difficult to rationalize if we focus strictly on the composition of the initial solution, whereas recognizing that alkalinity is associated with changes in speciation provides a simple explanation for the result. Example 8.9 A highly diluted leachate solution from a landfill is at pH 6.6 and contains 2 ⇥ 10 3 M of a group of organic acids whose acid/base behavior is essentially identical to that of acetic acid. The solution also contains 1.5 ⇥ 10 3 M T OT CO3 . What is the alkalinity of the solution, and what is the contribution of the organic acids to the alkalinity? Compare the alkalinity contributed by these acids to their total molar concentration and explain the difference. Solution The alkalinity can be computed as the change in T OT H when the solution is titrated to the alkalinity endpoint, i.e., from pH 6.6 to 4.5. Treating the organic acids as acetic acid (HAc), this calculation is ⇥ ⇤pH 4.5 ALK = H+ OH + HCO3 + 2 H2 CO3 + (HAc) pH 6.6
To carry out calculations using Visual Minteq, we first input T OT CO3 and T OT Ac. Then, we use the “Multi-problem/sweep” feature to instruct the program to carry out two runs at fixed pH’s, with the initial run at pH 6.6 and an increment of 2.1 pH units between runs. The reported initial and final values of T OT H are 2.07 ⇥ 10 3 M and 4.30 ⇥ 10 3 M, respectively, so the alkalinity is 2.23 ⇥ 10 3 equiv/L. pH 4.5 The contribution of the organic acids to the alkalinity equals (HAc)pH 6.6 . The HAc concentrations are reported by Visual Minteq to be 2.83 ⇥ 10 5 M and 1.29 ⇥ 10 3 M at pH 4.5 and 6.6, respectively, so the contribution of the organic acids to the alkalinity is 1.26 ⇥ 10 3 equiv/L, or 63% of T OT Ac. The explanation for this fraction lies in the a0 values at the two pH’s: HAc accounts for 1.4% of T OT Ac at the initial pH (6.6) and 64.4% of T OT Ac at the end of the titration. Therefore, 63% of the organic acids become protonated during the alkalinity titration, consuming an amount of H+ equal to 63% of their total concentration.
429
Since the CO2 system is of such overriding importance in most natural and many engineered water systems, a number of approaches have been developed to show the relationships among pH, T OT CO3 , and ALK in systems where the carbonate system provides essentially all the alkalinity. In such systems, only two independent variables are needed to completely define the system. For instance, knowing pH and T OT CO3 , we can compute the concentration of all species and ALK. Knowing – the concentrations of CO2– 3 and ALK, or of HCO3 and H2 CO3 , we could do the same (assuming, as usual, either that the solutes behave ideally or that their activity coefficients are known). Log C–pH graphs are very useful for quickly assessing the effect of a pH change on speciation in a system with fixed T OT CO3 , but they are not particularly convenient for evaluating the effect of changes in other system variables. For instance, to show the effect on solution composition of adding Na2 CO3 , we would have to prepare a new log C–pH diagram for each Na2 CO3 addition, because each system would have a different T OT CO3 . Using the same set of equations as are used to prepare log C–pH graphs, Deffeyes and others have prepared graphs which are easy to use to determine changes in solution composition in response to the addition or removal of a known quantity of H+ , OH– , any of the species of the carbonate acid/base group, or T OT CO3 . These graphs, which are simply a different way of representing the mass balance, T OT H, and equilibrium equations, are easier to use than log C–pH diagrams for some applications, but they have the disadvantage that they are not applicable if any other weak acid/base group affects the solution acid/base behavior. An example of such a graph is shown in Figure 8.19.
Additions of acids or bases to adjust the pH are common steps in water and wastewater treatment plants, either to improve the performance of a treatment process or to meet discharge guidelines. This image shows the controls in front of a large tank of NaOH at a wastewater plant.
430
Figure 8.19 Relationships among pH, T OT CO3 , and ALK for solutions whose acid/base properties are dominated by the carbonate system. Unlike log C–pH diagrams, these diagrams are easy to apply to systems with changing T OT CO3 . The broken lines and white circle in figure (a) are explained in the solution to Example 8.10. Adapted from Deffeyes (1965).9
Example 8.10 Photosynthesis can lead to substantial increases in the pH of natural waters during the day, due to removal of H2 CO3 from the water and its conversion to new biomass. The reverse reaction occurs at night, when respiration causes organic matter to decay and H2 CO3 to be released back into solution. The relatively large pH swings in one system are illustrated in Figure 8.20. 9 Deffeyes,
K.S. (1965) “Carbonate equilibria: a graphic and algebraic approach.” Limnology and Oceanography, 10, 412-426.
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Figure 8.20 pH fluctuations in the White River near Enumclaw, WA, during a two-week period in late September to early October, 2002. The pH increases during the day and reaches a peak in late afternoon, due to depletion of the H2 CO3 concentration by photosynthesis. It then drops overnight as bacterial respiration replenishes the H2 CO3 . Data from Ebbert (2003).10
(a) In Example 8.5, we considered a pond at pH 8.10 containing 7.42 ⇥ 10 4 M T OT CO3 and found that the carbonic acid equivalence point ( f = 0.0) was at pH 4.75. What is the alkalinity of the pond water, if the alkalinity is defined by titration to the equivalence point? (b) During the course of the day, algal photosynthesis removes 75% of the T OT CO3 by converting H2 CO3 and H2 O to algal biomass and O2 .11 What are the alkalinity and pH of the resulting solution? Assume that the algal biomass and O2 do not participate in any acid/base reactions. Solution (a) Because we know T OT CO3 and the pH of the initial solution, we can determine the concentrations of the individual carbonate species in that solution. Then, using H+ and CO2– 3 as components, we can compute T OT H as T OT H = 2(H2 CO3 ) + (HCO3 ) + (OH ) 10 “Water
(H+ )
temperature, specific conductance, pH, and dissolved-oxygen concentrations in the lower White River and the Puyallup River estuary, Washington, August-October 2002.” Water-Resources Investigations Report 03-4177, USGS, 2003. 11 The consumption of H CO and H O is the net result of this process, which might involve several 3 2 2 steps. For instance, the organisms might take up HCO–3 and Na+ initially, and then pump out Na+ and take up H+ when the HCO–3 is converted to organic matter.
432
Alternatively, we could input the pH and T OT CO3 into Visual Minteq and allow the program to calculate T OT H. In either case, we find T OT H in the initial solution to be 7.49 ⇥ 10 4 M. In Example 8.5, we noted that T OT H when f = 0 was 2(T OT CO3 ), or 1.484 ⇥ 10 3 M. The alkalinity is the difference in these two values, or 7.35 ⇥ 10 4 M. (b) The determination of the alkalinity in the evening is trivial, because the changes occurring during the day are removal of H2 CO3 , which has no effect on ALK, and generation of algal biomass and O2 , which we are told undergo no acid/base reactions and therefore also do not affect ALK. As a result, the alkalinity of the solution in the evening will be the same as in the morning, i.e., 7.35 ⇥ 10 4 equiv/L. Although the loss of three-fourths of the T OT CO3 does not alter the alkalinity, it does affect the solution pH. Here, we present three different approaches for determining what the final pH is. Approach 1. Based on the initial value and the fractional loss, T OT CO3 at the end of the day is 1.85 ⇥ 10 4 M. We found in part (a) that, if we use 4 H+ and CO2– 3 as components, T OT Hmorning is 7.49 ⇥ 10 M. Using these components, when H2 CO3 is removed from solution, the decline in T OT H is twice the decline in T OT CO3 , so we can find T OT Hevening as T OT Hevening = T OT Hmorning = 7.49 ⇥ 10
4
2DT OT CO3 2(75% ⇥ 7.42 ⇥ 10 4 ) = 3.66 ⇥ 10
4
The known values of T OT CO3 and T OT H can then be input into Visual Minteq to find the final pH, which the output indicates is 10.63. Approach 2. Because ALKevening is known, and we are modeling the solution as though it contains only water and carbonate species, we can apply Equation (8.21b) to write ⇥ ⇤ ALKevening = (H+ ) + (OH ) + T OT CO3 (a1 + 2a2 ) evening 7.35 ⇥ 10
4
= 1.85 ⇥ 10 4 (a1 + 2a2 ) + (OH )
(H+ )
(8.23)
The variables on the right side of Equation (8.23) all depend only on pH, so the pH at which the equation is satisfied can be determined. In essence, in this approach, we are writing an equation for the constancy of alkalinity that can be used in place of the T OT H equation to solve for the equilibrium solution composition. Like the T OT H equation, this equation for alkalinity can be solved either numerically or in conjunction with a log C-pH diagram. In the latter case, we draw a curve representing ALK on a log C-pH diagram for the 433
solution, as shown in Figure 8.21. The pH in the evening is the value at which the ALK computed based on the speciation [i.e., the RHS of Equation (8.23)] equals the known value of 7.35 ⇥ 10 4 equiv/L [the LHS of Equation (8.23)]. Once again, we find the evening pH to be 10.63.
Figure 8.21 A log C-pH diagram for a system with T OT CO3 fixed at 1.85 ⇥ 10 4 M, and with a curve shown for the alkalinity, based on the speciation at each pH. A line is also shown for a fixed value of 7.35⇥10 4 , which is the alkalinity (in equiv/L) of the solution of interest.
Approach 3. Finally, the same result can be obtained without any calculations at all by using the Deffeyes diagram (Figure 8.19a). To do so, we note that in the morning, the solution is at pH 8.10 (given in Example 8.5) and contains 7.42⇥10 4 M (i.e., 10 3.13 M) T OT CO3 . We can locate this point on the figure and estimate the alkalinity by reading the value on the ordinate (although we don’t actually need to know this value to answer the question). We know that T OT CO3 in the evening is 25% of the value in the morning, or 1.85⇥10 4 M (i.e., 10 3.73 M). Because ALK remains constant during the day, we can find the point on the diagram that corresponds to the evening condition by drawing a horizontal (constant ALK) line to the left from the “morning” point to a value on the abscissa equal to the known T OT CO3 in the evening. We can then estimate the evening pH from this point, by interpolation between the constant-pH isopleths shown on the plot. The whole sequence is shown by the broken lines in Figure 8.19a and yields an estimated evening pH of ⇠10.5. While this approach is not as accurate as the calculations in parts (a) and (b), it is very quick and yields a good estimate of the final pH. The accuracy could be improved by using an expanded version of the Deffeyes diagram.
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8.9.7
Alkalinity Calculations Embedded in Visual Minteq
Under some circumstances, Visual Minteq is able to estimate T OT CO3 and solution composition based on input values for ALK, the noncarbonate contributors to ALK, and the solution pH. This feature, which is accessed via the “Specify alkalinity” option under the “Parameters” menu on the home page, is most useful if one has experimental data for the alkalinity and pH of a solution and believes that the alkalinity is attributable primarily to carbonate species. In such cases, the program uses the principle that alkalinity is conservative to compute T OT CO3 . For instance, consider a solution at pH 8.7 with a measured alkalinity of 3.15 meq/L. If the pH is entered on the program’s home page, and the alkalinity is entered on the alkalinity menu as shown in Figure 8.22a, CO2– 3 is automatically added to the list of components, and the alkalinity is shown in the list of input concentrations. To emphasize that the concentration shown in the input summary is ALK and not T OT CO3 , the component name for this value is shown as CO3-2**** (Figure 8.22b). When the program is run, the value of T OT CO3 that causes the solution to have the given alkalinity and pH is calculated. In this case, the program reports that T OT CO3 is 3.09 ⇥ 10 3 M (on the “Equilibrated mass distribution” page).
Figure 8.22 Visual Minteq screens when the alkalinity is specified to be 3.15 meq/L.
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In carrying out these calculations, the program accounts for the alkalinity contributions of any weak acid/base groups that have been specified in the input based on ‘alkalinity factors’ (AF’s) that are associated with each component and species in the database. For example, the AF for PO3– 4 is 2, meaning that when a solution containing phosphate is titrated to the alkalinity endpoint, each PO3– 4 ion in the original solution consumes two H+ ions (because, at the endpoint, T OT PO4 is present almost entirely – as H2 PO–4 ). The corresponding alkalinity factors for HPO2– 4 and H2 PO4 are 1 and 0, respectively (and that for H3 PO4 is 1). If the inputs to Visual Minteq specify the pH and T OT PO4 in a solution, the program calculates the contribution of T OT PO4 to the alkalinity based on the speciation of T OT PO4 at that pH and reduces the alkalinity contribution from T OT CO3 accordingly. More generally, the program calculates the alkalinity attributable to CO3 species as ALKCO3 ⌘ [HCO3 ] + 2[CO23 ] = ALK
(8.24a)
 (AFi ) ci
(8.24b)
where ci is the concentration of a noncarbonate species i in solution with the given pH and T OTi, and AFi is the alkalinity factor of that species. After determining ALKCO3 from Equation (8.24b), the program calculates T OT CO3 from Equation (8.16). Example 8.11 An ideal solution at pH 7.7 contains 4 ⇥ 10 4 M T OT PO4 and has an alkalinity of 4.7 meq/L. Assuming that the only species contributing to ALK are those in the carbonate and phosphate groups, determine T OT CO3 in the solution both manually and using Visual Minteq. Solution Using the equations developed in Chapter 5 to determine the a values for the phosphate acid/base group at pH 7.7, we can compute the phosphate speciation in the 4 solution. The result is that [H2 PO–4 ] = 9.58 ⇥ 10 5 M, [HPO2– 4 ] = 3.04 ⇥ 10 M, and [H3 PO4 ] and [PO3– 4 ] are negligible by comparison. Inserting these values along with – + [H ] and [OH ] into Equation (8.24b), we find n o ALKCO3 = ALK AFH PO cH PO + AFHPO2 cHPO2 + AFH+ cH+ + AFOH cOH 2
= 4.7⇥10
3
= 4.4⇥10
3
4
2
4
4
4
(0)(9.58⇥10 5 )+(1)(3.04⇥10 4 )+( 1)(10
436
7.7
)+(1)(10
6.3
)
At pH 7.7, a1 and a2 for carbonate species are 0.955 and 0.002, respectively. Using these values and the preceding result for ALKCO3 in Equation (8.16) yields T OT CO3 =
4.4 ⇥ 10 3 = 4.58 ⇥ 10 0.955 + (2)(0.002)
3
Visual Minteq carries out essentially the same calculations if we input the solution pH and T OT PO4 on the program’s home page, and the alkalinity on the “Parameters/Specify alkalinity” page. When the program is run, the result that T OT CO3 is 4.58 ⇥ 10 3 can be found on the “Equilibrated mass distribution” page. Unfortunately, as of this writing, the Visual Minteq database includes nonzero 3+ AF values only for species in the PO3– 4 , H4 SiO4 , B(III), and Al component groups; all other components and species are assigned AF’s of zero. Therefore, if one wishes to use this feature of the program for a solution in which other groups contribute a significant portion of the alkalinity, it is necessary to assign an appropriate AF to the corresponding components and species. The procedure for determining AF’s and adding them to the Visual Minteq database is described in Appendix 8A.
8.9.8
Summary of the Alkalinity Concept
Alkalinity is a simple, widely used parameter to characterize the capacity of a solution to accept acid inputs without becoming “too acidic”. The definition of “too acidic” typically refers to a cutoff near pH 4.5 and is based on both the chemistry of the carbonate acid/base group and empirical observations about the ecology of natural systems. In many natural waters and process waters, the alkalinity can be approximated as the summation of the bicarbonate concentration and twice the carbonate concentration. Although this approximation is often valid, in any real system, weak bases other than those that comprise the carbonate system also contribute to the alkalinity. The pH and alkalinity both provide information about the acid/base properties of a solution, and although the two parameters are related, they are independent. Specifically, since the pH characterizes the activity of H+ in solution, it is an indicator of the current state of the system. By contrast, the alkalinity characterizes the ability of the system to maintain pH at 4.5 when acids are added. Unlike pH, alkalinity is a conservative parameter, so it is quite easy to compute the alkalinity of a solution prepared by mixing other solutions or by adding known chemicals to an existing solution. One consequence of this property is that addition of H2 CO3 or CO2 to solution, or its removal from solution, does not affect the alkalinity at all, because H2 CO3 has no alkalinity itself; i.e., it cannot combine with H+ , and at pH 4.5 it does not release H+ to a significant extent. 437
8.10
ACID/BASE TITRATIONS OF NATURAL ORGANIC MATTER
Aside from carbonate species, the most important acid/base groups in most natural waters are associated with the collection of molecules known as natural organic matter (NOM). These molecules, whose general chemistry was described in Chapter 5, play a major role in controlling the behavior of dissolved metals, suspended colloids and solids, and coagulants that are commonly added to water during treatment operations. Interpretation of the data from acid/base titrations of NOM is complicated by the fact that the NOM is a composite of many different molecules. Most dissolved NOM molecules are small polymers that contain a variety of acidic functional groups, of which carboxyl (COOH) groups are the most numerous. The carboxyl groups exist in a variety of chemical environments, and their Ka values can vary over a few orders of magnitude, typically centering around pKa ⇠ 4. The molecules typically contain other acidic groups with very different acid strengths, as well as basic (usually amine [NH2 ]) groups. As a result, titration curves of NOM are usually relatively featureless, with few sections that are dramatically steeper or flatter than others. A set of curves showing the change in anionic charge of NOM molecules due to protonation/deprotonation reactions during a titration is presented in Figure 8.23.
Figure 8.23 Titration of a sample of NOM isolated from a “pristine” groundwater source in Denmark. The anionic charge on the molecules gradually increases as the pH increases and the molecules deprotonate. The different curves are for titrations in solutions of different ionic strengths. Adapted from Christensen et al. (1998).12 12 Christensen,
J.B., Tipping, E., Kinniburgh, D.G., Groen, T., and Christensen, T.H. (1998) “Proton binding by groundwater fulvic acids of different age, origins, and structure modeled with the Model V and NICA-Donnan model.” Environ. Sci. Technol. 32, 3346-335.
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Titration curves of natural waters are sometimes categorized in terms of strong acidity, defined as the amount of base needed to titrate the sample from the initial pH to around pH 8, and weak acidity, the amount of base needed to titrate the sample from pH 8 to around pH 11 (the cutoffs are defined differently by different investigators). Frequently, the strong acidity is attributed to carboxyl groups and the weak acidity to phenolic groups, but these generalizations should not be taken as definitive. Many properties of NOM, including its acid/base behavior, have been comprehensively reviewed by Leenheer.13 Some efforts to model the acid/base behavior of NOM have attempted to capture the idea that the molecules contain one or distinct groups of sites, with each group characterized by a range of pKa values rather than a single, well-defined value. For example, one model represents each group as having a continuous, Gaussian distribution of pKa values around a user-specified mean value. The mean is defined such that, if pH = pKa , the activities of the acid and base are equal, as would be the case for a conventional acid/base pair. Under these conditions, the sites that are protonated represent a balanced mixture of those with higher and lower pKa values than pKa,mean . Representing the protonated and deprotonated sites as HNOM and NOM , respectively, we can define an apparent pKa that reflects the overall state of protonation of all the sites as follows: pKa,app ⌘
 (NOM ) (H+ )  (HNOM)
(8.25)
where the terms (NOM ) and (HNOM) are shown as summations because they include a range of acid/base sites with different pKa values. By definition, pKa,mean is the value of pKa,app under conditions where  (NOM ) =  (HNOM). As we would expect for a collection of molecules with a distribution of pKa values, the stronger binding sites (the more basic sites) are protonated first. Therefore, as the pH drops and more sites are protonated, the protonation occurs on sites that are progressively less basic (i.e., more acidic), and pKa,app decreases steadily. This decline in pKa,app with decreasing pH is the characteristic feature of NOM (or any collection of molecules containing acid/base sites with a distribution of acidities) that distinguishes it from conventional acids. A few other approaches have been proposed for incorporating a distribution of pKa values into the analysis of NOM acid/base behavior, and Visual Minteq includes options for utilizing several such models. In the program, DOM (an acronym for Dissolved Organic Matter) is used as the default name for the component that repre13 Leenheer,
J.A. (2009) “Systematic approaches to comprehensive analyses of natural organic matter.” Annals of Environmental Science, 3, 1-130 (on-line version in pdf format published with open access by Northeastern University at http://pubs.usgs.gov/sir/2007/5048/pdf/SIR07-5048 508.pdf).
439
sents NOM molecules. The reactive site-types on these molecules are referred to as subcomponents and are named DOM1, DOM2, etc. Readers are encouraged to explore these modeling options on their own; here, we consider only one example to illustrate how the distribution of pKa values in NOM molecules affects titration curves of samples that contain this material.
Example 8.12 A solution contains NOM at a concentration of 9 mg/L TOC. The NOM molecules have two major acid/base groups that are present at concentrations of 8 µmol/mg C and 5 µmol/mg C, and that have mean acidity constants of pKa1 = 3.9 and pKa2 = 8.9, respectively. (a) Simulate titrations and plot titration curves for the solution based on two scenarios. In one simulation, assume that all the acid/base sites in each group are identical, as they would be in a conventional diprotic acid; in the other, assume that each group has a Gaussian distribution of pKa values with a standard deviation of 1.8 pK units. (b) Plot pKa,app as a function of pH for each of the acid/base groups in the scenario where the actual pKa values have a Gaussian distribution. Solution (a) We will carry out both simulations using the Gaussian NOM feature in Visual Minteq, changing only the standard deviation between the two runs. To use this feature, we click on the “Show organic components” box on the home page, select “DOC (Gaussian DOM)” as the only component, and specify a total organic carbon concentration of 9 mg/L (by choosing “mg/L” in the “Concentration unit” box). When we click the “Add to list” button, a box opens showing that we have chosen to use the “Gaussian DOM distribution”, and a button appears allowing us to edit the DOC parameters. If we click on this button, a window opens that confirms the input concentration (which is always shown in mg/L units as DOC, regardless of the concentration units used on the home page). This window also shows DOM1 as the default (and only) subcomponent of the overall DOM. If we wished to model the DOM as having only one subcomponent (i.e., one group of acid/base site-types), we could accept this information as is and return to the home page. For this problem, however, we want to model NOM molecules that have two groups of sites, so we select DOM2 to represent the second group of sites, and then save the information. (Note that, to include three groups of sites, we would have to save after selecting DOM2 and then return to the page to select DOM3 as well.) The remaining information that we need to edit is accessed via the “Database management/ DOM complexes (Gaussian)” link. After clicking on that link, we specify the acid/base characteristics of DOM1 by inputting the properties of the species HDOM1, H2DOM1, etc., all of which can form (in principle) by protonation of the 440
subcomponent DOM1. In this particular simulation, we are assuming that DOM1 is monoprotic, so we select HDOM1 as the “Complex name” and edit the “log K”, “log K std dev”, and “Binding site density” boxes to input the appropriate values, as specified in the problem statement. We could also add new species that include the DOM1 subcomponent if we desired. For example, if we wanted to model this site-type as diprotic, we could add a new species H2 DOM1 by clicking on the “Add new species” button, filling in the stoichiometric information for formation of that species from the components, clicking the “Suggest ID number and mol. wt” button, and inputting the corresponding information for the mean and standard deviation of log K. In this problem, we are treating the DOM1 sites as monoprotic, so we need not add any new species. An image of the page after the proper information has been input for HDOM1 is shown in Figure 8.24, for the case where the DOM1 sites have a Gaussian distribution of pKa values. After saving these changes to the database, we repeat the process for HDOM2, save once again, and return to the main menu. We next select the “Multi-problem/ Sweep” parameters and specify the range parameters to explore. The titration can be carried out either by gradually changing
Figure 8.24 The DOM complexation database page, showing the input information for formation of the HDOM1 complex.
441
T OT H and letting the program compute the corresponding pH values of by gradually changing the (fixed) pH value and using the program output to find the corresponding T OT H. Once the program is run and the output is exported to Excel, the dose of strong base (e.g., NaOH) required to bring the solution from the initial pH of 4.0 to any other pH can be computed as the difference in T OT H (i.e., “Total dissolved H+ ”) between those two conditions. The same program can then be rerun after changing the standard deviations for the log Ka values in the database to zero to simulate titration of a conventional diprotic acid. The results for both simulations are shown in Figure 8.25. As we might expect, representing the acid/base groups in the NOM molecules as having a spread of pKa values causes the protonation/deprotonation reactions to proceed over a wider pH range and therefore flattens the titration curve.
Figure 8.25 Simulated titrations of NOM with two types of acid/base groups, considering each group to have either a single pKa or a distribution of pKa values.
8.11
BUFFERS
8.11.1
Definition of the Buffer Intensity
Because the presence of a weak acid/base group makes it more difficult to change the pH when the pH is near the pKa , such a group is said to buffer the pH near the pKa . Thus, ammonia is a good pH buffer near pH 9 but not at pH 5, and acetate is a good buffer near pH 5 but not at pH 9. If one wished to maintain a solution at a pH near 5.0, addition of acetate species would be an effective approach for doing so: if protons subsequently entered the solution, some of the Ac– ions would combine with them, 442
and if OH– ions entered solution, some HAc molecules would dissociate and provide protons to neutralize some of the added OH– . Obviously, the larger the concentration of weak acid or base that is present, the more proton-donating and proton-accepting species would be available to respond to addition of hydroxide or protons, and the greater would be the resistance of the system to pH changes. The ability of a system to resist pH changes is quantified by a term called the buffer intensity, b , defined as b ⌘
dT OT Hstrong dpH
(8.26)
where the subscript on T OT H indicates that the change in T OT H arises from the addition of strong acid (or of strong base, which is equivalent to removal of strong acid). In words, the buffer intensity is the incremental amount of base that must be added to a solution per unit increase in pH. A large buffer intensity means that a large amount of base must be added (or, equivalently, a large amount of protons must be consumed) in order to increase solution pH even slightly. Typical units of buffer intensity are equivalents per liter per pH unit. If a titration curve is plotted with pH on the ordinate and the amount of strong acid added (in equiv/L) on the abscissa, Equation (8.26) indicates that the buffer intensity is the negative inverse slope of the titration curve. Similarly, if the titration is with strong base and the amount of base added is plotted on the abscissa, the buffer intensity is the positive inverse slope. In either case, the buffer intensity is large when the titration curve is flat (small slope) and small when the titration curve is steep (large slope).
8.11.2
The Buffer Intensity of Water and Weak Acid/Base Groups
To compute and/or interpret the buffer intensity of a solution, it is useful to consider the three possible fates of H+ ions added to a solution containing a weak acid/base couple: reaction with OH– to form water; reaction with the weak base to form the conjugate acid; or no reaction, thereby increasing the H+ concentration in solution (or, more correctly, reacting with water to increase the H3 O+ concentration). Although these three changes are all linked chemically, they can be evaluated independently. Consider, for example, titration with a strong acid of a solution that contains only water and a single weak acid/base pair. To make the scenario concrete, assume that the pKa of the weak acid is 6.5, the total concentration of the acid/base pair in the solution is 10 4 M, and the initial pH is 6.1. We can compute the amount of H+ required to titrate this solution to pH 6.0 by considering the amount of H+ that reacts in each of the three ways described above. The speciation of the system at the two pH values can be determined analytically, using the techniques developed previously, to give the results shown in Table 8.7. 443
Table 8.7 Changes in speciation when an ideal solution containing 10 pKa,HA = 6.5) is adjusted from pH 6.1 to 6.0
Species concentration (OH– ) (HA)
+
(H ) pH 6.1 pH 6.0 Change
10 6.1 10 6.0 2.06 ⇥ 10
7
10 7.9 10 8.0 2.59 ⇥ 10
0.715 ⇥ 10 0.760 ⇥ 10 0.045 ⇥ 10
9
4M
T OT A (with
(A– ) 4 4 4
0.285 ⇥ 10 4 0.240 ⇥ 10 4 0.045 ⇥ 10 4
The expression used to calculate T OT H depends on whether HA or A– is chosen as a component. If HA is chosen as a component, then A– is ‘made’ by subtracting one H+ from HA. The tableau and T OT H equation for this scenario are as follows. H2 O
H+
HA
H2 O
1
0
0
0.00
H
0
1
0
0.00
OH–
1
1
0
14.00
HA
0
0
1
0.00
–
0
1
1
6.50
+
A
T OT H = H+
OH
A
(8.27)
By contrast, if A– is chosen as a component, then HA is considered to be formed by the combination of one A– with one H+ , and the T OT H equation is T OT H = H+
OH
(8.28)
+ (HA)
Differentiating Equations (8.27) and (8.28) to obtain expressions for dT OT H, we obtain HA as component: dT OT H = d H+ –
A as component: dT OT H = d H
+
d OH
d A
(8.29a)
d OH
+ d (HA)
(8.29b)
Because the sum of (HA) and (A– ) is constant at T OT A, d(HA) = d(A ), so the two expressions for d T OT H in Equations (8.29a) and (8.29b) are identical. That is, even though T OT H depends on which A–containing species is chosen as a component, d T OT H does not. For this reason, the computed buffer intensity is the same regardless of the choice of components. 444
The changes in the H+ and OH– concentrations shown in Table 8.7 would be the same when the pH of any solution was changed from 6.1 to 6.0, regardless of which (if any) weak acid/base groups were present. That is, whenever the pH of any solution is changed from 6.1 to 6.0, 2.59 ⇥ 10 9 M H+ is consumed by reacting with OH– to form H2 O, and the concentration of H+ in solution increases by 2.06 ⇥ 10 7 M. The amount of H+ that must be added to meet these requirements is the sum of these two values, or 2.09 ⇥ 10 7 M. Because these calculations depend only on Kw and are independent of any other acids or bases in the system, this sum is sometimes referred to as the amount of H+ needed to titrate the water. H+ needed to titrate water = D H+
D OH
(8.30)
Applying this equation when the changes in (H+ ) and (OH– ) are differentially small (i.e., for a differentially small change in pH), we can compute the contribution of these two processes (neutralization of OH– and accumulation of H+ ) to the buffer intensity of the solution. This contribution is designated bw and is referred to as the buffer intensity of the water. bw ⌘ bw =
d (H+ needed to titrate water) dpH d (H+ )
d OH dpH
=
d (H+ ) d OH + dpH dpH
(8.31)
Application of this equation over a wide range of pH’s yields the result shown by the broken curve in Figure 8.26. The curve is symmetric around pH 7.0, where it passes through a minimum. The steep rise in bw at both low and high pH corresponds to the flat portions of all titration curves at pH greater than approximately 10 and lower than approximately 4 regardless of which (if any) weak acid/base pairs are present in solution. In addition to the changes in (H+ ) and (OH– ), Table 8.7 shows that 0.045⇥10 4 M + H is needed to convert a portion of the A– to HA when the pH is changed from pH 6.1 to 6.0.14 Whereas the computation of bw depends only on Kw , this calculation depends on pKa of the HA/A– conjugate pair and on T OT A. The calculation for the amount of base needed for this conversion can be written generically as follows: calculation is listed twice in Table 8.7. Addition of 0.045 ⇥ 10 4 M H+ causes the HA concentration to increase by 0.045 ⇥ 10 4 and the A– concentration to decrease by the same amount; i.e., as noted previously, d(HA) = d(A ). Since both changes are part of the same process, the amount of base consumed is 0.045 ⇥ 10 4 M, not twice this value. By contrast, the calculations for the neutralization of H+ and the increase in OH– are additive, because those reactions are for two different conjugate acid/base pairs (H3 O+ /H2 O and H2 O/OH– ). 14 This
445
Figure 8.26 Buffer intensity contributed by water only and by HA only, for an acid HA with pKa = 6.5 present at T OT A = 3 ⇥ 10 3 M.
H+ required to convert A to HA = T OT A a0,pH6.0 when pH changes from 6.1 to 6.0
a0,pH6.1
= T OT A a1,pH 6.0
a1,pH 6.1
(8.32a) (8.32b)
Extrapolating this result to a pH change between any two pH values, we find H+ required to convert A to HA = T OT A a0,pHfinal when pH changes from pHinit to pHfinal = T OT A a1,pHfinal
a0,pHinit a1,pHinit
(8.33a) (8.33b)
As we did to compute bw , we can apply this expression to a differential change in pH to compute the buffer intensity contributed by the weak acid, bHA : bHA ⌘ =
d (H+ needed to titrate HA/A ) dpH
(8.34a)
d (a0 T OT A) dpH
(8.34b)
= T OT A = T OT A
da0 dpH
da1 dpH
(8.34c) (8.34d)
446
Applying these results across a wide pH range yields the curve labeled bHA in Figure 8.26. The curve is symmetric around pH = pKa,HA and passes through a maximum at that point, indicating, as expected, that the tendency for a weak acid to resist pH changes is greatest at a pH equal to its pKa . Equations (8.34c) and (8.34d) indicate that increasing T OT A increases bHA proportionately. Furthermore, since plots of a1 versus pH for monoprotic acids are all identical except for being offset from one another along the pH axis, a curve showing the buffer intensity contributed by a different acid (HB) would have an identical shape as the one for HA, but it would be shifted so that its peak was at pKa,HB . It should be apparent that the preceding calculations and discussion bear a strong similarity to those for the alkalinity of a solution containing a weak acid/base group. The similarity arises because the ideas underlying these two parameters are essentially the same; the only differences are that, to determine the buffer intensity, we consider small increments of acid or base added to the system instead of the total acid requirement to adjust the sample a specified pH endpoint, and we divide these increments by the change in pH. Combining the results shown in Equations (8.31) and (8.34), and noting that, in a system containing only one weak acid/base group, those equations account for all possible fates of H+ added to the solution, we can write the total buffer intensity of the solution as " # d (H+ ) d OH da0 b= + T OT A (8.35a) dpH dpH dpH " # d (H+ ) d OH da1 b= + + T OT A (8.35b) dpH dpH dpH b = bw + bHA
(8.35c)
By definition, pH = log (H+ ) = (1/2.303) ln (H+ ). Making this substitution, and noting that d(ln x) = dx/x, we can rewrite dpH as ✓ ◆ 1 1 d(H+ ) + dpH = d ln (H ) = (8.36) 2.303 2.303 (H+ ) Then, writing (OH– ) and a1 in terms of (H+ ) and substituting for dpH according to Equation (8.36), the three differentials on the right side of Equation (8.35b) can be expressed as d (H+ ) = dpH
d (H+ ) = 2.303 H+ 1 d (H+ ) 2.303 (H+ ) 447
(8.37)
d OH dpH
d [K w /(H+ ) ] ⇤ 1 ⇥ d H+ / H+ 2.303 h i 2 Kw /(H+ ) d (H+ ) = ⇤ 1 ⇥ d H+ / H+ 2.303 =
= 2.303
da1 = dpH
Kw = 2.303 OH (H+ )
(8.38)
Ka,HA Ka,HA + (H+ ) ⇤ 1 ⇥ d H+ / H+ 2.303 d
Ka,HA
=
d H+ [Ka,HA + (H+ )]2 ⇤ 1 ⇥ d H+ / H+ 2.303
= 2.303
(H+ ) Ka,HA
[Ka,HA + (H+ )]2
= 2.303a0 a1
(8.39)
Substituting Equations (8.37) and (8.38) into Equation (8.31), and Equation (8.39) into Equation (8.34d), we obtain the following concise expressions for the different contributors to the buffer intensity: ⇥ bw = 2.303 H+ + OH
bHA = 2.303 (T OT A) a0 a1
⇤
(8.40a) (8.40b)
Finally, inserting these results into Equation (8.35c), we find the overall value of b to be ⇥ b = 2.303 H+ + OH
+ (T OT A) a0 a1
⇤
(8.41)
This result establishes formally what we have already noted from visual inspection of buffer intensity curves: the buffer intensity becomes large when either (H+ ) or (OH– ) is large, regardless of whether or how much A is present, and the contribution of the HA/A– couple to the buffer intensity is directly proportional to T OT A. 448
Equation (8.41) can be extended to systems containing two or more weak monoprotic acids simply by adding terms for bHB , bHC etc. Each such term would have the same form as that for bHA , i.e., bHB = T OT B
da1,HB = 2.303 (T OT B) a0,HB a1,HB dpH
2
6 + b = 2.303 6 4 H + OH
Â
= bw +
X=A,B,C,...
+
Â
all HX/X pairs
(8.42) 3
7 (T OT X) a0, HX a1, HX 7 5
(8.43a)
bHX
(8.43b)
Equation (8.43b) indicates that the buffer intensity is a conservative quantity; that is, the total buffer intensity of a solution containing several acid/base groups is the summation of the buffer intensities attributable to the individual groups. Example 8.13 Estimate the buffer intensity of a solution of 5 ⇥ 10 3 M NH4 Cl and 2 ⇥ 10 3 M sodium benzoate (a common food preservative) at pH 9.24 (the pKa of NH+4 ). Benzoate is the conjugate base of benzoic acid, a monoprotic acid with pKa = 4.20. Solution At pH= 9.24, the benzoic acid/benzoate pair (which we will represent as HBz/Bz– ) is almost completely deprotonated, while the NH+4 /NH3 (aq) pair is split evenly between its two forms. Since the pH is 5.05 units above pKa for HBz/Bz– , we can estimate the concentrations of the four weak acid/base species as (Bz ) ⇡ T OT Bz = 10
2.70
(HBz) = 10
5.05
⇥ (Bz ) = 10
NH+ 4 = NH3 (aq) = 2.5 ⇥ 10
3
Inserting these values into Equation (8.43a), we obtain: ⇣ ⌘ h b = 2.303 10 9.25 + 10 4.75 + 10 2.7 10 5.05 (1.0) + 10 = 2.88 ⇥ 10
3
equiv/L pH unit
7.75
2.3
i (0.5) (0.5)
Evaluation of the terms in the buffer intensity equation indicates that more than 99% of the overall buffer intensity is contributed by the NH+4 /NH3 (aq) couple. 449
A titration curve and a plot of b versus pH for the solution described in Example 8.13 are shown in Figure 8.27. Note that the pH and the magnitude of the peaks in the buffer intensity curve are a good deal easier to evaluate than are the corresponding flat regions of the titration curve. For this reason, analysis of buffer intensity sometimes can provide information about the types and concentrations of weak acids in a sample that is almost impossible to ascertain from visual inspection of titration curves. In general, acids present at total concentrations somewhat larger than 10 4 M can be detected as peaks in plots of buffer intensity, with the ease of detection increasing as the pKa value of the acid gets closer to 7.
Figure 8.27 Response of the solution in Example 8.13 to addition of strong acid or base. (a) Titration curve; (b) Buffer intensity as a function of pH.
450
8.11.3
The Buffer Intensity of Multiprotic Acids
The buffer intensity in a system with multiprotic acids and bases can be analyzed by using essentially the same approach as for monoprotic groups; the results for mono-, di-, and tri-protic systems are shown in Table 8.8, along with a generic expression applicable to any acid/base group, regardless of how many protons it can release or acquire. A derivation of this equation is provided in Appendix 8B. The contributions of multiple multiprotic acids to the buffer intensity are additive, just as the contributions of monoprotic acids are. Table 8.8 Expressions for the contribution of weak acids to the buffer intensity Buffer intensity contributed by the Hn A acid/base group (bA ) Monoprotic, HA
2.3(T OT A)a0 a1
(8.44a)
Diprotic, H2 A
2.3(T OT A)(a0 a1 + 4a0 a2 + a1 a2 )
(8.44b)
Triprotic, H3 A
2.3(T OT A)(a0 a1 + 4a0 a2 + 9a0 a3 + a1 a2 + 4a1 a3 + a2 a3 )
(8.44c)
Any multiprotic, Hn A
2.3(T OT A) Â Â ( j
n n 1
j>i i=0
i)2 ai a j
(8.45)
The equations shown in Table 8.8 indicate that the buffer intensity contributed by an acid/base group is proportional to its total concentration and to a summation that includes all possible pairwise products of the a values. In almost all cases, only one or at most two of these pairwise products (the ones corresponding to the two most dominant species and to the most dominant and third most dominant species) are significant. As a result, although the complete summation might include many terms, it can be approximated almost exactly by considering only one or two terms. The buffer intensity of a solution containing 0.002 M T OT CO3 is shown in Figure 8.28. For this system, b is similar to that for a system containing two independent monoprotic acids, both present at a total concentration T OT A. In the case of carbonate species, the pKa ’s for the sequential deprotonation reactions are sufficiently separated that the two peaks in the buffer intensity curve are easily distinguished from one another. However, the peak associated with pKa2 blends with the contribution of bw , so that peak is not easily seen in the b curve.
451
Figure 8.28 Buffer intensity of a solution containing 2 ⇥ 10 3 M T OT CO3 , showing the individual contributions of water and carbonate species as well as the overall buffer intensity.
Example 8.14 (a) Based on qualitative reasoning, assess whether a solution containing 10 2 M ammonium phosphate, (NH4 )3 PO4 , would be more strongly buffered at pH 7.20 or pH 9.24, or whether the buffer intensity would be about the same at these two values. (b) Test your reasoning in part (a) by computing b at the two pH values. Solution (a) The system contains acid/base couples that have pKa values at both pH’s of interest. However, the acid/base pair providing the buffering at pH 9.24 (ammonia) is three times as concentrated as the one providing the buffering at pH 7.20 (phosphate). Therefore, we expect the solution to be more highly buffered at pH 9.24 than at pH 7.20. (b) The buffer intensity can be estimated by using the equations in Table 8.8. T OT A and a values of the various species at pH 7.20 are summarized in the following table. The table also shows the contributions of water, T OT NH3 , and T OT PO4 to the buffer intensity at each pH, based on the following equations:
452
⇥ ⇤ bw = 2.3 (H+ ) + (OH )
bNH3 = 2.3(3 ⇥ 10 2 )a0 a1 bPO4 = 2.3(10
2.0
)(a0 a1 + a1 a2 + +a2 a3 4a0 a2 + 4a1 a3 + 9a0 a3 )
The total buffer intensities at pH 7.20 and 9.24 are 6.43 ⇥ 10 3 and 1.75 ⇥ 10 2 equiv/L-pH unit, respectively, confirming our qualitative evaluation. The phosphate group accounts for approximately 89% of b at pH 7.20, and the ammonia group accounts for approximately 98% at pH 9.24. H2 O pH T OTi a0 a1 a2 a3 bi
8.12
7.20
5.10 ⇥ 10
PO4 9.24
7
4.09 ⇥ 10
5
7.20 10 2 10 5.3 0.5 0.5 10 5.4 5.75 ⇥ 10
3
9.24 10 2 10 9.1 0.01 0.99 10 3.1 2.47 ⇥ 10
NH3 7.20 3 ⇥ 10 0.99 0.01
4
9.24 3 ⇥ 10 0.5 0.5
2
6.83 ⇥ 10
4
2
1.73 ⇥ 10
2
SUMMARY
This chapter explores the qualitative and quantitative characteristics of acid/base titration curves. In a titration, known amounts of known acids or bases are added to solutions of unknown composition, and the pH response of the solutions can be used to infer the types and quantities of acids and bases in them. When H+ ions are added to a solution containing a strong base, some of those ions react with the OH– that was already in solution, forming H2 O, and others remain in solution as H+ , increasing the H+ concentration so that Kw is satisfied (since some OH– is consumed, the H concentration must increase). In such cases, the pH changes slowly until the amount of acid added is close to the total amount of base initially present. It then decreases extremely rapidly, passing through neutrality to an acidic value. If a weak base A– is present in the initial solution, some of the added H+ has another fate: it protonates the A– in order to maintain the HA/A– equilibrium in response to the increase in H+ . In this case, the rapid decrease in pH is mitigated, especially in the pH region near the pKa of the weak acid. The pH region where the rate of pH change is mitigated and the extent of the mitigation can be used to infer information about the strength of the dissolved weak acid/base couple (i.e., its pKa value) and its concentration. 453
Titrating a solution with a weak acid or base causes the pH to change less drastically than it would if a strong acid or base were added. However, the effective strength of an acid or base depends on the solution to which it is added. For instance, sodium carbonate acts as a strong base (with each carbonate ion acquiring two protons) at pH 5.0, but it acts as a weaker base at pH 8 (each carbonate acquiring one proton) and an even weaker one at pH 10. The amount of acid that must be added to a water sample to reach a pH near 4.5 is called the alkalinity. Alkalinity is conservative, and in systems dominated by the carbonate group, it can be approximated as the summation of the bicarbonate and twice the carbonate concentrations. Despite the frequent validity of this approximation, it is important to recognize that the definition of alkalinity is based on the amount of acid needed to titrate the solution to a given endpoint, and not on the amounts of specific acid/base groups present in the initial solution. Solutions are said to be buffered in regions where relatively large amounts of strong acids or bases are required to induce significant pH changes. Weak acid/base pairs buffer solutions near their pKa values. Therefore, we can buffer a solution (i.e., maintain the pH in a fairly narrow band) by adding large amounts of an acid/base pair whose pKa is in the pH region of interest. The carbonate acid/base system provides significant buffering to natural aquatic systems, reducing the likelihood that their pH will decrease below approximately 5.0 in response to the addition of moderate amounts of acid. If enough acid is added to a natural water to overcome this buffering and lower the pH substantially below 5.0, ecological consequences tend to be significant. The buffer intensity, defined as the incremental dose of base required per unit increase in pH, is a quantitative measure of the pH stability of a solution. Buffer intensity is always large at low and high pH due to the large concentrations of, respectively, H+ and OH– in such solutions. This contribution to the buffer intensity is referred to as the buffer intensity of water, bw . The buffer intensity contributed by weak acid/base couples passes through a maximum at the pH’s of their pKa ’s. The presence of a weak acid in solution is often easier to detect based on the solution’s buffer intensity than by inspection of unmodified titration curves (pH versus dose of strong acid or base added). The analysis of titration data represents a synthesis of many of the key concepts developed in prior chapters: chemical equilibrium, the relative strengths of acids and bases, the relationship of acid/base speciation to solution pH, and the idea that solutions have no memory. The analysis also reinforces the usefulness of both log C–pH diagrams and specialized software such as Visual Minteq as tools for understanding speciation. Once these concepts are mastered, we can explain behavior that, at first glance, is very complex (e.g., the shape of titration curves in systems containing multiple acid/base groups). In subsequent chapters, these same principles are applied to several other types of environmentally important chemical reactions. 454
8.13
APPENDIX 8A: DETERMINING ALKALINITY FACTORS AND INSERTING THEM INTO THE VISUAL MINTEQ DATABASE
The alkalinity factor (AF) of a species i is defined in Visual Minteq as the average number of H+ ions that a molecule of i acquires when a differential amount of i is added to a solution at the pH of the alkalinity endpoint (4.5 for the US definition of alkalinity, and 5.4 for the European definition). We can use Visual Minteq to compute the AF by carrying out a run in which the species of interest is the only input, the pH is fixed at the desired endpoint, and the ionic strength is fixed at zero. The AF for the species using pH 4.5 as the alkalinity endpoint can then be calculated as AFi =
T OT HpH 4.5 T OT Hinput [i]input
(8.46)
For example, to find AFPO2 , we could simulate a solution made by adding only 4
3– + HPO2– 4 to water by specifying inputs of 0.1 M each of PO4 and H . If we then ran the program with a fixed pH of 4.5 and I = 0, we would find from the output that T OT H is 0.2003 M. From this, we would compute that
AFHPO2 = 4
(0.2003 0.100) M = 1.00 0.100 M
(8.47)
If the AF for any species i in an acid/base group has been determined, the AF’s for any other species j in the group can be computed as AF j = AFi
DnH, i j
(8.48)
where DnH, i j is the number of H+ ions that are required to convert species i into – species j. Thus, for example, if i is HPO2– 4 and j is H2 PO4 , DnH, i j is 1 and AFH2 PO4 is 0. AF’s for components can be added to Visual Minteq only by editing the ALK.vdb database. To do this, the database should be opened with a text editor, and a line should be added with the component’s ID number and name (both of which can be found in the comp 2008.vdb database), followed by the appropriate AF. AF’s for species can be added by defining a problem that includes that species, migrating to the “Database management/ Aqueous species” page, typing the appropriate value in the “Total alkalinity factor” box, and saving the changes. Example 8.16 Oxalic acid (H2 C2 O4 ) is diprotic, with pKa1 = 1.25 and pKa2 = 4.27. Determine the alkalinity factors for all three species in the oxalic acid/base group.
455
Solution Following the procedures described in the preceding discussion, we can determine AFC O2 by investigating the speciation in an ideal solution containing an input 2
4
of C2 O2– 4 and nothing else, at a fixed pH of 4.5. When we simulate such a solution in Visual Minteq (where the component oxalate is shown as Oxalate-2), with an arbitrary concentration of, say, 0.1 M T OT C2 O4 , the program output indicates that T OT H is 3.69 ⇥ 10 2 . Inserting this information into Equation (8.45), we find AFC
3.69 ⇥ 10 2 0 = 0.369 0.1 M and AFH2 C2 O4 with Equation (8.47) as =
2 2 O4
We can then find AFHC
2 O4
AFHC
= 0.369
1 = 0.631
AFH2 C2 O4 = 0.369
2 = 1.631
2 O4
8.14
APPENDIX 8B: DERIVATION OF THE BUFFER INTENSITY OF MULTIPROTIC ACIDS
Consider a solution containing a multiprotic acid Hn A at a total concentration of T OT A. Choosing Hn A as the component, the upper portion of the tableau and the equation for T OT H in the equilibrium solution are as shown below. We do not know how the initial solution was prepared, so we leave out the bottom (input) portion of the table. +
H Hn A OH– Hn i A
H2 O
H+
Hn A
log K
0 0 1 0
1 0 1 i
0 1 0 1
0.00 0.00 14.00 log(K0i )(a)
K0i is the equilibrium constant for the reaction: Hn A )* Hn i + iH+ . The full tableau would contain one row for each deprotonated species (e.g., three rows [and three species] if Hn A could release one, two, or three protons). (a)
⇥ ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ ⇤ T OT Heq = H+ OH Hn 1 A 2 Hn 2 A2 ⇥ ⇤ ⇥ ⇤ i Hn i Ai · · · n An ⇥ ⇤ = H+
⇥
OH
⇤
n
Âi
i=1
456
⇥
Hn i Ai
⇤
...
(8.49)
If we titrate the solution from pHinit to pHfinal with strong acid, the amount of acid added (DT OT Hstrong ) must equal the change in T OT H of the equilibrium solution, i.e., DT OT Hstrong = DT OT Heq (8.50) Substituting the expression for T OT H from Equation (8.51) into Equation (8.52), we obtain ⇥ ⇤final DT OT Hstrong = H+ init
⇥ ⇤final OH init
n
Âi
i=1
⇥ ⇤final Hn i Ai init
(8.51)
Dividing both sides of Equation (8.53) by DpH (i.e., pHfinal pHinit ), multiplying by 1, and shrinking DpH to a differential yields an expression for b , as follows:
b⌘
DT OT Hstrong = DpH
⇥ ⇤final n ⇥ ⇤final final [H+ ]init + OH init + Â i Hn i Ai init
dT OT Hstrong = dpH
⇥ ⇤ ⇥ ⇤ ⇥ ⇤ i n d H d H+ d OH n iA + +Âi dpH dpH dpH i=1
DpH
i=1
(8.52a)
(8.52b)
Equation (8.54b) can be manipulated in essentially the same way as in the analysis of b for a monoprotic system shown in the main chapter, using the identities log x = ln x/ ln 10 and d ln x = dx/x. The algebra gets a bit long, but it is straightforward, leading to the results shown in Table 8.8.
8.15
PROBLEMS
Note: Many of the following problems can be solved either manually or using software like Visual Minteq. Your instructor will indicate whether you should use a particular approach, or whether the choice is up to you. In some cases, a manual solution is a practical option only if some simplifications are made (e.g., ignoring some of the species that might form). Again, your instructor will provide guidance on this issue. If chemical equilibrium software is used, present your solution with an explanation of how the problem was set up and solved with the program, give the results in a well formatted and titled table, and include appropriate axis labels and a caption for all graphs. Program output can be appended to your solution, but it is not acceptable to turn in only raw computer output. In all cases, assume ideal behavior unless the ionic strength is given explicitly or is to be computed as part of the problem. 457
1. The following solutions have been prepared: Solution 1: Solution 2: Solution 3:
H2 O plus sufficient NaOH to adjust the pH to 8.0. 10 3 M T OT OCl plus sufficient HCl and/or NaOH to adjust the pH to 8.0. 10 2 M T OT OCl plus sufficient HCl and/or NaOH to adjust the pH to 8.0.
(a) In Solutions 2 and 3, one-half of the T OT OCl was added as HOCl and one-half as NaOCl. Find the amount of HCl or NaOH that was needed to adjust the pH to 8.0 in each solution. (b) Compute and compare the pH when 10 of each of the three solution.
3
moles of HCl is added to 1 L
2. Plot a log C-pH diagram for a system containing 10 2 M T OT PO4 . Prepare a table showing the pH and the concentration of all phosphate species for HCl additions of 0, 5, 10, 15, 20, 25 and 30 meq/L to a solution of 10 2 M Na3 PO4 . 3. A solution of 0.1 N (i.e., 0.1 equiv/L) NaOH is added to 70 mL of 0.1 M HAc. What will the solution pH be after addition of (a) 35, (b) 70, and (c) 105 mL of the titrant? Assume ideal solution behavior, and take into account the dilution of the original solution by the titrant. 4. A solution contains 10
2.8 M
T OT CO3 and is at pH 8.1.
(a) Calculate and plot a titration curve for incremental additions of HCl until the pH is lowered to 4.0. (b) How much HCl (moles per liter of solution) is required to titrate the solution to pH 7.0? to the CO2 equivalence point? 5. A solution contains only one acid/base group, HA/A , at a total concentration of 2 ⇥ 10 4 M. pKa for HA is 9.0. If the solution is titrated from pH 9.0 to 10.0, what fraction of the NaOH added is attributable to reactions of the T OT A? Put another way, what is the ratio of the amount of base is needed to carry out the titration with the HA/A species present to the amount that would be required if they were absent? 6. A water sample at pH 6.5 has been titrated with both strong acid (HCl) and strong base (NaOH). The amount of titrant added at increments of 0.2 pH unit is summarized below, where the values represent the cumulative acid or base addition at each pH value. The sample is thought to contain just one or two weak acid/base groups. Analyze the titration curve to determine the number of acid/base groups present and the total concentration and pKa of each group. Hint: try to isolate the effect of the weak acid/base groups by separating their 458
contribution to the overall curve from the contribution needed just to titrate pure water. pH Base (meq/L) 6.5 0.000 6.7 0.004 6.9 0.006 7.1 0.008 7.3 0.009 7.5 0.010 7.7 0.011 7.9 0.012 8.1 0.013 8.3 0.014 8.5 0.017 8.7 0.021
pH Base (meq/L) 8.9 0.028 9.1 0.038 9.3 0.054 9.5 0.079 9.7 0.119 9.9 0.175 10.1 0.262 10.3 0.390 10.5 0.571 10.7 0.820 10.9 1.19 11.1 1.72
pH Acid (meq/L) 6.5 0.000 6.3 0.006 6.1 0.015 5.9 0.028 5.7 0.049 5.5 0.080 5.3 0.122 5.1 0.177 4.9 0.246
pH Acid (meq/L) 4.7 0.327 4.5 0.409 4.3 0.493 4.1 0.576 3.9 0.661 3.7 0.766 3.5 0.901 3.3 1.10 3.1 1.39
7. Calculate the initial pH and the pH at both equivalence points for titration of 10 2 M Na2 CO3 with strong acid. Next, based on your qualitative understanding of titrations, sketch the titration curve and buffer intensity curve for such a titration. Finally, use a spreadsheet or Visual Minteq to determine the two curves accurately. Compare your sketches to the calculated curves and describe the reasons that led to the discrepancies in your sketches. 8. Colorimetric indicators are convenient for evaluating the pH of a solution to within a few tenths of a unit. They are also very useful for indicating the endpoint of a titration. Consider a model indicator “In” which can exist as H2 In+ , HIn , or In– , with pKa1 = 6.0 and pKa2 = 9.0. H2 In+ is red, HIn is colorless, and In– is blue, with the red or blue color becoming noticeable when the solution contains at least 10 7 M of the respective species. (a) What are the pH and color of a solution made by adding 5 ⇥ 10 H2 InCl to water?
7M
(b) A solution contains 10 3 M ammonium carbonate, (NH4 )2 CO3 , and 5 ⇥ 10 4 M lime, Ca(OH)2 . Then, 5 ⇥ 10 7 M H2 InCl is added. What are the color and pH of the system? (c) Nitric acid (HNO3 ) is added to the solution in part (b) until the solution turns red. What is the pH? How much nitric acid was added? (d) If the solution in part (b) were adjusted to pH 12.0 with strong base and were then titrated with HNO3 to pH 4.0, would the blue-to-colorless or colorless-to-red change be sharper? That is, which would occur over a 459
smaller range of acid added? Why? Assume that both color changes occur over a pH range of about 0.1 pH unit.
9.
(a) How much does the alkalinity of a water sample change upon addition of 10 4 M of the following chemicals? Assume alkalinity is determined by titration to pH 4.5, that the solutions are ideal, and that the water is initially at pH 7.5 due to the presence of an unknown mixture of weak acids and bases. Briefly explain your reasoning. (i) (ii) (iii) (iv) (v) (vi) (vii)
HCl NaOH Na2 CO3 NaHCO3 CO2 NaH2 Cit FeCl3
(b) How would your answers to part (a) be different if the solutions were initially at pH 8.5? You need not do any calculations; just explain the reasoning that leads you to conclude whether the change in the alkalinity would be larger, smaller, or the same. 10. You have prepared a solution by adding enough Na2 CO3 to 2 L of deionized water to provide an alkalinity of 160 mg/L as CaCO3 . However, you subsequently find out that that bottle had previously contained 4 M HAc, and you are not sure that it was thoroughly cleaned before you used it. You measure the pH of your solution and find that it is 10.1. Determine whether any HAc was left in the bottle when you began your efforts and, if so, estimate its volume. 11. You wish to make a solution that has an alkalinity of 1.5 meq/L, with equal portions of the alkalinity contributed by NaOH, NaHCO3 , and Na2 CO3 . How much of each chemical should you add, in mg/L? 12. A solution at pH 8.0 contains 10 3 M T OT Ac and has an alkalinity of 1.5 meq/L. If the non-acetate alkalinity is contributed by carbonate species, what is T OT CO3 ? 13. A water contains 10 3 M Na2 CO3 and 10 equiv/L and mg/L as CaCO3 ?
3M
HCl. What is the alkalinity in
14. Estimate the alkalinity of a solution containing the following major species, based on titration with HCl to an endpoint of pH 4.5. Use the Davies equation to estimate activity coefficients.
460
T OT Na = 2 ⇥ 10 T OT Ca = 2 ⇥ 10
3
M
3
M
T OT CO3 = 2 ⇥ 10 T OT PO4 = 2 ⇥ 10
pH = 7.2
T OT Cl = 1 ⇥ 10
3
3
M
3
M
M
15. Calculate the alkalinity of aqueous solutions with the following characteristics: (a) pH= 3.5, T OT CO3 = 10
3M
(b) pH= 7.5, T OT CO3 = 10
3M
(c) pH= 11.5, T OT CO3 = 10
3M
16. An organic acid has been added to a solution with an alkalinity of 50 mg/L as CaCO3 . The alkalinity was added entirely as NaOH. If the total concentration of the organic acid added was 10 2.9 M and the final pH was 5.2, what is pKa of the acid? 17. An ideal solution has an alkalinity of 120 mg/L as CaCO3 . One-fourth of the alkalinity is provided by hydroxide ions, and the rest by carbonate species. Determine the pH of the solution and T OT CO3 . 18. An anaerobic biological treatment process is being designed to treat a solution containing 4000 mg/L HAc. The bacteria can convert acetic acid to carbon dioxide and methane by the following reactions, but only if the pH is above 5.2. CH3 COOH + H2 O ! CH4 + H2 CO3 CH3 COO + H2 O ! CH4 + HCO3
(a) Prepare a log C-pH diagram for the solution prior to treatment. What is the initial pH? (b) What concentration of NaOH must be added to adjust the solution to pH 5.2? What concentration of Na2 CO3 could be used as an alternative? (c) It is proposed to add 6 ⇥ 10 2 M NaOH to the original solution, after which the reaction is expected to convert 80% of the T OT Ac to end products. Prepare a log C-pH diagram for the resultant solution and find the pH before and after the biological conversion. 19. A log C-pH diagram for a solution containing 7 ⇥ 10 3 M of a triprotic acid, H3 A, is shown below. Note that HA2– is never a significant species compared to the other forms of the acid. Also shown is a base titration curve for a solution initially at pH 6.8 containing an unknown amount of T OT A. 461
(a) What is T OT A in the system being titrated? (b) Hydrochloric acid (HCl) is gradually added to the pH 12.0 solution until the pH decreases to 4.5. On a copy of the preceding titration curve, sketch a plot of pH vs. HCl added, starting at the point where the previous titration ended, i.e., at (19, 12.0). The amount of acid added can be represented as negative additions of base. 20. A solution has a pH of 7.3 and an alkalinity of 25 mg/L as CaCO3 . Essentially all the alkalinity is contributed by the carbonate system. The water also contains 10 4 M Fe2+ (5.5 mg Fe2+ /L). To remove the iron, oxygen is injected into the solution, converting the Fe2+ to Fe3+ ; the Fe3+ then precipitates as ferric hydroxide according to the following reaction: 1 Fe2+ + O2 (aq) + 2.5 H2 O )* Fe(OH)3 (s) + 2 H+ 4 Calculate the pH of the water after the reactions are complete. 462
21. Two small streams, one of which drains an abandoned mine and is quite acidic, combine to form a larger stream. Some characteristics of the streams are provided below. Use a charge balance to estimate the DIC in Stream B, and then determine the pH, alkalinity, and DIC concentration in the combined flow. Parameter Na+ Ca2+ Cl– SO2– 4 DIC pH Flowrate
Stream A 10 2.8 M 10 4.9 M 10 3.0 M 10 3.3 M 10 5.0 M 3.4 50 L/min
Stream B 10 4.0 M 10 4.1 M 10 4.0 M 10 5.1 M not analyzed 7.7 250 L/min
22. A solution at pH 7.5 contains 2 ⇥ 10 3 M T OT Ac, essentially all of which is consumed in a biological treatment process according to the following reactions: C2 H3 O2 + 2 O2 ! 2 HCO3 + H+ C2 H4 O2 + 2 O2 ! 2 HCO3 + 2 H+
(a) Determine the change in alkalinity, if any, that occurs if all the acetate undergoes the reaction. Express your answer in mg/L as CaCO3 . Assume that the solution behaves ideally. (b) If the ionic strength of the solution in part (a) were high enough to cause the activity coefficient for species with charge of ±1 to be 0.7, how would T OT Ac be distributed between HAc and Ac– in the original solution and at the endpoint of the alkalinity titration? (c) Repeat part (a), if the reaction occurred in the solution described in part (b). Assume that T OT CO3 is present entirely as H2 CO3 at the titration endpoint, regardless of the ionic strength. Keep in mind that pH is defined as log aH+ . 23. As noted in the chapter, many NOM molecules are acidic. As a result, water that contains NOM and low concentrations of dissolved inorganic species can be quite corrosive to both cementitious and metallic components of drinking water distribution systems. One approach for overcoming this problem is to contact the water with calcite (CaCO3 ) to neutralize some of the acidity. Assume that the NOM molecules in a particular water supply can be represented as diprotic acids with pKa1 = 4.8 and pKa2 = 7.7, and with an average molecular weight of 250. The water, at pH 6.1, contains 5 mg/L Ca2+ and 463
10 4 mol/L T OT CO3 initially, and is treated in a limestone (calcite) contactor. The remaining solutes (other than the NOM) are salt ions that are neither acids nor bases. The water exiting the contactor contains 9.5 mg/L Ca2+ and is at pH 7.9. (a) Using the species that are dominant in the treated water as components, write the T OT H equation for the influent solution. (Note: you will not be able to assign known values to all the terms in this expression at this point; don’t let that concern you, but do fill in as many values as you can.) (b) Using the same conventions as in part (a), write and solve a T OT H equation for the treated water to determine the NOM concentration, in mg NOM/L. 24. A treated water supply at pH 9.54 has an alkalinity of 20 mg/L as CaCO3 and contains 8 mg/L T OT Si as SiO2 , present primarily as H4 SiO4 and H3 SiO–4 . (a) Assuming that the solution is ideal, what fraction of the alkalinity is contributed by the silica species? (b) Explain briefly how your answer to part (a) would change, if at all, if you discovered that the ionic strength of the solution was high enough to cause the activity coefficient of H3 SiO–3 to be 0.8. Because it is neutral, the activity coefficient of H4 SiO4 can still be assumed to be close to 1.0. 25. A solution containing 125 mg/L butyric acid (C4 H8 O2 , MW 88, pKa = 4.73) is adjusted to pH 7.0 by addition of Na2 CO3 and then is treated in a biological process that converts essentially all of the butyrate to carbonate species via the following reaction: CH3 (CH2 )2 COO + 5 O2 ! 4 HCO3 + 3 H+ (a) Estimate the alkalinity (defined by titration to pH 4.5) after the neutralization step, but before biological treatment. (b) Estimate the buffer intensity of the solution in part (a) by computing the required amount of strong base to increase the pH to 7.1. Also compute the buffer intensity exactly, by using the equation given in the text. (c) What is the pH of the water after biological treatment? 26. A solution is made by adding 3 ⇥ 10 4 M Ca(OCl)2 , 4 ⇥ 10 4 M CaAc2 , 7 ⇥ 10 4 M NH4 Ac, and 2 ⇥ 10 4 M H2 SO4 to water. In the pH range of 4 to 10, at what pH do you expect the buffer intensity to be largest? Explain briefly. 27. Attempt to arrange the following solutions in order of increasing buffer intensity without carrying out any calculations, and then compute the buffer intensities numerically. 464
(a) 10
3M
T OT NH3 at pH 7.04
(b) 10
3M
T OT NH3 at pH 9.24
(c) 10
3M
T OT CO3 at pH 8.35
(d) 10
3M
T OT CO3 at pH 6.35
28. Determine the pH and buffer intensity of aqueous solutions made by dissolving the following chemicals in pure water: (a) 0.1 M HAc plus 0.1 M NaAc (b) 0.19 M HAc plus 0.01 M NaAc (c) 0.02 M HAc plus 0.18 M NaAc 29. Calculate the buffer intensity of a solution prepared by adding 0.2 M acetic acid and 0.1 M sodium acetate to water, and then adjusting the pH to 5.0 by addition of NaOH. 30. Plot a curve showing the buffer intensity of a solution containing 0.1 M total oxalate over the range from pH 2 to 12. Under what conditions does the oxalate contribute >90% to the total buffer intensity? 31.
(a) Calculate the buffer intensity at pH 6, 8, and 10 for a solution containing 10 2.8 M T OT CO3 . (b) For each of the solutions in part (a), use the calculated buffer intensity to estimate the final pH if 10 5 M NaOH is added.
32. In the metal finishing industry, “pickling” refers to the process of exposing metal parts to a strong acid solution to remove surface deposits, thereby preparing the surface for painting or other modification. An industrial wastewater from a pickling process contains 0.4 M T OT SO4 and 0.1 M T OT Fe at pH 0.8. Iron that is incorporated into dissolved species can have a charge of either +2 or +3 (referred to as ferrous and ferric species, respectively); all the Fe in the wastewater is ferric iron. Other ions might be present in solution, but they are not acids or bases. If you use Visual Minteq in your analysis, instruct the program to ignore species that contain more than one Fe3+ ion, both an Fe3+ ion 2+ 2– and an SO2– 4 ion, or both a Ca ion and an SO4 ion. (a) Compute the amount of lime (Ca(OH)2 ), in mol/L, needed to bring the solution to pH 5.5 for discharge. (b) Develop a titration curve (pH vs. meq/L of lime added) for titration of the solution to pH 10. Report the pH after 25%, 50%, and 75% of the ultimate lime dose has been added. Ignore possible formation of CaSOo4 . 465
(c) Compute the buffer intensity and indicate the major reactions responsible for the buffer intensity in the initial solution and at the four conditions investigated in part (b). 33. A waste solution containing 0.1 M sulfuric acid is to be discharged into a body of water having a pH of 7.5 and a carbonate alkalinity of 2 ⇥ 10 3 equiv/L. To what extent must the waste be diluted (i.e., what volume of river water must be combined with a unit volume of the waste) to assure that the pH of the mixture does not drop below 6.0? 34. A titration curve is shown below for a solution with an initial pH of 7.4. (a) What is the alkalinity of the solution, in both meq/L and mg/L as CaCO3 . (b) For the pH range shown, at approximately what pH is the buffer intensity maximized? (c) A species B(aq) that is present in the solution is a diprotic base with pKb1 = 4.0 and pKb2 = 12.0. Write out the reactions corresponding to these pK values. Note that, for acids, the numbering of acidity constants (Ka1 , Ka2 , . . .) starts with the most acidic species. Similarly, the numbering of basicity constants (Kb1 , Kb2 , . . .) starts with the most basic species. (d) When the solution equilibrates with a gas phase, 2 ⇥ 10 4 M B(aq) transfers out of solution and into the gas. By approximately how much does the alkalinity of the solution change, and in what direction? Explain your reasoning briefly. (Hint: It might be helpful to convert the Kb values to Ka ’s and determine the dominant species at different pH values.)
35. You want to prepare a solution in which the pH will remain in the range 9.8 ± 0.1 when up to 10 2 M strong acid or base is added. Noting that the target pH is close to pKa of NH+4 , you decide to prepare the solution by adding a mixture 466
of NH3 and NH4 Cl to water. What amounts of these chemicals should you add if you want to achieve the desired buffering using the minimum T OT NH4 ? 36. An anaerobic solution in a stagnant portion of a sewer can be represented as a mixture of 5 ⇥ 10 3 M Na(NH4 )2 PO4 and 5 ⇥ 10 4 M NaHS. (a) Sulfuric acid has formed on the upper surface of the pipe and is dripping into the wastewater. Prepare a plot of pH versus H2 SO4 added for additions of 0 to 6 ⇥ 10 3 M H2 SO4 . Ignore dilution of the wastewater by the dripping acid, and assume that, over the time frame of interest, sulfate and sulfide species are not interconvertible. (b) Prepare a plot of buffer intensity versus pH for the range of H2 SO4 additions considered in part (a). For completeness, also show the data for addition of base to pH 11. At what pH in the range investigated is the solution most highly buffered? (c) How much acid can be added before the H2 S concentration reaches 1 ⇥ 10 4 M? 37. The acid sulfite pulping process uses sulfurous acid (H2 SO3 ) to attack the lignin that holds the wood fibers together. The fibers are then released and processed into paper products, and a hot solution containing acetic acid, sulfurous acid, and some larger organic molecules is generated as a waste. This waste solution is partially evaporated and then condensed, with the smaller molecules being transferred into the evaporator condensate and the larger ones remaining behind in an organic-rich solution that can be burned to recover its energy content. The condensate can be treated in an anaerobic biological treatment process, but only if the pH is maintained near neutrality (pH 7). When the process is operating successfully, the microorganisms mediate the following reactions (CH4 is methane): CH3 COOH + H2 O )* CH4 + H2 CO3 4 H2 SO3 + 3 CH3 COOH )* 4 H2 S + 6 H2 CO3 (Note: The Visual Minteq database includes the species S2 O2– 5 , which can form from the combination of two HSO–3 ions and release of a water molecule. In answering the following questions, if you use Visual Minteq, exclude this species from the analysis so that your result is similar to that for a manual analysis that does not consider the species.) (a) Find the pH of a solution of 5000 mg/L HAc and 300 mg/L H2 SO3 S. (b) What would the solution pH be if all the HAc and all the H2 SO3 were destroyed according to the above reactions, and if all the H2 CO3 that was 467
generated remained in solution? CH4 undergoes no acid/base reactions. (In reality, much of the H2 CO3 dehydrates to CO2 , which then forms bubbles along with the CH4 and exits the solution.) (c) How much NaOH, in mol/L, would be needed to bring the pH of the initial solution [as characterized in part (a)] to 7.0? How much Na2 CO3 would be required to accomplish the same result? Comment on the relative requirements of these two bases, considering their relative strengths and the number of protons each can accept. (d) If the solution pH were increased to 7.0 with Na2 CO3 and the biological reaction then proceeded to destroy one-half of the sulfite and all the acetate according to the reactions shown, what would the final pH be?
468
9
GAS/LIQUID EQUILIBRIUM
469
Contents
9.1 9.2
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . BASIC CONCEPTS AND TERMINOLOGY FOR GAS/LIQUID EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 EFFECT OF GAS/LIQUID EQUILIBRATION ON GAS-PHASE COMPOSITION . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 FACTORS AFFECTING VOLATILITY AND HENRY’S CONSTANT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 HENRY’S LAW AND AQUEOUS-PHASE SPECIATION . . . . . 9.6 THE pH OF SOLUTIONS IN EQUILIBRIUM WITH ACIDIC OR BASIC GASES . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 GAS-LIQUID EQUILIBRIUM IN VISUAL MINTEQ . . . . . . . 9.8 BUFFERING IN SOLUTIONS EQUILIBRATED WITH WEAK ACID AND WEAK BASE GASES . . . . . . . . . . . . . . . . . 9.9 CO2 DISSOLUTION, ALKALINITY, AND ACIDITY . . . . . . . 9.10 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.11 APPENDIX 9A. THE LIQUID-PHASE ACTIVITY COEFFICIENT IN HENRY’S LAW USING THE RAOULT’S LAW STANDARD STATE CONVENTION . . . . . . 9.12 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1
470 471 482 483 490 494 502 508 512 515
516 519
INTRODUCTION
Equilibrium between gases and aqueous solutions drives a wide range of environmental processes, including the exchange of oxygen and carbon dioxide between the atmosphere and waters at the earth’s surface (both natural bodies of water and waters undergoing aerobic biological treatment); the release of methane and hydrogen sulfide from anaerobic sediments, wastewaters, and sludges; and the removal of sulfur 470
dioxide from the gaseous products of fossil fuel combustion, to name just a few. In addition, such exchanges play a central role in the global cycling of carbon, nitrogen, oxygen, and many other elements. In many cases, gaseous species undergo additional reactions once they dissolve, altering the solution pH, the solubility and behavior of metals, and biological productivity. Gas transfer into solution is commonly referred to as gas absorption or dissolution, and transfer out of solution is referred to as volatilization or gas stripping. Qualitatively, the tendency of a species to transfer into in the gas phase is referred to as its volatility. Systems in which molecules can transfer between a gas phase and solution are referred to as open systems, and those where such exchange is precluded are called closed systems. In this chapter, we explore equilibrium between gases and aqueous solutions, with special emphasis on how gases affect the pH of the solutions.
9.2 9.2.1
BASIC CONCEPTS AND TERMINOLOGY FOR GAS/LIQUID EQUILIBRIUM The Ideal Gas Law
Under normal environmental conditions, the behavior of all gases conforms closely to that described by the ideal gas law. As a result, all gaseous species share the following relationships among their molar- and mass-based concentrations, the pressure that they exert on their surroundings, and the ambient temperature: cG,i =
c0G,i MWi
=
ni Pi = VG RT
(9.1)
where cG,i and c0G,i are the gas-phase concentrations of i in moles and grams per unit volume, respectively; MWi is the molecular weight of i; ni is the number of moles of i in the gas phase; VG is the gas volume; Pi is the pressure exerted by i in the gas phase (the partial pressure of i); T is absolute temperature; and R is the universal gas constant. At 25 C, RT is 24.47 L·atm/mol.1 According to Equation (9.1), the partial pressure exerted by any ideal gas i, regardless of its chemical identity, is given by Pi = RT cG,i
(9.2)
This equation indicates that the partial pressure of i can serve as a surrogate for the gas-phase concentration of i if the temperature is specified. Pi is frequently used in that way in the literature, often with an implicit assumption that the temperature is 25 C, and it will be used in that way here as well. 1 The
SI unit for expressing pressure is the pascal (Pa). However, historically, pressures have been reported in the environmental and other scientific literature in atmospheres (atm). The metric unit that approximates one atm is one bar, which equals 105 Pa (often expressed as 100 kPa); 105 Pa actually equals 0.9869 atm.
471
The partial pressures exerted by different species in a gas phase are additive, so the total pressure is the sum of the partial pressures. Therefore, for a gas phase containing N different species Ptot = Pi = Ptot
N
N
 Pj = RT
 cG, j
j=1
RT cG,i RT
N
(9.3)
j=1
=
 cG,i
j=1
cG,i N
= yi
(9.4)
 cG, j
j=1
where yi is the dimensionless mole fraction of i in the gas phase, defined as yi ⌘
9.2.2
mole of i in gas phase total moles of all species in gas phase
(9.5)
Henry’s Law and Raoult’s Law
The transfer of a species X from an aqueous solution into a gas phase can be represented by the following reaction: X(aq) )* X(g)
(9.6)
where X(aq) represents molecules of X dissolved in the aqueous solution, and X(g) represents molecules of X in the gas phase. In some cases, dissolved molecules form strong enough bonds with water that a water molecule is considered to be part of the dissolved species. For instance, CO2 (aq) and SO2 (aq) are often written as H2 CO3 and H2 SO3 , respectively, so that the reactions characterizing volatilization of these species are H2 CO3 (aq) )* CO2 (g) + H2 O
(9.7)
H2 SO3 (aq) )* SO2 (g) + H2 O
(9.8)
Although, at times, it might be important to know whether a dissolved gas molecule is chemically bonded or just adjacent to a water molecule, the distinction between these two scenarios is usually unimportant for our purposes, so we can consider the corresponding pairs of species (e.g., CO2 (aq) and H2 CO3 (aq)) to be identical. In some literature, the combination of the two forms of such molecules is designated by adding an asterisk to the formula (e.g., the combination of CO2 (aq) and H2 CO3 (aq) is designated H2 CO3 *). The relationship at equilibrium between the liquid- and gas-phase concentrations of a volatile species is typically like that shown in Figure 9.1. The linear portions of 472
Bubbling a gas through water can be an e↵ective way to transfer some chemicals into solution. In this photograph, ozone is being bubbled through potable water in a laboratory-scale test. The ozone that dissolves acts as a powerful disinfectant. (MWH Global.)
this plot indicate that cG,i is directly proportional to cL,i in two limiting cases: one that applies when the volatile species is very dilute, and the other that applies when the liquid phase is almost pure i. As indicated, these regions of the plot are referred to as the Henry’s law and Raoult’s law regions, respectively. The proportionality constant in the Henry’s law region is usually designated H and is referred to as the Henry’s law constant, or simply the Henry’s constant, so we can write Henry’s Law:
cG,i = HcL,i
(9.9)
Raoult’s law, like Henry’s law, can be written using any selected units for cL,i and cG,i . However, in practice, it is most often written using the mole fraction of 473
i (i.e., xi ) to express the concentration in the liquid phase, and the partial pressure of i (i.e., Pi ) to express the concentration in the gas. In these units, the endpoints of the extrapolated line representing Raoult’s law in Figure 9.1 correspond to Pi = 0 (zero partial pressure of i) when i is absent from the solution (xi = 0), and Pi = Pvap,i (partial pressure of i equal to its vapor pressure) when the liquid is pure i (xi = 1.0). As a result, the relationship is commonly written as Raoult’s Law:
PG,i = Pvap,i xi
(9.10)
Figure 9.1 Gas-liquid equilibrium, demonstrating the Henry’s law and Raoult’s law regions.
Although the figure shows a continuum between the two limiting cases, the volatile species and water might not form a homogeneous solution at all mixing ratios. That is, at certain mixing ratios, the constituents might split into two different liquid phases – one consisting of nearly pure water with a small concentration of dissolved i, and the other consisting of nearly pure i with a small concentration of dissolved water. This scenario is commonly observed for mixtures of water with volatile, hydrophobic organic species such as benzene, toluene, and trichloroethylene. In such cases, the curve in Figure 9.1 is discontinuous, because a portion of the middle of the curve represents unattainable conditions for a single liquid phase. The curve in Figure 9.1 and Equations (9.9) and (9.10) characterize the distribution of i between gaseous and liquid phases at equilibrium, but that distribution does not always apply. Equilibration of gases with solutions typically is much slower than equilibration of acid/base reactions or of many other reactions that involve only dissolved species. This difference is due in part to the fact that the equilibration rate in gas/liquid systems depends not only on chemical interactions in the solution phase, but also on physical factors such as the amount of interfacial area available for gas transfer and the rate at which the gas and liquid near the interface are mixed with 474
the bulk fluids. We will not concern ourselves here with the rate at which gas/liquid systems approach equilibrium, but it is important to appreciate the fact that such systems are often not equilibrated. If cL,i is larger than the equilibrium value for the extant value of cG,i (i.e., if the system is characterized by a point below and to the right of the curve in Figure 9.1), the species will transfer out of the solution and into the gas. In such a case, the aqueous solution is said to be supersaturated with the gas. Correspondingly, if cL,i is less than the equilibrium value (characterized by a point above and to the left of the curve), the solution is said to be undersaturated with the gas, and the species will be absorbed from the gas into solution. Finally, if cL,i is equilibrated with cG,i , the solution is said to be saturated with the volatile species. Although some environmental systems are in the Raoult’s law region of the figure (e.g., underground aquifers that have been contaminated by a spill or leakage from a storage tank and contain a nearly pure, nonaqueous phase liquid [NAPL]), the primary interest in the vast majority of environmental engineering and science applications is in the Henry’s law region. Indeed, some volatile solutes (e.g., O2 , CO2 , H2 S) never exist as pure liquids at normal temperatures and pressures, in which case the Raoult’s law portion of the graph is never relevant. In the remainder of this chapter, we consider only dilute aqueous solutions that fall in the Henry’s law region of Figure 9.1. According to Equation (9.9), Henry’s constant can be defined as the ratio of the gas-phase concentration to the aqueous-phase concentration of a volatile species i, when the system is at equilibrium and the solution is almost infinitely dilute (almost pure water). The ratio of a constituent’s concentrations in two phases at equilibrium is often referred to as a distribution coefficient or partition coefficient. Thus, Henry’s constant can be described as the partition coefficient for a species i between the gas and a dilute aqueous phase. A related, but subtly different, definition of Henry’s constant identifies it as the equilibrium constant for the volatilization reaction shown as Reaction (9.6), in which case it represents a ratio of activities, not concentrations. Using the symbol H 0 for this definition, we can write aG,i (9.11) H0 ⌘ aL,i where aG,i and aL,i are the chemical activities of i in the gas and liquid phases, respectively. Like Equation (9.9), Equation (9.11) is understood to apply only when the gas exchange reaction is at equilibrium. As explained in Chapter 2, the chemical activity of a species i in any phase can be expressed as the product of an activity coefficient (gi ) and the concentration of i normalized to the standard state concentration. ai = gi 475
ci ci
(9.12)
Substituting this relationship for both activity terms in Equation (9.11), we obtain
H0 =
gG,i gL,i
cG,i cG,i cL,i cL,i
1 H = gL,i
gG,i
0
1 H = gL,i
gG,i
0
(9.13a)
cL,i cG,i cL,i cG,i
!
cG,i cL,i
(9.13b)
!
H
(9.13c)
The standard state concentrations in both phases (cG,i and cL,i ) are always assigned values of 1.0, with whatever units are being used to express the concentrations in the actual system, so the ratio of these concentrations can be written as cL,i cG,i
= 1.0
units of cL,i units of cG,i
(9.14)
Also, recall that the activity coefficient is an indicator of how the interactions of i with surrounding molecules compare to those in a reference environment; if the interactions are similar in the two environments, gi is close to 1.0, and if they differ substantially, gi is far from 1.0. The reference environment for gaseous species is always chosen to be that in an ideal gas. Correspondingly, because at normal temperatures and pressures, the behavior of any real gas actually does conform closely to that of an ideal gas, gG,i is always very close to 1.0 under environmental conditions. Therefore, we can rewrite Equation (9.13c) as ✓ ◆ units of cL,i 1 0 H = 1.0 H (9.15) gL,i units of cG,i Two different choices are commonly made for the reference environment of volatile species in aqueous solutions: infinite dilution (the Henry’s law reference state), and a pure phase of the species (the Raoult’s law reference state). In dilute aqueous solutions, the environment in the actual solution is similar to that in the Henry’s law reference state, and gL,i is likely to be close to 1.0 if that reference state is chosen. In such a case, the magnitudes of H 0 and H are almost identical. On the other hand, if the Raoult’s law reference state is adopted and applied to solutes in a dilute solution, gG,i can be far different from 1.0, and H and H 0 can have very different magnitudes. This choice for the reference state is made fairly commonly in the literature dealing with gas/liquid equilibrium of volatile organic compounds. In that case, gL,i can be approximated as the inverse of the mole fraction 476
Intense mixing is often used to contact air with water in biological wastewater treatment processes, where oxygen is consumed and CO2 is generated by the microorganisms as they consume organic matter. The aeration facilitates dissolution of more oxygen and stripping of the CO2 so that the process can continue. (Michael K. Stenstrom.)
of i in the aqueous solution, when that solution is equilibrated with pure i.2 For example, if water is equilibrated with pure liquid chloroform at 20 C, the mole fraction of chloroform in the aqueous solution is 1.2 ⇥ 10 3 . Therefore, if the Raoult’s law reference state were adopted, the activity coefficient of chloroform in the aqueous solution would be approximately 1/(1.2 ⇥ 10 3 ), or 830. Although the definitions of Henry’s constant given in Equations (9.9) and (9.11) are the ones used most often in environmental engineering and science, at least one other definition is in fairly common use. This definition is based on the reverse of Reaction (9.6), so that the expressions for the partition and equilibrium constants are inverted. When this definition is used, the constant is often written as KH , i.e.: KH =
cL,i 1 = H cG,i
(9.16a)
The corresponding equation if the constant is expressed as a ratio of activities instead of concentrations is ✓ ◆ units of cG,i 1 1 0 KH = 0 = g L,i 1.0 (9.16b) H units of cL,i H 2 This
relationship is derived in Appendix 9A.
477
This definition of the Henry’s constant is frequently used by researchers studying natural aquatic systems, especially when describing the equilibrium of atmospheric gases such as O2 and CO2 . Throughout this text, unless otherwise stated, we adopt the definition of Henry’s constant as a distribution coefficient [H, as given in Equation (9.9)]. With this definition, H has units that correspond to those of cG,i /cL,i , and higher values of H imply higher volatility (or, equivalently, lower solubility). If the infinite dilution convention is used and the solute behaves ideally (gL,i = 1.0), H has the same magnitude as H 0 , and the only difference between the two values is that H has units and H 0 is dimensionless. Some typical sets of units used to report H are shown in Table 9.1, and values of H for some environmentally significant gases are listed in Table 9.2. Table 9.1 Units commonly used for Henry’s constants.
Symbol(a)
Units of H
Mass concentration (mg i/L of solution)
Hcc
–
Molar concentration (mol i/L of gas)
Molar concentration (mol i/L of solution)
Hmm
–
Partial pressure (atm)
Mass concentration (mg i/L of solution)
Hpc
atm-L mg i
Partial pressure (atm)
Molar concentration (mol i/L of solution)
Hpm
atm-L mol i
Partial pressure (atm)
Mole fraction (mol i/total moles of solution)
Hpx
atm
Mole fraction(b)
Molar concentraion (mol i/L of solution)
Hym
L mol of gas
Mole fraction(b)
Mole fraction (mol i/total moles of solution)
Hyx
–
Dimensions used for gas phase concentration
Dimensions used for aqueous phase concentration
Mass concentration (mg i/L of gas)
(a)
These designations for the Henry’s constant are convenient ways of conveying the parameters being used to define H, but they have not yet been widely adopted. (b) When the units used for the gas-phase concentration are mole fractions, the value of H depends on the total pressure in the gas phase. Tabulated values of Hym and Hyx invariably assume Ptot = 1.0 atm. Ptot,system Values applicable for other pressures can be computed as Hsystem = Htabulated . Ptot,tabulated (= 1.0 atm)
478
Table 9.2 Henry’s constants of some environmentally important gases(a)
Compound Nitrogen Hydrogen Carbon monoxide Oxygen Methane Cyclohexane Radon Ozone Carbon dioxide Carbon tetrachloride Tetrachloroetheylene (PCE) Chlorine Trichloroethylene (TCE) Hydrogen sulfide Ethylbenzene Toluene Benzene o-Xylene Chloroform Chlorine dioxide Sulfur dioxide Bromoform 2,20 ,5,50 Tetrachlorobiphenyl Ethyl acetate Hydrogen cyanide Acetone Ammonia Methanol Phenol Acetic acid
Hpm (atm/(mol/L)) 1513 1280 1050 769 769 182 108 90.9 29.4 29.4 17.5 10.8 10.0 9.77 8.33 6.67 5.56 4.55 4.00 1.00 0.813 0.556 0.250 0.143 0.100 0.040 0.0171 4.6 ⇥ 10 3 5.0 ⇥ 10 4 2.0 ⇥ 10 4
(a) Values
Hcc or Hmm (b) – 61.8 52.3 42.9 31.4 31.4 7.44 4.41 3.72 1.20 1.20 0.715 0.441 0.409 0.399 0.340 0.273 0.227 0.186 0.163 0.0409 0.0332 0.0227 0.0102 5.84 ⇥ 10 3 4.09 ⇥ 10 3 1.64 ⇥ 10 3 6.99 ⇥ 10 4 1.88 ⇥ 10 4 2.04 ⇥ 10 5 8.17 ⇥ 10 6
are for 25 C and are from the Visual Minteq database or the compilation by Sander [Sander, R. (1999) Compilation of Henry’s Law Constants for Inorganic and Organic Species of Potential Importance in Environmental Chemistry (Version 3). http://www.henrys-law.org. Accessed June 19, 2006]. (b) Values in this column are valid for any units of concentration, provided that the same units are used for both the liquid and gas phases.
479
Example 9.1 Henry’s constant for oxygen at 20 C is 0.73 atm/(mol/m3 ). Express this constant in the following forms: (a) Hpm ; (b) Hpc ; (c) Hcc . (See Table 9.1 for the meaning of the subscripts.) Solution (a) The conversion of the given Henry’s constant to Hpm requires only a change in the liquid-phase concentration units from mol/m3 to mol/L. The calculation is ✓ ◆✓ ◆ atm L atm Hpm = 0.73 1000 3 = 730 3 m mol/L mol/m (b) Hpc is the Henry’s constant with the gas-phase concentration expressed as a partial pressure and the solution-phase concentration expressed in mol/L. Combining the result from part (a) with the conversions in Equation (9.1), we can find Hpc as atm Hpm Pi Pi atm mol/L Hpc = = = = = 0.023 0 cL,i (cL,i ) (MWi ) MWi 32, 000 mg/mol mg/L 730
(c) Hcc is the Henry’s constant with both the gas- and solution-phase concentrations expressed in mg/L. Thus, Hcc =
cG,i 0 Pi (MWi )/RT MWi = = Hpc 0 0 cL,i cL,i RT
= 0.023
atm 32, 000 mg/mol ✓ ◆ = 30 atm-L mg/L 0.082 (293 K) mol-K
The value of Henry’s constant expressed as Hcc is particularly useful for acquiring an intuitive feel for the solubility or volatility of a chemical. Consider, for example, a closed vial containing volumes VL of liquid and VG of gas, with a volatile species present in the two phases at concentrations of cL and cG , respectively. If these two phases are at equilibrium, then the concentration ratio equals Henry’s constant: cG mG /VG = = Hcc cL mL /VL 480
(9.17)
If we specify further that mG = mL (i.e., that the same mass of the volatile species is in each phase), then Hcc = VL /VG . That is, the Henry’s constant expressed as Hcc equals the liquid-to-gas volume ratio that causes equal masses of the constituent to be in the two phases at equilibrium. For example, Hcc for benzene at 25 C is 0.227. According to the preceding discussion, this value implies that, at equilibrium, the mass of benzene in 0.227 L of solution would be the same as that in 1.0 L of gas. On the other hand, Hcc for oxygen is 31.4, meaning that (again, at equilibrium) 31.4 L of solution is needed to hold the same mass of O2 as is present in 1.0 L of gas. This way of thinking about volatility and Henry’s constant is shown schematically in Figure 9.2a. A similar comparison can be made for a closed system that has equal volumes of liquid and gas (i.e., VL /VG = 1.0 L/L). In that case, at equilibrium, Hcc = mG /mL . That is, for a system with equal volumes of liquid and gas, Hcc equals the ratio of the species’ mass in the gas phase to that in solution. Thus, in an equilibrium system with VL = VG , benzene will be distributed in a ratio of 0.227 grams in the gas phase per gram in solution, whereas O2 will be distributed in a ratio of 31.4 grams in the gas per gram in solution (Figure 9.2b). Based on either of these ways of comparing the two species, we conclude that oxygen is far more volatile (i.e., less soluble) than benzene, consistent with the understanding that a higher Henry’s constant corresponds to greater volatility.
Figure 9.2 Various ways of interpreting H. (a) The solution-to-gas volume ratios needed to have an equal mass of benzene or oxygen in each phase. (b) Distribution of benzene and oxygen between the gas and solution in an equilibrium system with VL = VG .
The equilibrium distribution of a species between the gas and solution in a closed vial is taken advantage of when headspace analysis is used in gas chromatography. In this application, the gas phase concentration is measured, and the liquid phase concentration is inferred, assuming gas/liquid equilibrium and taking into account the ratio of volumes of liquid and gas in the vial. 481
9.3
EFFECT OF GAS/LIQUID EQUILIBRATION ON GAS-PHASE COMPOSITION
In many systems (e.g., when the gas phase of interest is the bulk atmosphere), transfer of volatile molecules between the two phases has virtually no effect on the composition of the gas phase. This approximation is sometimes described as an assumption that the gas phase is infinite (so its composition is unchangeable in response to finite additions or subtractions of a volatile species). There are, of course, also situations in which the assumption of an invariant gas-phase composition is not valid. For instance, gases present at extremely low concentrations in the atmosphere, such as SO2 in areas where human influence is minimal, can be substantially removed from the gas phase by contact with cloud droplets that have formed by condensation of water onto alkaline salt crystals.3 Similarly, the concentrations of oxygen, carbon dioxide, and trace gases in a bubble rising through the water column of a biological treatment process might change significantly as the bubble ascends. In such a case, one can take the change in gas-phase composition into account by writing a mass balance on the gas phase and linking that to a mass balance on the solution as part of the analysis, as shown in the following example. Example 9.2 You have received a sealed vial containing a water sample that is to be analyzed for cyclohexane. The volume of the vial is 10 mL, and the temperature is 25 C. You notice that the vial contains an air bubble, whose volume you estimate to be 0.25 mL. (a) If the sample originally contained 10 7 mol/L of the analyte, what concentration will you measure in the solution, assuming that it has equilibrated with the air bubble? (b) What are the concentrations of cyclohexane (MW = 84) in the gas and liquid phases in the bottle, in µg/L?
Solution (a) We can determine the final system composition by writing a mass balance on cyclohexane in the entire system (water plus air) and combining it with the Henry’s law expression. Assuming that the air bubble initially contained no cyclohexane, the total number of moles of cyclohexane (ntot,i ) in the vial is (0.00975 L) (10 7.0 mol/L), or 9.75 ⇥ 10 10 mol. Once the solution and air bubble have equilibrated, the cyclohexane will be distributed in accordance 3 This phenomenon, and the effect of gas/liquid equilibrium on the pH of precipitation, was addressed
in a classic paper in 1982 [Charlson, R.J., and Rodhe, H. (1982) “Factors controlling the acidity of natural rainwater,” Nature 295, 683-685.]
482
with Henry’s law, and its molar concentration in the gas phase will be consistent with the ideal gas law, so (representing cyclohexane as i): n tot,i = nG,i + nL,i = cG,iVG + cL,iVL n tot = Pi =
=
Pi Pi VG + VL RT Hpm ntot,i VG VL + RT Hpm 9.75 ⇥ 10 10 mol = 1.52 ⇥ 10 0.00025 L 0.00975 L ✓ ◆ + atm-L atm-L 0.082 (293 K) 182 mol mol-K
5
atm
The equilibrium concentration in solution can be found by combining this result with Hpm : cL,i =
1.52 ⇥ 10 5 atm Pi = = 8.4 ⇥ 10 atm Hpm,i 182 mol/L
8
mol/L
The solution originally contained 10 7 mol/L cyclohexane, so the calculation indicates that approximately 16% of this species volatilized when the solution equilibrated with the air bubble. This loss of the analyte would cause a corresponding 16% error in the concentration that would be reported if the analysis did not take the volatilization into account. (b) The value of cL,i in µg/L can be computed as the product of the molar concentration and the molecular weight, and cG,i can then be computed using Hcc,i : 84 ⇥ 106 µg mol/L = 7.0 µg/L ✓ ◆ mol L cG,i = Hcc,i cL,i = 7.44 (7.0 µg/L) = 52 µg/L L cL,i = 8.4 ⇥ 10
9.4
8
FACTORS AFFECTING VOLATILITY AND HENRY’S CONSTANT
The major factors controlling the Henry’s constant of a compound are its chemical structure and the temperature of the system. In the gas phase, the space separating 483
molecules is so large (on average) that attractive or repulsive interactions among the molecules are negligible, regardless of their identity. As a result, the molecular interactions that control volatility are those in the aqueous phase: the more favorable the interactions of the solute with water (i.e., the more hydrophilic the molecule), the less volatile it is. Correspondingly, increasing hydrophilicity (or, equivalently, decreasing hydrophobicity) is associated with decreasing values of H. Hydrophobic or hydrophilic nature of solute. The hydrophilicity of a solute is controlled primarily by two factors: bonding between the solute and water molecules, and the mobility of the water molecules surrounding the solute. Whenever a volatile molecule transfers from the gas phase to an aqueous solution, some bonds between adjacent water molecules (hydrogen bonds) must break to create space for the volatile molecule to occupy, and new bonds form between the solute and water. The net effect of these changes is quantified by the enthalpy of dissolution, which is negative (favorable) for dissolution of any volatile species. In other words, although the need to break water-water bonds opposes dissolution, the new solute-water bonds that form are favorable, and their effect always dominates the former one (Israelachvili, 1985).4 The reason for this somewhat surprising outcome is that, although some water-water bonds do break when the solute dissolves, the water molecules can rearrange themselves to form new bonds and thereby minimize the effect, as shown schematically in Figure 9.3.
Figure 9.3 Schematic of hydrogen-bonded water molecules surrounding a hydrophobic solute. Reprinted from Intermolecular and Surface Forces, by J. Israelachvili, c 1985, with permission from Elsevier (http://www.elsevier.com).
In addition to altering the intermolecular bonding in solution, dissolved molecules orient the nearby water molecules, reducing the freedom of movement of those molecules. If the solute is ionic or highly polar, the adjacent water molecules are 4 Israelachvili,
J. Intermolecular and Surface Forces, Academic Press, London (1985).
484
oriented to maximize favorable electrical interactions with the solute, and if it is nonpolar, they arrange themselves in a cage-like structure around the solute, as in Figure 9.3. The decreased freedom of motion of the water molecules represents a decrease in entropy, which opposes dissolution. The balance between this (unfavorable) decrease in freedom of motion and the (favorable) formation of solute-water bonds determines the volatility of any given compound. Bonds between ions and water are particularly strong, making all ions very hydrophilic and nonvolatile, i.e., Henry’s constant is essentially zero for all ions. Bonds between neutral molecules and water are weaker than those between ions and water, with the strength of the interaction increasing with increasing polarity of the solute.5 As a result, Henry’s constants of neutral molecules decrease with increasing polarity, other factors (e.g., molecular size) being equal. As indicated in Table 9.2, values of Hpm for relatively small, nonpolar molecules such as O2 , H2 , N2 , CO, and CH4 are on the order of 1000 atm L/mol. Molecules that are more polar (e.g., H2 S) or that are nonpolar but nevertheless have localized regions of finite charge (e.g., CO2 ) have Hpm values that are one to two orders of magnitude smaller, and NH3 , which is quite polar, has a Henry’s constant of only 0.017 atm L/mol. For a given molecular structure (e.g., a series of straight-chain hydrocarbons), increasing molecular size steadily increases both the favorable solute-water interactions and the unfavorable immobilization of water molecules. Invariably, increasing molecular size magnifies the favorable interactions more than the unfavorable ones (Israelachvili, 1985), so that an increase in molecular size increases solubility and decreases the Henry’s constant. Vapor pressure of pure compound. The preceding discussion focuses on the transfer of volatile molecules directly between an aqueous solution and the gas phase. An alternative and useful interpretation of the effects of molecular structure on volatility takes the pure compound (not an aqueous solution of the compound) as the starting point, i.e., it considers volatility from a Raoult’s law perspective. The volatility of pure compounds is quantified by their vapor pressure. Because intermolecular interactions in the gas phase are always negligible, the vapor pressure is a direct indicator of the interactions in the pure (liquid or solid) phase – the stronger the interactions, the lower the volatility and the vapor pressure. Therefore, if the interactions of a volatile species with water molecules in an aqueous solution were identical to those with other molecules of the same species in a pure phase, the trend in Henry’s constants for a group of compounds would be the same as the trend in their vapor pressures. However, most neutral compounds interact less favorably with water than with other like molecules. As a result, the compounds are more likely to volatilize from an aqueous solution than from the pure phase, as reflected in the greater slope of the curve in the Henry’s law region than the Raoult’s law region in Figure 9.1. 5 These bonds arise from so-called London - van der Waals interactions, in which resonance between
the electronic vibrations in adjacent molecules generates attractive forces between the molecules.
485
Thus, from this perspective, a Henry’s constant can be viewed as reflecting two factors – the inherent volatility of the pure compound, as indicated by its vapor pressure at the temperature of interest, and the activity coefficient of the compound when it is dissolved in water (using the pure compound as the reference state), which indicates the stability of the molecule when surrounded by water molecules. Some compounds, such as methanol, are quite stable when surrounded by water, but nevertheless have relatively high Henry’s constants because of their inherent volatility (high vapor pressure). Compounds such as polychlorinated biphenyls (PCBs), on the other hand, have very low inherent volatility as pure compounds, but their volatility when dissolved in water is nevertheless significant, because their interactions with water molecules are so unfavorable. Molecular structure. In recent years, a number of attempts have been made to predict several important molecular properties, including Henry’s constants, based on molecular structure. These so-called quantitative structure-activity relationships (QSARs) have been quite successful in some cases and less so in others. The predictions of QSARs for Henry’s constants of organic compounds are often quite good. Compilations of these predictions have been provided by Nirmalakhandan et al. (1997) and Schwarzenbach et al. (2002).6 Temperature. Henry’s constants also depend on temperature. This effect is typically modeled using the van’t Hoff equation, which describes the effect of temperature on any equilibrium constant and was presented earlier in Chapters 2, 3, and 4. The van’t Hoff equaiton is ln
HT2 DH r = HT1 R
✓
1 T1
1 T2
◆
(9.18)
where HT1 and HT2 are the Henry’s constants at temperatures T1 and T2 , respectively, DH r is the enthalpy change per mole of stoichiometric reaction under standard state conditions, and R is the universal gas constant.7 As noted previously, transferring a molecule from solution to the gas phase always requires energy to break the bonds between the solute and neighboring (mostly water) molecules, so the enthalpy of volatilization is always positive (i.e., the reaction is endothermic). As a result, volatility (and therefore Henry’s constant) always increases with increasing temperature. This trend is demonstrated for several organic compounds in Figure 9.4. 6 Nirmalakhandan, N., Brennan, R.A., and Speece, R.E. (1997) “Predicting Henrys law constant and the effect of temperature on Henrys law constant.” Water Research, 31, 1471-1481. Schwarzenbach, R.P., Gschwend, P.M., and Imboden, D.M. Environmental Organic Chemistry, 2nd ed., Wiley Interscience. New York, NY (2002). 7 As explained in Chapter 4, the van’t Hoff equation is based on the assumption that the molar enthalpy of reaction under the conditions of interest equals the molar enthalpy under standard state conditions. This assumption is generally acceptable for environmental systems. Also, note that H is the universally adopted symbol for standard molar enthalpy; it is not a Henry’s constant.
486
Figure 9.4 The temperature dependence of Henry’s constant for several halogenated VOCs. 1: Bromoform; 2: Hexachloroethane; 3: Chloroform; 4: Trichloroethylene; 5: 1,1,1-Trichloroethane; 6: Tetrachloroethylene; 7: Carbon tetrachloride; 8: Dichlorodifluoromethane. Based on data in Munz and Roberts (1987).8
Solution composition. Finally, the volatility of a molecule is affected not only by its own properties, but also by the presence of other species in solution. One commonly used expression for estimating the effect of salts on gL,i is log gL,i = ki I
(9.19)
where ki is a constant that is typically between 0.02 and 0.2 (but in some cases is < 0), and I is the ionic strength of the solution. For positive values of ki , this equation indicates that the activity coefficient is > 1, meaning that solubility of the gas declines as the ionic strength of the solution increases; this effect is sometimes referred to as salting out of the gas. Equation (9.19) is identical to Equation (2.8), which was introduced as a generic equation for the activity coefficient of any neutral solute, regardless of its volatility. As noted in Chapter 7, Visual Minteq uses this equation to compute the activity coefficients of such solutes, with ki assumed to be 0.10 in all cases. Because of the overwhelming importance of dissolved oxygen as a water quality parameter, its solubility as a function of temperature and salt content has been studied more than that of any other gaseous species. This relationship is shown for a solution in equilibrium with the atmosphere (PO2 = 0.21 atm) in Figure 9.5, which was prepared from the following empirical correlation (APHA, 2012):9 8 Munz,
C. and Roberts, P.V. (1987) “Air-water phase equilibria of volatile organic solutes.” Journal AWWA, 79, 5, 62-69. 9 APHA, AWWA, and WEF. Standard Methods for the Evaluation of Water and Wastewater, 13th ed., APHA, New York, NY (2012).
487
ln cO2 |0.21 atm =
139.344 + +
1.575701 ⇥ 105 T
6.642308 ⇥ 107 T2
1.2438 ⇥ 1010 8.621949 ⇥ 1011 T3 T4 19.428 3.8673 ⇥ 103 Chl 3.1929 ⇥ 10 2 + T T2
(9.20)
where cO2 |0.21 atm is the equilibrium concentration of dissolved oxygen at a partial pressure of 0.21 atm, T is temperature in Kelvins, and Chl is the chlorinity of the solution, defined as the chloride concentration in g/kg. The overall salt content of the solution is assumed to include all the salt ions that are in seawater that has been diluted or concentrated until it has the specified chlorinity.10
Figure 9.5 The solubility of oxygen in water as a function of temperature and salt content. The numbers on the graph refer to the chloride concentration in grams per kilogram of solution. For reference, the Cl– concentration in seawater is 19.3 g/kg. The data are for solutions containing the concentration of Cl– indicated, and other salts in the same ratio to Cl– as the ratio in seawater.
Solutes other than salts can also affect the solubility of a gas. In the waste treatment literature, it is conventional to represent the ratio of the oxygen saturation concentration in a particular water to the corresponding concentration in clean water at the same temperature by the symbol beta (b ): cO2 ,sat,actual = b cO2 ,sat,clean 10 When
(9.21)
Equation (9.20) is applied at normal temperatures, the absolute values of the individual terms on the right side are on the order of hundreds, but the result of the summation is ⇠ 2. As a result, calculations must use coefficients with a large number of significant figures to yield a correct result.
488
In this case, the effect on the Henry’s constant is given by HO2 ,actual =
HO2 ,clean b
(9.22)
Except in highly saline waters, b tends to be a fairly minor correction. For instance, in activated sludge systems treating domestic wastewater, b values are usually between 0.7 and 0.98, and a value of 0.95 is frequently assumed (Metcalf and Eddy, 2003).11
If the concentration of a dissolved species exceeds that in equilibrium with the gas phase, the constituent will be driven into the gas phase. This fact is used to advantage in analysis of dissolved organic carbon (DOC) (upper image). In the analysis, organic molecules are oxidized to CO2 , which then enters the gas phase, where its concentration is easily measured. Similarly, the lower image shows equipment used to strip radioactive, gaseous radon from groundwater to prevent the radon from escaping from tapwater into the air inside a building.
11 Metcalf and Eddy, Inc.
Wastewater Engineering, 4th ed., G. Tchobanoglous, F.L. Burton, and H.D. Stensel, eds., McGraw-Hill, New York (2003).
489
Example 9.3 Henry’s constant for oxygen is 588 atm·L/mol at 10 C. (a) What is the concentration of oxygen dissolved in the epilimnion (surface layer) of a lake at that temperature if the water has equilibrated with the atmosphere? (b) Estimate the standard enthalpy of volatilization of oxygen based on a comparison of the result from part (a) with the value for HO2 (g) at 20 C given in Example 9.1. Solution (a) The partial pressure of O2 (g) in the atmosphere is 0.21 atm, so the equilibrium dissolved oxygen concentration at 10 C can be computed from Henry’s law as [O2 (aq)] =
PO2 (g) Hpc,O2
=
0.21 atm = 3.57 ⇥ 10 588 atm/(mol/L)
4 mol
L
= 11.4
mg L
(b) The enthalpy of volatilization can be estimated by applying the van’t Hoff equation [Equation (9.18)], as follows: ✓ ◆ HT DH r 1 1 ln 2 = HT1 R T1 T2 HT T1 T2 ln 2 T2 T1 HT1 ✓ ◆ (293K) (283K) 588 atm-L/mol kJ 3 kJ DH r = 6.7 ⇥ 10 ln = 12.0 mol-K 283K 293K 730 atm-L/mol mol DH r = R
9.5
HENRY’S LAW AND AQUEOUS-PHASE SPECIATION
As has been noted, only neutral molecules can exist in the gas phase at any substantial concentration, so Henry’s law describes equilibrium between those molecules and their dissolved neutral counterparts. For example, Henry’s law describes the equilibria between CO2 (aq) (or, equivalently, H2 CO3 ) and CO2 (g), H2 S(aq) and H2 S(g), and NH3 (aq) and NH3 (g). However, it does not describe the relationship between gaseous CO2 and species that are generated when dissolved CO2 undergoes a reaction, at least not directly (e.g., it does not directly relate CO2 (g) to HCO–3 or T OT CO3 ). One can, of course, derive such relationships if one wishes. For example, the equilibrium relationship between CO2 (g) and HCO–3 is given by 490
CO2 (g) + H2 O )* H2 CO3 (aq)
K(9.23) = 1/H
(9.23)
H2 CO3 (aq) )* HCO–3 + H+
K(9.24) = Ka1
(9.24)
CO2 (g) + H2 O )* HCO–3 + H+
K(9.25) = Ka1 /H
(9.25)
Reaction (9.25) represents the overall gas transfer reaction that dominates when solution pH is in the range 6.35 < pH < 10.33, and K(9.25) is a valid equilibrium constant for that reaction. However, K(9.25) would not properly be referred to as a Henry’s constant.
9.5.1
Effect of Ionization of Dissolved Gases on Gas/Liquid Equilibrium
Next, we consider the implications of Henry’s law for a gaseous species such as CO2 , which can undergo acid/base reactions in solution. The partial pressure of CO2 in the atmosphere is approximately 3.8 ⇥ 10 4 atm (10 3.42 atm), and its Henry’s constant at 25 C is 29.41 atm·L/mol. As a result, any ideal solution at 25 C in equilibrium with the normal atmosphere has a dissolved CO2 concentration of PCO2 /Hpm,CO2 , or 1.29 ⇥ 10 5 mol/L (i.e., 10 4.89 mol/L).12 Solutions that contain more than 1.29 ⇥ 10 5 mol/L H2 CO3 , such as carbonated beverages, are supersaturated with respect to atmospheric CO2 , causing them to release carbon dioxide when they are exposed to the atmosphere. Similarly, solutions containing less than 1.29 ⇥ 10 5 mol/L H2 CO3 are undersaturated and absorb CO2 if they are in contact with the atmosphere. We next develop the log C-pH diagram for an open system consisting of pure water in equilibrium with atmospheric CO2 to explore how the presence of the gas phase affects the solution composition. As was just noted, Henry’s law dictates that an ideal solution in equilibrium with the atmosphere must contain 10 4.89 M H2 CO3 . This requirement is independent of pH, so the H2 CO3 concentration in such a solution is represented on a log C-pH diagram by a horizontal line at log C = 4.89. However, we know that in any solution containing H2 CO3 , the deprotonated species HCO–3 and CO2– 3 must also be present, in concentrations that satisfy the Ka expressions. When we combine those expressions with the fixed activity of H2 CO3 , we obtain the following equations for the activities of the deprotonated species:
(HCO3 ) =
Ka1 (H2 CO3 ) 10 6.35 10 = (H+ ) (H+ )
log (HCO3 ) = 11.24 12 In
4.89
=
10 11.24 (H+ )
log (H+ ) = 11.24 + pH
(9.26)
the remainder of the chapter, all systems will be assumed to be at 25 C unless otherwise indi-
cated.
491
(CO23 ) =
Ka1 Ka2 (H2 CO3 ) 10 = (H+ )2
log (CO23 ) = 16.68
4.89
6.35 10 10.33 10 4.89
(H+ )2
2 log (H+ ) = 21.57 + 2 pH
(9.27)
Equations (9.26) and (9.27) indicate that HCO–3 and CO2– 3 can be represented on the log C-pH diagram by straight lines with slopes of +1 and +2, respectively. The diagram, including lines for H+ and OH– , is plotted in Figure 9.6. Note that the lines for H2 CO3 and HCO–3 intersect at pH = pKa1 = 6.35. This result could have been anticipated, because the equality of the concentrations of these two species at pH = pKa1 is a consequence of the equilibrium between them and must apply regardless of whether the solution is in equilibrium with a gas phase (as in this diagram) or not (as in diagrams that we considered previously). Similarly, the HCO–3 and CO2– 3 lines intersect at pH = pKa2 (= 10.33).
Figure 9.6 Log C-pH diagram for a solution in equilibrium with atmospheric CO2 (g).
Also, as in systems in which gas/liquid equilibrium is not considered, carbonic acid is the dominant carbonate-containing species at pH < 6.35, bicarbonate is dominant at 6.35 < pH < 10.33, and carbonate is dominant at pH > 10.33, and the slopes of the lines are always related by d(CO23 ) d(HCO3 ) d(H2 CO3 ) = +1 = +2 dpH dpH dpH
(9.28)
Now consider the total concentration of dissolved carbonate-containing species (T OT CO3 ) as a function of pH. This total concentration equals the sum of the concentrations of the three carbon-containing species and is approximately equal to the concentration of the dominant species at each pH. Unlike previous systems we have 492
considered, the total dissolved carbon varies tremendously as a function of pH. This result is an unavoidable consequence of the requirement that all the equilibria be satisfied simultaneously. Since the H2 CO3 concentration does not depend on pH, the only way for the constraints of these equilibria to be met (i.e., for H2 CO3 to be the dominant carbonate species at low pH and nondominant at higher pH) is for both of the other carbonate-containing species to be present at concentrations less than 1.29 ⇥ 10 5 M at pH < 6.35, and for one or both to be present at concentrations greater than 1.29 ⇥ 10 5 M at pH > 6.35. Put another way, in the systems considered in previous chapters, the total concentration of all carbonate species was fixed, so an increase in the concentration of any of the species required a decrease in the concentration of at least one of the others. By contrast, in the system equilibrated with an infinite gas phase, the concentration of an individual species (H2 CO3 ) is fixed, so an increase in the concentrations of the other species requires that the total concentration of dissolved carbon increase. In a real system that includes a solution that is physically in contact with the atmosphere, this increase is made possible by the dissolution of more and more CO2 as the solution pH is increased. As noted previously, Henry’s constants always relate the concentration of dissolved neutral species to their concentration in the gas. In the case of the carbonate system, the neutral dissolved species is an acid which can dissociate to form its conjugate base and a second, more basic species. The Ka expressions require that the concentrations of these bases increase as pH increases. If the neutral, volatile species is a base, then it can be converted to more acidic species upon dissolution, and the concentrations of these acidic species decrease with increasing pH. For instance, a solution in equilibrium with gaseous ammonia (Hpm,NH3 = 0.0171 atm·L/mol) at a partial pressure of 10 7 atm would contain NH3 (aq) and NH+4 at concentrations given by the following equations: NH3 (aq) =
PNH3 10 7 atm = = 5.9 ⇥ 10 HNH3 0.0171 atm/(mol/L)
6 mol
L
log NH3 (aq) = 5.23 (NH+ 4)=
(NH3 (aq))(H+ ) 10 5.23 (H+ ) = = 10+4.01 (H+ ) Ka 10 9.24
+ log (NH+ 4 ) = 4.01 + log (H ) = 4.01
pH
(9.29)
A log C-pH diagram for this system is shown in Figure 9.7. It has some of the same features as the diagram characterizing equilibrium with carbon dioxide, but the total concentration of dissolved ammonia increases with decreasing pH instead of with increasing pH. As noted, this result is a consequence of the fact that NH3 is a base, whereas H2 CO3 is an acid. 493
Figure 9.7 Log C-pH diagram for a solution in equilibrium with NH3 (g) at a partial pressure of 10 7 atm.
9.6
THE pH OF SOLUTIONS IN EQUILIBRIUM WITH ACIDIC OR BASIC GASES
The necessary and sufficient condition for determining the equilibrium speciation of a solution is that the equations expressing the mass balances, the equilibrium relationships, and the charge balance, proton condition, or T OT H equation be satisfied. In problems that we have solved previously, we have typically been able to write a mass balance for each weak acid/base group in solution and insert a known value of T OTi. If a solution is in equilibrium with a gas phase containing an acidic or basic gas at a fixed partial pressure, we have the same number of unknowns as if the gas were not there (one unknown corresponding to the concentration of each species in an ideal solution). However, it is not possible to write a mass balance specifying the total dissolved concentration of the volatile species, because we do not know how much gas will dissolve into or evolve out of the solution as the system equilibrates. This situation means we have one fewer mass balance than in the past. However, we also have one additional equilibrium expression that must be satisfied: the equilibrium specified by Henry’s law. As a result, we still have equal numbers of equations and unknowns, and the equations can be solved to determine the solution composition at equilibrium. For example, say we are interested in computing the solution pH and speciation for a system containing pure water equilibrated with atmospheric CO2 . As we did when solving acid/base problems in the absence of a gas phase, we can prepare a system tableau to summarize the relevant information for the problem. The tableau 494
is shown in Table 9.3, using the default components from Visual Minteq. Note that, even though the gas phase plays an important role in controlling the equilibrium solution composition, CO2 (g) does not appear in the upper portion of the tableau; it appears explicitly only in the lower portion of the tableau, as an input with an unknown dose. Nevertheless, the influence of CO2 (g) is accounted for implicitly in the upper portion of the tableau, via the fixed, known, equilibrium concentration shown for H2 CO3 in the final column. Table 9.3 Tableau for equilibration of pure water with atmospheric CO2 , using CO2– 3 as a component
Stoichiometric Coefficient H2 O H+ CO2– 3 Species H2 O H+ CO2– 3 OH– HCO–3 H2 CO3
1 0 0 1 0 0
0 1 0 1 1 2
0 0 1 0 1 1
Inputs CO2 (g)
1
2
1
log K 0.00 0.00 0.00 14.00 10.33 16.68
Conc’n
1.29 ⇥ 10
5
?
Although the analysis appears to be straightforward, we recognize a problem when we write the T OT H equation: (H+ )
(OH ) + (HCO3 ) + 2(H2 CO3 ) = 2[CO2 (g)]in =?
(9.30)
Even though we can write an expression equating T OT H in the equilibrium solution with twice the input of CO2 (g) (in moles of CO2 that dissolve per liter of solution), we do not know how much CO2 (g) enters solution, so the equation provides us no useful information. At first glance, it appears that we have to know how much CO2 dissolves before we can write a useful T OT H equation; yet we have to solve the equation to know how much CO2 dissolves! This dilemma can be resolved by reformulating the tableau using dissolved CO2 (g) (i.e., H2 CO3 ) as a component. By doing this, we ensure that the stoichiometry for forming the species H2 CO3 is simply one unit of the component H2 CO3 and nothing else. Then, because H+ does not appear in the reaction forming the species H2 CO3 , the concentration of H2 CO3 does not appear in the T OT H equation. The modified tableau using H2 CO3 as a component is shown in Table 9.4. The T OT H equation based on this tableau is: 495
(H+ )
(OH )
(HCO3 )
2(CO23 ) = 0[CO2 (g)]in = 0
(9.31)
Table 9.4 Tableau for equilibration of pure water with atmospheric CO2 , using H2 CO3 as a component
Stoichiometric Coefficient H2 O H+ H2 CO3 Species H2 O H+ H2 CO3 OH– HCO–3 CO2– 3
1 0 0 1 0 0
0 1 0 1 1 2
0 0 1 0 1 1
Inputs CO2 (g)
1
0
1
log K
Conc’n
0.00 0.00 0.00 14.00 6.35 16.68
1.29 ⇥ 10
5
?
The key new feature of the T OT H equation when H2 CO3 is used as a component is that even though we still do not know how much H2 CO3 enters solution, we have made our ignorance of that value irrelevant, because the coefficient applied to [CO2 (g)]in in the equation is 0. As a result, the equation can be converted into a PC that can be used in conjunction with the log C-pH diagram to determine the pH at equilibrium. When we rearrange Equation (9.31) to convert it to a PC, the result turns out to be identical to the charge balance: (H+ ) = (OH ) + (HCO3 ) + 2(CO23 )
(9.32)
The log C-pH diagram for the system is shown in Figure 9.8, with the LHS and RHS of Equation (9.32) highlighted. The equilibrium pH is 5.62, and the equilib4.89 M, 10 5.62 M, and rium concentrations of H2 CO3 , and HCO–3 , and CO2– 3 are 10 10.33 10 M, respectively. Most of the dissolved carbonate is present as H2 CO3 , and T OT CO3 is 1.53 ⇥ 10 5 M. The key insight from this analysis is that the composition of solutions in equilibrium with a gas phase can be determined following a procedure very similar to that for solutions in which no gas/liquid exchange is considered. The main difference is that, if we wish to generate a useful T OT H equation or PC, we need to eliminate all 496
Figure 9.8 LogC-pH diagram for a solution in equilibrium with atmospheric CO2 (g). The RHS and LHS of the charge balance are shown in bold for a solution that has equilibrated with the gas phase and that contains no other acids or bases.
terms from that equation that account for exchange of gaseous constituents with the solution, because we do not know in advance how much gas exchange occurs. We can accomplish this by choosing the volatile species as components. Once a T OT H equation is written consistent with this guideline, the problem can be solved just like those that do not involve a gas phase. Example 9.4 Compute the equilibrium composition of a solution containing 10 3 M NaAc that equilibrates with atmospheric CO2 . Ignore possible volatilization of HAc. Solution Based on the preceding discussion, we should prepare the tableau for this system using H2 CO3 as a component. Choosing Na+ , Ac– , and, of course, H+ as the other components, we can generate the tableau shown on the next page, which leads to the following T OT H and PC equations.
T OT H:
(H+ )
(OH ) + (HAc)
(HCO3 )
2(CO23 )
= 0[NaAc]in + 0[CO2 (g)]in = 0 PC:
(H+ ) + (HAc) = (OH ) + (HCO3 ) + 2(CO23 ) 497
(9.33) (9.34)
Stoichiometric Coefficient H2 O
H+
H2 CO3
Ac–
Na+
log K
H2 O
1
0
0
0
0
0.00
H
0
1
0
0
0
0.00
H2 CO3
0
0
1
0
0
0.00
Ac–
0
0
0
1
0
0.00
+
0
0
0
0
1
0.00
–
1
1
0
0
0
14.00
HAc
0
1
0
1
0
4.74
HCO–3 CO2– 3
0
1
1
0
0
6.35
0
2
1
0
0
16.68
NaAc
0
0
0
1
1
10
CO2 (g)
1
0
1
0
0
?
Conc’n
Species +
Na
OH
1.29 ⇥ 10
5
Inputs 3
A log C-pH diagram for the system is shown in Figure 9.9. The lines for the carbonate species are the same as in the preceding example, and lines for the acetate species have been added. The RHS and LHS of Equation (9.34) are highlighted, indicating that the equilibrium pH is 6.50. The solution speciation is as follows:
Figure 9.9 LogC-pH diagram for 10 3 M T OT Ac that is equilibrated with atmospheric CO2 (g). The RHS and LHS refer to Equation (9.34), which is the PC for a system in which the acetate is added as NaAc.
498
(H+ ) = 10
6.50
(H2 CO3 ) = 10 (HAc) = 10
(OH ) = 10 4.89
4.75
7.50
(HCO3 ) = 10 (Ac ) = 10
4.74
3.01
(CO23 ) = 10
8.57
The total dissolved carbonate in this solution (3.10 ⇥ 10 5 M) is more than twice as much as in the system equilibrated with the atmosphere but containing no acetate. The reason for this is that sodium acetate is a base, so when it is added, the pH of the solution increases. This pH increase, in turn, causes more CO2 (g) to dissolve. Despite this dissolution process, the concentration of H2 CO3 at equilibrium is identical to that when no NaAc is added. Rather than increase [H2 CO3 ], the dissolution provides a way for [HCO–3 ] and [CO2– 3 ] to increase, so that Ka1 and Ka2 are satisfied. The dissociation of H2 CO3 releases H+ , so the pH does not increase as much as it would if the solution did not re-equilibrate with the atmosphere. Example 9.5 Compute the equilibrium composition of the solution in the previous example if it contacts air containing not only CO2 at its normal partial pressure, but also NH3 at a partial pressure of 10 7 atm. Solution The concentration of dissolved NH3 (aq) at equilibrium in this system is given by PNH3 (g) /Hpm,NH3 , or 10 5.23 M(= 5.88 ⇥ 10 6 M) at all pH values. A horizontal line at this value can be added to the log C-pH diagram to represent NH3 (aq). Then a straight line that has a slope of 1 and that intersects the NH3 (aq) line at pH=pKa = 9.24 can be drawn to represent NH+4 . The diagram is shown in Figure 9.10.
Figure 9.10 LogC-pH diagram for a solution that contains 10 3 M T OT Ac and is in equilibrium with atmospheric CO2 (g) and NH3 at a partial pressure of 10 7 atm.
499
Once again, we can solve for the equilibrium pH by writing the T OT H equation, using the dissolved volatile constituents (both NH3 and H2 CO3 in this case) as components. The system tableau, T OT H equation, and PC are as follows. Stoichiometric Coefficient H2 O
H+
H2 CO3
Ac–
Na+
NH3
log K
1
0
0
0
0
0
0.00
0
1
0
0
0
0
0.00
Conc’n
Species H2 O H
+
H2 CO3
0
0
1
0
0
0
0.00
–
0
0
0
1
0
0
0.00
+
0
0
0
0
1
0
0.00
NH3
0
0
0
0
0
1
0.00
OH–
1
1
0
0
0
0
14.00
HCO–3 CO2– 3
0
1
1
0
0
0
6.35
0
2
1
0
0
0
16.68
HAc
0
1
0
1
0
0
4.74
NH+4
0
1
0
0
0
1
9.24
CO2 (g)
1
0
1
0
0
0
?
NaAc
0
0
0
1
1
0
10
NH3 (g)
0
0
0
0
0
1
?
Ac Na
Inputs
T OT H: PC:
(H+ )
(OH )
(HCO3 )
3
2(CO23 ) + (HAc) + (NH+ 4)=0
2 (H+ ) + (HAc) + (NH+ 4 ) = (OH ) + (HCO3 ) + 2(CO3 )
The LHS and RHS of the PC are included on the log C-pH diagram in Figure 9.10. The LHS is dominated by (NH+4 ) over the whole pH range shown, and the RHS is dominated by (HCO–3 ) over most of that range, so the equation is satisfied when those two species are approximately equal to one another. As would be expected considering that NH3 is a base, the equilibrium pH is higher than in the previous example, equaling 7.62. Concentrations of all the dissolved species at equilibrium are as follows:
500
(H+ ) = 10
7.63
(H2 CO3 ) = 10
(OH ) = 10 4.89
(HAc) = 10
5.88
(NH3 ) = 10
5.23
6.37
(HCO3 ) = 10 (Ac ) = 10 (NH+ 4 ) = 10
3.62
3.01
(CO23 ) = 10
6.32
3.62
In this case, T OT CO3 is 2.64 ⇥ 10 4 M, more than an order of magnitude greater than in the system without the bases (NaAc and NH3 ). Example 9.6 The solution described in Example 9.5 is amended with 10 3 M Na2 CO3 while remaining in contact with the gas phase. What is the composition of the equilibrated solution? Does CO2 (g) enter the solution, leave it, or not exchange with the atmosphere in response to this perturbation? Solution We can use the same log C-pH diagram to solve for the equilibrium speciation in this system as we used in Example 9.5. The lines representing the species’ concentrations in that figure are not affected by the addition of Na2 CO3 , because this addition does not alter any of the equilibria or mass balances conveyed by the diagram. However, when we include Na2 CO3 in the Inputs section of the tableau, it contributes ( 2)(10 3 ) mol/L to T OT Hin (because the stoichiometric coefficient for H+ is 2 in the reaction forming NaCO3 from the components). The T OT H equation and PC are therefore T OT H: PC:
(H+ )
(OH )
(HCO3 )
2(CO23 ) + (HAc) + (NH+ 4 ) = 2 ⇥ 10
(H+ ) + (HAc) + (NH+ 4 ) + 2 ⇥ 10
3
3
= (OH ) + (HCO3 ) + 2(CO23 )
The equilibrium solution composition in this case is (H+ ) = 10
8.51
(H2 CO3 ) = 10
(OH ) = 10 4.89
(HAc) = 10
6.78
(NH3 ) = 10
5.23
5.49
(HCO3 ) = 10 (Ac ) = 10 (NH+ 4 ) = 10
2.73
3.02
(CO23 ) = 10
4.54
4.50
The equilibrium value of T OT CO3 is now 2.01 ⇥ 10 3 M, representing an increase of 1.75 ⇥ 10 3 M compared to the previous example. Of this increase, 1 ⇥ 10 3 M is attributable to the Na2 CO3 that was added, and the remainder entered by dissolution of yet more atmospheric CO2 (g). The increase in pH causes NH3 to exit the solution, as evidenced by the fact that T OT NH3 is less than before the Na2 CO3 was added. Thus, in response to the addition of a base (Na2 CO3 ), an acidic gas (CO2 ) 501
was absorbed into solution and a basic gas (NH3 ) exited. These gas exchanges partially counteract the effect of the sodium carbonate on solution pH; in the absence of these gaseous constituents, the pH change due to addition of Na2 CO3 would have been much more substantial. The pH’s of solutions of pure water equilibrated with various partial pressures of CO2 (g), SO2 (g), and NH3 (g) are shown in Figure 9.11. The effect of these gases on solution depends on both their acid/base behavior and their solubility. SO2 is both much more acidic and more soluble than CO2 , so it has a much greater effect on pH, even when its gas-phase pressure is low. Because ammonia is a base, it drives the pH of water up, whereas CO2 and SO2 cause the pH to decline. It is therefore appropriate to compare the base strength of ammonia (as reflected in its pKb value, 4.76) with the acid strengths of CO2 (pKa1 = 6.35) and SO2 (pKa1 = 1.85). The basicity of ammonia is intermediate between the acidities of CO2 and SO2 , as is its solubility (based on comparison of the Henry’s constants). Correspondingly, at a given partial pressure, NH3 affects the pH more than CO2 and less than SO2 .
Figure 9.11 The effect of dissolution and acid/base reactions of various gases on the pH of pure water.
9.7
GAS-LIQUID EQUILIBRIUM IN VISUAL MINTEQ
Software for computing the equilibrium speciation of aqueous solutions can incorporate gas/liquid equilibrium in a variety of ways. Currently, most programs (including Visual Minteq) can compute the speciation in solutions that are equilibrated with a gas that has a fixed composition, but they do not allow for the possibility that equilibration of the two phases will substantially change the gas-phase composition; 502
modifications to allow that possibility are relatively straightforward and will undoubtedly be incorporated in the future. Like dissolved species, gaseous species are specified in Visual Minteq by inputting the reaction stoichiometry for forming the species from components, and the corresponding equilibrium constant. For instance, the Visual Minteq components used to form NH3 (g) are H+ and NH+4 , so the reaction is * 1 H+ + 1 NH+ 4 ) NH3 (g)
(9.35)
As is true for dissolved species, information about reaction stoichiometry and thermodynamics is already in the Visual Minteq database for most gaseous species of interest. Whenever a volatile species is equilibrated between a gas and solution, the activities of the species in the two phases are linked by the Henry’s constant written as an equilibrium constant. Furthermore, because the species in the two phases have the same chemical composition (except, perhaps, for the inclusion of a water molecule when the species is dissolved), they are formed by identical combinations of components. As a result, the reaction forming the gaseous species can be generated by summing the reaction forming the dissolved species and the volatilization reaction. For example, Reaction (9.35) can be generated as follows:
Sulfur dioxide emitted from the burning of fossil fuels can react with atmospheric oxygen and water to form sulfuric acid, which can later fall to the earth as acid precipitation. Such precipitation can dramatically increase the corrosion of statues, buildings, and other structures, in addition to damaging forests and disrupting poorly bu↵ered aquatic systems. (Photo: Monika Andersson/Stock Boston.)
503
1 H+ + 1 NH+4 )* NH3 (aq)
KNH3 (aq)
(9.36)
0 HNH
(9.37)
NH3 (aq) )* NH3 (g) 1 H+ + 1 NH+4 )* NH3 (g)
3
0 KNH3 (aq) HNH
3
(9.35)
In Visual Minteq, concentrations of gaseous species are quantified via the species’ partial pressures, in units of atmospheres, and the standard-state concentration for all gaseous species is chosen to be 1.0 atmosphere. Solute concentrations are quantified in molal units (i.e., moles per kilogram of water),13 so the appropriate version of the 0 . The equilibrium constant equilibrium constant for the volatilization reaction is Hpm used in Visual Minteq for Reaction (9.35) is therefore 0 K(9.35) = KNH3 (aq) Hpm,NH = 10 3
9.24
10
1.77
= 10
11.01
(9.38)
Generalizing this result, the equilibrium constant for formation of any gaseous species A(g) in Visual Minteq is 0 KA(g) = KA(aq) Hpm,A
log KA(g) = log
0 KA(aq) + log Hpm,A
(9.39a) (9.39b)
where KA(aq) is the equilibrium constant for forming the dissolved, volatile species. If the equilibrium constants (i.e., KA(g) values) for the gaseous species of interest are included in the Visual Minteq database, the only input required from the user is the partial pressure(s) of those species. The screen for inputting this information is accessed via the “Gases” link at the top of the main menu and is shown in Figure 9.12.
Figure 9.12 Visual Minteq input screen for gases. 13 Recall
that, for solutions of environmental interest, molal units are essentially identical to molar units (moles per liter of solution). Also, as we have seen, solute concentrations can be input in a variety of units. Visual Minteq always converts those values to molal units prior to the numerical processing step.
504
Because equilibrium with CO2 is so frequently of interest, a separate option for entering PCO2 is provided, showing the partial pressure of CO2 (g) in the normal atmosphere as a default value. If the actual PCO2 of interest is different from this, that value can be entered either directly or as a multiplier of the atmospheric value. The multiplier is shown in a separate box below the default PCO2 ; the correct result will be obtained as long as the product of the values in these two boxes equals the desired PCO2 . For other gases in the database, clicking the button to “Choose a gas” adds some fields to the screen, including a pulldown menu for choosing the gas to be considered. If a volatile species other than CO2 (g) is selected on this page, log K for its formation from the components and the enthalpy of the corresponding reaction appear. The user can then input the partial pressure of that species, click the “Add” button, and either use the pulldown menu to begin the process for another gaseous species or use the buttons at the bottom of the screen to navigate to another screen. If desired, the stoichiometry and log K value for forming the gaseous species can be viewed on the “Database management/Gases” page, as shown in Figure 9.13.
Figure 9.13 Visual Minteq screen showing the information in the database for formation of NH3 (g).
505
Example 9.7 Derive log KCO2 (g) for use in Visual Minteq. Solution The log K value can be found by a direct application of Equation (9.39b). In this case, the dissolved species that has (essentially) the same composition as the gaseous species is H2 CO3 . The reaction forming this species from Visual Minteq components, and the corresponding equilibrium constant, are 2 H+ + 1 CO23 + 1 H2 O )* H2 CO3
log K = 16.68
0 for H CO is 1.47, so Log Hpm 3 2 0 log KCO2 (g) = log KH2 CO3 (aq) + log Hpm, CO2 = 16.68 + 1.47 = 18.15
This value is indeed the one in the Visual Minteq database, as can be confirmed on the “Database management/Gases” page. If we want the program to consider a gaseous species that is not in the database, Equation (9.39b) can be used to compute log K for its formation. That information can be input on the “Database management/Gases” page, and the species’ stoichiometry can be specified via the “Add new species” button. Note that the components needed to form the new species must be input on the home page before the stoichiometry for the species can be input, and the species name must end with “(g)”. Also, because temperature typically has a much larger effect on Henry’s constants than on the equilibrium constants for reactions that take place entirely in solution, it is important to input a value for the enthalpy of the volatilization reaction (shown on the screen as dHr) if we wish to analyze systems at temperatures other than 25 C. Once the partial pressure of the gaseous species has been input, the program can be run as usual. The numerical processing that is carried out is essentially identical to that described in previous chapters for scenarios in which the activity of a dissolved species is fixed. The only difference is that, in the current case, the activity of the gaseous species is not included when the total concentrations of the components are computed. Once the program has satisfied all the relevant equilibrium constants and mass balance equations, the problem is solved. The amount of the gas that entered or exited solution during the equilibration process can then be computed based on the mass balance (i.e., by comparing T OT Xfinal with T OT Xin ) for any of the components that are used to form the volatile species. Example 9.8 In drinking water treatment operations, “hard” water containing an unacceptably large concentration of calcium is often “softened” by adding enough base and carbonate to precipitate much of the calcium as the mineral calcite, CaCO3 (s). The 506
operation is carried out at high pH, after which the solid is separated from solution. Often, the solution pH is then lowered by dissolving CO2 (g) into the solution, in a process called recarbonation. Consider a softened solution at pH 10.5 that contains 7.40 ⇥ 10 5 M each of T OT CO3 and T OT Ca. Use Visual Minteq to determine the final pH if this solution is equilibrated with air in a recarbonation step. Let the program account for nonideal solute behavior. Solution The inputs to Visual Minteq required to answer this question include the total component concentrations (T OT Ca, T OT CO3 , T OT H) in the initial solution and the partial pressure of CO2 with which the solution equilibrates. Of these, T OT Ca and T OT CO3 in the initial solution and PCO2 (g) are known. Furthermore, T OT H in the initial solution can be determined by running the program using T OT Ca and T OT CO3 and the initial pH of 10.5 as inputs. Doing so, T OT H is found to be 2.99 ⇥ 10 4 M. We can then use that as an input value, switch the calculation options so that pH is calculated rather than fixed, and run the program with a fixed CO2 partial pressure at its normal atmospheric value. The output page, shown in Figure 9.14, indicates that the final pH is 7.88.
Figure 9.14 Visual Minteq output page for Example 9.8.
The amount of CO2 that dissolves during the recarbonation step can then be computed from the mass balance on T OT CO3 . T OT CO3 in the solution after recarbonation is 4.57 ⇥ 10 4 M (found in Visual Minteq on the “Equilibrated mass distribution” page). Since the initial solution contained only 7.4 ⇥ 10 5 M T OT CO3 , the amount of CO2 that dissolves is 507
CO2 entering during recarbonation = T OT CO3 ,final = 4.57 ⇥ 10
9.8
4
T OT CO3 ,init 7.4 ⇥ 10
5
= 3.83 ⇥ 10
4
M
BUFFERING IN SOLUTIONS EQUILIBRATED WITH WEAK ACID AND WEAK BASE GASES
The preceding examples indicate that addition of acids to a solution that is in contact with a gas phase causes acidic gases to exit solution and basic gases to enter. The response to addition of a base is exactly the opposite. As a result, acidic or basic gases can serve as pH buffers. Furthermore, gases can provide buffering over a much wider pH range than nonvolatile species do, because they can enter or exit the solution (and thereby provide buffering capacity) in significant amounts under any condition where the equilibrium concentration of total dissolved gas is a strong function of pH. By contrast, nonvolatile species can be effective buffers only at pH values near their pKa . For instance, in a system equilibrated with atmospheric CO2 , T OT CO3 is sizable and depends strongly on pH at all pH values greater than about 7.5. Therefore, this gas serves as a good pH buffer at any pH> 7.5. Similarly, NH3 (g) can serve as a good buffer at any pH< 9.0 if PNH3 is large enough to cause T OT NH3 to be significant. Keep in mind, though, that transfer of gas into or out of solution tends to be a much slower process than protonation/deprotonation of dissolved species, so the efficacy of buffering a system by a gas might not be as great as is implied by equilibrium considerations. The buffer intensity of solutions equilibrated with a gas phase can be computed using the same concepts as described in Chapter 8, but some terms in the calculation must be modified, as follows. The buffer intensity can be thought of as a way to quantify a stimulus-response sequence, where the stimulus is the addition of a differential amount of strong acid to the sample and the response is the change in pH. In Equation (8.26), we quantified the amount of acid added as dT OT Hstrong , with the implicit understanding that the term referred to acid that was intentionally added as a stimulus. In the scenarios considered in that chapter, the intentional addition of acid was the only change that affected T OT H, so dT OT Hstrong could be interpreted simply as dT OT H. However, if the solution is in contact with a gas phase, dissolution or evolution of the gas could also alter T OT H. Therefore, if Equation (8.26) is used in an open system, dT OT Hstrong must be interpreted as the amount of strong acid intentionally added, and not as the overall change in T OT H between the initial and final states.
508
The possibility the total dissolved concentration of A will change in response to acid addition also requires modifications to the expressions for buffer intensity that include that term. Specifically, although the generic, defining expressions for bHA [Equation (8.33a and 8.33b)] remain the same, the more detailed equations for calculating that value must include an additional term, as follows: bHA ⌘ =
d (H+ needed to titrate HA/A ) dpH
(9.40a)
d (a0 T OT A) dpH
(9.40b)
= T OT A = T OT A
da0 dpH
a0
dT OT A dpH
(9.40c)
da1 dT OT A + a1 dpH dpH
(9.40d)
Using these equations as the starting point, the same algebra as is shown in Chapter 8 can be applied to derive expressions analogous to Equations (8.34) through (8.42), but that apply to open systems. Example 9.9 An ideal solution at pH 8.4 is in equilibrium with the atmosphere and contains no weak acids or bases in significant concentrations other than carbonate species. The solution then receives an input of 10 3 mol/L HCl. Compare the pH change in a closed system (i.e., one that undergoes no further CO2 exchange with the atmosphere) with that in an open system. What are the buffer intensities of the closed and open systems at pH 8.4? Solution Because the initial solution is ideal and in equilibrium with the atmosphere, the H2 CO3 activity in the solution is 10 4.89 , and the H2 CO3 concentration is 10 4.89 M. We also know the pH of the solution, so we can compute the activities and concentrations of the remaining solutes as follows: (OH ) =
Kw 10 = + (H ) 10
(HCO3 ) = (CO23 ) =
14.0 8.4
= 10
5.6
Ka1 (H2 CO3 ) 10 6.35 10 = (H+ ) 10 8.4
Ka2 (HCO3 ) 10 10.33 10 = (H+ ) 10 8.4 509
4.89
2.84
= 10
2.84
= 10
4.77
The system tableau and the values of T OT H and T OT CO3 in this solution are shown below. Stoichiometric Coefficient H2 O
H+
CO2– 3
log K
H2 O
1
0
0
0.00
H
0
1
0
0.00
CO2– 3 –
0
0
1
0.00
OH
1
1
0
14.00
HCO–3
0
1
1
10.33
H2 CO3
0
2
1
16.68
H+
0
1
0
10
8.4
OH–
1
1
0
10
5.6
CO2– 3
0
0
1
10
4.77
HCO–3
0
1
1
10
2.84
H2 CO3
0
2
1
10
4.89
Conc’n
Species +
Inputs
T OT Hinit = (H+ ) = 10
8.4
(OH ) + (HCO3 ) + 2(H2 CO3 ) 5.6
10
+ 10
2.84
+ 2(10
4.89
) = 1.47 ⇥ 10
3
T OT CO3 ,init = (CO23 ) + (HCO3 ) + (H2 CO3 ) = 10
4.77
+ 10
2.84
4.89
+ 10
= 1.48 ⇥ 10
3
The addition of 10 3 M HCl increases T OT H by 10 3 , so the T OT H equation after the acid is added but before any CO2 transfer is T OT Hinit+HCl = (H+ )
(OH ) + (HCO3 ) + 2(H2 CO3 ) + 10
= 1.47 ⇥ 10
3
+ 10
3
= 2.47 ⇥ 10
3
3
In the closed system, we can use this value of T OT H along with the initial T OT CO3 (which is also the final T OT CO3 in that system) as inputs to Visual Minteq and allow the program to solve for the equilibrium pH. The result, shown in the output table in Figure 9.15a, is that pH= 6.05.
510
Figure 9.15 Visual Minteq output for addition of 10 4 M HCl to a pH 8.4 solution that is initially in equilibrium with atmospheric CO2 (g). (a) Closed system; (b) Open system.
For the open system, we use the same input values of T OT H and T OT CO3 , but we indicate that the solution is equilibrated with atmospheric CO2 . The program output, shown in Figure 9.15b, indicates that the equilibrium pH is 7.93, representing a much smaller decline than in the closed system. In both systems, the addition of HCl causes a substantial fraction of the HCO–3 in the initial solution to protonate to H2 CO3 . However, in the open system, the increase in the H2 CO3 concentration causes the solution to become supersaturated with respect to CO2 in the overlying gas phase, so CO2 exits the solution, in effect carrying some of the added H+ with it. This process continues until the Henry’s constant, the Ka
511
expressions, and the mass balances are all satisfied simultaneously. As a result of this process, T OT CO3 in the equilibrated, open system is 5.40 ⇥ 10 4 M, or about one-third of that originally present. The buffer intensities of the solutions in the closed and open systems can be determined by using Visual Minteq to determine the pH change that accompanies a small change in the input value of T OT H (which accounts for the intentional addition of strong acid, but not for any change in T OT H induced by gas exchange), and then computing the buffer intensity as DT OT H/DpH. The results of the simulations using Visual Minteq with a DT OT H of 10 5 mol/L are summarized in the following table. (Because we are focusing on small differences in the computed values, an extra significant figure has been added to the pH values.) Consistent with the prior discussion, the buffer intensity is far greater in the open system. Inputs
T OT CO3
Initial
⇤
Closed
Initial+10
5M
H+
Open
Initial+10
5M
H+
T OT H
1.47 ⇥ 10
3
0.94 ⇥ 10
4
1.47 ⇥ 10
3
pH
1.48 ⇥ 10
3
8.394
3
8.051
1.52 ⇥ 10
3
8.391
1.49 ⇥ 10
b⇤ 2.92 ⇥ 10 3.33 ⇥ 10
5 3
Computed as DT OT H/DpH, and expressed in equiv per liter per pH unit
9.9
CO2 DISSOLUTION, ALKALINITY, AND ACIDITY
Recall that the alkalinity of a solution is defined as the concentration of H+ required to titrate the solution to some defined endpoint near pH 4.5, when the H+ is added as a strong acid, and that carbonate species are often the dominant contributors to the alkalinity of a solution. However, as shown in Chapter 8, the alkalinity changes negligibly when H2 CO3 is added to or removed from a solution. The minimal effect of H2 CO3 addition or removal on alkalinity reflects the fact that only species that consume or release H+ when the solution is titrated to pH 4.5 contribute to the alkalinity, and H2 CO3 does neither. Because H2 CO3 can be thought of as just a combination of a molecule of CO2 with water, exchange of CO2 (g) between a solution and a gas phase also has a negligible effect on the solution alkalinity. This idea can perhaps best be understood by writing the dominant, net reaction that occurs when CO2 (g) enters solution and the solution is titrated to pH 4.5: CO2 (g) + H2 O
pH 4.5
! H2 CO3
(9.41)
Reaction (9.41) characterizes the fate of the overwhelming majority of CO2 (g) molecules in a solution that has been titrated to the alkalinity endpoint. Because the 512
net reaction neither consumes nor releases H+ , it has no effect on the alkalinity.14 The effect on alkalinity when other gases enter or leave solution depends on whether H+ appears in the net reaction analogous to Reaction (9.41) for those gases. If it does, then the addition or removal of the gas changes the number of H+ ions needed to carry out the alkalinity titration, so the gas exchange affects the alkalinity; if H+ does not appear in the reaction, then the gas exchange does not affect the alkalinity. Example 9.10 Use the constancy of ALK in response to CO2 exchange to find the pH of the open system in Example 9.9. Solution The speciation of the initial solution in Example 9.9 is given in the solution to that example. The alkalinity of the solution can be computed as ALKinit = (HCO3 ) + 2(CO23 ) + (OH ) = 10
2.84
+ 2(10
4.77
) + 10
5.6
(H+ ) 10
8.4
(9.42a) = 1.48 ⇥ 10 3 equiv/L
(9.42b)
Because alkalinity is conservative, the addition of 10 3 M HCl to this solution decreases ALK by 10 3 equiv/L, to 4.8 ⇥ 10 4 equiv/L. Equation (9.42a) applies generically to any solution where carbonate species comprise the only weak acid/base group, so it still applies after the HCl addition. However, at that point, ALK is known and the terms on the right side of the equation are not. That is, we can write ALKinit+HCl = 4.8 ⇥ 10
4
= (HCO3 ) + 2(CO23 ) + (OH )
(H+ )
(9.43)
This equation, like the one we developed in Example 9.9 for closed systems, can be used in place of the T OT H equation to solve for the equilibrium solution composition. To do so, we can prepare a log C-pH diagram for an open system in equilibrium with the atmosphere, and add a curve showing the solution alkalinity, computed as the sum on the right side of Equation (9.43). Such a diagram is shown in Figure 9.16. The equilibrium pH is at the point where ALK= 4.8 ⇥ 10 4 equiv/L, which occurs at pH 7.92, consistent with the results we obtained using other solution approaches in Example 9.9. 14 As
was explained in Chapter 8, if the carbonate group were the only acid/base group in a sample, the alkalinity could be defined by titration to the pH corresponding to f = 0, rather than to the arbitrary endpoint of pH 4.5. In such a case, dissolution or volatilization of CO2 (g) would have absolutely no effect on the alkalinity. As a practical matter, other acids and bases are invariably present, and the alkalinity endpoint is usually chosen to be a fixed pH, rather than f = 0. For these reasons, the experimentally measured alkalinity does change slightly in response to addition or removal of CO2 (g), but the change is small.
513
Figure 9.16 A log C-pH diagram for an open system in equilibrium with atmospheric CO2 (g), with a curve showing the solution alkalinity as a function of pH and a line showing an alkalinaity of 4.8 ⇥ 10 4 equiv/L.
Example 9.11 In Example 8.10, we considered the composition of a pond subject to pH swings caused by photosynthesis during the day and respiration at night. In that example, we treated the pond as a closed system and found that the pH at the end of the day would be 10.63. What would the pH of the pond be at the end of the day if the water equilibrated with the atmosphere (i.e., if the pond behaved as an open system)? Solution We determined in the solution to Example 8.10 that the alkalinity of the pond was 7.35 ⇥ 10 4 equiv/L, and we have seen that exchange of CO2 between a solution and the atmosphere does not affect the alkalinity. We prepared a log C-pH diagram showing the alkalinity of a solution in equilibrium with atmospheric CO2 (g) in Example 9.10 (Figure 9.16), and the curves for the concentrations of individual species and the alkalinity in that diagram apply equally well to the current system of interest; the only difference is that the known alkalinity in the current case is a bit higher than in that example. The diagram is shown again in Figure 9.17, with the line for the known alkalinity adjusted to the value in the pond. We conclude that the equilibrium pH in the pond at the end of the day is 8.10, dramatically lower than in the closed system. Thus, we see once again the important role that exchange of CO2 (g) can play in buffering solution pH. Example 8.10 and the current example characterize the two extremes of no CO2 (g) exchange and full equilibration of the solution with the atmosphere. As has been noted, the actual amount of exchange in a system like this one would depend on fac514
Figure 9.17 The log C-pH diagram from Figure 9.16, for a system with ALK= 7.35 ⇥ 10 4 equiv/L.
tors such as the temperature, wind speed, water depth, and mixing intensity in the solution. Typically, some CO2 (g) exchange occurs, but the rate of exchange is not rapid enough to maintain gas/liquid equilibrium.
9.10
SUMMARY
Several important chemical species that can be present in solution can also exist in gas phases. The exchange of such species between the two phases can be represented as a chemical reaction, and their partitioning at equilibrium can be characterized by an equilibrium constant known as Henry’s constant. All chemical species tend to become more volatile (less soluble, corresponding to higher Henry’s constant) with increasing temperature. Henry’s constant also tends to increase with decreasing molecular weight and decreasing ionic or polar character of the molecule. Although Henry’s constant, like all equilibrium constants, is a ratio of activities, the constant is commonly reported as a partition coefficient, i.e., as a ratio of concentrations. Acidic and basic volatile species affect the acid/base balance when they enter or leave a solution. If a solution is equilibrated with an acidic or basic gas, the soluble concentration of the neutral form of the acid/base group (e.g., NH3 or H2 CO3 ) is independent of pH. The dissolved concentrations of species that have more protons than the neutral species (e.g., NH+4 ) decrease with increasing pH, while those of species formed by deprotonation of the neutral molecule (e.g., HCO–3 and CO2– 3 ) increase with increasing pH. As a result, the total dissolved concentration of a volatile acidic or basic gas can be very sensitive to solution pH. 515
Acidic gases tend to enter solution and basic gases to leave solution when some other reaction raises the pH, thereby mitigating the effect of the other reaction. Because of this, gases can serve as excellent pH buffers. Also, the nature of gas/liquid equilibrium allows acidic and basic gases to serve as buffers at any pH where the neutral species is not the dominant dissolved species of the acid/base group. The farther the pH is from the region where the neutral species is dominant, the better the gas can buffer the solution. However, the rate of equilibration of gases with solutions depends on the intensity of mixing in each phase and the amount of interfacial area across which the species can transfer, and such reactions often proceed much more slowly than reactions among dissolved species. Therefore, the predicted effect of a gas on solution composition at equilibrium might not be observed in the time frame of interest. Although acidic and basic gases can alter solution pH, they might or might not alter the alkalinity of the solution. In particular, CO2 has a negligible effect on the alkalinity when it enters or exits solution. Systems equilibrated with gases can be modeled using chemical equilibrium software. In Visual Minteq, the input value of log K for the gaseous species equals the sum of the log K value for the corresponding neutral, dissolved species and 0 , where log H 0 is the equilibrium constant for the volatilization reaction, log Hpm pm using 1.0 atm and 1.0 mol/L as the standard state concentrations in the gas and liquid phases, respectively. Once the equilibrium composition of the solution has been determined, the amount of the gaseous species that had to enter or leave solution to reach equilibrium can be computed.
9.11
APPENDIX 9A. THE LIQUID-PHASE ACTIVITY COEFFICIENT IN HENRY’S LAW USING THE RAOULT’S LAW STANDARD STATE CONVENTION
Figure 9.1 shows the partial pressure of a solute i as a function of the liquid-phase concentration of i. If the liquid phase is pure i, the equilibrium partial pressure of i is, by definition, the vapor pressure of i, Pvap,i . Raoult’s law is based on the assumption that the equilibrium partial pressure of i is directly proportional to the liquid-phase mole fraction of i, xi , but the diagram shows that, especially at low values of xi , the actual partial pressure of i is larger than xi Pvap,i . As indicated in the chapter, this result reflects the fact that when the volatile species is present as a dilute solute in water, it is less ‘comfortable’ than when it is surrounded primarily or exclusively by other molecules of i in a pure liquid. As a result, in an aqueous solution, i is less stable than in pure i, and it is more likely to escape into the gas phase. This tendency is quantified by the liquid-phase activity coefficient of i. Specifically, if the Raoult’s law reference state is adopted, g L,i equals 1.0 for pure i and is >1 for i dissolved in 516
an aqueous solution; on the other hand, if the Henry’s law reference state is adopted, g L,i equals 1.0 in a dilute aqueous solution and is <1 for pure i. We can express gas/liquid equilibrium of a volatile species i in a given system using either Henry’s law or Raoult’s law. Assuming that the activity coefficient of gaseous i is 1.0, the activity of gaseous i can be defined in either the Henry’s law or the Raoult’s law expression as the ratio of the partial pressure of i to the standardstate pressure (1.0 atm). However, the activity of aqueous i is defined differently (i.e., using different standard-state conditions) in these two laws. Therefore, we can write aRG,i =
Pvap,i Pi R = gL,i xi Pi Pi
(9.44)
aH G,i =
0 Pi H ci = Hpc,i gL,i Pi ci
(9.45)
where the superscripts R and H indicate the reference state for aqueous i. Because the activity of gas-phase i is defined identically in the two laws, the two preceding expressions can be equated to one another. Doing so and inserting values for Pi and ci yields Pvap,i 0 R H ci = Hpc,i g L,i (9.46) g L,i xi Pi ci R g L,i H g L,i
0
= Hpc,i 0
= Hpc,i
ci 1 Pi ci xi Pvap,i
(9.47)
ci 1 1.0 atm Pvap,i xi 1.0 mol/L
(9.48)
If we consider a dilute aqueous solution in the Henry’s law region, we can assume that the molar density of the solution is approximately the same as in pure water (r 0 = 55.6 mol/L) and can write ci as r 0 xi . In addition, in this region, gL,H j = 1.0, so R gL,i 0 r 0 xi 1 1.0 atm = Hpc,i 1.0 Pvap,i xi 1.0 mol/L R gL,i
0
Hpc,i r 0 1.0 atm = Pvap,i 1.0 mol/L
0
(9.49a) (9.49b)
The product Hpc,i r 0 is the slope of the Henrys law portion of the curve in Figure 9.1, and Pvap,i is the slope of the Raoult’s law portion. Therefore, Equation (9.49b) indicates that, when the Raoult’s law standard-state convention is used for aqueous i, the activity coefficient for dissolved i in the Henry’s law region is the ratio of the slopes of the Henry’s law and Raoult’s law curves. Now consider a scenario in which pure i is mixed with pure water and the system reaches equilibrium. In this process, some i will dissolve into the water, and some 517
water will dissolved into the (initally) pure i. Furthermore, assume that the volume ratio of the two liquids is such that neither liquid completely dissolves into the other phase; i.e., when the system reaches equilibrium, both liquid phases are still present. If i is volatile and a gas phase is present, then when the gas reaches equilibrium with the liquids, the partial pressure of i must be in equilibrium with both phases. For many such mixtures, the two liquids are sufficiently insoluble in one another that the aqueous phase is characterized by Henry’s law, and the phase that is predominantly i has a mole fraction of i that is very close to 1.0. For such a scenario, we can use Equation (9.45) to express the equilibrium with the aqueous phase as 0
Pi = Hpc,i
ci,sat0 n Pi ci
(9.50)
where ci,sat0 n is the ‘saturation concentration of i’ (i.e., the concentration of i in equi[H] librium with [nearly] pure i), and g L,i has been assigned a value of 1.0 because we stipulated that the solution is in the Henry’s law region. Assuming that the molar density of the solution is the same as that in the standard state, we can replace ci,sat0 n 0 with r xi,sat0 n to obtain 0 r xi,sat0 n 0 Pi = Hpc,i Pi (9.51) ci However, as noted above, the gas phase in this scenario must be in equilibrium with (nearly) pure i, so the partial pressure of i must be Pvap . Making this substitution, we can rearrange Equation (9.51) as follows:
0
Pvap,i = Hpc,i 1 xi,sat0 n
0
0
r xi,sat0 n Pi ci
(9.52a)
0
0
0
Hpc,i r Pi Hpc,i r 1.0 atm = = Pvap,i ci Pvap,i 1.0 mol/L
(9.52b)
Finally, comparison of Equations (9.49b) and (9.52b) shows that R g L,i =
1 xi,sat0 n
(9.53)
Equation (9.53) indicates that the activity coefficient for a slightly soluble volatile species is the inverse of the solubility of that species in water, when the solubility is expressed as a mole fraction. This result applies if, when a pure phase of the species of interest equilibrates with water, the aqueous solution is characterized by Henry’s law and the nonaqueous solution is almost pure i. These conditions apply to many solvents and other nonaqueous phase liquids.
518
9.12
PROBLEMS
Please re-read the note preceding the Chapter 8 Problems on p.457. The same guidelines apply to the following problems. 1. Calculate the pH of water in a cloud that has equilibrated with SO2 (g) at a concentration of 1.0 ppm by volume. 2. A solution contains 2 mg/L CN– and is at pH 8.5. (a) What is the concentration of HCN(aq) in solution? (b) If the partial pressure of HCN(g) above the solution is 10 6 atm, would HCN transfer from the gas phase into solution or from the solution into the gas, or is the system at equilibrium? 3. A solution is at pH 8.35 and has reached equilibrium with the atmosphere. (a) How much H2 SO4 (mol/L) must be added to lower the pH to 6.0, if no gas/liquid exchange occurs? (b) Repeat part (a) if the solution remains in equilibrium with the atmosphere as the acid is added. 4. Soil water with 10 PCO2 = 10 1 atm.
3N
alkalinity is in equilibrium with a soil gas in which
(a) Calculate the pH and total dissolved inorganic carbon concentration (M). (b) If the soil water is pumped from the ground and exposed to air, what will the equilibrium pH be? 5. A soft drink has a pH around 2.7 due partially to its high H2 CO3 content. (a) If all the acidity in the beverage was generated by bubbling it with pressurized CO2 , what PCO2 would have been required in the bubbles? (b) What would the pH of the solution be if it was in equilibrium with a PCO2 of 4 atm? (c) The pH of the solution is actually kept low by a combination of dissolved CO2 plus phosphoric acid. Assuming PCO2 is 4 atm in the bottle, what concentration of H3 PO4 must be added to solution to reach pH 2.7? 6. Calculate the pH, alkalinity, and composition of a solution in equilibrium with a gas phase having PCO2 = 10 2 atm and PH2 S = 10 4 atm. 7. A groundwater at 25 C is initially at pH 8.0 and contains 2 ⇥ 10 alkalinity. 519
3
equiv/L
(a) If the alkalinity is all attributable to the carbonate system, determine whether the water is undersaturated or supersaturated with respect to atmospheric CO2 . (b) What will the pH and alkalinity of the solution be if it equilibrates with CO2 in the atmosphere? (c) How much CO2 dissolves into or volatilizes out of solution as the system equilibrates with the atmosphere? 8. The surface water of a lake is at pH = 7.9 and contains 10 inorganic carbon (i.e., T OT CO3 ).
2.9 M
dissolved
(a) What is the alkalinity of this water? (b) Is the water undersaturated or oversaturated with atmospheric CO2 ? (c) If the water equilibrates with the atmosphere, what will the final alkalinity, pH, and dissolved inorganic carbon concentration be? 9. Hydrogen sulfide gas at a partial pressure of 10 solution.
4.82 atm
is in contact with a
(a) Prepare a log C-pH diagram showing the concentrations of dissolved H2 S, HS– , and S2– in equilibrium with the gas phase over the range 4< pH <12. (b) If the solution pH is 8.5 and the S-containing species are the only contributors to alkalinity, what is the alkalinity of the solution? Assume that the solution is closed during the alkalinity titration. (c) Assuming that the system in part (b) remains in continuous equilibrium with the gas phase, find the pH after 5 ⇥ 10 3 M HCl has been added to the solution. What is the change in total dissolved sulfide concentration when the HCl is added? 10. One hundred mL of air is contacted with 1 L of water at 10 C in a closed vessel. The water is initially devoid of oxygen. What are the concentration of dissolved oxygen and the partial pressure of oxygen in the air once the system comes to equilibrium? 11. What is the effect of dissolution of 10 4 M NH3 (g) and 2 ⇥ 10 5 M H2 S(g) on the alkalinity of a solution that is initially at pH 8.0, if alkalinity is defined by a titration to pH 4.5? 12. The raw water supply for a community contains 18 mg/L total particulate matter and is to be treated by addition of 60 mg alum (Al2 (SO4 )3 · 18 H2 O) per liter of water. 520
(a) For a total flow of 8000 m3 /d, compute the daily alum requirement and the concentration of solids in the water following alum addition, assuming that all the alum precipitates as Al(OH)3 (s). (b) The water is initially at pH 7.5 and has [ALK] = 40 mg/L as CaCO3 . It is desired to maintain solution pH at 6.5 or higher. Will the pH be in the acceptable range after alum addition and Al(OH)3 (s) precipitation? (c) What will the pH be if the treated solution is bubbled with air, so that it reaches equilibrium with atmospheric CO2 (g)? 13. A wastewater at pH 7.5 contains 45 mg/L SO4 S, which can be converted to sulfide species (H2 S or HS– ) under anaerobic conditions. What fraction of the sulfate can be reduced to sulfide without causing the partial pressure of H2 S in equilibrium with the solution to exceed 10 6 atm? Assume the solution is well-buffered at pH 7.5. 14.
(a) Compare the buffer intensities of closed solutions at pH’s of 3, 8, and 11, if all the solutions contain 10 3 M T OT CO3 . (b) Repeat part (a) for open systems in equilibrium with the atmosphere (and therefore containing different concentrations of T OT CO3 ).
15. At a Superfund site, concentrated NaOH has leaked into the groundwater, increasing the pH to 12.1 and dissolving large amounts of Si and Al from the soil (12,000 mg/L T OT Si and 1400 mg/L T OT Al). The water also contains 0.003 M T OT CO3 . One remediation strategy that is being considered is injection of CO2 into the groundwater to lower the pH. A log C-pH diagram showing the speciation of Al and Si in the groundwater is provided below. (a) Prepare a system tableau that includes all the species that will be present in the equilibrium solution after CO2 injection. Use the dominant Si- and Al-containing species in the original solution (H3 SiO–3 and Al(OH)–4 ) as components, and choose a component from the carbonate group, keeping in mind that an unknown amount of CO2 will dissolve when CO2 (g) is injected. (Hint: If you have trouble filling in any of the cells in the tableau, writing the reaction for forming the species from the components should help.) (b) Determine T OT H in the initial solution (before CO2 (g) injection). (c) Write the T OT H equation for the solution after it reaches equilibrium with air, assuming that CO2 is the only gas of interest that dissolves. (d) Add lines for H2 CO3 , HCO–3 , and CO2– 3 to the log C-pH diagram for the solution after it has equilibrated with air.
521
(e) Finally, use the log C-pH diagram in conjunction with the T OT H equation to solve for the equilibrium pH and determine how much CO2 would dissolve into the solution (moles per liter of solution) if it reached equilibrium with air.
16. After water has been ‘softened’ to remove hardness ions like Ca2+ and Mg2+ , it is typically at high pH, and must be partially neutralized before being distributed. The neutralization is sometimes carried out by bubbling the solution with air. (a) Imagine such a water that is at pH 11.33 and contains 10 4 M each of T OT Ca and T OT Mg, and 2 ⇥ 10 4 M T OT CO3 . Prepare a log C-pH diagram for the solution after it equilibrates with air, assuming that the solutes behave ideally. Consider Ca2+ , Mg2+ , CaOH+ , MgOH+ , and species of the carbonate group as possible solutes. (b) Write the T OT H equation for the solution after it equilibrates with air, and determine the equilibrium pH. 17. The water in a sewer is at pH 8.20, has an ALK of 2.50 meq/L, and contains 2.25 mmol/L T OT CO3 . All of the alkalinity is contributed by the carbonate and sulfide acid/base groups. (a) What would the partial pressures of CO2 and H2 S be in a gas phase that has equilibrated with the solution? (b) What is T OT H in the solution, if H+ , HCO–3 and HS– are chosen as components? (c) As the water enters a sewage treatment plant, it is dosed with HOCl, converting all the sulfide species to sulfate. The reactions of H2 S and HS– with HOCl are as follows: 522
H2 S + 4 HOCl ! SO24 + 4 Cl + 6 H+
HS + 4 HOCl ! SO24 + 4 Cl + 5 H+
If the HOCl dose is just enough to react with all the T OT S(II), what will T OT H be before any re-equilibration with the gas phase occurs? Use the same components as in part (b), along with Cl– and SO2– 4 . (Hint: determine the net change in T OT H that occurs for each mole of H2 S or HS– that reacts, and then apply that information to determine the overall change in T OT H when the H2 S and HS– in the original solution undergo the reactions shown.) 18. The micro-organisms in a waste treatment pond are exposed to 35 mg/L of easily degradable organic compounds with an average composition of C5 H10 O5 . The standard molar Gibbs energy of formation of these compounds has been estimated as 2797 kJ/mol. The pond is at pH 7.3, its temperature is 25 C, and its alkalinity is 6 ⇥ 10 3 equiv/L. The overall reaction for oxidation of the organic matter is: C5 H10 O5 + 5 O2 ! 5 H2 CO3 (a) Due to oxygen consumption by the above reaction, the dissolved oxygen concentration in the water is only 2 mg/L. What is the maximum amount of energy that an organism could obtain by oxidizing 1 mole of the organic matter? (b) The operator of the treatment facility has suggested that the treatment efficiency could be improved by bubbling air through the water to encourage biological activity. Would the organisms find the environment in the untreated water more favorable, less favorable, or identical (from an energy standpoint) after bubbling with air, if the process causes the solution to equilibrate with both O2 (g) and CO2 (g)? (c) Determine the pH of the system after it has been bubbled with air. 19. Two waters at 25 C are in equilibrium with air, one at pH 9.5 and one at pH 7.3. These waters are then mixed in various proportions. (a) Assuming that carbonate species dominate the acid/base behavior of both solutions, and that no gas transfer occurs when they are mixed, determine the alkalinity and pH of the mixture as a function of the fraction of the high-pH solution in the mixture. Consider mixing proportions such that the high-pH solution accounts for zero to 10% of the total in 0.5% increments, and then from 10% to 100% of the total in 10% increments.
523
(b) For each mixture analyzed in part (a), determine whether the solution will be undersaturated, supersaturated, or in equilibrium with atmospheric CO2 . For the 1:1 mixture, how much CO2 will dissolve into or volatilize out of the solution as it equilibrates with the atmosphere? Express your answer in moles CO2 transferred per liter of solution. 20. A water supply at pH 7.5 has ALK = 40 mg/L as CaCO3 . Essentially all the alkalinity is attributable to the carbonate acid/base system. The water is treated by addition of alum, causing Al(OH)3 (s) to precipitate and aiding in the removal of colloidal matter from suspension. The precipitation of the alum generates acid by the following reaction: Al2 (SO4 )3 · 14 H2 O ! 2 Al(OH)3 (s) + 3 SO24 + 6 H+ + 8 H2 O The production of acid by this reaction lowers the solution pH. In reality, as the pH declines, the likelihood that the added Al will precipitate declines as well. Nevertheless, for this question, assume that all of the added alum undergoes the reaction shown, releasing six moles of H+ per mole of alum added. (a) Prepare a titration curve showing the pH as a function of alum added, for alum additions of 0 to 100 mg/L. (Note: you need not consider the addition of alum or the formation of Al(OH)3 (s) explicitly. Rather, determine the pH as a function of H+ added, and use the alum reaction only to convert each value of H+ added to a corresponding alum dose.) (b) It is determined that the optimal alum dose for coagulation is 60 mg/L. To minimize corrosion, it is desired to maintain pH 6.5 in the treated water. Will chemical (base) addition be necessary when the optimal alum dose is added? If so, how much? Assume that gas exchange during the various treatment steps is slow, so that the solution can be treated as a closed system. (c) If no base is added to the solution, will it be possible to reach the target pH by bubbling the treated water with air after adding the optimum alum dose? How much CO2 must be stripped out of solution to reach the target pH? (Hint: recall what happens to alkalinity when CO2 is added to or removed from a solution. Also, note that the solution might reach the target pH after some CO2 has been stripped out of the solution, but before the solution reaches equilibrium with the air.) 21. Sulfur dioxide, SO2 , is a major air pollutant emitted during the burning of sulfur-bearing fossil fuels. Consider a stack gas containing 15% CO2 and 2000 ppm SO2 . Treatment must lower the SO2 concentration to 40 ppm SO2 . It is proposed to absorb the SO2 into an alkaline solution containing 0.1 mol/L 524
Ca(OH)2 . The operation is counter-current, as shown in the diagram below, so the liquid effluent is in contact with the untreated influent gas. Assume the gas and liquid phases come to equilibrium rapidly, so that the water exiting the column is in equilibrium with the gas phase that enters it. (However, the gas and liquid are not equilibrated at the top, where Ca(OH)2 containing no CO2 or SO2 enters.) Answer the following questions to explore the feasibility and some of the potential problems associated with this process.
(a) What is the pH of the influent water? Do not consider Ca(OH)2 to be a strong base for this part of the problem. (b) What are the pH and composition of the water as it exits the reactor, i.e., in equilibrium with the influent gas? (Note that Visual Minteq accounts for the formation of S2 O2– 3 from two H2 SO3 molecules. If you use Visual Minteq to model the solution, instruct the program to ignore this reaction.) (c) Do a mass balance on sulfur (flow rate in ⇥ concentration in = flow rate out ⇥ concentration out) to compute the required ratio of gas to liquid flow rates (QG /QL ) in liters of gas per liter of liquid. Assume T = 25 C and Ptot = 1.0 atm. (d) Assuming T OT CO3 in the influent water is negligible, do another mass balance to determine the fraction of the CO2 that will be absorbed from the gas phase. (e) The point of the process is to remove SO2 from the gas. If most of the base added is used to neutralize CO2 rather than SO2 , the process may be considered inefficient, at least in terms of reagent utilization. Of the total hydroxide consumed in the reactor, what fraction neutralizes SO2 , and what fraction neutralizes CO2 ? 525
526
10
CHEMISTRY OF METALS IN AQUEOUS SYSTEMS 1. SOLUTION-PHASE REACTIONS
527
Contents 10.1 10.2 10.3 10.4 10.5 10.6
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . METAL OXIDATION STATES . . . . . . . . . . . . . . . . . FORMATION AND STRUCTURE OF METAL COMPLEXES COMPLEXES WITH HYDROXIDE IONS . . . . . . . . . . . LOG C-pH DIAGRAMS FOR DISSOLVED METALS . . . . . COMPLEXES WITH LIGANDS OTHER THAN WATER AND HYDROXIDE . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 MIXED LIGAND COMPLEXES AND CHELATING AGENTS 10.8 METAL SPECIATION IN SYSTEMS CONTAINING COMPLEXING LIGANDS . . . . . . . . . . . . . . . . . . . . 10.9 METAL ION BUFFERS . . . . . . . . . . . . . . . . . . . . . 10.10PREDOMINANCE AREA DIAGRAMS . . . . . . . . . . . . . 10.11SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.12PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.1
. . . . .
. . . . .
528 529 529 530 536
. . 538 . . 539 . . . . .
. . . . .
541 553 556 566 567
INTRODUCTION
As we are all aware, metals are used in and are central to virtually every aspect of modern life, from buildings and vehicles to water collection and distribution systems, machinery and tools, and computer chips. Unfortunately, though, the associated mining and refining activities and the careless disposal of metallic objects have, in many cases, greatly accelerated the biogeochemical cycling of metals and exposed organisms or entire ecosystems to unhealthy, and at times lethal, levels of these chemicals. No definition distinguishing metals from nonmetals is universally accepted. Any such definition identifies as metals those elements that are refined, used in commerce, and referred to colloquially as such (e.g., Cu, Fe, Ag), but virtually all elements in the third and lower rows of the periodic table (sodium and all elements with atomic numbers larger than that of sodium) other than the halogens and noble gases share some 528
metallic properties. In this chapter, we focus on the reactions that are commonly associated with metallic behavior in aquatic environments. One key characteristic of metals in such systems is their ability to exist as either dissolved or solid species and to combine with numerous other species. That is, in contrast to acetate, which is likely to be present in only two forms (acetate ion or acetic acid) in aquatic systems, copper might be present as a constituent in dozens of different dissolved and solid species. The speciation of metals has a profound effect on their physical and chemical behavior, their bioavailability, and their toxicity. The primary goal of this chapter is to provide a context for predicting and interpreting the speciation of dissolved metals; reactions involving metals in solid phases are considered in Chapter 11.
10.2
METAL OXIDATION STATES
The form in which we are most familiar with metals is as uncharged, nearly pure elements (e.g., copper wires, zinc coatings on galvanized pipes, or nickel or chromium plating to give solid surfaces the desired color or texture). However, metals in aqueous solutions exist primarily in ionic form. The ionic charge on a metal is called its oxidation number or oxidation state. Many metal ions can be stable in two or more oxidation states (e.g., Fe2+ or Fe3+ , Cr3+ or Cr6+ ). In such cases, it is common to refer to all the species in which the metal has a given charge by that charge (e.g., divalent or trivalent iron; trivalent or hexavalent chromium). The same concept is often represented by writing the charge on the metal ion in Roman numerals after the metal’s atomic symbol [e.g., Fe(II) or Fe(III); Cr(III) or Cr(VI)]. The behavior of a given metal can change enormously when it undergoes a change in oxidation state. In this chapter and Chapter 11, we consider only reactions in which metals remain in a given oxidation state; conversion of metals between oxidation states is addressed in Chapter 12.
10.3
FORMATION AND STRUCTURE OF METAL COMPLEXES
As noted in previous chapters, even though water molecules are electrically neutral, their charges are not distributed symmetrically. Rather, they are dipoles, with a small amount of positive charge residing near the hydrogen atoms and a balancing negative charge (associated with unshared pairs of electrons) on the opposite side of the oxygen atom. Water molecules surrounding a (positively charged) metal ion orient themselves with the unshared electrons adjacent to the metal. Some of the electron density and associated negative charge in these orbitals is shared with the metal ion, forming a chemical bond. 529
Depending on the size of the metal ion, four to eight oriented water molecules may surround it. Since these water molecules are bonded to the metal ion (albeit often weakly), the water plus the metal can be thought of as a single entity, and a reasonable representation of the metal in solution would be Me(OH2 )n+ x , where 4 x 8. The group of water molecules immediately surrounding the metal is called the inner hydration sphere or inner coordination sphere. Depending on the strength and extent of the electric field generated by the metal ion, a second sphere of water molecules (an outer hydration sphere) may also be bonded to the complex. In general, the smaller and more highly charged the central metal ion is, the more strongly it is able to bond to and orient water molecules around it, and the larger the complex becomes. Schematics (not to scale) of a Cr3+ ion surrounded by six water molecules and of a Cr6+ ion surrounded by four water molecules are shown in Figures 10.1a and 10.1b, respectively. As is described subsequently in the chapter, a variety of other molecules can replace the water molecules in the hydration spheres. All such molecules are called ligands, and the entire cluster is called a metal-ligand complex or simply a complex. Complexes in which some of the waters of hydration around a Cr6+ ion have been replaced by OH– or NH3 are shown in Figures 10.1c and 10.1d, respectively. Fully hydrated metal ions like the Cr3+ ion in Figure 10.1a are often referred to as free metal ions or uncomplexed ions even though, in a formal sense, they are complexed with water molecules. The number of molecules to which a dissolved metal ion binds directly (either water molecules or other ligands) is referred to as the metal’s coordination number. A few inorganic ligands of importance and some metals to which they bind are listed in Table 10.1. The exchange of water molecules in the hydration sphere with other molecules from the bulk solution is generally fast, so that the formation of complexes proceeds rapidly and can be considered to reach equilibrium within seconds in most cases. However, in a few cases (especially for more highly charged metals), the waters of hydration exchange more slowly, and equilibration with ligands in solution might take much longer. For instance, the characteristic residence time of water in the hydration sphere of Cr3+ ranges from several seconds to several days, depending on the solution pH.
10.4
COMPLEXES WITH HYDROXIDE IONS
The attraction of the central metal ion for the unshared electron pairs on water molecules in the inner hydration sphere causes electrons to be withdrawn from the water’s O H orbitals, weakening those bonds. As a result, the H+ ions that are part of the water molecules in a metal ion’s hydration sphere are more easily dissociated than those on water molecules in bulk solution, causing metal ions to act as acids. This process is shown schematically in Figure 10.2. 530
Figure 10.1 Chromium can exist in both trivalent (Cr3+ ) and hexavalent (Cr6+ ) oxidation states. (a) A fully hydrated Cr3+ ion, i.e., a Cr(H2 O)3+ 6 complex. (b) A fully hydrated Cr(H2 O)6+ ion in a tetrahedral complex. This ion is theoretically plau4 sible, but it is so acidic that it is never found at significant concentrations in aqueous solutions. (c) A Cr(H2 O)4 (OH)+2 complex, i.e., a species similar to that in (a), except that one H+ ion has been lost by each of two of the waters of hydration. (d) A Cr3+ ion in which all six waters of hydration have been replaced by ammonia molecules, forming the complex Cr(NH3 )3+ 6 . (e) A fully deprotonated chromate ion (CrO2– ), which is the dominant form of Cr(VI) in neutral or 4 alkaline solutions.
531
Figure 10.2 Schematic representation of the shifts in electron density between a metal ion and a water molecule in the inner hydration sphere. The unshared electrons in the oxygen atom’s orbitals are attracted to the metal ion, causing the electrons that form the O H bonds to shift toward the oxygen. This process forms a bond between the oxygen and metal ion and simultaneously weakens the O H bonds, so that the H ions are more readily released (i.e., the water becomes more acidic).
Although the actual reaction taking place is best represented as shown in Equation (10.1), it is most often abbreviated as in Equation (10.2) or (10.3). Regardless of how the reaction is written, the net effect is replacement of a water molecule by a hydroxide ion as one of the ligands in the inner hydration sphere. Me(OH2 )x n+ )* Me(OH2 )x 1 (OH)n Men+ + H2 O )* Me(OH)n Me
n+
1
1
+ H+
+ H+
n 1
+ OH )* MeOH
(10.1) (10.2) (10.3)
Table 10.1 Some important inorganic ligands and some metals to which they bind.
Ligand
Metals
Environment/Application
H2 O
All
Any aquatic system
OH–
Virtually all
Any aquatic system
F–
Fe3+ , Al3+
Some natural systems; industrial systems where HF is used to treat metal surfaces
Cl–
Cu2+ , Cu+ , Pb2+ , Cd2+
Estuaries, seawater; corrosion of metals
CN–
Fe3+ , Fe2+ , Cu+ , Cu2+ , Ni2+ , Ag+
Metal plating
NH3
Cu+ , Cu2+ , Cd2+ , Ni2+
Metal finishing
S2 O2– 3
Ag+
Photofinishing
5– P2 O4– 7 , P3 O10
Ca2+ , Mn2+ , Fe3+
Detergents, corrosion inhibitors
532
The equilibrium constant written for a reaction in which a metal-ligand complex forms is called a stability constant or formation constant. In the above reactions, the stability constant K(10.2) is also the acidity constant Ka for a reaction in which the free metal ion Men+ is the acid and Me(OH)n–1 is the conjugate base; similarly, K(10.3) is the inverse of a basicity constant. Many metal ions can act as multiprotic acids, releasing protons from several of the water molecules in their inner hydration sphere. For example, Figure 10.1c shows a Cr3+ ion after two of its waters of hydration have released an H+ ion. In some cases, once one H+ has been released from each water of hydration, the remaining protons are bonded very tightly to the oxygen atoms and are essentially non-releasable. For 3+ instance, Fe(OH)3– ion surrounded 6 is a stable species consisting of a central Fe + by six water molecules, each of which has lost one H . This species can act as a base by acquiring an H+ to form the conjugate acid Fe(OH)2– 5 , but it cannot act as an acid. In other cases, the protons that are part of the OH– ligands in the inner hydration sphere can be released, leaving the metal ion bonded to one or more oxide (O2– ) ions. Examples of metal ions that undergo this type of reaction include As5+ , Se4+ , and Cr6+ , which can induce the surrounding water molecules to deprotonate 2– 2– 2– completely to form AsO3– 4 , SeO3 , and CrO4 ions, respectively; the species CrO4 2– is shown in Figure 10.1e. In these types of molecules, the Me O bond is typically too strong to be broken under normal conditions. In such cases, rather than Men+ and O2– being considered to form a complex, they are considered to form a stable molecule, sometimes referred to as a metal oxyanion. For most metals, no more than four acid/base species are of practical significance. Thus, although the fully protonated species Cr(OH2 )6+ 4 shown in Figure 10.1b is theoretically plausible, it is so acidic as to be unimportant in all systems of interest; rather, when one is considering hydrolyzed forms of hexavalent Cr, only CrO2 (OH)2 , CrO3 (OH)– , and CrO2– 4 are considered. (The first two of these species are chromic acid and bichromate ion, commonly represented as H2 CrO4 and HCrO–4 , respectively.) Just as the equilibrium constants for deprotonation of an acid (Ka ) and for release of a gas from solution (H) are given special designations for convenience, conventions have been adopted for representing the stability constants of metal-ligand complexes. These designations can be summarized as follows: Ki :
Equilibrium constant for a reaction between two dissolved species, adding one ligand to a complex with i 1 ligands, to form a complex with i ligands.
bi :
Equilibrium constant for a reaction between an uncomplexed ion (i.e., an aquo-complexed ion) and i ligands to form a complex of the type MeLi . 533
bi j : ⇤K i
or ⇤ bi :
Equilibrium constant for a reaction between MeLi and j form a complex of the type MeL j .
i ligands to
Same as Ki or bi except that the reaction is written with the ligand in a protonated form, usually requiring inclusion of H+ on the product side.
Reactions for the formation of several hydroxo complexes of cadmium are shown below, along with the conventional way of representing their stability constants. Cd2+ + OH– )* CdOH+
log K1 = 3.90
(10.4)
CdOH+ + OH– )* Cd(OH)2
log K2 = 3.81
(10.5)
Cd(OH)2 + OH– )* Cd(OH)–3
log K3 = 0.99
(10.6)
Cd(OH)–3 + OH– )* Cd(OH)2– 4
log K4 = 0.01
(10.7)
Cd2+ + 2 OH– )* Cd(OH)2
log b2 = 7.71
(10.8)
Cd2+ + 4 OH– )* Cd(OH)2– 4
log b4 = 8.71
(10.9)
CdOH+ + H2 O )* Cd(OH)2 + H+
log⇤ K2 = 10.19
(10.10)
Cd2+ + 3 H2 O )* Cd(OH)–3 + 3 H+
log⇤ b3 = 33.30
(10.11)
If there is any chance of confusion about which ligand is being considered, its identity can be added to the subscript; e.g., K1 might be written as KOH,1 . Not all the stability constants listed above are independent. For instance, b values are related to the various K’s by n
bn = K1 K2 K3 ...Kn = ’ Ki
(10.12)
i=1
and b4 and ⇤ b4 are related through the following reactions: Cd2+ + 4 OH– )* Cd(OH)2– 4
b4
4 H2 O )* 4 H+ + 4 OH–
Kw4
+ Cd2+ + 4 H2 O )* Cd(OH)2– 4 +4H
⇤b 4
(10.13)
Thus, ⇤ b4 = b4 Kw4 or, in general, for formation of hydroxide complexes bi = bi Kwi
(10.14)
Ki = Ki Kw
(10.15)
⇤
and ⇤
534
A list of stability constants written in various ways for complexation of several metals by OH– is provided in Table 10.2. Table 10.2 Stability constants for complexation of metals by OH– .
i 1 2
log Ki 2.00 2.00
log⇤ Ki 12.00 12.00
log bi 2.00 4.00
log⇤ bi 12.00 24.00
Al3+
1 2 3 4
9.00 8.71 7.60 7.69
5.00 5.29 6.40 6.31
9.00 17.71 25.31 33.00
5.00 10.29 16.69 23.00
Ca2+
1
1.30
12.70
1.30
12.70
Cd2+
1 2 3 4
3.90 3.81 0.99 0.01
10.10 10.19 13.01 13.99
3.90 7.71 8.70 8.71
10.10 20.29 33.30 47.29
Co2+
1 2 3
4.30 4.91 1.30
9.70 9.09 12.70
4.30 9.21 10.51
9.70 18.79 31.49
Cr3+
1 2 3 4
10.43 7.73 7.65 2.54
3.57 6.27 6.35 11.46
10.43 18.16 25.81 28.35
3.57 9.84 16.19 27.65
Cu2+
1 2 3 4
6.50 5.27 3.59 0.91
7.50 8.73 10.41 13.09
6.50 11.77 15.36 16.27
7.50 16.23 26.64 39.73
Fe2+
1 2 3
4.60 2.91 3.50
9.40 11.19 10.50
4.60 7.51 11.01
9.40 20.49 30.99
Fe3+
1 2 3 4
11.98 10.27 4.75 6.30
2.02 3.73 9.25 7.70
11.98 22.25 27.00 33.30
2.02 5.75 15.00 22.70
Ag
+
535
Table 10.2 – Continued from previous page i 1 2 1
log Ki 10.60 11.24 2.58
log⇤ Ki 3.40 2.76 11.42
log bi 10.60 21.84 2.58
log⇤ bi 3.40 6.16 11.42
Ni2+
1 2 3
4.11 4.90 3.00
9.89 9.10 11.00
4.11 9.01 12.01
9.89 18.99 29.99
Pb2+
1 2 3 1 2 1 2 3 4
6.40 4.51 3.00 4.16 1.80 5.00 6.11 2.50 1.20
7.60 9.49 11.00 9.84 12.20 9.00 7.89 11.50 12.80
6.40 10.91 13.91 4.16 5.96 5.00 11.11 13.61 14.81
7.60 17.09 28.09 9.84 22.04 9.00 16.89 28.39 41.19
Hg2+ Mg2+
H4 SiO4 Zn2+
10.5
LOG C-pH DIAGRAMS FOR DISSOLVED METALS
Recognizing metals as polyprotic acids, we can develop log C-pH diagrams showing the relationships among various Me(OH)x species, using the same procedures we used previously for other acids. For instance, consider a system containing 10 4 M T OT Cd. Up to four protons can be released from the waters of hydration in the free aquo Cd2+ complex; i.e., Cd2+ is a tetraprotic acid, with Ka values as follows.1 Cd2+ + H2 O )* Cd(OH)+ + H+
log Ka1 = log ⇤ K1 = 10.10
CdOH+ + H2 O )* Cd(OH)2 + H+
log Ka2 = log ⇤ K2 = 10.19
Cd(OH)2 + H2 O )* Cd(OH)–3 + H+
log Ka3 = log ⇤ K3 = 13.01
+ Cd(OH)–3 + H2 O )* Cd(OH)2– 4 +H
log Ka4 = log ⇤ K4 = 13.99
Other than the fact that the Kai values are written as ⇤ Ki , these relationships are identical to those in any other system with a tetraprotic acid. The corresponding log C-pH diagram is shown in Figure 10.3. 1 Neutral
complexes are commonly designated with a superscript o or 0 , or by the suffix (aq) to distinguish them from solids that might have the same chemical formula.
536
Figure 10.3 LogC-pH diagram for Cd in system with 10
4M
T OT Cd.
Assuming that the solutes behave ideally, the concentration of any Cd OH complex can be written in terms of the concentration of free Cd2+ and either the b or ⇤ b stability constant, as follows: (Cd(OH)2i
i
) = bi Cd2+ OH ⇤b i = Cd2+ i + (H )
i
(10.16a) (10.16b)
If Cd2+ and Cd OH complexes are the only Cd-containing species in the solution, these equations allow us to write a mass balance on Cd that expresses T OT Cd strictly in terms of the concentration of free Cd2+ , the stability constants, and the OH– or H+ concentration, as follows: T OT Cd = (Cd2+ )+(CdOH+ )+(Cd(OH)2 )+(Cd(OH)3 )+(Cd(OH)24 ) (10.17a) "
(CdOH+ ) (Cd(OH)2 ) (Cd(OH)3 ) (Cd(OH)24 ) = (Cd ) 1 + + + + (Cd2+ ) (Cd2+ ) (Cd2+ ) (Cd2+ ) 2+
= (Cd2+ ) 1 + K1 (OH )1 + b2 (OH )2 + b3 (OH )3 + b4 (OH )4 4
= (Cd2+ ) 1 + Â bi (OH )i i=1
!
4
= (Cd2+ ) 1 + Â
i=1
537
⇤b i + (H )i
!
#
(10.17b) (10.17c) (10.17d)
The fraction of the total dissolved metal that is present as a given species is designated by a, as in other acid/base systems. Thus, for example, aCd2+ is aCd2+ ⌘
[Cd2+ ] T OT Cd
(10.18a)
4
=
1 + Â bi (OH )
i
i=1 4
=
1+ Â
i=1
10.6
⇤b i + (H )i
!
!
1
(10.18b)
1
(10.18c)
COMPLEXES WITH LIGANDS OTHER THAN WATER AND HYDROXIDE
In addition to water and hydroxide, other dissolved species can act as ligands and form soluble complexes with metal ions. In all such cases, at least a portion of the ligand molecule must have a local excess of electronic (negative) charge which it donates or shares with the metal ion, thereby forming a chemical bond. For instance, ammonia can form strong complexes with some metal ions because the nitrogen atom in an ammonia molecule has an orbital containing an unshared pair of electrons (see Figure 5.8). These electrons, which cause the nitrogen atom to carry a slight negative charge in free NH3 molecules (much as the oxygen atom does in water molecules), can be shared with metal ions to form Me NH3 bonds (as in Figure 10.1d). The most common donor atoms in environmental systems (i.e., ligand atoms that can donate electronic charge and thereby bond to metal ions) are O, N, and S. If the bonds formed between a ligand and a metal ion are strong enough that the ligand replaces water in the inner hydration sphere, the reaction is referred to as ligand exchange, and the ligand is said to form an inner sphere complex, e.g., * Me(H2 O)n+ x + NH3 ) Me(H2 O)x
n+ + H2 O 1 (NH3 )
(10.19)
The above reaction is more commonly written as Men+ + NH3 )* Me(NH3 )n+
(10.20)
Yet other ligands bond to the metal ion but remain separated from it by a water molecule. These outer sphere complexes are relatively weak and are called ion pairs. The rules for naming stability constants for ligands other than hydroxide are essentially identical to those for hydroxide. For instance, for ammonia complexes of Zn2+ , the reactions and labels assigned to the various stability constants are as follows: 538
Zn2+ + NH3 )* Zn(NH3 )2+
log K1 = 2.214
(10.21)
Zn(NH3 )2+ + NH3 )* Zn(NH3 )2+ 2
log K2 = 2.284
(10.22)
2+ Zn(NH3 )2+ 2 + NH3 )* Zn(NH3 )3
log K3 = 2.364
(10.23)
2+ Zn(NH3 )2+ 3 + NH3 )* Zn(NH3 )4
log K4 = 2.024
(10.24)
Zn2+ + 2 NH3 )* Zn(NH3 )2+ 2
log b2 = 4.498
(10.25)
Zn2+ + 3 NH3 )* Zn(NH3 )2+ 3
log b3 = 6.862
(10.26)
Zn2+ + NH+4 )* Zn(NH3 )2+ + H+
log ⇤ K1 = 7.030
(10.27)
Note that, in this case, the ligand is neutral, so all the complexes are divalent (as is the case when water is the ligand).
10.7
MIXED LIGAND COMPLEXES AND CHELATING AGENTS
Complexes that contain more than one type of ligand (other than water) can also form. Such complexes are called mixed-ligand complexes. For instance, formation of a mixed, ammonia-chloro complex might be represented as * Me(H2 O)n+ x + Cl + NH3 ) Me(H2 O)x
n 1 + 2 H2 O 2 (NH3 )Cl
(10.28)
Also, some ligands have more than one donor atom, in which case one ligand molecule might expel more than one water molecule from the inner hydration sphere and form multiple bonds to the metal ion. Such ligands are called multidentate ligands, and they tend to form strong complexes. If the complexes are very strong, they are called chelates, and the ligands are called chelators or chelating agents. Commercially, the most important chelating agent is the tetraprotic acid EDTA (ethylene diamine tetra-acetic acid), which, in its most dissociated form, has the structure shown in Figure 10.4. EDTA can bind to a metal to form a hexadentate chelate. To do so, it replaces the six water molecules in the hydration sphere and forms six new bonds with the metal ion by sharing electrons associated with the two nitrogen atoms and the four carboxylate groups. Thus, a single EDTA molecule can displace six individual monodentate ligands. The binding of EDTA to many metals is so strong that, under many circumstances, EDTA Me complexes form until either the free EDTA or free metal is essentially 100% complexed. Thus, adding excess EDTA to a solution is an effective way to maintain the activity of free (i.e., uncomplexed) metal ions at extremely low levels. 539
Figure 10.4 (a) The bond structure of the strong chelating agent EDTA, in its fully deprotonated form (EDTA4 ). (b) The structure of a (Pb-EDTA)2 chelate. In part (b), the circle at the center is the Pb2+ ion, and the remaining circles represent oxygen, nitrogen, and carbon atoms; hydrogen atoms have been omitted to minimize clutter. The six bonds between the EDTA and Pb are shown in bold.
In addition to applications in industry, EDTA is used extensively as a food additive, at least in part to “tie up” metal ions that might come in contact with the food and prevent them from participating in other reactions. It is also used medically in the treatment of acute metal poisoning, as a way of collecting metal ions from organs and holding them in solution so that they can be excreted. The fact that metal-ligand complexes form in a system indicates that the potential energy of the system is lower (i.e., the system is more stable) with those complexes present than if the metal and ligand remained separate. Therefore, it is not surprising that the formation of complexes reduces the likelihood that the metal will undergo other reactions (at least from a thermodynamic perspective; the kinetics might be a different story). Mathematically, this effect is obvious from the fact that when complexes form, the concentration, activity, and molar Gibbs energy of the free metal ion in the solution all decrease. The stronger the complex, the greater the reduction in free metal ion activity, so chelates can reduce metal reactivity very substantially. One consequence of the lower reactivity of complexed metals compared to the same concentration of free metal is that complexation or chelation tends to stabilize metals in solution at the expense of metals associated with solids. For instance, complexation reduces the tendency for metals to precipitate out of solution as solids or to bind (adsorb) to the surfaces of solids that are in contact with the solution. The former fact is exploited extensively in industry, where chelating agents are used to prepare solutions with very high concentrations of metals under conditions where most of the metal would precipitate if the chelating agent were not present. 540
Similarly, adsorption of metals tends to diminish when they become complexed (Figure 10.5), although there are circumstances (discussed in Chapter 13) under which complexation enhances metal adsorption.2 The same effect (lower chemical activity, and therefore lower reactivity) generally causes uptake of chelated metals by organisms to be much less efficient than uptake of free metal ions, so that the chelated metals are less toxic at a given total metal concentration. Stability constants for a variety of metals and ligands are listed in Table 10.3.
Figure 10.5 Adsorption of Cd2+ to SiO2 (s) is diminished when Cl– or SO2– 4 is added to the solution and free Cd2+ ions are converted to Cd Cl or Cd SO4 complexes. The systems characterized in this figure all contained 30 g/L SiO2 (s) and 5 ⇥ 10 7 M T OT Cd and had an ionic strength of 0.7 M [from Benjamin and Leckie (1982)2 ].
10.8
METAL SPECIATION IN SYSTEMS CONTAINING COMPLEXING LIGANDS
The equilibrium speciation of metals in systems containing ligands bears a close connection to the speciation of simple acids and bases and can be represented on log C-log L (i.e., ligand) diagrams that bear many similarities to log C-pH plots. Consider, for instance, the speciation of Cd over a range of Cl– activities in a system containing some known total cadmium concentration. In the preliminary analysis, we will assume that the pH is low enough that OH– complexes of Cd2+ can be ignored. Reactions and stability constants for the formation of the two known Cd Cl complexes are as follows: Cd2+ + Cl– )* CdCl+ CdCl+ + Cl– )* CdCl2 (aq)
log K1 = 1.98 log K2 = 0.62
2 Benjamin,
(10.29) (10.30)
M.M., and Leckie, J.O. (1982). “Effects of complexation by Cl, SO4 and S2 O3 on adsorption behavior of Cd on oxide surfaces. Environ. Sci. Technol. 16, 162-170.
541
542
Co2+
Cd
2+
Ca2+
Al3+
Ag+
3.22
3.84
4.28
CoHL 12.22
CoL
CoL 2.30
CdL2 3.50
CdL2 7.23
CdHL 11.83
CdL 2.37
CaL 2.36
AlL2 5.58
AlL
AgL3 5.20
AgL2 5.25
AgL 1.30
SO2– 4
4.37
CdL
CaHL 11.43
CaL
CO2– 3
1.98
0.40
0.39
CoL
0.35
CdL2 2.60
CdL
CaL
AlL
AgL 3.31
Cl–
7.01
CoL 1.40
CdL 1.20
CaL 1.14
AlL4 19.40
AlL3 16.70
AlL2 12.63
AlL
AgL2
AgL 0.40
F–
2.55
0.11
0.20
2.03 CoL2 3.49
CoL
CdL4 6.80
CdL3 5.89
CdL2 4.55
CdL
CaL2
CaL
7.21
AgL 3.31
NH3
6.46
CoHL 15.43
CdHL 16.08
CaH2 L 20.92
CaHL 15.04
CaL
Al2 L 18.98
AlHL 20.01
PO3– 4 8.05
18.10
15.97
12.44
21.78
18.96
21.70
18.16 CoHL 21.59
CoL
CdH2 L 23.23
CdHL 21.43
CdL
CaHL
CaL
AlHL
AlL
AgL3
AgHL 14.74
AgL
EDTA
20.48
CdL4
CdL3
CdL2
CdL
17.92
15.65
11.12
6.01
AgH-1 L2 8.59
AgH-1 L 5.30
AgL2
CN–
CoL
CdL4
CdL3
CdL2
CdL
AgL2
AgL
HS–
5.20
19.31
17.11
15.31
8.01
17.91
13.82
Table 10.3 Stability constants for some metal–ligand complexes. Values correspond to log b for formation of the complex from the free metal, the ligand, H+ , and H2 O.
543
Fe3+
Fe2+
Cu2+
Cr
3+
FeHL 11.43
CuHL 12.13
4.25
2.39
FeL2 5.38
FeL
FeL
CuHL 2.34
CuL2 10.20
3.10
CuL 2.36
CrL
SO2– 4
6.77
CuL
CO2– 3
FeL
1.48
0.20
4.59
CuL4 FeL
2.29
0.26
0.30
0.54
CuL3
CuL2
CuL
CrL
Cl
–
5.20
6.04
1.21
FeL3 13.70
FeL2 10.66
FeL
FeL
CuL 1.70
CrL3 11.71
CrL2 9.04
CrL
F–
4.40
4.02
FeL4 2.73 FeH2 L 23.85
FeHL 22.28
FeH2 L 22.27
FeL2 2.24 FeL3 2.67
FeHL 15.98
CuHL 16.50
CrH2 L 22.07
CrHL 16.16
PO3– 4
1.40
FeL
CuL4 12.28
CuL3 10.19
CuL2 7.41
CuL
CrL2 4.10
CrL
CoL5 5.11
CoL4 5.05
CoL3 4.42
NH3
Table 10.3 – continued from previous page
20.49
FeHL
FeL
FeHL
FeL
29.17
27.66
19.05
16.01
CuH2 L 26.23
CuHL 24.02
CuL
CoH2 L 23.49
EDTA
39.71
35.40
23.00
14.31
Fe2 L6
FeL6
47.64
43.60
FeH2 L6 42.11
FeHL6
FeL6
CoL5
CoL3
CN–
FeL
HS–
5.62
544
2+
Zn
Pb
2+
Ni2+
Hg2+
4.57
PbL2 3.47
PbL2 9.94
ZnL2 3.28
ZnL2 7.30
ZnHL 11.83
ZnL 2.34
4.76
ZnL
PbHL 13.23
PbL 2.69
NiL2 0.82
2.30
6.53
PbL
NiHL 12.42
NiL
NiL
HgL2 3.48
HgL2 15.58
HgHL 16.35
HgL 2.47
SO2– 4
HgL 12.13
CO2– 3
1.56
1.89
0.43
2.15
1.30
2.21
0.46
ZnL3 6.85 ZnL4 8.87
ZnL4 0.20
ZnL2 4.49
ZnL3 0.50
ZnL2 0.45
ZnL
ZnHL 15.69
PbH2 L 21.07
PbHL 15.48
23.24
23.64
20.11
PbHL
PbL
22.54
19.71
NiH2 L 24.74
NiHL
NiL
HgH2 L 29.17
HgHL 26.87
HgL
EDTA
21.43
18.00 ZnH2 L2 22.83
ZnHL
ZnL
PbH3 L 25.64 ZnL
NiL6 8.27
NiL5 8.31
NiL4 7.65
NiL3 6.53
NiH2 L 20.50
NiHL 15.33
PO3– 4
PbL4 1.38 ZnL 1.30
2.72
NiL2 4.87
NiL
HgL3 10.04
HgL2 17.79
NH3
PbH2 L 24.44
PbL2 3.24
PbL
NiL
HgL 1.60
F–
Table 10.3 – continued from previous page
PbL3 1.80
PbL2 1.90
PbL
NiL2
NiL
HgL4 15.63
HgL3 15.03
HgL2 14.03
HgL 7.33
Cl
–
36.03
30.20
38.97
36.31
32.75
17.00
ZnL4
ZnL3
ZnL2
19.62
16.05
11.07
NiH3 L4 43.34
NiH2 L4 40.74
NiHL4
NiL4
HgL4
HgL3
HgL2
HgL
CN–
38.42
ZnH-2 L2
PbL3
PbL2
NiL
1.43
16.57
15.27
5.49
HgH-2 L2 23.22
HgH-1 L2 31.93
HgL2
HS–
By writing the equilibrium expression for formation of CdCl+ as (CdCl+ )/(Cd2+ ) = K1 (Cl ), we can derive the relationships in Table 10.4. The table includes analogous expressions for the HAc/Ac– acid/base pair for comparison. Table 10.4 Comparison of relationships between an acid/base and a metal/complex system.
Acid/Base Pair (HA/A– )
Metal/Complex Pair (Me/MeL) (CdCl+ ) = (Cd2+ )
at
pCl– = log K1
(HAc) = (Ac )
at
pH = pKa
(CdCl+ ) > (Cd2+ )
at
pCl– < log K1
(HAc) > (Ac )
at
pH< pKa
(CdCl+ ) < (Cd2+ )
at
pCl– > log K1
(HAc) < (Ac )
at
pH> pKa
The only significant difference between the equations for the acid/base and metal/ complex systems is that the crossover point is at pH = pKa in the acid/base system and at pCl = log K1 = pK1 in the metal/complex system. This difference in sign arises because Ka is defined as the equilibrium constant for a dissociation reaction, whereas K1 is defined for an association reaction. The total dissolved cadmium concentration in this system can be represented as a function of only (Cd2+ ), (Cl– ), and the stability constants, as follows: 2
T OT Cd = (Cd2+ ) + Â (CdCli ) i=1
⇥ ⇤ = (Cd2+ ) 1 + bCl,1 (Cl )1 + b2 (Cl )2 " # n
= (Cd2+ ) 1 + Â bCl,i (Cl )i
(10.31)
i=1
The fraction of T OT Cd that is in a given form can then be computed. For instance, the fraction of T OT Cd present as free Cd2+ ion is " # 1 2 (Cd2+ ) i aCd2+ = = 1 + Â bCl,i (Cl ) (10.32) T OT Cd i=1 Using a spreadsheet in conjunction with Equation (10.31) and the equilibrium constant expressions for Reactions (10.29) and (10.30), or a chemical equilibrium program like Visual Minteq, we can determine the concentrations of all Cd species at any given value of (Cl– ), much as we did previously to develop the log C-pH diagram for a multiprotic acid. Such a plot summarizes the results from simultaneous solutions of the two equilibrium-constant expressions for formation of Cd Cl complexes and the mass balance on Cd. Plotting the corresponding values against log (Cl ) for a system with 10 4 M T OT Cd yields the log C-log (Cl ) diagram shown in Figure 10.6. 545
Figure 10.6 LogC-log (Cl ) diagram for a system containing 10 4 M T OT Cd, at a pH low enough that formation of Cd OH complexes is negligible.
The activity of Cl– in seawater is approximately 0.5. Inserting this value and the appropriate values for bCl,i into Equation (10.32), we find aCd2+ = 5.57 ⇥ 10 3 . In other words, only about 0.6% of the dissolved Cd in seawater is present as Cd2+ ions. (In truth, aCd2+ in seawater is even less than this value due to formation of complexes with other ligands in the solution, as described shortly.3 ) The same approach can be used to compute a values of the other Cd species in the system. Next, consider a system in which complexation by both chloride and hydroxide might be significant. Specifically, say we wish to analyze the Cd speciation in a solution containing 10 4 M T OT Cd in which (Cl ) is fixed at 0.5 M, but the pH is variable. Such a system can be represented by a log C-pH diagram similar to Figure 10.6, but differing because of the presence of Cl– . We note first that the relative concentrations of Cd2+ and the Cd OH complexes at any given pH must satisfy the equilibrium relationships among them [e.g., Equations (10.4)-(10.11)], and that those relationships are independent of (Cl– ). Therefore, the relative concentrations of those species at a given pH are the same as when no Cl– is present. For example, (Cd2+ ) = (CdOH+ ) at pH = p⇤ KOH,1 = 10.10, and (CdOH+ ) = (Cd(OH)2 ) at pH = p⇤ KOH,2 = 10.19. Similarly, because the reactions among Cd2+ and the Cd Cl complexes do not involve exchange of H+ or OH– , the relative concentrations of these species are independent of pH. This inference can be demonstrated formally by writing an expression for the activity of a generic Cd Cl complex, as follows: 3 In
addition, the calculation is based on the unrealistic assumption that the ions behave ideally in a solution with the high ionic strength of seawater. The effect of ionic strength can easily be taken into account when chemical equilibrium software is used to carry out the calculations.
546
2 Cd2+ + iCl )* CdCli
bCl,i =
i
(10.33)
(CdCl2i i ) (Cd2+ )(Cl )i
(10.34a)
(CdCl2i i ) = bCl,i (Cd2+ )(Cl )i
(10.34b)
log(CdCl2i i ) = log bCl,i + log(Cd2+ ) + i log(Cl )
(10.34c)
Based on Equation (10.34c), if (Cl– ) is fixed (as in the system of interest), the activity of each CdCl2i i complex can be written as log(CdCl2i i ) = log(Cd2+ ) + constant
(10.35)
where the constant equals log bCl,i + i log(Cl ). Equation (10.35) indicates that the curve representing any CdCl2i i species on a log C-pH diagram will be identical to the curve for Cd2+ , except that it will be offset vertically from the Cd2+ curve by this constant amount. We can now write a mass balance on Cd using Equation (10.16a) or (10.16b) to account for the Cd OH complexes, and Equation (10.34b) to account for the Cd Cl complexes, as follows: T OT Cd = (Cd2+ ) + (CdOH+ ) + (Cd(OH)2 ) + (Cd(OH)3 ) + (Cd(OH)24 ) + (CdCl+ ) + (CdCl2 ) "
4
(10.36a) 2
= (Cd ) 1 + Â bOH,i (OH ) + Â bCl,i (Cl ) 2+
i
i=1
i=1
i
#
(10.36b)
In the system of interest, T OT Cd (10 4 M), the activity of Cl– (0.5 M), and all the b values are known, so (Cd2+ ) can be computed at any pH. The concentration of each complex can then be computed from (Cd2+ ) and the appropriate stability constant. For example, at pH 9.0, we find (Cd2+ ) = 6.74 ⇥ 10
(CdOH+ ) = 5.61 ⇥ 10
(Cd(OH)2 ) = 3.01 ⇥ 10 (Cd(OH)3 ) = 3.38 ⇥ 10
(Cd(OH)24 ) = 3.01 ⇥ 10
7 8 9 13
(CdCl+ ) = 3.21 ⇥ 10
(CdCl2 ) = 6.71 ⇥ 10
5 5
18
By carrying out analogous calculations, we can determine the speciation of the system at other pH values and prepare the log C-pH diagram, which is shown in 547
Figure 10.7. Consistent with the preceding discussion, the curves for CdCl+ and CdCl2 are parallel to that for Cd2+ and to one another, and the various acid/base conjugate pairs of the Cd OH system have equal activities at the pH corresponding p⇤ KOH,i . However, at these intersection points, each of the two acid/base species does not necessarily account for 50% of T OT Cd. For instance, at pH 10.08, (Cd2+ ) = (CdOH+ ), but each of these species represents only about 0.67% of T OT Cd. The reason for this result is, of course, that the vast majority of the Cd in the system at that pH is present as Cd Cl complexes.
Figure 10.7 LogC-pH diagram for a system containing 10 tivity of 0.5.
4M
T OT Cd and a fixed Cl– ac-
Comparing Figures 10.3 and 10.7, we see that the presence of Cl– at a fixed activity of 0.5 narrows the pH region where Cd(OH)2 is dominant and prevents CdOH+ from becoming dominant at any pH. More generally, the addition of Cl– and the formation of Cd Cl complexes ‘ties up’ some of the T OT Cd and thereby reduces the concentration of all other Cd species. We could, of course, develop the same log C-pH diagram using Visual Minteq. To input the fixed activity of 0.5 for Cl– , we click the “Fixed activity” radio button when Cl– is selected as one of the input components. Then, to determine the speciation at different pH values, we use the “Multi-problem/Sweep” option to scan through the pH range of interest. Because the fixed Cl– activity is quite high, allowing Visual Minteq to compute and apply activity coefficients yields a significantly more accurate prediction for the speciation than if we made the unrealistic assumption that the solution behaves ideally. To summarize the key points of the preceding discussion, the presence of complexing ligands in a solution containing a weak acid metal group forces us to consider more equilibrium expressions and more terms in the metal mass balance than if those ligands were absent. Correspondingly, the log C-pH diagram contains curves 548
The acidic waters that drain from mines can contain large concentrations of metals. When these waters mix with others or are neutralized by some other process, large amounts of metals can precipitate. (Laurie Balistrieri [USGS].)
for more species. However, the relationships among those species (i.e., the stability constant relationships) are the same as in simpler systems; the only relationship that changes when additional complexes are considered is the mass balance on total metal. Generalizing based on Equation (10.36b), we can write the mass balance for a generic system with m different ligands concisely as follows: ! m
n
T OT Me = (Mex+ ) 1 + Â Â bL j ,i (L j )i
(10.37)
j=1 i=1
Example 10.1 Use Visual Minteq to determine the speciation of dissolved Zn(II) in seawater. Assume that T OT Zn(II) is 2.0 µg/L and that the ionic strength is 0.7 M, and con2– sider OH– , Cl– , HCO–3 , CO2– 3 , and SO4 as possible complexing ligands. Use the concentrations of these ligands given in Table 1.1. What is the dominant dissolved Zn species? What is aZn2+ ? Solution Table 1.1 gives the pH of seawater as 7.9, so we input that value and the given ionic strength on the Visual Minteq home page. The concentrations of T OT Cl and 549
T OT SO4 are given in Table 1.1 in units of mg/L. We can input those values directly by selecting mg/L as the “Concentration unit” on the home page. We can also input T OT Zn using those units, as 0.002 mg/L. Table 1.1 does not give T OT CO3 explicitly, but we can infer this value based on the given alkalinity of 120 mg/L as CaCO3 , if we assume that all the alkalinity that is not accounted for by other given constituents is attributable to carbonate species. The carbonate species can be included in the Visual Minteq analysis by inputting the given alkalinity on the “Parameters/Specify alkalinity” menu. After doing this, returning to the home page, and switching the concentration units to molal, we can click on the “View/edit list” button to see the summary list of components, which is displayed in Figure 10.8a. The corresponding tableau, shown in Figure 10.8b, includes 14 Zn ligand complexes.4
Figure 10.8 Visual Minteq summary of (a) components and (b) species for analysis of Zn speciation in seawater. 4 The
maximum number of species that Visual Minteq displays on a single screen depends on the graphics settings of the computer. In the system of interest, the program’s database includes 18 species that can form from the components, 16 of which are shown. The image shown here includes all the Zn complexes that are expected to be present; the two species that are not shown are H2 CO3 and HCO–3 , but those species are nevertheless included in the calculations.
550
After running the program, we can click the “View species distribution” button on the main output page to see a summary of the speciation, component by component. This summary, shown in Figure 10.9, indicates that the dominant form of zinc is the free aquo ion, but that ion nevertheless accounts for only 38% of T OT Zn (i.e., aZn2+ = 0.38). ZnSO4 and ZnCl+ account, respectively, for 20% and 19% of T OT Zn, and a wide variety of other complexes account for the rest. Needless to say, solving for the speciation at this level of detail without a program like Visual Minteq would be impractical.
Figure 10.9 Visual Minteq output summary for Zn speciation in seawater.
Example 10.2 As noted in Chapter 8, most natural organic matter (NOM) molecules contain acidic functional groups (thought to be primarily carboxyl and phenolic groups) with pKa values that fall into two broad ranges centered near pH 4 and 9. These same functional groups are able to bind with metal ions to form soluble complexes that, in many environments, account for the majority of the total dissolved metal concentration. A lake contains NOM like that characterized in Example 8.12. The acidic functional groups in this NOM were represented by two Gaussian distributions of pKa ’s, with mean values of 3.9 and 8.4. Assume that these same functional groups form complexes with Cu2+ , and that in all cases, log K = 4.9 for reactions of the type Cu2+ + DOMi )* Cu DOMi 551
where i can be either 1 or 2 (i.e., the binding strength is the same with both sets of functional groups). Determine the distribution of dissolved copper in the solution described in Example 8.12 (containing NOM at a concentration of 9 mg TOC/L) over the pH range from 3 to 11, for T OT Cu equal to 50 µg/L. Solution The solution can be modeled using Visual Minteq by opening the saved file from Example 8.12 and specifying 50 µg/L Cu2+ as an additional input. We then go to the “Database management/ DOM complexes (Gaussian)” page, where we see that the default value of log K for formation of both Cu DOM1 and Cu DOM2 complexes is 4.9; we therefore do not need to make any modifications to that page. After indicating the Cu species that we wish to have reported in the “Selected sweep” results, we can run the program, export the output, and plot the results, which are shown in Figure 10.10.
Figure 10.10 Copper speciation in the example solution containing 50 µg/L T OT Cu and NOM at a concentration of 9 mg DOC/L.
The calculations indicate that Cu DOM complexes are the dominant dissolved form of Cu over the range from pH⇠3.5 to ⇠8.25. In this range, Cu DOM1 is present at a considerably higher concentration than Cu DOM2, although the latter complex does account for a nontrivial portion of T OT Cu at neutral and slightly alkaline pH. The dominance of Cu DOM1 over Cu DOM2, even though they have identical log K values, is partially caused by the fact that DOM1 groups are more concentrated than DOM2 groups in the NOM molecules. However, the major cause of this difference is that DOM1, which represents the carboxyl groups in the molecules, is considerably more acidic than DOM2, which represents the phenolic groups. As a result, a much larger fraction of the DOM2 remains protonated at acidic pH values; 552
put another way, H+ is a much more effective competitor with Cu2+ for DOM2 sites than it is for DOM1 sites. At pH greater than 8.25, DOM2 becomes almost as extensively deprotonated as DOM1, so that the difference in their abilities to complex Cu is primarily due to their different concentrations. However, those are also the conditions under which CuOH+ and more hydrolyzed species become dominant over Cu2+ , so that OH– begins to outcompete both groups of organic sites for Cu2+ . As a result, the dominant forms of dissolved copper become Cu OH complexes.
10.9
METAL ION BUFFERS
As shown in Chapter 8, an understanding of acid/base chemistry can help us choose appropriate chemicals to add to a solution if we wish to buffer the pH at certain values. Given the similarities between acid/base and metal/complex chemistry, it is not surprising that the activity of dissolved metal ions can be buffered just as the activity of H+ can be. Buffering of the activity of free metal ions is especially important in biological systems, because this activity can control many metabolic reactions and because free metal ions are thought to be more toxic than most complexes. As a result, organisms have developed the capability to maintain the activity of Men+ within narrow bounds, even if the concentration of total Me in the system changes significantly. The critical feature of a good acid/base buffering system is the presence of a species that can readily combine with or release H+ . Furthermore, that species must be present at a concentration that is much greater than the desired activities of free H+ and OH– . Optimal buffering is achieved when half of the buffering species is bound to an H+ ion, while the other half is not. In this way, the protonated (acidic) form of the buffer serves as a reservoir of H+ that can be released if the H+ activity in solution declines, and the unprotonated (basic) form serves as a reservoir of sites where H+ can attach if more H+ enters solution. Similarly, ligands are good buffers for the activity of free Men+ when they are present at a much higher concentration than the desired concentration of free Men+ and can combine with or release Men+ readily if the total Me concentration in solution changes. Just as a given acid group is an effective buffer only over one or more limited pH ranges (near the pKa values), we expect that a given ligand will serve as an effective buffer only for certain limited ranges of Men+ activity – specifically, under conditions where two species containing different numbers of ligands are present at approximately equal concentrations. For instance, for a metal and ligand that form only a single complex MeL, the buffering is expected to be maximized when one-half of the ligand is free L and the other half is complexed with the metal. This situation applies when (Men+ ) = 1/K1 [i.e., log(Men+ ) = pK1 ]. 553
Consider, for example, the potential for ethylene diamine (EN) to buffer the activity of free, dissolved Cu2+ in a pH 7 solution containing 10 3 M T OT EN. EN contains the structural core of EDTA, but lacks the four carboxyl groups that make EDTA such a strong chelator. Nevertheless, it is a relatively strong complexing agent for many metals. The amine (ammonia-like) groups in EN can protonate, so EN is a weak base. The corresponding diprotic acid, H2 EN2+ , has pKa1 and pKa2 values of 6.85 and 9.93, respectively. The complexes CuEN2+ and Cu(EN)2+ 2 have log KEN,1 = 10.50 and log bEN,2 = 19.58, respectively. The analog of a log C-pH diagram for this system is a plot of log C-log (Cu2+ ). Such a diagram can be prepared in Visual Minteq by inputting the known values of pH and T OT EN, and systematically varying log (Cu2+ ) using the “Multi-problem/ sweep” option. (EN is shown in the Visual Minteq component list as EtDiAm.) The diagram is shown in Figure 10.11.
Figure 10.11 A log C-log (Cu2+ ) diagram for a system at pH 7 and containing 10 T OT EN.
3M
The Cu2+ activity in the system will be well buffered under conditions where changes in the Cu2+ concentration can be substantially counteracted by formation or dissociation of Cu EN complexes, corresponding to conditions where the number of Cu2+ ions associated with the dominant EN species is changing. The log Clog (Cu2+ ) diagram indicates that this criterion is met near pCu2+ values of 10.1 and 4.9, and a titration curve showing changes in log (Cu2+ ) as a function of T OT Cu (Figure 10.12) confirms that the EN is a good buffer near these pCu2+ values. Consider why the titration curve has the shape it does. At log (Cu2+ ) less than 11 (very low Cu2+ activities), the vast majority of T OT Cu is complexed (aCu2+ is only ⇠ 10 7.2 ). On the other hand, most of the EN is protonated but not complexed with Cu2+ , so there is a large pool of uncomplexed EN (i.e., unused Cu-complexing capacity) in the system. When log (Cu2+ ) increases to approximately 11, this pool 554
Figure 10.12 A titration curve showing the response of log (Cu2+ ) to changes in T OT Cu, for the system characterized in Figure 10.11.
begins to be used, and once log (Cu2+ ) has increased to 9, most of the EN is present in Cu(EN)2+ 2 complexes. This transition consumes one Cu ion for every two EN molecules in the system, so the titration curve has a flat portion with a width of approximately 0.5 T OT EN or 5 ⇥ 10 4 M. Once essentially all the EN is complexed, the dominant EN containing species 2+ remains Cu(EN)2+ 2 as the Cu activity increases through a few orders of magnitude. As a result, most of the Cu2+ that is added while the Cu2+ activity is in this range remains as free Cu2+ ions. The additional Cu2+ therefore has a substantial effect on the Cu2+ activity, and the titration curve steepens. This scenario applies until log (Cu2+ ) gets large enough to induce a second dramatic change in EN speciation, when most 2+ of the EN in solution is converted from Cu(EN)2+ 2 to Cu(EN) , consuming an additional 5 ⇥ 10 4 M T OT Cu. Because most of the Cu2+ added as this transition occurs is converted to a complex rather than remaining as free ions, the Cu2+ activity increases more gradually again, generating a second flat region in the titration curve near log (Cu2+ ) = 4.9. At still higher log (Cu2+ ) values (> 4), the capacity of EN molecules to complex Cu2+ ions has been fully utilized, and Cu species that are not complexed to EN (CuOH+ and Cu2+ ) become the dominant Cu containing species. Under these conditions, the presence of the EN becomes irrelevant. The titration curve in this region would be flat regardless of the presence or absence of EN for the same reason that acid titration curves tend to be flat at very high H+ activity (low pH). As in the case of pH buffers, a plot of the metal buffering intensity (the absolute value of the slope of the titration curve) often provides a more dramatic illustration of the sensitivity of buffering to the free ion activity. Such a plot is shown for the Cu2+ EN system in Figure 10.13 and highlights both the buffering of the EN 555
near log (Cu2+ ) of 10.1 and the fact that any buffering provided by the EN near log (Cu2+ ) of 4.9 blends with the non-EN buffering at higher Cu2+ activities. Thus, the Cu activity is well buffered under conditions where the amount of Cu bound to EN species is undergoing a significant transition, but the EN provides minimal buffering under conditions where its speciation is relatively stable. The similarities to acid/base buffering are clear.
Figure 10.13 Buffer intensity provided by EN over a range of Cu2+ activities.
As this example illustrates, strong complexing agents can buffer the activity of free metal ions at extremely low values–the stronger the complexing agent, the lower the Men+ activity that it buffers–making the ligands powerful metabolic tools. That is, by producing an appropriate complexing agent, organisms can retain a pool of an essential metal ion, and yet maintain the activity of the free metal ion at a very low and stable level. If, for some reason, the organism acquires additional metal, the complexant binds it and prevents the free metal ion activity from increasing to potentially toxic levels. By the same token, if the environment becomes depleted in metals, the complexes dissociate to release free metal ions and reduce the danger that the organism will suffer from a metal ion deficiency. As is the case for acid/base buffers, the buffering intensity provided by a complexing agent is directly proportional to its concentration.
10.10
PREDOMINANCE AREA DIAGRAMS
In this and previous chapters, we have displayed information about the speciation of a system primarily on log C-pH diagrams. However, any such diagram can only reflect a single set of conditions for all variables other than pH. For instance, the log C-pH diagrams for Cd speciation shown in Figures 10.3 and 10.7 are for Cl– activities of 0 and 0.5, respectively. Similarly, Figure 10.6 displays the speciation of Cd as a 556
function of Cl– activity, but it is applicable only for pH values at which the activity of free Cd2+ is much greater than that of any Cd OH complex. If we wished to display speciation graphically as a function of two independent variables, we could do so by using a three-dimensional plot with, for instance, pH and log (Cl ) varying along the x and y axes, respectively, and log C plotted on the z (vertical) axis. Any vertical plane slicing through such a plot at a fixed value of Cl– activity would be a conventional log C-pH diagram for that Cl– activity, and any vertical plane at a fixed pH would be a conventional log C–log (Cl ) diagram for that pH (Figure 10.14).
Figure 10.14 Two-dimensional slices through a three-dimensional log C-pH–log{L} diagram.
While such three-dimensional diagrams are conceptually very informative, as a practical matter they are difficult to prepare, visualize, and use. On the other hand, drawing a separate diagram for each value of pH and/or Cl– concentration of interest also has obvious drawbacks. As an alternative, plots can be prepared that display a horizontal slice through such a three-dimensional diagram, at a fixed value of log C 557
equal to log T OT Cd. Such a slice, represented by the top surface of Figure 10.14, has axes representing values of log (Cl ) and pH. In contrast to the diagrams we have used previously, the body of such a plot does not indicate the concentrations of various species in the system. Rather, at each point on the plot (i.e., at each point in log (Cl )-pH space), the diagram identifies only the species present at a concentration approximately equal to T OT Cd; i.e., it identifies the dominant Cd-containing species in the system. Such diagrams are called predominance area diagrams. Although predominance area diagrams lack detailed information about the concentrations of nondominant species in the system, they can display information over a large range of activities of two different ligands, identifying the species accounting for most of the mass of the metal (or other constituent of interest) anywhere in that two-dimensional space. In this section, we develop a predominance area diagram for a model metal-ligand system. As an introduction, however, it is useful to consider the analogous, but much simpler, one-dimensional predominance diagram that characterizes speciation of an acid/base group as a function of pH. For instance, to characterize dominance in the carbonate acid/base group, we need consider only three species. The three pairwise comparisons among these species are summarized in the three right-hand columns of Table 10.5. Each comparison divides the pH scale into two portions (summarized in the bottom two rows of the table) that differ with respect to which of the species being compared is dominant over the other. The two regions are separated by a single point at which the activities of the two species are equal. Note that the transition point at which dominance switches between H2 CO3 and CO2– 3 (pH 8.34) can be determined by writing the equilibrium constant for the reaction between these species (K02 = Ka1 Ka2 ), even though neither of these species is dominant at that pH. Consideration of all three pairwise comparisons indicates the pH region where each species is dominant. For instance, the activity of HCO–3 is greater than that of H2 CO3 at all pH’s greater than 6.35, and it is greater than that of CO2– 3 at all pH’s less than 10.33, so it is dominant over all (i.e., both) other possible species if and only if the pH is between these two values. Similar considerations for the other two species allow us to prepare what might be called a predominance line diagram for the system, as shown in Figure 10.15.
Figure 10.15 A ‘predominance line diagram’ showing the pH ranges for dominance of various carbonate species.
558
559
(H2 CO3 ) > (CO23 ); at pH > 8.34, (H2 CO3 ) < (CO23 )
(HCO3 ) > (CO23 ); at pH > 10.33, (HCO3 ) < (CO23 )
(H2 CO3 ) > (HCO3 ) at pH > 6.35, (H2 CO3 ) < (HCO3 )
pKa1 + pKa2 = 8.34 2 at pH < 8.34,
pH =
at pH < 10.33,
pH = pKa2 = 10.33
at pH < 6.35,
{CO23 }{H+ }2 {H2 CO3 }
Comparison of species’ activities at various pH values
Ka1 Ka2 =
pH = pKa1 = 6.35
{CO23 }{H+ } {HCO3 }
H2 CO3 , CO2– 3
Conditions when species have equal activities
Ka2 =
HCO–3 , CO2– 3
Ka1 =
{HCO3 }{H+ } {H2 CO3 }
H2 CO3 , HCO–3
Equilibrium relationship
Species being compared
Table 10.5 Predominance relationships among carbonate species.
To extend the above analysis to systems in which two variables are considered (e.g., pH and chloride activity), we note that the conditions under which predominance switches between any two species can be determined by evaluating the equilibrium expression relating that pair of species. The region where a particular species is dominant over all the other species can then be identified by considering, simultaneously, the comparisons between that species and all others. For example, the reactions characterizing a system containing Pb(II) at some fixed but unspecified total concentration, and two complexing ligands (Cl– and OH– ) whose activities might vary widely are Pb2+ + H2 O )* Pb(OH)+ + H+
⇤
KOH,1 = 10
7.60
PbOH+ + H2 O )* Pb(OH)2 + H+
⇤
KOH,2 = 10
9.49
Pb(OH)2 + H2 O )* Pb(OH)3 + H+
⇤
KOH,3 = 10
11.00
Pb2+ + Cl )* PbCl+ PbCl+ + Cl )* PbCl2
KCl,1 = 101.56
PbCl2 + Cl )* PbCl3 PbCl3 + Cl )* PbCl24
KCl,3 = 10
0.10
KCl,4 = 10
0.42
KCl,2 = 100.34
The reactions and K values relating Pb OH and Pb Cl complexes to one another can be derived by various combinations of the preceding reactions. For example, the reaction for conversion of PbOH+ to PbCl2 can be generated by adding the reverse of the first reaction to the fourth and fifth reactions, with K = KCl,1 KCl,2 /⇤ KOH,1 . The equilibrium constant expression for any of these reactions can be manipulated to yield the equation of a line in pH-log (Cl ) space along which the two Pb– containing species are present at equal activity, with one species dominant on one side of the line and the other dominant on the other side. Such a reaction and equilibrium constant expression can be written generically for conversion of a Pb Cl complex to a Pb OH complex as follows: PbCl2y y + xH2 O )* Pb(OH)2x x + xH+ + yCl ⇣ ⌘ Pb(OH)2x x (H+ )x (Cl )y K= (PbCl2y y )
(10.38) (10.39)
If the species being compared are present at equal activity, then (PbCl2y (Pb(OH)2x x ), so K = (H+ )x (Cl )y log K = log (H+ )x + log (Cl )y = x pH + y log(Cl ) log (Cl ) = (log K + x pH)/y
y
)=
(10.40)
The results obtained by applying similar analyses to all possible pairs of Pb-containing species are collected in Table 10.6. 560
561
⌘ K(2) ⌘ K(3)
⇤K OH,2 ⇤K OH,3
KCl,1 ⌘ K(4)
K(1) /K(4) K(8) K(2) K(9) K(3) K(1) /(K(4) K(5) ) K(11) K(2) K(12) K(3)
PbOH+ /Pb(OH)2 Pb(OH)2 /Pb(OH)–3 Pb2+ /PbCl+ PbCl+ /PbCl2 PbCl2 /PbCl–3 PbCl–3 /PbCl2– 4 PbCl+ /PbOH+ PbCl+ /Pb(OH)2 PbCl+ /Pb(OH)–3 PbCl2 /PbOH+ PbCl2 /Pb(OH)2 PbCl2 /Pb(OH)–3
2
3
4
5
6
7
8
9
10
11
12
13
KCl,4 ⌘ K(7)
KCl,3 ⌘ K(6)
KCl,2 ⌘ K(5)
⌘ K(1)
⇤K OH,1
Pb2+ /PbOH+
1
(a)
Expression for K
Species compared (Reactant/Product)
Line #
29.99
18.99
9.50
29.65
18.65
9.16
0.42
0.10
0.34
1.56
11.00
9.49
7.60
Log K
log(Cl ) = ( 29.99 + 3pH)/2
log(Cl ) = ( 18.99 + 2pH)/2
log(Cl ) = ( 9.50 + pH)/2
log(Cl ) = 29.65 + 3pH
log(Cl ) = 18.65 + 2pH
log(Cl ) = 9.16 + pH
log(Cl ) = 0.42
log(Cl ) = 0.10
log(Cl ) = 0.34
log(Cl ) = 1.56
pH 11.00
pH 9.49
pH 7.60
species have equal activities
Equation of line along which
Table 10.6 Equilibrium relationships used to develop a predominance area diagram for the Pb2+ -Cl– -OH– system
562
K(15) K(3) K(1) /(K(4) K(5) K(6) K(7) ) K(17) K(2) K(18) K(3)
PbCl–3 /Pb(OH)–3 + PbCl2– 4 /PbOH
PbCl2– 4 /Pb(OH)2 – PbCl2– 4 /Pb(OH)3
16
17
18
19 K( j) refers to the equilibrium constant from line # j.
K(14) K(2)
PbCl–3 /Pb(OH)2
15
(a)
K(1) /(K(4) K(5) K(6) )
PbCl–3 /PbOH+
14
Line #
29.47
18.47
8.98
29.89
18.89
9.40
log(Cl ) = ( 29.47 + 3pH)/4
log(Cl ) = ( 18.47 + 2pH)/4
log(Cl ) = ( 8.98 + pH)/4
log(Cl ) = ( 29.89 + 3pH)/3
log(Cl ) = ( 18.89 + 2pH)/3
log(Cl ) = ( 9.40 + pH)/3
Table 10.6 – continued from previous page Expression Species compared Log K Equation of line along which (Reactant/Product) for K (a) species have equal activities
Table 10.6 contains all the information we need to prepare the predominance area diagram. We begin by preparing a plot with pH as the abscissa and log (Cl ) as the ordinate, on which we draw lines along which two Pb OH or two Pb Cl species are present at equal activity (Figure 10.16). On such a diagram, the lines relating any two Pb OH species are independent of the Cl– activity (because Cl– does not appear in the reaction or the equilibrium relationship) and therefore vertical. Similarly, those relating any two Pb Cl species are horizontal. On the lower-pH side of each vertical line, the more acidic of the two species is dominant over the more basic species, and on the higher-pH side, the opposite is true. Correspondingly, below each horizontal line, the less-chlorinated species is dominant over the more-chlorinated species, and above each such line, the more-chlorinated species is dominant.
Figure 10.16 Lines along which either two Pb OH or two Pb Cl species are present at equal activity. On one side of each line, one of the species is present at a higher activity than the other, and on the other side of the line, the other species is present at the higher activity.
The seven lines divide the graph into 20 rectangular zones. In the zone at the lower left (zone A, corresponding to the lowest pH values and lowest chloride activities), Pb2+ is present at a higher activity than any of the other species under consideration and is therefore the dominant Pb-containing species in solution. In the zones to the right of that one (B, C, and D), the activity of Pb2+ is still higher than that of any of the Pb Cl complexes (because these zones are below the Pb2+ /PbCl+ line), but one or more of the Pb OH complexes has a higher activity than Pb2+ . For instance, in zone C, Pb(OH)2 has the highest activity of any Pb OH complex. Since that complex has a higher activity than Pb2+ , and Pb2+ has a higher activity than any of the 563
Pb Cl complexes, Pb(OH)2 also has a higher activity than any of those complexes. As a result, it must be the dominant Pb-containing species in zone C. A similar analysis allows us to identify the dominant species in zones B and D, and also in the zones directly above zone A (where the dominant species are Pb Cl complexes). In the zones that do not lie along the perimeter of the plot, it is not immediately clear whether the dominant species is a Pb OH or Pb Cl complex, but the same type of reasoning allows us to limit the potential dominant species to one of two possibilities. When these inferences about the dominant species in each zone are added to the diagram, we obtain Figure 10.17.
Figure 10.17 Figure 10.16, modified to identify the Pb-containing species that are plausibly dominant in each rectangular region.
The preceding logic fully establishes the region where Pb2+ is the dominant Pbcontaining species (zone A). It also establishes, for example, that PbOH+ is dominant in zone B, and leaves open the possibility that this species might be dominant in parts or all of the zones directly above zone B. Similar statements apply to all the other Pb OH and Pb Cl complexes. To assess which species is dominant in regions where this issue remains uncertain, we compare the activities of the two plausibly dominant species in each zone, using the information in Table 10.6. For example, Figure 10.17 includes a vertical line along which (Pb2+ ) = (PbOH+ ) and a horizontal line along which (Pb2+ ) = (PbCl+ ). Therefore, it must be the case that (PbOH+ ) = (PbCl+ ) at the intersection of these two lines (indicated by a circle in Figure 10.17). Furthermore, according to Equation (10.40), the slope of the line defining the boundary between PbOH+ and PbCl+ on the diagram must be 1 [since for these species, x = y = 1 in Reaction (10.38) and Equation (10.40)]. Since we 564
have now identified the slope of the PbOH+ /PbCl+ line and one point that the line passes through, we can draw the line directly. PbOH+ is present at a higher activity than PbCl+ everywhere below and to the right of this line, and the opposite is true everywhere above and to the left. However, for our purposes, the line is relevant only in zone F, so we extend it only until it hits one of the boundaries of this zone. The boundary that it hits first is the upper one, where PbCl2 becomes dominant over PbCl+ . Once this line is drawn, we know which species is dominant at any point in zone F. Knowing that PbCl+ is dominant both in zone E and the left part of zone F, we can eliminate the line segment that previously separated those two areas, establishing a continuous region where PbCl+ is the dominant species. We can do the same with the line segment at the bottom of zone F to establish a region where PbOH+ is dominant, yielding Figure 10.18.
Figure 10.18 Figure 10.17, modified to include consideration of the relative activities of PbOH+ and PbCl+ in zone F.
We can then apply the same approach to zone J. Specifically, the line we just drew identified points where the activities of PbOH+ and PbCl+ were equal, and we terminated that line at the upper boundary of zone F, where the activities of PbCl+ and PbCl2 are equal. Therefore, at that intersection point, the activities of PbOH+ and PbCl2 must be equal. We also know that the slope of the PbOH+ /PbCl2 line is 0.5, based on the equation for that line given in Table 10.6. We can draw that line on the predominance area diagram, extending it until it hits one of the boundaries of zone J, and thereby establish the portions of zone J in which either PbOH+ or PbCl2 is dominant. By repeating this process, we can identify which of the two possible species is dominant in each portion of each of the other zones, ultimately obtaining the complete predominance area diagram shown in Figure 10.19. 565
Figure 10.19 Final predominance area diagram for Pb(II) considering possible formation of Pb OH and Pb Cl complexes.
The preceding discussion identifies a simple algorithm for drawing predominance area diagrams. It also leads to the useful observation that whenever two lines intersect on a predominance area diagram (say, one for the species A/species B comparison and the other for the species B/species C comparison), a third line must pass through the same intersection point (the line for the species A/species C comparison). Thus, once all the relevant comparisons are made, three line segments will emanate from any such intersection point. These line segments divide the space immediately surrounding the intersection point into three regions, in each of which a different species (A, B, or C) is dominant over the other two. Equation (10.40) indicates that the three line segments must all have different slopes. Review of Figure 10.19 confirms that these rules apply to all the intersection points shown.
10.11
SUMMARY
This concludes our exploration of metal chemistry in systems that contain only dissolved species. The key points of the discussion can be summarized as follows. The cationic charge on dissolved metal ions causes these ions to be attracted to the unshared electrons in the orbitals of the oxygen atom of water molecules. This attraction constitutes an Me H2 O bond that can range from very weak to very strong. The formation of the bond causes the electrons that are shared between the oxygen and the hydrogens in the water molecule to shift toward the oxygen, weakening the O H bonds, so that the acidity of these water molecules is enhanced. In some cases, the Me O bond is so strong that both hydrogen ions are released from the water 566
molecule. Although hydrolysis reactions occurring adjacent to a metal ion are just that – the splitting of H2 O – they are often represented as the bonding of a free metal ion (Men+ ) to a free hydroxide ion, obscuring the fact that OH– ions reside in the hydration sphere with other unhydrolyzed water molecules. Other ligands can replace water molecules or hydroxide ions in the coordination sphere of a metal ion. Because the coordination sphere is always fully occupied, these reactions always involve ligand exchange. Again, however, the waters of hydration are not usually shown explicitly when these reactions are written, so the reactions are typically represented as the binding of the ligand to the free metal. Almost all ligands that are environmentally significant bind to metal ions through atoms that have unshared electrons – primarily, oxygen, nitrogen, and sulfur. Ligands that bind metals especially strongly and do so via several binding groups are called chelating agents or chelators. Chelators play an important role in the bioenvironmental chemistry of metals, since evidence suggests that chelated metals are often unavailable to biota. Because the reactions relating metals and ligands to complexes are direct analogs of those relating protons and bases to conjugate acids, metal complexation reactions can be modeled using the same algorithms and the same software (like Visual Minteq). The analogy between metal/ligand and acid/base systems can be extended to the idea of metal ion buffering as well. Often, we are interested in metal speciation as a function of both pH and the concentration of another ligand in a system. The speciation could be represented on a three-dimensional log C-pH-log L diagram, but since such diagrams are difficult to prepare, we sometimes choose to display only a horizontal slice through the diagram, indicating the species that are dominant in different regions of pH-log L space. Such diagrams are referred to as predominance area diagrams. Next, in Chapter 11, we turn our attention to the characteristic ability of metals to form solids in addition to soluble complexes.
10.12
PROBLEMS
Please re-read the note preceding the Chapter 8 Problems on p.457. The same guidelines apply to the following problems. 1. Calculate the speciation (both concentration and activity) of zinc in seawater, based on the typical seawater composition given in Table 1.1. Assume that T OT Zn = 5 ⇥ 10 8 M. 2. Prepare a log C-pH diagram for a solution containing 10 4 M T OT Pb if the only Pb complexes that form are with OH– . Compare the diagram with the corresponding diagram for Cd in Figure 10.3 and comment on the relative acidities of Pb2+ and Cd2+ . 567
3. A solution at pH 7.0 contains Cu2+ and NH3 . (a) What is the ratio (CuOH+ )/(Cu2+ )? (b) What is the ratio (NH+ 4 )/(NH3 )? (c) Are there any hydroxo complexes of Cu present at concentrations greater than that of CuOH+ ? (d) Use the data in the Visual Minteq database to infer values of log bi for Cu(NH3 )i complexes with 1 i 4.
(e) Estimate the slopes of the following lines on a log C-log (NH3 ) diagram for a fixed pH of 7.0, under conditions where almost all the copper is present as Cu(NH3 )2+ 4 : (i) (ii) (iii) (iv)
Cu2+ CuOH+ Cu(NH3 )2+ Cu(NH3 )2+ 3
4. Use the data in Tables 10.2 and 10.3 to determine equilibrium constants for the following reactions. (a) Cr3+ + 2 OH– )* Cr(OH)+2 (b) Cr(OH)+2 + OH– )* Cr(OH)3 (c) CrOH2+ + H2 O )* Cr(OH)+2 + H+ (d) Pb2+ + 2 Cl– )* PbCl2 (e) PbCl2 + Cl– )* PbCl–3 5. Determine the speciation of Cu, Zn, and a ligand L in a solution at pH 6.0, if the solution contains 10 6 M each of T OT Zn, T OT Cu, and T OT L, and if L participates in the following reactions:
6.
Cu2+ + HL )* CuL+ + H+
⇤K 1
Zn2+ + L– )* ZnL+
K1 = 107.0
HL )* H+ + L–
Ka = 10
= 106.0 3.0
(a) What is the speciation of mercury in an ideal solution of 10 4 M HgCl2 ? What is the concentration of free Cl– in the solution? (If you use Visual Minteq to answer the question, note that the default input component for mercuric species is Hg(OH)o2 .) (b) Repeat part (a) for a solution at pH 7.0. (c) Repeat part (a) for a solution with a fixed Cl– activity of 2 ⇥ 10 4 . 568
7. Determine the speciation of Ca, Cd, and the chelating agent EDTA in a pH 7.9 solution that contains 10 6 M T OT Cd, 10 2 M T OT Ca and 10 4 M T OT EDTA. 8. Prepare a plot of log Cd2+ vs PH2 S for solutions at pH 8.0 and with T OT Cd = 10 6 M, for a range of PH2 S from 10 6 to 10 3.0 atm. 9. Prepare a log C-pH diagram for an aqueous solution containing 10 6 M T OT Zn and 10 4 M T OT NH3 , over the pH range from 4 to 10. What is the equilibrium partial pressure of NH3 (g) in the system at pH 8.0? 10. A solution containing 0.1 M each of HgSO4 , NaOH, and NaCl is prepared. The ionic strength of the solution is approximately 0.25 M. (a) Write mass balances for total Hg, SO4 , and Cl. Consider all the complexes that are included in the Visual Minteq database. (b) Determine the effective equilibrium constant, K1,eff , for HgSO4 in the solution (i.e., the equilibrium constant applicable for the given ionic strength, if the terms for the solutes are expressed as concentrations rather than activities). (c) Write the T OT H equation for the solution, using Visual Minteq defaults as components. (d) Use Visual Minteq to determine the solution pH and composition at equilibrium. What is the dominant Hg species? What fraction of T OT Hg is in complexes with Cl– ? What fraction of T OT Cl is complexed with Hg2+ ? 11. In general, the fractional distribution of metal ions among their acid/base forms (i.e., Men+ , MeOHn 1 , Me(OH)n2 2 , etc.) depends only on the solution pH, and not on T OT Me. However, this generalization does not apply if polymeric species form. Explore the consequences of this fact by plotting log C-pH diagrams for solutions containing total dissolved ferric iron [T OT Fe(III)] concentrations of 10 5 , 10 3 , or 10 1 M, considering both the monomeric and polymeric species that are included in the Visual Minteq database. At what pH does the concentrations of Fe(OH)o3 equal that of Fe2 (OH)4+ 2 in each solution? 12.
(a) Use Visual Minteq to prepare a log C-pH diagram over the pH range from 3 to 12 showing Cu species in a solution of 3 ⇥ 10 4 M CuCl2 , 10 3 M NH4 Cl, and 5 ⇥ 10 4 M Ca(OH)2 , if the ionic strength of the solution is 0.01 M. Show all Cu-containing species that are present at concentrations greater than 10 6 M. (b) In the slightly alkaline pH range, the diagram indicates that the sum of the concentrations of all Cu-containing species is less than 3 ⇥ 10 4 M. Why is this the case? 569
(c) What is the pH of the solution in part (a)? (d) What partial pressure of ammonia would be in equilibrium with the solution in part (a)? 13. Prepare a log C-pH diagram for the solution in Problem 12 if it equilibrates with a gas phase containing ammonia at a partial pressure of 10 6.5 atm. Consider pH’s between 6 and 12, and continue to apply the assumption that the ionic strength is 0.01 M. What are the two Cu species present at the highest concentrations at pH 6, 9, and 12? 14. Thiosulfate ion, S2 O2– 3 , is a complexing agent used extensively in industry. Use Visual Minteq to prepare a log C-log (S2 O23 ) diagram showing silver speciation in systems containing 10 5 M T OT Ag at pH 9.0 and pH 13.0, over the range 12 < log(S2 O23 ) < 3. In the simulation, analyze the system composition at equal increments of log(S2 O23 ) (not T OT S2 O3 ), and show all species that are present at concentrations >10 12 M. (a) What are the values of (S2 O23 ) at the points where (Ag+ ) = (AgS2 O3 ) and where (Ag+ ) = Ag(S2 O3 )32 , for the system at pH 9? What are the corresponding values at pH 13? Explain this result. (b) The log C-log (S2 O23 ) diagram should indicate that AgS2 O–3 is dominant over a range of values of (S2 O23 ) at each pH. Explain briefly why AgS2 O–3 is dominant over a smaller range of log (S2 O23 ) at pH 13 than at pH 9. (c) A solution buffered at pH 9 and containing 10 5 M T OT Ag is titrated with Na2 S2 O3 . Plot the concentration of S2 O3 -complexed Ag (i.e., the sum of the concentrations of all Ag(S2 O3 )1–2x complexes) as T OT S2 O3 x 5 increases from 0 to 10 M. What are the concentrations of all species at the end of the titration, i.e., when T OT Ag = T OT S2 O3 = 10 5 M? Based on this result, is thiosulfate a relatively strong or weak complexing agent for silver? 15. The owner of an industry that generates wastewater containing both Ag+ and Hg2+ is considering two options for discharge of the water after treatment. One option is to discharge the water through an ocean outfall, in which case the initial dilution is predicted to yield concentrations of 0.04 mg/L T OT Hg and 2.5 mg/L T OT Ag. The alternative is to discharge to a large river, in which the dilution would be faster and more extensive, so the total concentrations of the metals near the outfall would be lower by a factor of 10. Both Ag+ and Hg2+ are more toxic as free ions than as complexes. The major complexing ligands in the two potential receiving waters are Cl– (19,400 mg/L 570
in the ocean, 35 mg/L in the river) and SO2– 4 (2710 mg/L in the ocean, 12 mg/L in the river). Both solutions are at pH 8.0, and the ionic strengths are 0.62 M in the ocean and 0.01 M in the river. Which disposal option would you recommend to minimize toxicity near the outfall? 16. Groundwater in a formation rich in gypsum has the following composition: [Na+ ] = 35 mg/L [SO24
] = 110 mg/L
[Cl ] = 48 mg/L 2+
[Ca ] = 116 mg/L
DIC= 65 mg/L T OT Zn = 65 µg/L
pH= 7.7 (a) Calculate the alkalinity (meq/L) of the solution. (b) What are the three most abundant species of Zn and Ca in the solution, and what fraction of T OTi do these species account for? (c) The water is pumped to the surface and placed in an an open tank, where it is dosed with 5 ⇥ 10 4 M HCl and aerated. Determine the pH, DIC (mg/L) and ALK (meq/L) of the water after it equilibrates with air. 17. An aqueous solution at pH 7.3 contains 10 5 M T OT CO3 and natural organic matter (NOM) at a concentration of 10 mg/L as DOC. The NOM is modeled as a mixture of two discrete ligands: L–I , with a charge density of 7.7 meq per gram of DOC, and L–II , with a charge density of 3.9 meq per gram of DOC. The two ligands are monoprotic acids, with pKa ’s of 4.0 and 8.9, respectively. (a) How much alkalinity does the NOM contribute to the solution? (b) The solution also contains 10 9 M T OT Pu(VI), which, in the uncomplexed form, exists as the divalent cation PuO2+ 2 . This ion forms complexes with OH– and CO2– , several of which are included in the Visual 3 Minteq database. Determine the speciation of the plutonium, ignoring the possible formation of complexes with the NOM ligands. (c) Now assume that PuO2+ 2 also forms complexes with the NOM ligands, as characterized by the following reactions: + – PuO2+ 2 + LI )* PuO2 LI
K = 106.8
+ – PuO2+ 2 + LII )* PuO2 LII
K = 107.4
Determine the speciation of plutonium considering these reactions in addition to the ones considered in part (b). Note that, because the pH is known and T OT CO3 is much greater than T OT Pu(VI), the concentration ratios of inorganic Pu complexes to free PuO2+ 2 remain the same as in part (b), but the absolute values of the concentrations change when complexation by the organics is taken into account. Note also that the total concentrations of the organic ligands are much larger than T OT Pu(VI). 571
18.
(a) Prepare the following plots showing the speciation of mercury in solutions containing 10 6 M T OT Hg(II). (i) A log C-pH diagram for a solution containing 2 ⇥ 10 4 M T OT Cl, a typical concentration in low-mineral fresh waters. Consider the pH range from 2 to 12. (ii) Same as part (a), but for 5 ⇥ 10 3 M T OT Cl, a typical concentration in brackish water. (iii) Same as part (a), but for 0.5 M T OT Cl, a typical seawater concentration. (iv) A log C log {Cl } diagram for a solution at pH 3.0, for log{Cl } values from 5 to 0. (v) Same as part (iv), but for pH 7.0. (vi) Same as part (iv), but for pH 10.0. (b) What is the lowest pH at which a complex containing OH– becomes the dominant Hg species in parts (a)(i-iii)? Why is this trend observed? (c) What is the lowest log(Cl ) at which a complex containing Cl– becomes the dominant Hg species in parts (a)(iv-vi)? Why is this trend observed?
19. A log C-pH diagram is provided below for an ideal solution containing 10 4 M total ferrous iron [T OT Fe(II)], 2 ⇥ 10 5 M total sulfide [T OT S(II)], and 3 ⇥ 10 4 M T OT CO3 . (a) If the solution is at pH 9.0, what is the total contribution (mol/L) of the six Fe-containing species to T OT H, if FeOH+ , HS– , and CO2– 3 are chosen as components? (b) What is the total contribution of the six Fe-containing species to the alkalinity of the solution, in equiv/L? (c) Using only the information provided in the plot, determine ⇤ b3 for formation of Fe OH complexes.
572
20. Consider the predominance area diagram for the Fe(II)/Cl– /H2 O system in Figure 10.19, and the equations used to develop it. (a) What is the predominant form of ferrous iron in a system at pH 9.2, if Cl = 0.1? (b) If the chloride activity in part (a) is increased to 0.6 while the pH is held at 9.2, what will the predominant form of ferrous iron be? (c) If the pH of the system in part (b) is increased to 10.0, what will the predominant form of ferrous iron be? (d) Repeat part (c), but for a final pH of 13. 21. In a study investigating the capacity of sewage effluent to form complexes with cupric ion (Cu2+ ), increasing amounts of CuSO4 were added to a sample of the effluent, and the activity of Cu2+ was analyzed using an ion-selective electrode. The following graph shows some of the results. Assume that the ligands in the effluent can be modeled as a single species, L2– , present at a concentration T OT L, and that only a single Cu L complex forms. Use the results shown in the graph to estimate T OT L and the stability constant K1 for formation of CuL(aq). The pH was low enough that Cu OH complexes could be ignored.
573
574
11
CHEMISTRY OF METALS IN AQUEOUS SYSTEMS
2. PRECIPITATION AND DISSOLUTION REACTIONS
575
Contents
11.1 11.2 11.3 11.4 11.5 11.6
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . FORMATION OF SOLIDS CONTAINING METAL IONS . . . . . THE CHEMICAL ACTIVITY OF PURE SOLIDS . . . . . . . . . THE SOLUBILITY PRODUCT . . . . . . . . . . . . . . . . . . . PRECIPITATION OF SOLIDS OTHER THAN HYDROXIDES . . FORMATION OF NONHYDROXO COMPLEXES IN SYSTEMS WITH SOLIDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 ADDING NEW SOLIDS TO THE VISUAL MINTEQ DATABASE 11.8 DETERMINING WHETHER A SOLID WILL PRECIPITATE UNDER GIVEN CONDITIONS . . . . . . . . . . . . . . . . . . . 11.9 THE GIBBS PHASE RULE . . . . . . . . . . . . . . . . . . . . . 11.10PREDOMINANCE AREA DIAGRAMS CONSIDERING POSSIBLE PRECIPITATION OF SOLIDS . . . . . . . . . . . . . 11.11SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.12PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.1
576 577 578 579 593 599 604 608 629 641 647 649
INTRODUCTION
In this chapter, we continue our exploration of the chemistry of metals in aqueous systems, now focusing on the precipitation of metal-containing solids. Just as we used analogies with acid/base chemistry to understand and model the formation of metal complexes, analogies between solid/solution equilibrium and gas/solution
576
equilibrium will help us understand the reactions in these systems. The major differences between these two phenomena is the added complexity in the metal precipitation systems due to the large number of reactions in which metals can participate.
11.2
FORMATION OF SOLIDS CONTAINING METAL IONS
If the concentration of metal and/or hydroxyl ions in a solution is steadily increased, substantial concentrations of complexes containing multiple metal ions begin to appear. For instance, in the case of Fe(III), well-defined molecules containing two Fe3+ ions (dimers) may form as shown in Reaction (11.1a), which is commonly written in the shorthand form shown in Reaction (11.1b). 2 Fe(H2 O)5 OH2+ )* (H2 O)5 Fe O Fe(H2 O)4+ 5 + H2 O 2 FeOH
2+
)*
Fe2 (OH)4+ 2
(11.1a) (11.1b)
Trimers, tetramers, and higher-order complexes may form as well. In each step, an additional Fe-containing subunit (a monomer) is added. For some metals, the existence of fairly large complexes (e.g., Al8 (OH)4+ 20 ) has been well established. Molecules made of several monomers are called polymers. In complexes containing a single metal ion, the metal ion can be thought of as the core or nucleus of the complex. Correspondingly, complexes containing more than one metal ion are sometimes called polynuclear complexes. If the polymerization process continues, eventually a three-dimensional molecular network builds up, and many of the ions on the interior lose contact with the bulk aqueous solution. In other words, the polymer becomes a separate phase from the water. When the new phase first forms, it may not be visible or easily separated from the bulk solution, complicating any effort to identify the exact conditions when precipitation begins. However, in many cases, the system composition changes fairly abruptly from a mixture of small, soluble species to a collection of distinct particles, so the initiation of precipitation is relatively unambiguous. As is the case for soluble complexes, metals can form solids with a variety of ligands. In fact, the ligands that can form soluble complexes are often the same ones that can combine with the metals to form solids. For instance, many metals can form both soluble complexes and solids with hydroxide, carbonate, phosphate, and sulfide ions. We now turn to the task of identifying and solving the equations (mass balances, equilibrium equations, etc.) that characterize systems in which solids might precipitate. As always, analysis of the equilibrium system composition involves finding a set of concentrations that meets the constraints imposed by these equations. We begin by noting that any equations that describe relationships among soluble species apply regardless of whether or not a solid is present. That is, in any solution 577
containing a metal ion and a ligand (Me, L), if soluble complexes between Me and L can form, they will do so, and the activities of the complexes will be related to the activities of free (uncomplexed) metal and ligand species by the relevant stability constants. The complexes might be present at very low concentrations in some cases, but they will always be present at some nonzero concentration. By contrast, the fact that a solid could form between Me and L (as indicated by the availability of an equilibrium constant for the precipitation reaction) does not ensure that such a solid will form and be present at equilibrium. Solids sometimes precipitate in systems that initially contain only dissolved species, while in other systems solids that are present initially dissolve completely before equilibrium is attained. Obviously, if a certain solid is not present, the equilibrium expressions involving it are irrelevant (just as we do not concern ourselves with gas/liquid equilibrium if no gas phase is present). The uncertainty about whether a particular solid will be present at equilibrium distinguishes these systems from those that we have considered previously. In light of this uncertainty, the only way to conduct the analysis is to make an assumption about whether the solid will be present once the system equilibrates and to solve the system of equations based on this assumption. The computed equilibrium condition can then be checked for consistency with the assumption; if the assumption turns out to be wrong, the alternative assumption is made, and the process must be repeated. In the following section, we consider how the set of equations characterizing the system differs depending on which assumption is made. First, though, we review material from earlier chapters on how the chemical activity of the solid is defined, and we define some terms applicable to systems where precipitation must be considered.
11.3
THE CHEMICAL ACTIVITY OF PURE SOLIDS
Recall from Chapter 2 that the activity of any species is defined as the product of two terms, one expressing the concentration of the constituent relative to its concentration in the standard state, and the other characterizing the chemical environment relative to the environment in the reference state. Recall also that concentrations of constituents in solids are conventionally expressed in terms of mole fractions, and that the standard state concentration is a mole fraction of 1.0, i.e., pure solid. The reference state environment for molecules in a solid phase is also conventionally defined based on the pure solid. Therefore, if we are actually dealing with a pure solid of, say, Cd(OH)2 (s), both the concentration (i.e., mole fraction) of Cd(OH)2 molecules in the solid and the environment of those molecules correspond to standard state conditions, so the activity of the Cd(OH)2 (s) is 1.0. In this text, we will consider only solids that are in their standard state. However, it is important to rec578
ognize that if a solid phase contains a mixture of solid species [e.g., Cd(OH)2 (s) and CdCO3 (s)], the activity of each constituent in the phase is likely to be different from 1.0, both because the dilution of each constituent by the other causes their concentrations to be lower than in the standard state, and because, in such a solid, the molecular environment near each molecule is different from that in the reference state. Another point that was emphasized in Chapter 2 is the importance of distinguishing between the activity of a pure solid (1.0, dimensionless) that is dispersed in a solution and the concentration of that solid in the dispersion (mass of solid per unit volume of the solution or of the whole suspension in, say, milligrams per liter). A justification for the constancy of a solid’s activity, independent of its concentration in a suspension, was provided in Chapter 3.
11.4
THE SOLUBILITY PRODUCT
The reaction for dissolution of a metal hydroxide solid and the corresponding equilibrium constant expression can be written as follows: Me(OH)n (s) )* Men+ + n OH Ks0 =
(Men+ ) (OH )n (Me(OH)n (s))
= (Men+ )(OH )n eq
(11.2)
eq
(11.3)
where the second equality in Equation (11.3) applies if the precipitate is a pure solid. Conventionally, equilibrium constants involving solids are written with the solid as a reactant, and the symbol Ksi is used to represent the equilibrium constant for such a reaction when a dissolved complex containing i ligands is the product. Additional species such as uncomplexed ligands might appear on either side of the reaction, as required by stoichiometry. For example, Ksi values for equilibrium with Cd(OH)2 (s) are1 Cd(OH)2 (s) )* Cd2+ + 2 OH–
log Ks0 = 14.36
(11.4)
Cd(OH)2 (s) )* CdOH+ + OH–
log Ks1 = 10.45
(11.5)
Cd(OH)2 (s) )* Cd(OH)2
log Ks2 = 6.65
(11.6)
Cd(OH)2 (s) + OH– )* Cd(OH)–3
log Ks3 = 5.66
(11.7)
Cd(OH)2 (s) + 2 OH– )* Cd(OH)–4
log Ks4 = 5.64
(11.8)
1 Note
that although Cd(OH)2 (s) has the same chemical formula as the dissolved dihydroxo complex Cd(OH)2 , they are different species. The former is a solid whose activity is 1.0 regardless of its concentration in a solution (assuming it is a pure solid), while the latter is a dissolved species whose activity can vary widely, depending on its concentration and the ionic composition of the solution.
579
An alternate designation for the equilibrium constant Ks0 is Ksp , where ‘sp’ stands for solubility product. Recall that the activity quotient Q for a reaction is defined as the same ratio of activities as appears in the equilibrium constant, K, but that Q is used to represent that ratio under any conditions, whereas K is the value of the ratio only when the reaction is at equilibrium. For reactions involving solids, Q is always computed assuming that the pure solid is present, so that asolid = 1.0. As a result, for a solid Me(OH)n (s), Qs0 equals (Men+ )(OH )n under any (equilibrium or nonequilibrium) conditions. Solutions in which Qs0 < Ks0 are said to be undersaturated with respect to the solid; in such solutions, if the solid is present, it will dissolve as the system equilibrates, and if no solid is present, none will form. Correspondingly, solutions in which Qs0 > Ks0 are said to be supersaturated with respect to the solid; in such solutions, the solid will tend to precipitate. Values of Ks0 for a few solids of interest are listed in Table 11.1. Example 11.1 The process of softening water by inducing the precipitation of calcite, CaCO3 (s), was described in Example 9.8. Typically, softening involves addition of lime (Ca(OH)2 ) to increase the solution pH and, in some cases, of sodium carbonate (commonly referred to as soda ash) to increase T OT CO3 . (It might seem paradoxical that lime is added in a process intended to remove Ca from solution, but the precipitation reaction actually removes enough of both the Ca originally present and the added Ca to achieve the treatment goal, and often this approach is cheaper than using a different base to raise the pH.) Historically, the lime and soda ash doses were often chosen based on a target of stoichiometric softening, meaning that the doses were chosen so that T OT Ca = T OT CO3 . Consider a solution that has been subjected to stoichiometric softening and has a final pH of 10.8. If the solution has reached equilibrium with calcite, what is the concentration of dissolved Ca2+ ? If the sum of the Ca2+ in the original solution and that added with the lime is 180 mg/L Ca2+ , how much calcite is generated per liter of water treated? Solution The softened solution is in equilibrium with CaCO3 (s), and the treatment process assures that, in this solution, T OT Ca = T OT CO3 . Ca2+ is a very weak acid (pKa = 12.70), so we can assume that, at pH 10.8, (Ca2+ ) = T OT Ca. Since a2 for H2 CO3 at this pH is 0.74, we can write for the softened water: Ks0 = (Ca2+ )(CO23 ) 10
8.48
= (T OT Cadiss )(0.74 ⇥ T OT CO3,diss ) = 0.74 (T OT CO3,diss )2
T OT Cadiss = T OT CO3,diss = 6.7 ⇥ 10 580
5
581
Ca2+
Al3+
Ag+
Metal
Ca5 (OH)(PO4 )3 (s)
CaSO4 (s)
Gypsum
Hydroxyapatite
CaO(s)
Ca(OH)2 (s)
Portlandite
Lime
CaCO3 (s)
Aragonite
58.33
4.61
4.70
5.30
8.34
8.48
22.50
AlPO4 (s) CaCO3 (s)
34.26
31.20
9.75
Al(OH)3 (s)
Calcite
Gibbsite
Al(OH)3 (am)
AgCl(s)
Cerargyrite
17.59
Ag3 PO4 (s) 53.62
11.09
Ag2 CO3 (s) Ag2 S(s)
7.70
log Ks0
AgOH(s)
Formula
Acanthite
Mineral Name
Cu+
Covellite
Malachite
CuCl(s)
CuS(s)
Cu3 (PO4 )2 · 3 (H2 O)(s)
CuCO3 (s)
Cu2 (OH)2 CO3 (s)
CuO(am)
Cu(OH)2 (s)
Cu2+
CoCO3 (s)
Co(OH)2 (am)
Cd3 (PO4 )2 (s)
CdS(s)
CdCO3 (s)
Cd(OH)2 (s)
Formula
Cr(OH)3 (s) Tenorite
Greenockite
Otavite
Mineral Name
Cr3+
Co2+
Cd2+
Metal
Table 11.1 The Ks0 values of some solids of interest.
6.76
39.62
35.12
11.50
33.47
19.51
18.71
32.65
11.20
14.91
32.60
31.42
12.01
14.36
log Ks0
582
Mn2+
Hg2+
Fe3+
Fe2+
Metal
Fe3 (PO4 )2 · 8 H2 O(s)
Vivianite
Pyrochroite
Cinnabar
Montroydite 39.28
Hg(CN)2 (s) Mn(OH)2 (s)
12.81
56.52
22.52
HgCO3 (s) HgS(s)
25.45
25.33
77.61
HgO(s)
Hg(OH)2 (s)
Fe2 O3 (s)
Maghemite
85.42
40.63
g-FeOOH(s)
Lepidocrocite Fe2 O3 (s)
41.51
a-FeOOH(s)
Goethite
Hematite
38.80
Fe(OH)3 (s)
20.35
37.76
10.59
14.51
log Ks0
Ferrihydrite
FeS(ppt)
FeCO3 (s)
Fe(OH)2 (am)
Formula
Siderite
Mineral Name
Zn2+
Si4+
Pb2+
Wurtzite
Zincite
Silica
Galena
Cerussite
Hydrocerussite
Massicot
ZnS(s)
ZnO(s)
Zn3 (PO4 )2 · 4 H2 O(s)
ZnCO3 · H2 O(s)
Zn(OH)2 (am)
SiO2 (am, ppt)
Pb3 (PO4 )2 (s)
PbS(s)
PbCO3 (s)
Pb3 (CO3 )2 (OH)2 (s)
PbO(s)
Ni3 (PO4 )2 (s)
NiCO3 (s)
Ni(OH)2 (am)
Ni
MnCO3 (am)
Formula
2+
Mineral Name
Mn2+
Metal
Table 11.1 – Continued from previous page
26.02
16.77
35.42
10.26
15.53
2.74
43.53
32.32
13.20
46.76
15.11
31.30
11.20
17.11
10.50
log Ks0
where the subscript ‘diss’ stands for ‘dissolved’. The total concentration of Ca in the system (considering both the solution and the solids that precipitate) is given as 180 mg/L, which equals 4.5 ⇥ 10 3 M. Since the preceding calculations indicate that T OT Ca is only 6.7 ⇥ 10 5 M, ⇠98.5% of the Ca in the system must be converted to CaCO3 (s). The molecular weight of CaCO3 (s) is 100, so the mass of solid generated is 0.985 (4.5 ⇥ 10 3 M)(100,000 mg/mol), or approximately 443 mg/L.
11.4.1
Metal Speciation in Solutions Equilibrated with a Solid
As with other equilibria that involve H+ or OH– ions, a great deal of information about the solubility of Me(OH)n (s) solids can be displayed on log C-pH diagrams. We can develop such diagrams by deriving the equation for the activity of each species as a function of pH, subject to the constraint that all the equations be consistent with the expression for solid/liquid equilibrium. For instance, we can derive the equation for the line representing (Men+ ) on such a diagram by adding n times the reverse of the water dissociation reaction to the solid dissolution reaction, as follows: Me(OH)n (s) )* Men+ + n OH–
Ks0
n H+ + n OH– )* n H2 O
Kw n
Me(OH)n (s) + n H+ )* Men+ + n H2 O
⇤K s0
⇤
Ks0 =
(11.2)
(11.9)
n ⇠⇠ (Men+ )⇠ (H⇠ Ks0 2 O) = ( ( + n Kwn (Me(OH) (((( ( n (s))(H )
log⇤ Ks0 = log (Men+ ) + n pH log (Men+ ) = log⇤ Ks0
n pH
(11.10)
Note that, consistent with the convention for stability constants of soluble metalligand complexes, a solubility constant written for a reaction involving the protonated ligand (H2 O in this case) is designated by an asterisk before the K. Equation (11.10) describes the activity of Men+ in a solution that has equilibrated with Me(OH)2 (s). Although Me(OH)2 (s) does not appear explicitly in the equation, the presence of the solid is implicit, because its activity was assumed to be 1.0 as part of the derivation. In any system that contains Men+ and OH– , Me OH complexes are bound to form in accordance with their respective equilibrium expressions. We have already analyzed these types of equilibria in systems that do not contain solids, and those relationships must still be satisfied when a solid is present. 583
The equilibrium reaction between a metal hydroxide solid and a soluble hydroxo complex of the metal can be written as follows: Me(OH)n (s) + (i
n) H2 O )* Me(OH)ni i + (i
n)H+
Manipulating the equilibrium constant expression for this reaction, we obtain the following generic equations for the activities of Me OH complexes in equilibrium with the solid: ⇤
Ksi =
(Me(OH)ni i )(H+ )i n Me(OH)n (s) (H2 O)i n
log ⇤ Ksi = log (Me(OH)ni i ) + (i = log (Me(OH)ni i )
(i
(11.11)
n) log (H+ )
(11.12a)
n) pH
(11.12b)
log (Me(OH)ni i ) = log ⇤ Ksi + (i
n)pH
(11.13)
Equation (11.13) indicates that, on a log C-pH diagram, a line representing the dissolved species Me(OH)ni i in equilibrium with Me(OH)n (s) has a slope equal to i n. Applying Equation (11.13) to a system in equilibrium with Cd(OH)2 (s), we derive the following equations for the lines of the various dissolved Cd species: log (Cd2+ ) = 13.64
2 pH
(11.14)
pH
(11.15)
log (CdOH+ ) = 3.55
log Cd(OH)2 (aq) = 6.65
(11.16)
log (Cd(OH)3 ) = 19.66 + pH
(11.17)
log (Cd(OH)24 ) = 33.64 + 2 pH
(11.18)
The total dissolved Cd (i.e., T OT Cddiss ) can be computed at any pH as the sum of the concentrations of the individual species. Alternatively, we can note that, because the presence of the solid does not alter the equilibrium relationship between dissolved species, the relationship between T OT Cddiss and the activity of free Cd2+ when Cd(OH)2 (s) is present must be the same as in the case developed in Chapter 10 for a completely soluble system [Equation (10.17d)]: 4
T OT Cddiss = (Cd2+ ) 1 + Â bOH,i (OH )i i=1 4
= (Cd ) 1 + Â 2+
i=1
584
⇤b
i (H+ )i
!
!
(11.19a)
(11.19b)
However, for a system in equilibrium with solid Cd(OH)2 (s), the Ks0 expression must be satisfied, and the equation establishing a fixed value of T OT Cddiss is no longer applicable. Substituting the Ks0 relationship for (Cd2+ ) into Equation (11.19), we obtain, for a solution in equilibrium with Cd(OH)2 (s), Ks0 T OT Cddiss = (OH )2 =
⇤ K (H+ )2 s0 Kw2
4
!
(11.20a)
!
(11.20b)
1 + Â bi (OH )i i=1
4
1+ Â
i=1
⇤b i (H+ )i
Either form of Equation (11.20) indicates the total dissolved Cd concentration that must be present in a solution in equilibrium with solid Cd(OH)2 (s), if the only complexes that form are between Cd2+ and OH– and the solutes behave ideally. The only variable in these equations is H+ . Thus, if the pH is known, the concentrations of all dissolved Cd species are established.
Precipitation and accumulation of solids on the insides of pipes can interfere with their ability to convey water. This photograph shows the substantial buildup of solids inside a pipe used to carry a mixture of wastewater and brine from desalination processes to an ocean outfall.
A log C-pH diagram showing Cd species in a solution in equilibrium with Cd(OH)2 (s) is shown in Figure 11.1. The lines representing the various Cd(OH)2i i species do not “bend” at all, and in this way they are similar to the lines for the dissolved species in solutions equilibrated with a gas phase. This similarity is not coincidental. Recall that in a solution equilibrated with a gas phase, the activity of the dissolved volatile species is independent of pH and therefore is represented by a 585
horizontal line on a log C-pH diagram. If the dissolved gas is an acid (e.g., H2 CO3 ), it equilibrates with its conjugate base (e.g., HCO–3 ). In that case, the line for the base must intersect the line for the acid at pH = pKa and have a slope of +1. The lines for species generated by further deprotonation (e.g., CO2– 3 ) have correspondingly larger slopes. The net result is that the concentration of total dissolved gas (e.g., T OT CO3 ) increases continuously with increasing pH. If the gaseous species is a base (e.g., NH3 ), the same reasoning applies, but the concentration of the conjugate acid, and therefore of the total dissolved gas, increases with decreasing pH.
Figure 11.1 LogC-pH diagram showing dissolved Cd species in equilibrium with Cd(OH)2 (s). The point identified as the initial condition is for an example presented later in this chapter.
Precipitation followed by settling and/or filtration of the solids is one of the most common approaches for removing dissolved metals from solution. (Courtesy of ASKO Processing, Inc.)
586
In systems equilibrated with a solid, the activity of the dissolved species that has the same chemical formula as the solid itself (Cd(OH)2 in the example system) is independent of pH, because that species is formed by dissolution of the solid without any consumption or release of H+ ions. The line representing the concentration of Cd(OH)2 in Figure 11.1 is therefore horizontal. Furthermore, because Cd(OH)2 can act either as an acid [by combining with additional OH– ions to form Cd(OH)–3 and – + 2+ Cd(OH)2– 4 ] or as a base (by releasing OH ions to form CdOH or Cd ), a log C-pH diagram for Cd(OH)2 (s) combines features that are characteristic of the diagrams for both CO2 (g) and NH3 (g). That is, the concentrations of some Cd(OH)2i i species increase with increasing pH, while those of others increase with decreasing pH. This result leads to the bowl-shaped curve for the total dissolved cadmium concentration as a function of pH shown in Figure 11.1. This shape is a common feature of the solubility diagrams for many metals.
11.4.2
Systems Containing Ligands Other than Hydroxide
If ligands other than OH– are present in a solution containing Cd(OH)2 (s), then Cd2+ will form complexes with those ligands as well. The extension of Equations (11.19) and (11.20) to such a situation is straightforward, since the presence of the solid has no effect on the stability constant relationships between Cd2+ and dissolved ligands. For instance, if complexation by either Cl– or OH– is possible, then, regardless of whether or not a solid is present, the total dissolved concentration of Cd is given by ! T OT Cddiss = Cd2+
4
1 + Â bOH,i OH i=1
i
2
+ Â bCl,i Cl
i
(11.21)
i=1
The only difference in the case where solid Cd(OH)2 (s) is present is that, as in Equation (11.20), the activity of the free, uncomplexed Cd2+ ion is related to pH via the solubility constant. In such a case, the total soluble cadmium concentration is ! 4 2 Ks0 i i T OT Cddiss = 1 + Â bOH,i OH + Â bCl,i Cl (11.22) 2 OH i=1 i=1 A log C-pH diagram for a system in equilibrium with Cd(OH)2 (s) and at a fixed Cl activity of 0.5 is shown in Figure 11.2. Note that, as in Figure 10.7, the lines for the CdCl2i i species are parallel to that for Cd2+ . The extension to cases with even more ligands present is straightforward, since each ligand forms complexes with the metal independently. –
11.4.3
Modeling Solid/Solution Equilibrium with Visual Minteq
Like the systems discussed in previous chapters, simulation and analysis of systems in which solid/solution equilibrium must be considered can be greatly facilitated by 587
Figure 11.2 LogC-pH diagram for a system in equilibrium with Cd(OH)2 (s) for a fixed Cl– activity of 0.5.
the use of chemical equilibrium software packages like Visual Minteq. The types of systems that can be modeled using Visual Minteq when solid/solution equilibrium is of interest are listed under the “Solid phases and excluded species” pulldown menu on the home page. To model a solution that is known to be in equilibrium with one or more solids, we select “Specify infinite solid phases”. This selection instructs the program to assume that an infinite amount of each of the specified solid phases is present, so that no matter how much dissolves, each solid remains present with an activity of 1.0. The solids to be included can be selected in two different ways. One option is to migrate to the “Specify infinite solid phases” page and choose the appropriate solids from the pulldown menu, which includes all the solids in the program’s database. Once a solid is selected from this menu, its components are automatically added to the component list. However, the database identifies some solids by their chemical formulas and others by their mineral names, which can present challenges to users who might not know the mineral names of the solids of interest. As an alternative, the user can add the components of the solid in the usual way (on the program’s home page), and then check the box at the top of the “Specify infinite solid phases” page so that the pulldown menu includes only those solids that can be made from the components that have been previously selected. In this way, the user can see all the solids that can be made from those components (including the solids that are identified only by mineral name), simplifying the selection. Note that, if the solution does not initially contain the constituents of the solids but only acquires those constituents when the solid dissolves, it is acceptable to specify that the components that make up the solid are present initially, but at input concentrations of zero. 588
Once the solid phases that are known to be present are specified, the program simulates dissolution of each solid by (mathematically) titrating the solution with the components that make up the solids until the solubility constant expression is satisfied. If a solution is initially supersaturated with the solid, a ‘reverse titration’ is conducted, with the solid components being removed from solution in their stoichiometric ratios until the solubility criterion is met. To illustrate this process, we can re-solve Example 11.1 using Visual Minteq. We begin by clicking “Solid phases and excluded species” and “Specify infinite solid phases” on the home page, choosing calcite from the “Name of solid” pulldown list. Once that is done, the stoichiometric and thermodynamic parameters for formation of the solid appear (Figure 11.3). The log Ks value that appears on this page is the logarithm of the equilibrium constant for the dissolution of the solid via a reaction in which all the reactants and products other than the solid are components in the Visual Minteq database.2 In the case of calcite dissolution, the reaction is CaCO3 (s) )* Ca2+ + CO23 so the equilibrium constant is Ks0 , or 10 8.48 . Clicking the “Add” button causes the components Ca2+ and CO2– 3 to be included in the “List of components” to be considered in the subsequent calculations and instructs the program to assure that the computed solution composition is in equilibrium with the solid.
Figure 11.3 Visual Minteq screen for specifying that the solution is in equilibrium with calcite. 2 Note
that this convention differs from the one used to input log K values for other species in Visual Minteq. In all other cases, log K corresponds to the reaction for formation of the species, whereas for solids, log K corresponds to the reaction for disappearance of the species. Oddly, the Visual Minteq convention for inputting the stoichiometry of solids is based on the reaction for formation of the solid. As a result, the stoichiometry and log K values associated with a given solid in Visual Minteq are for different reactions (the former for formation of the solid, and the latter for its dissolution).
589
We next note that, when stoichiometric softening is implemented, T OT Ca = T OT CO3 . According the the problem statement, T OT Ca in the system after addition of the lime was 180 mg/L, or 4.5 ⇥ 10 3 M, so that must be the initial concentration of T OT CO3 as well. We therefore input these values, along with the known, fixed pH (10.8). If we wish to confirm our inputs, we can review the list of components (H+ , + Ca2+ , CO2– 3 ) and the list of fixed species (H , calcite); assuming those are correct, we then run the program. The screen showing the composition of the equilibrated solution for a run in which the ionic strength was calculated by the program is shown in Figure 11.4. Note that, because the pH was fixed at an alkaline value, but no cation was input to balance the charge on OH– , the solution is reported to have a charge imbalance. Also, the ionic strength is unrealistically low because the only ions being considered are – + Ca2+ , CO2– 3 , H , and OH , and other species that can be generated by combinations of those ions; if we wanted to simulate a real solution more accurately, we would have to input at least the other major cations and anions (e.g., Na+ , Mg2+ , Cl– , SO2– 4 , etc.) or specify a fixed ionic strength closer to that in the real solution.
Figure 11.4 Output screen for stoichiometric softening at pH 10.8.
By clicking the “Equilibrated mass distribution” button, we see that T OT Ca and T OT CO3 are both 7.96 ⇥ 10 5 M, or about 20% more than we calculated in Example 11.1. A small portion of this difference can be attributed to the fact that ionic strength effects were taken into account in the Visual Minteq calculations. However, the majority of the difference reflects the fact that Visual Minteq accounts for the presence of the complexes CaCO3 , CaHCO+3 , and CaOH+ . Although we could have considered 590
these complexes in our manual analysis of the solution, that would have complicated the calculations considerably. Solving the problem using the software is thus not only easier, but also yields a more accurate result. Example 11.2 Prepare a log C-pH diagram showing the total soluble aluminum concentration in a solution in equilibrium with amorphous aluminum hydroxide [Al(OH)3 (am)] at total fluoride concentrations of 0, 1, and 10 mg/L, and for the range 5 pH 9.3 What fraction of the total dissolved Al is present as Al F complexes in the three solutions at pH 7.0? Assume the solution ionic strength is 0.01 M in all cases. Solution The diagram can be prepared using Visual Minteq by selecting Al(OH)3 (am) as a solid phase available at infinite concentration, specifying the T OT F concentration to be 0, 1, or 10 mg/L, and using the “Multi-problem/Sweep” option to explore pH values in the range of interest. The output can then be exported and converted to prepare the plot shown in Figure 11.5. The figure indicates that the presence of 1 mg/L T OT F dramatically increases aluminum solubility at pH’s between about 5.0 and 7.0, and the presence of 10 mg/L does so from pH less than 5.0 to about 7.5. Review of the numerical output at various pH’s reveals that Al F complexes are present at concentrations much greater than Al3+ in both of these solutions throughout the pH range shown. However, at pH > 6.5, Al(OH)–4 is also present at a concentration much greater than Al3+ . At sufficiently high pH, the concentration of Al(OH)–4 far exceeds that of the Al F complexes, so the Al F complexes do not have a large effect on the total dissolved aluminum, even though their concentration is much greater than that of Al3+ . In the pH ranges where the Al F complexes account for the majority of the dissolved Al, the curves have an odd, undulating shape. This shape is generated by the fact that the activity of free fluoride ion (F– ) changes significantly with pH in these ranges, due to both its acid/base behavior (pKa = 3.18) and its complexation with Al. This situation contrasts with that in Figure 11.2, where the effect of Cl– on the solubility of Cd(OH)2 (s) was characterized. The key distinction is that, in the earlier diagram, the activity of Cl– was held constant, whereas in this case T OT F, rather than the activity of free F– , is being held constant. As in prior examples, Visual Minteq 3 Many
pure solids have a well-defined crystal structure that can often be identified using an analytical tool known as x-ray diffraction, with different structures yielding different diffraction patterns. Some solids, especially if they have been precipitated rapidly from a highly supersaturated solution, consist of many small crystalline units (sometimes called crystallites) linked together in a less organized structure. Such solids often generate no clear pattern when subjected to x-ray diffraction and are therefore referred to as ‘amorphous’ solids.
591
Figure 11.5 Total dissolved aluminum in equilibrium with Al(OH)3 (s) at three different fluoride concentrations. The broken line indicates the concentration of Al(OH)–4 in all three solutions.
allows us to easily account for the complex, nonlinear effects associated with the changing F– activity, whereas doing so via manual calculations would be extremely complicated. It is important to recognize that the concentrations of Al3+ and of all the Al OH complexes are unaffected by the presence of fluoride in these systems. Those concentrations are determined solely by the Ksi values for formation of Al OH complexes and the solution pH, and the sum of these concentrations in all three solutions is represented by the curve for (F ) = 0 . The Al F complexes are then solely responsible for the deviation between this curve and the curves for the systems with F– present. For example, at pH 7.0, the values of T OT Al are 8.64 ⇥ 10 6 , 9.61 ⇥ 10 6 , and 9.23 ⇥ 10 5 M in the presence of 0, 1, and 10 mg/L T OT F, respectively. The concentration of Al F complexes in the latter two solutions can therefore be computed as 1 mg/L T OT F:
T OT (AlFi ) = 9.61 ⇥ 10
10 mg/L T OT F:
T OT (AlFi ) = 9.23 ⇥ 10
6 5
8.64 ⇥ 10 8.64 ⇥ 10
6 6
= 9.61 ⇥ 10 = 8.37 ⇥ 10
7 5
Thus, 10% of T OT Al is present as Al F complexes at pH 7.0 in the presence of 1 mg/L T OT F, and 91% is present as such complexes when T OT F is 10 mg/L. The same information can be inferred directly from Visual Minteq by selecting the problem number corresponding to pH 7.0 on the main output page and clicking the “View species distribution” button. When we do that for the run at pH 7.0 with 592
1 mg/L T OT F, the screen shown in Figure 11.6 appears. The concentrations of the Al F complexes can be summed and compared to T OT Al to find the fraction of the soluble Al present as AlF3i i species.
Figure 11.6 Distribution of Al and F species in a system at pH 7 containing 1 mg/L T OT F and equilibrated with Al(OH)3 (am).
11.5
PRECIPITATION OF SOLIDS OTHER THAN HYDROXIDES
We next extend the concepts developed in the preceding sections to solutions that are in equilibrium with nonhydroxide solids, many of which are important in natural systems or in water and wastewater treatment operations. For instance, CaCO3 (s) is the mineral limestone, and, as has been noted, precipitation of CaCO3 (s) is one of the most common approaches for removing hardness from drinking waters. Other metal carbonates are also important both naturally and in engineered processes. Because we will ultimately wish to compare the solubilities of metal carbonate and metal hydroxide solids, and we have already analyzed the composition of systems in equilibrium with Cd(OH)2 (s), we focus here on precipitation of CdCO3 (s) as an example system. We begin by deriving the relevant equations and drawing log C-pH diagrams for two solutions in equilibrium with CdCO3 (s): one contain593
ing a fixed concentration of total dissolved carbonate species (T OT CO3,diss ) and the other in equilibrium with atmospheric CO2 (g).
11.5.1
Speciation in a System with CdCO3 (s) and Fixed TOTCO3,diss
Cadmium carbonate (also known by the mineral name otavite) is a relatively insoluble solid with Ks0 = 10 12.01 . CdCO3 (s) )* Cd2+ + CO23
Ks0 = 10
12.01
(11.23)
Thus, in a solution in equilibrium with pure CdCO3 (s): (Cd2+ )(CO23 ) = 10
12.01
log (Cd2+ ) + log (CO23 ) = 12.01
(11.24)
Equation (11.24) points out that, in a solution equilibrated with CdCO3 (s), the value of (Cd2+ ) can be determined by knowing (CO2– 3 ), regardless of the value of – 2– 5.0 (HCO3 ) or (H2 CO3 ). That is, if (CO3 ) is 10 and the system is in equilibrium with CdCO3 (s), then (Cd2+ ) will be 10 7.01 , regardless of whether the CO2– 3 activity 5.0 5.0 of 10 is established by having ⇠10 M T OT CO3,diss in a solution at pH 12, or 3 ⇠10 M T OT CO3,diss in a solution at pH 8.3, or ⇠ 0.2 M T OT CO3,diss in a solution at pH 6.3. Furthermore, any change in log (Cd2+ ) must be mirrored exactly by a change in log (CO23 ) that is equivalent in magnitude but opposite in direction. Thus, for a fixed T OT CO3,diss , the shape of the (Cd2+ ) line on a log C-pH diagram is easy to 2+ deduce. At pH 11, the (CO2– 3 ) line is essentially horizontal, so the (Cd ) line must 2 2+ be horizontal also; only in this way will log (Cd ) + log (CO3 ) be constant. In the region between the two carbonate pKa ’s, the slope of the (CO2– 3 ) line is approximately 2+ +1, so the slope of the (Cd ) line must be approximately 1. Similarly, at pH < 2+ 6.35, the slope of the CO2– 3 line is approximately +2, so that of the Cd line will be approximately 2. Both log (Cd2+ ) and log (CO23 ) are shown over a wide pH range for a system with a constant T OT CO3,diss of 10 3 M in Figure 11.7.4 Although the solid of interest in this system is CdCO3 (s), the solution contains both Cd2+ and OH– , so we expect dissolved Cd(OH)2i i complexes to form, and we should include them on the log C-pH diagram for the system. There are a number of ways to evaluate the equilibrium concentrations of these complexes. Perhaps the 4 In most scenarios where CdCO (s) is likely to precipitate, T OT Cd is much less than T OT CO . 3 3 Precipitation of CdCO3 (s) would therefore have a negligible effect on the total dissolved carbonate concentration, and the assumption of a constant T OT CO3 would be acceptable. Systems in which the precipitation reaction causes the total concentrations of both the metal and ligand to change significantly are handled well by chemical equilibrium software and are considered later in the chapter.
594
Figure 11.7 Cd2+ and CO2– concentrations in an ideal solution containing 10 3 T OT CO3,diss and equilibrated with CdCO3 (s).
3M
easiest approach is to use the curve for the Cd2+ activity in conjunction with the equilibrium constant expressions relating the activities of the various complexes to that of Cd2+ . For example, because ⇤ K1 is the acid dissociation constant relating Cd2+ to CdOH+ , we know that the following equation applies in any solution containing Cd2+ , regardless of whether or not a solid is present: log CdOH+ = log Cd2+ + log ⇤ K1 + pH
(11.25)
Equation (11.25) is the basis for two facts that we already know about the curves representing (Cd2+ ) and (CdOH+ ) on the log C-pH diagram: the CdOH+ curve always has a slope that is greater by 1 than the Cd2+ curve, and (Cd2+ ) = (CdOH+ ) at pH = p⇤ K1 = 10.10. Using this information, we can draw the curve for CdOH+ directly; the lines for the other hydroxo complexes can be drawn by following similar logic. The resulting diagram is shown in Figure 11.8. The T OT Cddiss curve in the log C-pH diagram for this system has a bowl-like shape similar to that for equilibrium with Cd(OH)2 (s), and for essentially the same reason: At low pH, the activities of soluble species that are more acidic than the solid become very large, and at high pH, the same is true of soluble species that are more basic than the solid. The exact value of T OT Cddiss at any pH and the pH of minimum solubility differ for different solids, but the general pattern observed here is consistently observed. The equation describing T OT Cddiss as a function of pH can be written concisely as the sum of the concentrations of free Cd2+ and the various complexes. The form of this summation is identical for any solution in which Cd OH species are the only 595
Figure 11.8 The concentrations of Cd2+ and all Cd OH complexes in the system shown in – Figure 11.7. Dissolved complexes of Cd2+ with CO2– 3 or HCO3 might also be present, but they are not considered in this analysis.
complexes of significance and is given by Equation (11.19a) or (11.19b): 4
T OT Cddiss = (Cd ) 1 + Â bOH,i (OH ) 2+
i
i=1 4
= (Cd2+ ) 1 + Â
i=1
⇤b i + (H )i
!
!
(11.19a)
(11.19b)
As we did when analyzing a solution in equilibrium with Cd(OH)2 (s), we can substitute for (Cd2+ ) based on Ks0 for the solid that is present, in this case CdCO3 (s). Making this substitution, we find ! 4 Ks0,CdCO3 (s) i T OT Cddiss = 1 + Â bi OH (11.26a) CO23 i=1 =
Ks0,CdCO3 (s)
4
a2 T OT CO3
1 + Â bi OH i=1
i
!
(11.26b)
Next, consider another solution in equilibrium with CdCO3 (s), but this time with T OT CO3 ,diss two orders of magnitude larger, at 10 1 M. In such a system, the curve representing the CO2– 3 concentration on the log C-pH diagram would be two units 596
Massive amounts of CaCO3 (s) can be precipitated from hard water when the water is softened. This photo shows the precipitate forming in a water treatment plant in Kansas. The solids form in the center of the tank and then settle as the water flows slowly to the tank perimeter. The disposal of these solids, along with Al(OH)3 (s) and Fe(OH)3 (s) that form in other water treatment operations, is often a major component of the cost of water treatment. (Charles Weber.)
higher than in Figure 11.7, but would otherwise be identical. As a result, the curves representing the equilibrium Cd2+ concentration and that of each of the Cd OH complexes would be two log units lower. The log C-pH diagram for this scenario is shown in Figure 11.9.
11.5.2
Speciation in a System with CdCO3 (s) and Fixed PCO2
The next case we will consider is one in which the solution is open to the atmosphere. The analysis of this system is actually quite easy, because Henry’s law establishes a 597
Figure 11.9 Concentrations of Cd2+ and Cd OH complexes in an ideal solution containing 0.1 M T OT CO3diss and in equilibrium with CdCO3 (s). As in Figure 11.8, possible formation of Cd CO3 and Cd HCO3 complexes is ignored.
fixed value of (H2 CO3 ), independent of pH. As a result, the CO2– 3 concentration is represented by a straight line of slope +2 at all pH values. For such a solution to be in equilibrium with CdCO3 (s), the value of (Cd2+ ) must decrease by two log units for each unit increase in pH, because the Ks0 expression still requires that the sum of log (Cd2+ ) and log (CO23 ) be a constant (= log Ks0 ). As a result, (Cd2+ ) is represented on the diagram by a straight line of slope 2. The line representing (CdOH+ ) is also straight, with a slope of 1, intersecting the Cd2+ line at pH = p⇤ KOH,1 = 10.10 (Figure 11.10a). You should be able to derive the equations necessary to compute the concentrations of the other complexes in the system. The log C-pH diagram for a system with fixed PCO2 of 10 3.46 atm (the partial pressure of CO2 in the atmosphere) is shown in Figure 11.10b. Once again, the total dissolved Cd in an open system equilibrated with CdCO3 (s) is given by Equation (11.19). Applying that equation in conjunction with the specifics of the system of interest, we obtain T OT Cddiss =
=
Ks0,CdCO3 (s) CO23
4
1 + Â bi OH i=1
Ks0,CdCO3 (s)
i
!
(11.27a) 4
PCO2 /HCO2 (Ka1 Ka2 /(H+ )2 )
598
1 + Â bi OH i=1
i
!
(11.27b)
Figure 11.10 Concentrations of Cd2+ and Cd OH complexes in an ideal solution equilibrated with atmospheric CO2 (g) and with CdCO3 (s). (a) Concentrations of 2+ Cd2+ and CO2– 3 . (b) Concentrations of Cd and all Cd OH complexes. As in previous figures, Cd CO3 and Cd HCO3 complexes are ignored.
11.6
FORMATION OF NONHYDROXO COMPLEXES IN SYSTEMS WITH SOLIDS
As was suggested earlier, in any of the systems considered in the preceding sections, complexes can form not only with hydroxide, but also with other ligands that are present in the system. These ligands include the ones that are incorporated in any – nonhydroxide solids in the system (e.g., CO2– 3 and HCO3 in systems containing metal carbonate solids), as well as other ligands, such as Cl– . Each such complex can be represented by an additional curve on the relevant log C-pH diagrams. 599
A generic expression for T OT Mediss in a system containing m ligands can be written as follows: ! m
k
T OT Me = (Men+ )0 1 + Â Â bL j ,i (L j )i
(11.28)
j=1 i=1
where (Men+ )0 is the activity of the free, uncomplexed metal ion, computed from the solubility product and the activity of the free, uncomplexed ligand (e.g., as Ks0,Cd(OH)2 (s) /(OH )2 or Ks0,CdCO3 (s) /(CO23 ). Note that this equation is essentially identical to Equation (10.37), except that now we specify that the free metal ion activity is based on the solubility product. Example 11.3 The solid PbCO3 (s), known as cerussite, is thought to exert an important control over the release of soluble lead from plumbing materials (e.g., lead pipes, solder, and brass fittings) as water passes through distribution systems. When Pb2+ is released from these materials, if the pH and T OT CO3 are in certain ranges, cerussite can precipitate and remove the lead from solution. In addition to (possibly) forming this precipitate, Pb2+ and carbonate species can form various complexes. The stoichiometry and equilibrium constants for dissolution of PbCO3 (s) and for formation of Pb CO3 complexes are summarized below; stability constants for the formation of Pb OH complexes are given in Table 10.2. PbCO3 (s) )* Pb2+ + CO2– 3
log Ks0 = 13.20
(11.29)
Pb2+ + CO2– 3 )* PbCO3
log KCO3 ,1 = 6.53
(11.30)
2– Pb2+ + 2 CO2– 3 )* Pb(CO3 )2
log bCO3 ,2 = 9.94
(11.31)
Pb2+ + HCO–3 )* PbHCO+3
log KHCO3 ,1 = 2.90
(11.32)
Because of concerns over the effect of lead on the nervous system, and in particular its effect on the cognitive development of young children, the USEPA has established an ‘action level’ of 15 µg/L for lead in tap water. Prepare a log C–log (CO23 ) diagram for a system at pH 8.0 and in equilibrium with cerussite, showing all soluble Pb species that are present at concentrations >10 9 M (0.2 µg/L). Consider CO2– 3 concentrations from 10 7 to 10 3 M. Solution The Pb2+ concentration in equilibrium with cerussite at any CO2– 3 concentration can be determined directly from Ks0 : log (Pb2+ ) = log Ks0 600
log (CO23 )
(11.33)
In addition, since the pH is specified to be 8.0, the concentration of each Pb OH complex is a fixed multiple of the Pb2+ concentration, independent of the CO2– 3 concentration. For example, the concentration of PbOH+ is given by (Pb2+ )(⇤ KOH,1 ) (H+ )
(PbOH+ ) =
log (PbOH+ ) = log (Pb2+ ) + log (⇤ KOH,1 ) + pH = log (Pb2+ ) + ( 7.60) + 8.00 = log (Pb2+ ) + 0.40
(11.34)
Combining this result with Equation (11.33), we find log (PbOH+ ) = 12.80
log (CO23 )
(11.35)
Expressions for log C of the other Pb OH complexes can be found from similar calculations. In addition, expressions for log (PbCO3 ) and log (Pb(CO3 )22 ) can be obtained by combining the solubility reaction (11.29) with Reactions (11.30) and (11.31), respectively; the resulting equations are
log (PbCO3 ) = log Ks0 + log KCO3 ,1 = 13.2 + 6.53 = 6.67 ⇣ ⌘ log Pb(CO3 )22 = log Ks0 + log bCO3 ,2 + log (CO23 )
(11.36)
= 13.2 + 9.94 + log (CO23 ) = 3.26 + log (CO23 ) (11.37)
Finally, the equation describing the concentration of PbHCO+3 as a function of the CO2– 3 concentration can be derived by combining the stability constant expression for Reaction (11.32) with that for Reaction (11.29) and Ka2 for the carbonate acid/base group, yielding 2+ (PbHCO+ 3 ) = KHCO3 ,1 (Pb )(HCO3 )
= KHCO3 ,1
Ks0 (H+ )(CO23 ) KHCO3 ,1 Ks0 + = (H ) Ka2 Ka2 (CO23 )
log (PbHCO+ 3 ) = log KHCO3 ,1 + log Ks0 = 2.90 + ( 13.2)
log Ka2
( 10.33)
pH
(8.0) = 7.97
(11.38)
Equations (11.33) and (11.35) through (11.38) indicate that the concentrations of all five soluble Pb species can be expressed in the form log C = m log (CO23 ) + b 601
(11.39)
with m being negative in some cases and zero or positive in others. As a result, each species is represented by straight line on the log C-log (CO23 ) diagram, and the curve for T OT Pb is bowl-shaped (Figure 11.11). The same diagram can also be developed with minimal manual calculations by using Visual Minteq. In that case, the fixed partial pressure of CO2 (g) and the presence of the solid are specified as described previously, and the program accounts for all the soluble complexes automatically.
Figure 11.11 log C-log (CO23 ) diagram for a solution in equilibrium with cerussite [PbCO3 (s)] at pH 8.0.
The preceding example emphasizes that not only OH– , but also other ligands, can either increase or decrease the solubility of metals, depending on the exact solution conditions and, especially, the ligand concentration. As a result, by a judicious choice of which ligand(s) to add, and at what concentrations, we can induce precipitation to facilitate removal of a metal from solution, or induce formation of complexes to prevent precipitation. This duality of functions provides enormous flexibility in the industrial and commercial use of ligands. Example 11.4 Uranium can exist in a number of oxidation states, taking on a charge of +3 to +6. When in the +6 oxidation state, it is covalently bound to two oxygen ions to form the uranyl ion, UO2+ 2 . This cation can undergo hydrolysis reactions to form UO2 OH complexes or react with carbonate to form UO2 CO3 complexes, and it can also form y small polymeric complexes with compositions of the general form (UO2 )x (OH)2x y 2(x y) or (UO2 )x (CO3 )y . The Visual Minteq database includes 14 such UO2 OH and UO2 CO3 complexes, in addition to free UO2+ 2 . (a) Consider a carbonate-free solution that has an ionic strength of 0.004 M and is 602
in equilibrium with the uranyl hydroxide solid schoepite, UO2 (OH)2 ·H2 O(s). Use Visual Minteq to determine T OT UO2,diss and the distribution of T OT UO2,diss between free and OH complexed UO2 over the range 4 < pH < 10. (b) How does the presence of 0.002 M T OT CO3 affect the total dissolved uranium over this pH range? Identify the three species that account for the largest amounts of UO2 in the latter solution at pH 7.5. Solution (a) A log C-pH diagram for the carbonate-free system can be developed by inputting the known ionic strength, specifying that schoepite is an infinite solid phase, and using the “Multi-problem/Sweep” option to instruct the program to carry out calculations over the specified range of pH values. T OT UO2,diss is shown over the range 4 < pH < 10 in Figure 11.12, along with the concentrations of UO2+ 2 and UO2 OH complexes [computed as the difference T OT UO2 (UO2+ 2 )]. The T OT UO2 vs. pH curve is bowl-shaped and similar to the corresponding curves that we have seen previously. Free UO2+ 2 is not the dominant species anywhere in the pH range of interest (although it would be at even lower pH). (b) When 0.002 M T OT CO3 is added to the system, UO2 CO3 complexes dominate over UO2+ 2 and UO2 OH complexes at all pH values above ⇠5.4, and T OT UO2 is dramatically increased. Although the T OT UO2 curve still has a bowl-like shape, the bowl is much shallower (i.e., T OT UO2 is much less sensitive to pH). The relative contributions of different species to T OT UO2 at pH 7.5 in the solution containing carbonate can be identified by viewing the output for the pH 7.5 run (or running the program again at that pH, if none of the runs were at pH 7.5) and clicking the “View species distribution” button. The window for the run at pH 7.5 is shown below and indicates that > 95% of T OT UO2 is 4– present as either UO2 (CO3 )2– 2 or UO2 (CO3 )3 , with an additional 3.8% present as UO2 CO3 .
603
Figure 11.12 Soluble UO2 in equilibrium with schoepite in (a) a carbonate-free system, and (b) a solution containing 0.002 M T OT CO3 . Ionic strength 0.004 M in both solutions.
11.7
ADDING NEW SOLIDS TO THE VISUAL MINTEQ DATABASE
Although the Visual Minteq database includes information on an impressively large array of solids, there are bound to be situations in which a user wishes to explore the solubility of a solid that is not included. In these cases, the relevant information can be 604
added to the database following a straightforward algorithm, as is illustrated here for inclusion of the mineral struvite (Mg(NH4 )PO4 (s)·6 H2 O, Ks0 = 10 13.26 ). This solid often forms in or downstream of anaerobic digesters, in which microorganisms and other organic solids decay, releasing ammonia and phosphate into the water. In recent years, efforts have been made to recover these nutrients from digester solutions, both to reduce the release of the chemicals into receiving waters and because they have potential economic value when incorporated into fertilizers.5,6 As is the case for any species in Visual Minteq, the stoichiometry for struvite formation must be input based on a reaction in which struvite is a product and all of the reactants or products other than struvite are components, i.e., 3 * Mg2+ + NH+ 4 + PO4 + 6 H2 O ) Mg(NH4 )PO4 ·6 H2 O(s)
(11.40)
To add struvite to the database, we first choose Mg2+ , NH+4 , and PO3– 4 as components. The input concentrations of these components are irrelevant at this point, so we can leave them all as 0. We then go to the “Database management/Solid phases” page, click on “Add new species”, and proceed to specify the name of the solid and the stoichiometry for the solid formation reaction. The name ’struvite’ is input in the “Species name” box near the bottom of the window. The first component in the solid is then specified by using the pulldown list for “Component no. 1”, selecting “Mg+2”, and typing the corresponding stoichiometric coefficient (in this case, 1) in the box to the left. When the “Next” button is clicked, the information we have input appears higher in the window as the “Stoichiometry of 1st component”. We then repeat the process to specify the remaining components and stoichiometric coefficients (including H2 O, since the inclusion of water molecules in a solid is an important part of its identity). Once the stoichiometry for the final component has been input, the “Suggest ID number and mol. wt.” button can be clicked, and the program fills in the corresponding fields. The user must then fill in the remaining fields — most importantly, the log K for the reaction — at which point the changes to the database can be saved for use in subsequent problems. (Keep in mind that, in Visual Minteq, the stoichiometry that is input is for the solid formation reaction, but the equilibrium constant is for the dissolution reaction.) An image of this page after all the data for struvite have been input is shown in Figure 11.13. The standard molar enthalpy of formation of struvite is not known, so a value of 0 has been input to that field. 5 When this section was originally written, Visual Minteq did not include struvite in its default solids database. However, the solid was subsequently added to the database, so it can be considered in a simulation without inputting it as a new species. Nevertheless, the current discussion provides a template for how a new solid could be added to the database, if needed. 6 Visual Minteq lists some solids by their chemical formula and others by their mineralogical name. Therefore, when searching the list for a specific solid, it is important to look for the solid in both ways.
605
Figure 11.13 The Visual Minteq page for adding struvite to the database of available solids.
Example 11.5 Assume that struvite has formed on pipes downstream of a digester, and that the treatment process is then modified to eliminate the digestion step. As a result, the struvite is exposed to water containing 8 mg/L T OT Mg, 3 mg/L T OT NH4 N and 0.2 mg/L T OT PO4 P. The water is at pH 8.5, and it has an alkalinity of 600 mg/L as CaCO3 (all assumed to be attributable to carbonate and phosphate species) and an ionic strength of 0.03 M. How much struvite will dissolve in one liter of such a solution, and what will the final pH be, if this solution equilibrates with the solid? Assume that dissolution of struvite has a negligible effect on the ionic strength. Solution The values of T OT Mg, T OT NH4 and T OT PO4 in the water before it contacts the struvite are given in mg/L units, with the total ammonia and phosphate values expressed as mg N/L and mg P/L, respectively. This information can be input into Visual Minteq directly by choosing Mg2+ , N (NH+4 ), and P (PO4 ) as the components and selecting mg/L as the “Concentration unit” from the main menu. 606
Since the pH is known, and we are assuming that all the alkalinity is contributed by carbonate and phosphate species (both of which are taken into account in Alkalinity calculations within Visual Minteq), we can input the given alkalinity via the “Parameters/Specify alkalinity” option and let the program calculate T OT CO3 . After specifying that the alkalinity is 600 mg/L as CaCO3 , we close that window and then change the specification of the concentration units on the main page to molal. When we do this, the input concentrations are recomputed and appear in molal units (which we assume are equivalent to mol/L) when we view the components list. The values shown for Mg, NH+4 -N, and PO4 -P at this point are T OT Mg = 0.000329 T OT NH4 = 0.000214 T OT PO4 = 6.46 ⇥ 10
6
The component shown for the carbonate species in the list is designated “CO–2 3 ****”, with an assigned value of 0.011989. Recall from Chapter 8 that this value is not the concentration of T OT CO3 , but rather the alkalinity (in equiv/L) that the program attributes to carbonate species, H+ , and OH– . Since the equivalent weight of CaCO3 is 50, the given total alkalinity of 600 mg/L as CaCO3 corresponds to 12 meq/L (i.e., 0.012 equiv/L). The slightly smaller value of 0.011989 equiv/L reported by Visual Minteq accounts for just the contributions from the carbonate species, H+ , and OH– ; the remaining alkalinity (i.e., 0.000011 equiv/L) is provided by phosphate species, so that the total alkalinity in the solution simulated by Visual Minteq is, in fact, the correct value. Ultimately, we want to determine the solution composition when the original solution contacts and equilibrates with solid struvite. To do this, we need to know the complete composition of the original solution, represented as a combination of Visual Minteq components. At this point, we know T OT Mg, T OT NH4 , T OT PO4 , and T OT CO3 (or, at least, CO–2 3 ****, which Visual Minteq can use in conjunction with the pH to compute T OT CO3 ). The remaining piece of information we need is T OT H, which we can find based on the known pH of 8.5. Specifying that pH = 8.5 on the home page and inputting the known ionic strength of 0.03 M, we can run the program. The output indicates that T OT H is 1.1477 ⇥ 10 2 M (accessed on the “Equilibrated mass distribution” output page), implying that that much T OT H must have been in the original solution for the solution to be at pH 8.50. The output also indicates that T OT CO3 is 1.1727 ⇥ 10 2 M. To confirm these results, we can switch the pH option on the main page to indicate that the pH should be calculated rather than assigned a specified value, and input a value of 1.1477 ⇥ 10 2 M T OT H. We also ‘turn off’ the calculation of T OT CO3 from the alkalinity menu and instead input a T OT CO3 value of 1.1727 ⇥ 10 2 M.7 Running the program with these inputs confirms that the equilibrium pH is 8.50. 7 Keep
in mind that these values must be input in the current units for the Components list. Thus, if the units are millimolal, the input should be 11.477 for T OT H and 11.727 for T OT CO3 .
607
Finally, we can compute the solution composition after it equilibrates with struvite by using the values computed for T OT H and T OT CO3 in the original solution as inputs, specifying that struvite is present as an infinite solid phase, and allowing the program to determine the equilibrium pH. The output from such a run indicates that the equilibrium pH is 8.978 and that T OT Mg is 1.052 ⇥ 10 3 M. The difference between this value and T OT Mg in the original solution indicates the amount of struvite that dissolved per liter of water. Struvite dissolved = (1.052 ⇥ 10
11.8
3
3.29 ⇥ 10 4 )
mol = 7.23 ⇥ 10 L
4 mol
L
DETERMINING WHETHER A SOLID WILL PRECIPITATE UNDER GIVEN CONDITIONS
The discussion thus far has established an approach for determining the composition of solutions that are known to be in equilibrium with a solid. We next consider systems that are, initially, not in equilibrium with a solid, and we ask these questions: (1) Will a solid precipitate and/or dissolve as the system equilibrates, and if so, how much solid reacts? (2) If more than one solid can form, which one(s) will be present at equilibrium? (3) How will the equilibration process affect other water quality variables, such as pH and T OT CO3,diss ?
11.8.1
Systems in Which Only One Solid Need Be Considered
We begin the analysis by consideration of a system that is well buffered at pH 10.0 and contains 10 4 M T OT Cd, and in which we believe Cd(OH)2 (s) is the only solid likely to form. The T OT Cd might be present initially as soluble species, as solid Cd(OH)2 (s), or as a combination of soluble and particulate species. For simplicity, we assume that hydroxo complexes are the only ones likely to form. Our goal is to determine if solid Cd(OH)2 (s) will be present when the system reaches equilibrium, and if so, how much. The approach for determining the composition of the equilibrium solution involves making an initial assumption — either that the solid will or will not be present. If we assume that no solid is present at equilibrium, then the analysis proceeds by computation of the Cd speciation in a solution containing 10 4 M total dissolved Cd. The calculations are carried out by substituting 10 4 for both T OT Cddiss and (OH– ) (because pH = 10) in Equation (11.19) and solving for (Cd2+ ). Designating this value as (Cd2+ )0 , the calculation yields: 608
T OT Cddiss = Cd2+
0
4
0
10
= Cd2+
h 1 + K1 OH
⇥
1 + (103.92 )(10
+ (108.70 )(10 10
4
+ b2 OH 4.0
5
= 10
+ b3 OH
) + (107.65 )(10
4.0 3
) + (108.65 )(10
= (Cd2+ )0 (2.28)
(Cd2+ )0 = 4.39 ⇥ 10
2
4.36
4.0 2
)
4.0 4
)
3
+ b4 OH
4
i
⇤ (11.41)
Thus, if all the Cd in the system were soluble, (Cd2+ ) would be 10 4.36 . We can then calculate the product (Cd2+ )0 (OH– )2 (i.e., the value of Qs0 ) to determine whether the computed solution composition would be supersaturated with respect to Cd(OH)2 (s). For the current system, this product is (10 4.36 )(10 4.0 )2 = 10 12.36 . Since this product is greater than Ks0 (= 10 14.36 ), we conclude that, if all the T OT Cd remained soluble, the solution would be supersaturated with respect to Cd(OH)2 (s); correspondingly, our assumption that no solid would be present at equilibrium must be in error. We therefore abandon the initial assumption and recalculate the equilibrium composition, this time with the assumption (now confirmed) that the solid will be present. Given that the solution is well buffered, the final pH will be 10.0, and the equilibrium value of (Cd2+ ) will be the value in equilibrium with Cd(OH)2 (s), i.e., (Cd2+ ) = Ks0 /(OH )2 = 10 6.36 . Designating this value as (Cd2+ )00 and the corresponding the total soluble Cd concentration as T OT Cd00diss , we can then compute that ⇥ ⇤ T OT Cd00diss = (Cd2+ )00 1 + K1 (OH ) + b2 (OH )2 + b3 (OH )3 + b4 (OH )4 = (10
6.36
)(2.28) = 9.95 ⇥ 10
7
As expected, the total dissolved Cd concentration in equilibrium with the solid is less than the total initially present, confirming that some Cd is present as a solid once the system reaches equilibrium. This whole process is summarized by the sequence of steps 1-2a-3a-4a-3b-4b-5b in the flow chart shown in Figure 11.14. The difference between T OT Cd and T OT Cddiss is the amount of Cd that precipitates (or, if the solid was initially present, the amount that fails to dissolve) as the system approaches equilibrium. This difference is 10 4.0 9.95 ⇥ 10 7 , or 9.90 ⇥ 10 5 M; i.e., approximately 99% of the Cd in the system is present as the solid once the system equilibrates. The total concentration of the solid in the system is typically reported as its mass concentration (e.g., milligrams per liter), which includes the contribution of hydroxide ions. In this example, this concentration would be ⇣ ⌘⇣ mg ⌘ mg [Cd(OH)2 (s)] = 9.9 ⇥ 10 5 M 146, 000 = 14.5 (11.42) mol L 609
Figure 11.14 Flow chart for determining the equilibrium speciation in a system with known T OT Me, in which Me(OH)2 (s) might precipitate and only hydroxo complexes form. For other types of solids and other ligands, Equation (11.28) is used in place of (11.19).
The mass of solid that actually forms, as determined by weighing a dried sample of the solid, is often quite a bit more than indicated by calculations such as this one, because significant amounts of water can be incorporated into the solid when it forms and remain there during drying. If, in our initial calculations, (Cd2+ )0 had been less than Ks0 /(OH )2 , then we would have concluded that the solution containing 10 4 M T OT Cddiss was undersaturated; i.e., the assumption that all the Cd was soluble would have been confirmed, and the sequence would have proceeded from 4a to 5 in Figure 11.14. The concentrations of the Cd OH complexes could then be calculated based on (Cd2+ )0 and the stability constants. The calculations that we would carry out if the initial assumption had been that the solid was present at equilibrium are essentially identical to those described above, but they would be carried out in a different sequence. Specifically, we would first compute the Cd2+ activity in equilibrium with the solid [i.e., Ks0 /(OH)2 , which we defined earlier as (Cd2+ )00 ]. We would then compute the corresponding value of T OT Cd00diss from Equation (11.19) and compare that value with the actual total con610
centration of Cd in the system, T OT Cd. If T OT Cd00diss < T OT Cd (i.e., if the soluble Cd concentration based on the assumption that solid was present was less than the total amount of Cd known to actually be in the system), we would conclude that the assumption that some Cd(OH)2 (s) is present at equilibrium was correct; this outcome corresponds to sequence 1-2b-3b-4b-5 in Figure 11.14. On the other hand, if T OT Cd00diss > T OT Cd, we would conclude that the assumption was in error and that all the Cd in the system was soluble. In that case, the speciation would be recomputed by setting T OT Cddiss equal to T OT Cd and solving for the Cd2+ concentration from Equation (11.19) (sequence 1-2b-3b-4b-3a-4a-5). In the current example, of course, we would find that the assumption that the solid is present was correct. Thus, if we know the initial composition of a solution and we believe that only a single, specific solid can precipitate, we can determine whether that solid will be present at equilibrium by initially assuming either that the solid will be present or that it will be absent; the only penalty associated with making the wrong assumption is that a few additional calculations must be carried out. If a log C-pH graph describing the system is available, the process of determining whether the solid will be present at equilibrium is even simpler. On such a graph, the curve showing T OT Cddiss as a function of pH represents T OT Cd00diss ; i.e., it is the total dissolved Cd concentration in a solution that is in equilibrium with the solid. The point representing the initial pH and total concentration of Cd in the system of interest (which might or might not be an equilibrium system) can also be identified on the graph. If this point is above the T OT Cd00diss curve, then T OT Cdinitial > T OT Cd00diss , and the solid will be present when the system reaches equilibrium; if it is below the curve, then T OT Cdinitial < T OT Cd00diss , and the solid will not be present. The point representing the initial conditions in the example system just analyzed is shown in Figure 11.1. The fact that it is above the curve for total dissolved Cd indicates that the solid will be present when the system equilibrates, consistent with our conclusion from the numerical analysis. To address the same question using Visual Minteq, we follow essentially the same procedure as for a system in which a solid is known to be present, except that the solid is input using either the “Specify finite solid phases” (if some of the solid is present initially) or the “Specify possible solid phases” (if none of the solid is present in the initial solution). In the former case, in addition to identifying the solid that might be present at equilibrium, the user specifies on the input screen for the solid how much of the solid is initially present. Two Visual Minteq output screens for the example system (10 4 M T OT Cd, pH fixed at 10.0) are shown in Figure 11.15. The output information includes both the solution speciation and the molar concentration of the solid that is present at equilib-
611
rium (accessed via the “Amount of finite solids” button on the main output page).8 For this system, the equilibrium composition is fully determined by T OT Cd and the final pH. Because both of those values are fixed, the results are the same regardless of the form in which Cd is added. If the final pH was not specified, then that value and the rest of the solution composition would depend on what anions were added to the system along with the Cd, as is demonstrated in the following example.
Figure 11.15 Visual Minteq output for a solution prepared by adding 10 4 M Cd(OH)2 (s) to water with pH fixed at 10.0, assuming ideal solution behavior. (a) Solution composition. (b) Concentration of solid present at equilibrium. the simulation is carried out, Visual Minteq adds Cl– as a dummy component at a very low total concentration (10 16 M). This component is needed because the solution algorithm requires the total number of components to exceed the number of species present at fixed activity. In the current problem, the only components are Cd2+ and H+ . The pH is fixed by the user (so the species H+ has a fixed activity), and Cd(OH)2 (s) has a fixed activity of 1.0, so the requirement is not met. Adding the dummy component (and a corresponding species) overcomes this problem. These additions have no effect on the other calculations. 8 When
612
Example 11.6 A solution is made by adding NaOH to pure water until the pH is 10.0. Determine the system composition if 10 4 M T OT Cd is then added. Consider two cases: one where the Cd is added as Cd(NO3 )2 , and the other where it is added as Cd(OH)2 (s). Assume the solution behaves ideally both before and after the Cd addition. Solution If the solution behaves ideally, the pH of pure water can be raised to 10.0 by addition of 10 4 M NaOH. We can therefore use Visual Minteq to simulate the system in which Cd(NO3 )2 is added by inputting 10 4 M T OT Na and 10 4 M T OT H to account for the NaOH, and 10 4 M T OT Cd plus 2 ⇥ 10 4 M T OT NO3 to account for the Cd(NO3 )2 . In addition, we designate Cd(OH)2 (s) as a possible solid. The system made by adding Cd(OH)2 (s) is simulated by replacing the inputs of T OT Cd and T OT NO3 with an input of 10 4 M Cd(OH)2 (s) as a finite solid. (Note that a given solid may be included as only one of the “Solid phases and excluded species” options. Therefore, when Cd(OH)2 (s) is added as a finite solid, it must be deleted as a “possible solid”.) For both simulations, the ionic strength is fixed at 0 M (to model ideal solution behavior). The output from the two runs is summarized below. Chemical added
Equilibrium pH
[Cd(OH)2 (s)](M)
Cd(NO3 )2
8.691
4.32 ⇥ 10
Cd(OH)2 (s)
10.002
9.90 ⇥ 10
5 5
TOTCddiss (M) 5.68 ⇥ 10 1.01 ⇥ 10
5 6
In the run where Cd is added as Cd(NO3 )2 , the pH drops when the solid precipitates, because the precipitation reaction removes OH– along with Cd2+ ; this decline in pH causes Cd to be considerably more soluble in this system than in the one analyzed previously (where the pH was fixed at 10.0). By contrast, the pH increases (albeit only slightly) when Cd(OH)2 (s) is added, because OH– enters solution as a small amount of the solid dissolves. By invoking the “Specify possible solid phases” option in conjunction with the “Multi-problem/sweep” feature, we can use Visual Minteq to prepare a log C-pH diagram for a system containing 10 4 M T OT Cd, assessing whether Cd(OH)2 (s) will precipitate at each pH analyzed. The results of such an exercise are shown in Figure 11.16a. At pH values less than 8.83, the total dissolved Cd concentration is 10 4 M, implying that no solid is present at equilibrium. Under these conditions, the speciation of dissolved Cd is identical to that shown in Figure 10.3, which was developed without ever considering possible precipitation of solids. However, at higher pH values (at least up to pH 14), the solid is present in the equilibrium system, so its 613
activity is 1.0, and T OT Cddiss is <10 4 M. Under these conditions, the speciation of dissolved Cd is identical to that in Figure 11.1, which was developed by assuming that Cd(OH)2 (s) was always present. Visual Minteq also reports the concentration of the solid in these systems (computed as T OT Cd T OT Cddiss ). These values are accessed via the “Amount of finite solids” button on the output page for an individual run or by selecting Cd(OH)2 (s) as one of the sweep output parameters on the “Multiproblem” page; the results for the current example are shown in Figure 11.16b.
Figure 11.16 Speciation of Cd in a system containing 10 4 M T OT Cd, considering possible precipitation of Cd(OH)2 (s). (a) Dissolved species. (b) Concentration of solid.
Almost all the Cd in the system precipitates at pH values between approximately 10 and 13, but the concentration of dissolved Cd increases dramatically when the pH drops below 9.2 or increases above 14.2. Because of this, adjusting the solution pH to between 11 and 12 appears to be a good approach for removing Cd from solution in, 614
for example, an industrial water treatment process. The solid could then be filtered out and reused or disposed of, and the water could be discharged. In addition to determining whether, and how much of, a specified solid will precipitate in these systems, Visual Minteq computes and reports on the solubility ‘status’ of each potential precipitate via the ion activity product (IAP) and the saturation index. The IAP has the same meaning as the activity quotient, Q, for the dissolution reaction, when the reaction is written in the Visual Minteq format (i.e., with all reactants and products other than the solid as components). The saturation index is defined as the difference log IAP log K and equals log Q/K for the dissolution reaction. If this value is positive, the thermodynamic driving force favors conversion of products to reactants (i.e., precipitation of the solid); if it is negative, the reverse is true; and if it is zero, the solution is saturated with respect to the solid. In the simulation used to generate Figure 11.16, log K for the dissolution reaction is 13.644, and the program reports that the activity of Cd2+ at pH 7.0 is 9.99 ⇥ 10 4 . The relevant reaction and calculations for the IAP and saturation index of Cd(OH)2 (s) are therefore Cd(OH)2 (s) + 2 H+ )* Cd2+ + 2 H2 O
IAPCd(OH)2 (s) = = Sat.Index ⌘ log IAP
(11.43)
(Cd2+ )(H2 O)2 (Cd(OH)2 (s)) (H+ )2 (9.99 ⇥ 10 4 )(1.0)2 = 1011.00 (1.0)(10 7.0 )2
log K = 11.00
13.644 = 2.644
(11.44) (11.45)
This result not only confirms that Cd(OH)2 (s) is undersaturated under the specified conditions, but also gives an indication of how far the solution is from equilibrium with that solid. The log IAP and/or saturation index for Cd(OH)2 (s) can be displayed in the Visual Minteq output by clicking the “Display saturation indices” button or including these parameters in the “Selected sweep” output list. For the current system, the saturation index is positive at pH < 8.82 and zero at higher pH values, consistent with our earlier conclusion that the solution is undersaturated with respect to the solid at pH < 11.82 and that the solid precipitates (so the solution reaches equilibrium with it) at higher pH’s. The program computes the solubility status of all solids that are in its database and that could be made from the given inputs, regardless of whether the solids are allowed to form in a given simulation. For instance, if we ran a simulation of a solution containing 10 4 M T OT Cd at fixed pH values from 4 to 14 but did not include 615
Cd(OH)2 (s) as an infinite, finite, or possible solid, we could nevertheless instruct the program to report the IAP or saturation index for the solid in each simulation. In this case, the output would show that the saturation index was negative at pH < 8.82, positive at pH > 8.82, and zero at pH = 8.82, but the program would ignore the solid when computing the solution speciation and would therefore report that T OT Cd was all dissolved at all pH’s. We can summarize the preceding discussion of systems in which only one, specified solid need be considered, and that solid might or might not be present at equilibrium, as follows: • If the initial solution is undersaturated with respect to the possible solid (Qs0 < Ks0 ), and if the solid is not present initially, it will not precipitate. • If the initial solution is undersaturated with respect to the possible solid and the solid is present initially, it will dissolve until either the solubility product is satisfied or no solid remains. • If the initial solution is supersaturated with respect to the possible solid (Qs0 > Ks0 ), the solid will precipitate until the solubility product is satisfied. • A log C-pH diagram can be prepared for a system with known T OT Me, showing both regions where the solid is present and where is absent. The diagram is, essentially, a composite of those that we have prepared previously when considering only one of the two possibilities. At the transition point between the two regions, the curve for each species is continuous, but the slope changes abruptly. • We can instruct Visual Minteq to take solids into account or to ignore them. In either case, the program can provide information about the thermodynamic ‘status’ of existing or potential solids. If a given solid is allowed to precipitate in the simulation, the program will report whether it is indeed present at equilibrium. If the solid is not allowed to precipitate, the program will report whether the solution is undersaturated or supersaturated with respect to the solid once all the solution-phase reactions reach equilibrium without precipitation of the solid.
11.8.2
Systems in Which More Than One Solid Must Be Considered
At this point, we have computed the concentration of Cd2+ in equilibrium with Cd(OH)2 (s) and CdCO3 (s) for a number of scenarios. However, in each case we assumed that only one solid could be present and that we knew which one it was. Consider next – 2+ a solution containing both CO2– is steadily added. Will 3 and OH , to which Cd 616
CdCO3 (s) precipitate? Will Cd(OH)2 (s)? Which solid will precipitate first, or be present at the highest concentration once equilibrium is reached? We now turn our attention to answering these questions. Possible precipitation of Cd(OH)2 (s) and CdCO3 (s) in an example system. For any value of T OT Cd in a system where both OH– and CO2– 3 are present, we can identify four possible equilibrium scenarios with respect to the solids: neither solid is present; Cd(OH)2 (s) is present and CdCO3 (s) is not; CdCO3 (s) is present and Cd(OH)2 (s) is not; or both CdCO3 (s) and Cd(OH)2 (s) are present. If we made an initial assumption that neither solid was present at equilibrium, we could compute the equilibrium speciation using the techniques described earlier and synthesized on the left side of Figure 11.14. The validity of the assumption could then be checked by computing the products (Cd2+ )0 (OH )2 and (Cd2+ )0 (CO23 ). If both solids were undersaturated based on this calculation, then the assumption would be confirmed and the problem would be solved. Similarly, if such an analysis indicated that one of the two solubility products was exceeded but the other was not, we would conclude that the former solid would be the only one present at equilibrium. The problem could then be solved by following the procedure described previously for a situation where it is known that a particular solid is present at equilibrium, as summarized by the right side of Figure 11.14. The only other possible outcome of the initial analysis is that, if no solid precipitated and all the solution-phase equilibria were satisfied, both solids would be supersaturated. Our first guess might be that such an outcome indicates that both solids will be present when the solution reaches equilibrium; however, this inference is not necessarily correct. To understand why, consider an ideal solution that initially contains 5 ⇥ 10 4 M T OT Cddiss but no solid, is well buffered at pH 9, and is in equilibrium with atmospheric CO2 . In such a system, precipitation of either solid would lower T OT Cddiss and the concentration of each dissolved Cd species, but (OH ) and (CO23 ) would remain constant — (OH ) because the pH is well buffered, and (CO23 ) because of the combined effects of a constant pH and equilibration of the solution with atmospheric CO2 (g). Calculations using Visual Minteq that account for the formation of Cd OH, Cd HCO3 , and Cd CO3 complexes in such a system indicate that, if no solid precipitated, the concentration of Cd2+ [i.e., Cd2+ )0 ] would be 5.67⇥10 5 M. The corresponding values of IAP (i.e., Qs0 ) for Cd(OH)2 (s) and CdCO3 (s) when the reactions 13.753 and 10 7.816 , are written with Cd2+ , H+ , and CO2– 3 as components would be 10 respectively, and the saturation indices would be 0.109 and 4.194, respectively. Thus, if no precipitate formed, both solids would be supersaturated. Using Visual Minteq to simulate the equilibrium composition of this system, but allowing only one of the two solids to precipitate, we obtain the results summarized in Table 11.2. 617
Table 11.2 Composition of a solution containing 5 ⇥ 10 4 MT OT Cd at pH 9.0 in equilibrium with atmospheric CO2 , with possible precipitation of Cd(OH)2 (s) or CdCO3 (s)
Solid allowed to precipitate T OT Cddiss
Conc’n of solid
(Cd2+ )
Saturaton Index Cd(OH)2 (s) CdCO3 (s)
Cd(OH)2 (s) 3.89⇥10
4
4.41⇥10
5
1.11⇥10
CdCO3 (s)
8
3.62⇥10
9
4.9997⇥10
3.20⇥10
4
0 4
4.085
4.085 0
The calculations indicate that, for the given pH and CO2 (g) partial pressure, the T OT Cddiss required for the solution to be in equilibrium with pure Cd(OH)2 (s) is higher than that required for equilibrium with pure CdCO3 (s). Clearly, a single solution cannot simultaneously have both values of T OT Cddiss , so we conclude that it is impossible for a solution to be in equilibrium with both solids under the specified conditions. The table also shows that the same conclusion could be reached by comparison of the Cd2+ activities required for equilibrium with the two solids, and further exploration of the output shows that it could be reached by comparison of any of Cd species. What, then, will the equilibrium condition be? To answer this question, consider the sequence of events that might ensue as the system moves from its initial condition toward equilibrium. Since both solids are initially supersaturated, it is reasonable to assume that both will begin to precipitate when the solution is first prepared. As they do so, Cd2+ will be removed from solution while the activities of H+ and CO2– 3 remain constant, so the solubility quotients of both solids (i.e., Qs0,Cd(OH)2 (s) and Qs0,CdCO3 (s) ) will decrease. This process is shown schematically in Figure 11.17a. Precipitation of both solids is expected to continue until the solubility quotient of one of them is reduced to the point where Qs0 = Ks0 . In the example system, this point is reached first for Cd(OH)2 (s), when T OT Cddiss has been lowered to 3.89 ⇥ 10 4 M. At this point, approximately 22% of the dissolved Cd has precipitated, and (Cd2+ ) has declined to 4.41 ⇥ 10 5 M (i.e., 10 4.36 M). CdCO3 (s) is still supersaturated under these conditions, so it continues to precipitate. This scenario is shown in Figure 11.17b, where the double-headed arrows and equal signs are used to indicate equilibrium. Note that the form of the precipitated Cd [i.e., whether most of the Cd has been removed from solution as CdCO3 (s) or Cd(OH)2 (s)] is unimportant for this portion of the analysis; all that matters is that the Cd2+ activity has declined enough for the solution to be in equilibrium with Cd(OH)2 (s). When the system conditions are as shown in Figure 11.17b, CdCO3 (s) is precipitating, Cd(OH)2 (s) is neither precipitating nor dissolving, and the solution pH is being held steady at 9.0. As a result, both (Cd2+ ) and the product (Cd2+ )(OH )2 de618
Figure 11.17 Schematic of precipitation and dissolution processes in an example system. (a) Both Cd(OH)2 (s) and CdCO3 (s) are supersaturated in the initial solution, so both start to precipitate. (b) After the Cd2+ concentration has declined by ⇠ 22%, the solution is in equilibrium with Cd(OH)2 (s), so that solid stops precipitating. CdCO3 (s) is still supersaturated. (c) CdCO3 (s) continues to precipitate, removing Cd2+ from solution. This causes Cd(OH)2 (s) to become undersaturated, so that the solid begins to dissolve. (d) Eventually, enough Cd2+ has been removed from solution so that the solution is equilibrated with CdCO3 (s). At this Cd2+ concentration, Cd(OH)2 (s) is undersaturated, so it is completely dissolved.
crease, causing Cd(OH)2 (s) to become undersaturated and inducing the Cd(OH)2 (s) that precipitated previously to begin to dissolve. That is, under these conditions, CdCO3 (s) is precipitating to reduce its degree of supersaturation, and Cd(OH)2 (s) is simultaneously dissolving in an attempt to bring the solution back to saturation with respect to that solid (Figure 11.17c). 619
The CdCO3 (s) is bound to win this battle for control of the solution composition, because as the reactions progress, CdCO3 (s) accumulates and Cd(OH)2 (s) is depleted. Eventually, the last bit of Cd(OH)2 (s) dissolves, and CdCO3 (s) continues to precipitate until T OT Cddiss is lowered to 3.20 ⇥ 10 8 M, and (Cd2+ ) is 3.62 ⇥ 10 9 M (i.e., 10 8.44 M). At that point, the system contains and is equilibrated with CdCO3 (s), and it is undersaturated with respect to Cd(OH)2 (s), which is not present. Thus, for the given system conditions, even though both solids were supersaturated initially, only CdCO3 (s) is present at equilibrium (Figure 11.17d). General criteria for equilibrium with multiple solids. Generalizing the preceding result, we infer that if a solution is supersaturated with multiple solids, any or all of those solids may precipitate when the system begins to approach equilibrium. As these reactions proceed, the solution composition and the extent of disequilibrium of all the solids change. Once the ultimate equilibrium condition is reached, the solution can be saturated with respect to one or more of the solids (which can therefore be present in the system), and it can be undersaturated with respect to other (potential) solids (which are not present). However, the only way that multiple solids can be present is if they are all in equilibrium with the same solution composition. Although CdCO3 (s) is the only solid present at equilibrium for the scenario that was just explored, by altering the constraints on the system composition, it is possible to identify other conditions that cause only Cd(OH)2 (s), both solids, or neither solid to be present. We next explore the conditions that lead to these outcomes. For convenience, we initially retain the assumption that Cd(OH)2 (s) and CdCO3 (s) are the only two potential precipitates, but we later expand the analysis to consider more complex systems. If both Cd(OH)2 (s) and CdCO3 (s) were in equilibrium with the same solution, the solids would also have to be in equilibrium with one another. We can determine the conditions necessary for this scenario to apply by writing the equilibrium reaction and corresponding equilibrium constant for conversion of one solid to the other. For example, by adding the reaction for dissolution of CdCO3 (s) to that for precipitation of Cd(OH)2 (s), we obtain CdCO3 (s) )* Cd2+ + CO2– 3
Ks0,CdCO3 (s)
Cd2+ + 2 OH– )* Cd(OH)2 (s)
1 Ks0,Cd(OH)
2 (s)
CdCO3 (s) + 2 OH– )* Cd(OH)2 (s) + CO2– 3
K=
Ks0,CdCO3 (s) Ks0,Cd(OH)2 (s)
(11.46)
If both pure solids were present, the activity of each would be 1.0,9 and the equilibrium constant for the preceding reaction would be 9 The assignment of a value of 1.0 for the activity of each solid presumes that the solids are present as distinct, pure phases. That is, if a pure CdCO3 (s) solid phase is present, its mole fraction in that phase and its activity would both be 1.0. If a pure Cd(OH)2 (s) is in the same system, its activity would also be 1.0. This situation differs from that in which the two solids are present as a solid solution, in which case they are intimately mixed in a single solid phase, each with a mole fraction < 1.0.
620
K=
(Cd(OH)2 (s)) (CO23 ) (CO23 ) 10 = = (OH )2 10 CdCO3 (s) (OH )2
12.01 14.36
= 10+2.35
(11.47)
Equation (11.47) indicates that, in any system equilibrated with both solids, the ratio of carbonate ion activity to the square of the hydroxide ion activity must be 102.35 . If, in a system containing both solids, this activity quotient is greater than 102.35 , the reaction will tend to proceed to the left [Cd(OH)2 (s) dissolving and CdCO3 (s) precipitating], and if it is less than 102.35 , the reaction will proceed to the right. As with any equilibration process, these changes tend to reduce the degree of disequilibrium, and the reaction would proceed until Q = K, at which point the two solids would be equilibrated with each other. However, if all of one solid dissolved before equilibrium was achieved, the other solid would be the only one in the system, and the equilibrium reaction involving the two solids becomes irrelevant. In the example system considered previously, the combination of fixed pH and fixed PCO2 maintained (OH ) at 10 4.0 and (CO23 ) at 10 1.57 , no matter how much of each solid precipitated or dissolved. As a result, (CO23 )/(OH )2 was maintained at 106.53 , equilibrium between the two solids could never be achieved, and Cd(OH)2 (s) could not be present at equilibrium. An even simpler criterion for equilibrium between Cd(OH)2 (s) and CdCO3 (s) can be derived by writing the reaction between these solids in terms of the H2 CO3 activity rather than the activities of OH– and CO2– 3 , as follows: CdCO3 (s) + 2 OH– )* Cd(OH)2 (s) + CO2– 3 – CO2– 3 + 2 H2 O )* H2 CO3 + 2 OH
CdCO3 (s) + 2 H2 O )* Cd(OH)2 (s) + H2 CO3
K = 102.35 Kw2 K= = 10 Ka1 Ka2 K=102.3510
(11.39) 11.32
(11.48)
11.32 =10 8.97 (11.49)
The form of the equilibrium expression for the reaction written in this way is simply (H2 CO3 ) = 10 8.97 . In other words, the two solids can be in equilibrium with one another only if the dissolved H2 CO3 activity is 10 8.97 . In the example system, the fixed PCO2 maintains the H2 CO3 activity at 10 4.89 , leading once again to the conclusion that the two solids cannot be equilibrated with one another. In fact, this result indicates that it is impossible for Cd(OH)2 (s) and CdCO3 (s) to be present together in an equilibrium solution in contact with atmospheric CO2 (g) not only at pH 10, but at any pH. If a solution in equilibrium with the atmosphere initially contained both solids, the Cd(OH)2 (s) would always dissolve completely as equilibrium was approached. The same result can be inferred by preparing a log C-pH graph showing the total dissolved Cd concentration in equilibrium with Cd(OH)2 (s) or CdCO3 (s) for a 621
system in equilibrium with the atmosphere (Figure 11.18). In this plot, the curve for equilibrium with CdCO3 (s) is always below that for equilibrium with Cd(OH)2 (s), consistent with the idea that a solution that is in equilibrium with Cd(OH)2 (s) and with atmospheric CO2 will always have “too much” dissolved Cd for equilibrium with CdCO3 (s); that is, it will always be supersaturated with respect to CdCO3 (s). (The values shown for T OT Cddiss in this figure are higher than those in Figures 10.1 and 10.9 because only Cd OH complexes were considered in the earlier figures, whereas Cd HCO3 and Cd CO3 are included here.)
Figure 11.18 The total dissolved Cd concentration in equilibrium with Cd(OH)2 (s) or CdCO3 (s) in a system equilibrated with atmospheric CO2 (g). All the Cd OH, Cd HCO3 , and Cd CO3 complexes in the Visual Minteq database have been included in the calculations for both solids.
Consider now the situation that would apply in a closed system containing 10 2 M T OT CO3,diss . If we plot the total dissolved Cd concentration in equilibrium with each of the two possible solids on a single log C-pH diagram for this system (Figure 11.19), we see that, unlike the case for an open system, the solid that is in equilibrium with the lower T OT Cddiss depends on solution pH. Following the same logic as in the system in equilibrium with atmospheric CO2 , we infer that, at each pH, the solid that induces the lower value of T OT Cddiss is the only one that could be present at equilibrium; such a solution would always be undersaturated with respect to the other solid (which, correspondingly, would not be present). On the other hand, if the solution equilibrated with the solid that induced the higher T OT Cddiss , the other solid would be supersaturated, indicating that the system was not fully equilibrated. At the intersection of the two curves, and only at that condition, both solids are in equilibrium with the same solution, and so both could be present at equilibrium. For the current scenario, this condition occurs at pH 11.82 and T OT Cddiss equal to 4.28 ⇥ 10 7 M. Under those conditions, the ratio (CO23 )/(OH )2 equals 102.35 , 622
Figure 11.19 Curves showing T OT Cddiss in equilibrium with Cd(OH)2 (s) or CdCO3 (s), for a system with fixed T OT CO3,diss of 10 2 M. Each curve divides the space into a region where the solid is undersaturated and a region where it is supersaturated. The two curves together therefore define four regions where zero, one, or both solids are supersaturated. The circled point indicates the only condition where the same T OT Cddiss is in equilibrium with both solids.
and (H2 CO3 ) = 10 8.97 , consistent with calculations shown in Equations (11.47) and (11.49). The key results of this analysis are summarized in Figure 11.19. The system can be at equilibrium anywhere that no solid is supersaturated. This space includes any point on the lower of the two curves or any point below both curves. If the system is characterized by a point on the lower of the two curves, then the solid corresponding to that curve can be present, but the other solid cannot; if the system is characterized by a point below both curves, then neither solid can be present; only if the system is at the point where the curves intersect can both solids be present. Although the preceding analysis was based on identifying the pH and T OT Cddiss that allow the two solids to coexist in the same solution, the criterion for coexistence could equally well be based on the concentration of any individual Cd species, instead of the sum of the concentrations of all such species. For example, a log C-pH diagram for the same system showing Cd2+ instead of T OT Cd (Figure 11.20) indicates that the concentration of Cd2+ in equilibrium with CdCO3 (s) is less than that in equilibrium with Cd(OH)2 (s) at pH < 11.82, that the reverse is true at pH > 11.82, and that at pH = 11.82, the same Cd2+ concentration is in equilibrium with both solids; a similar result would be obtained for a comparison of any dissolved Cd species in these solutions. The concentrations of all Cd species in an equilibrium system containing 10 4 M T OT Cd and 10 2 M T OT CO3,diss are shown over the pH range 3-14 in Figure 11.21, 623
Figure 11.20 The concentration of Cd2+ in equilibrium with either Cd(OH)2 (s) or CdCO3 (s), for the same system as shown in Figure 11.19.
with dissolved species shown in Figure 11.21a and the solids in Figure 11.21b. All of the Cd is soluble at pH < 5, with Cd2+ being the dominant species. At pH above ⇠ 5.5, on the other hand, almost all the Cd is present in solids, but the identity of the solid containing the Cd changes abruptly at pH 11.82: at pH < 11.82, only CdCO3 (s) is present, and at pH > 11.82, only Cd(OH)2 (s) is present. Only if the pH is exactly 11.82 can both solids be present. At the high-pH end of the plot, it is clear that Cd(OH)2 (s) is dissolving with increasing pH. At even higher pH, all of the solid would dissolve, and the solution would again contain no solids at all. Note that, consistent with all the prior discussion and with the principles explained in prior chapters: • Cd2+ and the various Cd OH complexes comprise a group of acid/base conjugate pairs. The activity of each acid equals that of its conjugate base at pH = pKa , regardless of which solid (if any) is present; • The slope of the curve for any acid is always one less than that of its conjugate base; • All the curves in the region where a single solid or no solid is present are identical to the corresponding curves derived earlier for a similar system where only that solid or no solid was considered; • The curve for every species is continuous across the whole diagram, but the slope of each curve is discontinuous at transition points where a solid appears and/or disappears. 624
Figure 11.21 Cd speciation in a system containing 10 4 M T OT Cd and 10 2 M T OT CO3 , considering possible precipitation of Cd(OH)2 (s) or CdCO3 (s). (a) Soluble Cd species; (b) Cd-containing solids. Values on the y axis extend to negative number so that the conditions where the solids’ concentrations are zero can be seen more clearly.
Graphs like those shown in Figures 11.19 through 11.21 provide an overview of both the equilibrium and nonequilibrium possible states of the system, tell us which solids might precipitate in a given system, and help us identify the conditions that must be met for multiple solids to be present at equilibrium. In systems where the total concentration of only one component changes significantly as the system equilibrates, the graphs can also be used to determine the exact conditions that will apply at equilibrium. This was the case in the first example we considered, in which T OT Cd 625
was the only variable that could change [because the final pH was specified and the concentrations of the carbonate species were fixed by the known pH and the assumption of equilibrium with atmospheric CO2 (g)]. A similar situation often applies when considering precipitation of a metal or ligand that is present at a trace concentration, so that, even if the vast majority of the species precipitates, the precipitaton reaction has a negligible effect on the concentration of any other constituent of the solution.
11.8.3
Visual Minteq Modeling of Systems with Potential Solids
Typically, our interest is in more complicated systems than those considered in the preceding sections. For example, we might want to model systems in which the total concentrations of both the anion and cation that form a solid can change significantly as a result of precipitation, or one of the precipitating species is an acid or base and the precipitation reaction causes a nontrivial change in the solution pH. In these situations, log C-pH diagrams are useful for understanding the initial conditions and the possible changes as precipitation proceeds, but a program such as Visual Minteq is needed to predict the ultimate, equilibrium conditions. The use of Visual Minteq to solve these kinds of problems involves a straightforward extension of the analysis for cases where only one solid need be considered. We simply include all the solids to be considered in the list of possible solid phases, along with their initial concentrations, and the program solves for the equilibrium condition. Example 11.7 A wastewater at pH 7.5 and with an ionic strength of 0.01M contains the following constituents: Total ferrous iron [T OT Fe(II)]
2.0 mg/L
T OT Ca
65 mg/L
T OT PO4 P
10 mg/L
T OT NH4 N
140 mg/L
Alkalinity
2.5 meq/L
(a) What is T OT CO3 in the wastewater, assuming that any alkalinity not accounted for by the species listed is contributed by the carbonate group? (b) Find the ion activity products of any solids that might precipitate in the solution, based on the Visual Minteq database. Which of these solids are supersaturated in the given solution? (c) Which solid(s) will be present once the solution reaches equilibrium, and what will their concentration(s) be? (d) What fraction of the T OT Ca, T OT Fe, and T OT PO4 will remain in the equilibrated solution? 626
Solution (a) The alkalinity includes contributions from the phosphate, ammonia, and carbonate acid/base groups. Because the pH is 7.5, essentially all of the ammonia is in the form of NH+4 , which is the same form it is in at the end of the alkalinity titration. (A small amount of ammonia is present in its deprotonated form as part of Fe NH3 complexes, but given the low concentration of T OT Fe compared to that of T OT NH4 N, that NH3 can be assumed to make a negligible contribution to the alkalinity.) We can therefore estimate the alkalinity contribution from the carbonate species as the difference between the total alkalinity and the contribution from the phosphate species. This calculation can be carried out by Visual Minteq by inputting the total alkalinity on the Parameters page, along with the known pH and T OT PO4 on the main input page. If, in addition, we input the specified values for T OT Ca, T OT NH4 N, and ionic strength, we can run the program and find T OT CO3 reported as part of the output, on the “Equilibrated mass distribution” page. When we do so, we find T OT CO3 is 2.185 ⇥ 103 M. (b) In the output from the same run as described in part (a), we can look at the “Display saturation indices” page, which is reproduced below. The output indicates that Ca4 H(PO4 )3 · 3 H2 O(s), hydroxyapatite, siderite, and vivianite are all supersaturated in the initial solution, while calcite and Fe(OH)2 (am) are not.
(c) If we list the solids identified in the problem statement in part (b) as “Possible solid phases” and then run the program, we find that even though several solids are supersaturated in the initial solution, only two — hydroxyapatite and siderite — are actually present at equilibrium. The reason for this is that when these two solids precipitate, they change the composition of the solution enough that the other solids which were supersaturated initially are no longer 627
supersaturated. The concentrations of the two solids at equilibrium are displayed by clicking the “Amount of finite solids” button on the output page; the reported values are 1.08 ⇥ 10 4 M hydroxyapatite and 2.27 ⇥ 10 5 M siderite. (d) The fractional distribution of all the components between the solution and solids is displayed on the “Equilibrated mass distribution” page, a portion of which is shown below. About two-thirds of the T OT Ca and one-third of the T OT Fe remain in solution when the system reaches equilibrium, but more than 99.9% of the T OT PO4 precipitates as hydroxyapatite.
11.8.4
The Relative Solubilities of Solid Polymorphs
It is fairly common for several solids to have the same chemical formula and to differ from one another only in their crystal structure. Such solids are called polymorphs of one another. In other cases, the compositions of two solids differ only in the number of water molecules incorporated into their structures. In such cases, the form of the solubility product for the solids is essentially identical. For instance, the dissolution reactions and the solubility product expressions for four Fe(III) oxide or hydroxide solids are shown in Table 11.3. The final equalities in each row of the table indicate that, for any of these solids to be in equilibrium with a solution, the product (Fe3+ )(OH )3 must be a specific value. Obviously, since this product can have only one value in a given solution, no two of the solids can be equilibrated with the same solution. Furthermore, the relative values of Ks0 for the different solids indicate that the solubility decreases in the order ferrihydrite > lepidocrocite > goethite > hematite. As a result, any solution in equilibrium with one of the first three solids will be supersaturated with respect to hematite. The conclusion is that no solution that contains any of the first three solids can be an equilibrium solution; conversion of the solid to hematite will always be favored. Nevertheless, the solids listed in the upper rows of the table are significant in the natural environment, and they can be prepared and used in the laboratory. These solids are commonly referred to as being metastable, meaning that they do not normally convert to more stable solids in the time frame of interest. In such cases, we 628
can treat any of the solids as though it were the most stable one that could form in the system, ignoring its conversion to more stable polymorphs. This situation arises commonly when we wish to model the initial formation of solids, such as in cases where metals are precipitated as part of a water or wastewater treatment process. The multiplicity of potential solids that might form in such situations makes clear the importance of specifying which solids are being considered in any given analysis.
11.9
THE GIBBS PHASE RULE
Throughout this text, starting with the discussion of manual approaches for solving acid/base problems in Chapter 5, we have formulated the approach for determining the equilibrium composition of a solution as involving two steps: identifying a group of unknowns, and then identifying a set of simultaneous equations that can be solved to determine those unknowns. In each case, the number of equations had to match the number of unknowns, and a group of other parameter values had to be known in advance (things like equilibrium constants; the total amounts of certain chemicals in a system; the fixed activity of some species, etc.). The relationship among the total number of variables in a chemical equilibrium problem, the number of variables that can and/or must be specified, and the number of variables that can then be determined by solving the available equations is formalized in a statement known as the Gibbs phase rule (or, simply, the phase rule). This rule states that the number of degrees of freedom F in a chemical system equals the number of components C minus the number of phases P plus two: F =C P+2 (11.50) The number of degrees of freedom indicates how many independent constraints can be (and need to be) specified to completely define the system composition at equilibrium. These constraints can be system parameters, such as temperature or pressure, the activity of a particular chemical species, or the total concentration of a group of species (e.g., T OT CO3 ), or they can be other physical/chemical laws, such as the requirement that the total charge on the dissolved cations equal that on the dissolved anions. The term ‘components’ as used in the phase rule is defined in different ways by different people, but these differences are primarily semantic; in the end, the same set of equations must be solved regardless of how the term is defined. For our purposes, the most convenient definition of ‘components’ for use with the Gibbs phase rule is the same one we use when analyzing a problem using chemical equilibrium software; in fact, the phase rule was the basis for choosing the term ‘components’ for use in the software jargon. Therefore, the number of components in any system (i.e., the value of C in the phase rule) can be identified as the minimum number of chemicals that can be used to make all the species present in the equilibrium system, subject to the constraint that no component can be made solely by combining other components. 629
630
Dissolution Reaction
Ks0 = Ks0 = Ks0 =
a-FeOOH(s) + H2 O )* Fe3+ + 3 OH–
0.5 a-Fe2 O3 (s) + 1.5H2 O )* Fe3+ + 3 OH–
Goethite a-FeOOH
Hematite a-Fe2 O3
Ks0 =
{Fe3+ }{OH }3 = {Fe3+ }{OH }3 = 42.71 {a-Fe2 O3 (s)}0.5 {H2 O}1.5
{Fe3+ }{OH }3 = {Fe3+ }{OH }3 = 41.51 {a-FeOOH(s)}
{Fe3+ }{OH }3 = {Fe3+ }{OH }3 = 40.63 {g-FeOOH(s)}
{Fe3+ }{OH }3 = {Fe3+ }{OH }3 = 38.8 {Fe(OH)3 (s)}
Ks0 Expression
Lepidocrocite g-FeOOH(s) + H2 O )* Fe3+ + 3 OH– g-FeOOH
Ferrihydrite Fe(OH)3 (s) )* Fe3+ + 3 OH– (amorphous iron hydroxide)
Solid
Table 11.3 Equilibrium constants for dissolution of four forms of iron oxyhydroxide.
The phases in a chemical system might include a gas, one or more liquids, and a number of different solids. Phases are defined by the possibility of mixing at the molecular level and by the presence of an interface with any other phase. Thus, each distinct solid in a system (each solid with a different composition or crystal structure) represents a separate phase. For instance, a system containing metallic iron, an iron oxyhydroxide solid (e.g., goethite), and iron carbonate solid would have three solid phases. On the other hand, a solid that contains different ions more or less randomly distributed in the crystal matrix [e.g., a solid consisting of a CaCO3 (s) crystal structure in which 10% of the Ca2+ ions are randomly replaced by Cd2+ ions] would be a single phase. A well-ordered crystal containing repeating units or layers of SiO2 and Al2 O3 (e.g., aluminosilicate clays) is also treated as a single phase. Although multiple liquid phases exist in some environmental systems of interest (e.g., organic solvents in some contaminated groundwaters), the focus of this text is strictly on aqueous solutions, so we will only consider systems with one liquid phase. Because gases always mix at the molecular level and never form an interface with
In the past, wastes generated by industrial processes such as electroplating were a major source of metals to wastewater treatment plants and receiving waters. Improvements in process control and in onsite industrial wastewater treatment have dramatically reduced the metal load discharged to the environment from these sources. (Courtesy of ASKO Processing, Inc.)
631
one another, there can be only one gas phase in a given system at equilibrium. One of the simplest examples of the application of the phase rule is for a system containing only H2 O(l), H+ , and OH– . Any two of these three species can serve as components for the system for the purposes of the Gibbs phase rule; the remaining species cannot be a component because it could be made by a combination of the other two. Therefore, for this system, C = 2. The system consists of only one phase (P = 1), so according to Equation (11.50), F = 3. One of the degrees of freedom is associated with the requirement for electroneutrality, and the other two are normally used to specify the system’s temperature and pressure. Once these parameters are specified, the state of the system is completely defined. (The temperature and pressure establish the value of Kw , and that value plus the electroneutrality condition establish the equilibrium concentrations of H+ and OH– .) If, in a system containing only H2 O, some of the liquid water froze, the system would have two phases (ice and liquid water), and the number of degrees of freedom would decrease to two. In such a case, in addition to the electroneutrality requirement, we could arbitrarily specify the pressure or temperature, but not both. For example, if the pressure were 1.0 atm, the temperature would be 0 C; it is simply not possible to have an equilibrium system at 1.0 atm total pressure that contains both ice and water (and no other components) and that is at any temperature other than 0 C. Similarly, if we specified that all three phases of H2 O were present (P = 3) in a system containing only H2 O, H+ , and OH– , the system would have to be at the triple point of water; i.e., both temperature and pressure would be constrained to known values. In almost all systems of interest in this text, the temperature and total pressure of the system are assumed to be known (typically 25 C and 1.0 atm). Even in cases where values of these parameters are not specified explicitly, they are implicitly established when we state the equilibrium constants for the reactions that take place in the system. Furthermore, since we are always interested in systems that contain dissolved ions, the charge balance places another constraint on the system. Therefore, the systems of interest always have at least three prespecified constraints, so the number of remaining degrees of freedom (F) in these systems is C P 1. Although we have not considered the phase rule explicitly in previous chapters, the analyses we have conducted have always been consistent with it. For instance, if a solution is made by adding HAc to water, the system has three components. These components might be chosen to be H2 O, H+ , and Ac– , or H2 O, H+ , and HAc, or any of a variety of other combinations (including combinations of imaginary chemicals). However, no matter what component set is chosen for the system, it must comprise three chemicals. Therefore, for such a system, C = 3, P = 1, and F = 1. This degree of freedom is utilized when we specify that the total acetate concentration is 10 3 M or that the pH is 5.0, or make any other chemical assignment. Once this degree of freedom is utilized, the equilibrium composition of the solution is fully established. 632
It is not possible to determine a unique equilibrium composition for a system in which the number of parameters that have been specified is less than the number of degrees of freedom. An example would be to ask the question, What is the pH of a solution made by adding HAc to pure water? Obviously, any pH could be obtained, depending on the concentration added.10 Similarly, it is usually not possible to find any equilibrium composition that is consistent with more constraints than specified by the number of degrees of freedom. An example of this situation would be to ask the question: What is the concentration of Ac– in a system made by adding 10 3 M HAc to pure water if the pH is 6.5? The problem with this question is that the pH of a solution of 10 3 M HAc is fixed by the given addition of HAc, and one cannot cause such a solution to have a pH of 6.5 (except by adding a base, like NaOH, which then changes the number of components in the system). Although this particular violation of the phase rule is obvious and would be unlikely to go unnoticed, one might inadvertently describe a system that violates the phase rule when dealing with complex systems having lots of phases. Unfortunately, Gibbs phase rule violations are sometimes reported incorrectly by chemical equilibrium software. Conformance with the Gibbs phase rule is checked by most software packages before each iteration, and if a violation occurs, the program terminates. The user is then alerted to the problem and must correct the error. If the violation occurs on the first iteration, chances are that too many species were input with known, fixed activities, in which case the problem can be solved by reducing the number of such specifications. However, phase rule violations can sometimes be generated by the program in the midst of solving the problem, if it “guesses” that more solids are present than can truly be present at equilibrium. This type of phase rule violation is software-related and can be overcome by instructing the software to ignore some of the possible precipitates that might form. The user must then check the computed final equilibrium condition to see if the ignored solids might in fact precipitate (by comparing Qs0 with Ks0 ). Some strides have been made to overcome this problem, but it still can occur occasionally. Example 11.8 Apply the Gibbs phase rule to determine the number of degrees of freedom in the following solutions. Assume that the temperature and total pressure of each system are fixed. 10 The phase rule is, in effect, a statement of mathematical possibilities, not all of which are chemically realistic. Mathematically, one could meet all the constraints of the equilibrium constants and the mass balance and charge balance equations by specifying a negative total concentration of a component. In this way, one could even get the pH to an alkaline value by adding “negative” amounts of HAc. Thus, mathematically, it is possible to choose any value one wants for pH in the example system (i.e., the choice really is unconstrained), even though achieving that pH might not be possible in a realistic system.
633
(a) A mixture of sodium carbonate and sodium bicarbonate that has been adjusted to pH 7.0 by addition of acetic acid. (b) The solution in part (a), if it subsequently equilibrates with atmospheric CO2 (g). (c) A solution in equilibrium with both CdCO3 (s) and Cd(OH)2 (s), in which T OT CO3,diss is 10 2 M. The solution also contains some Na+ , added along with some of the carbonate. (d) The same solution as in part (c), but prepared by adding only H2 CO3 , CdCO3 (s), and Cd(OH)2 (s) to water, so that the solution contains no Na+ . Solution Each of the solutions of interest contains ions, so the electroneutrality condition in solution is a constraint in addition to the temperature and pressure. Therefore, the number of degrees of freedom is given by F = C P 1. This value of F in each system is then determined as follows. (a) The solution can be made by a combination of five components (e.g., H2 O, H+ , Na+ , H2 CO3 , and Ac– ), and it contains only one phase (the solution). Thus, it has three degrees of freedom. One degree of freedom is associated with the constraint that the pH is 7.0. We could use the other two degrees of freedom to specify, for example, the concentrations of sodium carbonate and sodium bicarbonate added, or the total concentrations of carbonate and acetate in the system, or any other two parameters that characterize the system. (b) The number of components in the system is the same as in part (a), because CO2 (g) can be made from the existing components. However, we do have an additional phase to consider (the gas phase), so the number of degrees of freedom is reduced to one. Thus, if we specified T OT Ac, for example, the system would be completely defined. (c) This system contains five components (e.g., H2 O, H+ , Na+ , Cd2+ , and CO2– 3 ) and three phases, so it has one degree of freedom (F = 5 3 1 = 1). This degree of freedom is ‘used’ when we impose the constraint on T OT CO3 ). Thus, the system composition at equilibrium is fully defined. This result is consistent with the conclusion we reached earlier, that a solution equilibrated with 10 2 M T OT CO3,diss can be in equilibrium with both CdCO3 (s) and Cd(OH)2 (s) only at pH = 11.82. Given the pH and known T OT CO3 concentration, the concentrations of all the Hx COx3 2 species are also established under these conditions. (d) The system contains only four components, but it still has three phases, and we are trying to impose the constraint that T OT CO3 is 10 2 M. This system is therefore overly constrained (F = 1) and violates the phase rule; no system can meet the specified conditions and be at equilibrium.
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11.9.1
Integrated Analysis of a Model System: Titration of a CarbonateContaining Solution with Zn(NO3 )2
To illustrate the integrated application of many of the concepts presented in this chapter, we next carry out a detailed analysis of an aqueous solution containing a variable concentration of Zn and a known initial concentration of total carbonate species. Specifically, we analyze the composition of a solution of 0.10 M ZnCl2 when it is mixed with another that contains 0.02 M Na2 CO3 in volume ratios ranging from 0.003:1 to 0.3:1. This scenario might be of interest, for instance, if an industry generated waste streams that could be reasonably represented by these simplified solutions, and engineers wanted to identify mixing ratios that would lead to the maximum possible precipitation of Zn (e.g., to meet a regulatory requirement that T OT Zn remain less than some specified value). In the analysis, we consider the interactions of Zn with hydroxide, carbonate and chloride ligands, the possible precipitation of Zn(OH)2 (s) and ZnCO3 (s), and the effects of these interactions on solution pH. We also compute the ionic strength for each mixture and use that information to estimate activity coefficients of all the solutes. The analysis can be carried out using the titration function in Visual Minteq. We can choose either of the solutions as the ‘main’ solution, and the other as the titrant. Choosing the Na2 CO3 solution as the main one, we begin by inputting the appropriate information for T OT Na and T OT CO3 . The solution pH and ionic strength will be calculated when the program runs, so we leave those selections at their default values. In order to assure that the tableau shows all the species that are expected to form once some of the ZnCl2 solution has been mixed into the Na2 CO3 solution, we also list Zn2+ and Cl– as components of the first solution, but we specify that both T OT Zn and T OT Cl are zero. When we do this, the tableau of aqueous species is as shown in Figure 11.22. Next, we specify that both Zn(OH)2 (s) and ZnCO3 (s) are possible solid phases. The database includes a number of solids with each of these stoichiometries. For this example, we assume that amorphous zinc hydroxide [shown in the list as Zn(OH)2 (am)] and unhydrated zinc carbonate [shown as ZnCO3 (s)] are the two solids that can form. We then move to the “Multi-problem/Sweep” menu, check the box labeled “Titration/mixing with a titrant with defined composition”, and click on the “Go to Titration Manager” button. In the Titration manager, we can input an arbitrary value for the volume of the main (Na2 CO3 ) solution to be titrated. Choosing a value of 1, we then enter the volume of titrant to be added in each step. Since the lowest volume ratio we wish to investigate is 0.003, we enter that value in the box for the “Volume of titrant”. It is also convenient to check the box specifying that no titrant is to be added until the second step, so that the first calculation is for the ‘pure’ Na2 CO3 solution. We next define the composition of the titrant solution, which in our case consists of only two components: 0.1 M Zn2+ and 0.2 M Cl– . After saving this information, we return to the “Multi-problem” menu and indicate that we wish to include 100 titration 635
Figure 11.22 Aqueous species considered by Visual Minteq in the simulated titration of 0.02 M Na2 CO3 with 0.1 M ZnCl2 .
points (100 additions of the ZnCl2 solution, each with a volume equal to 0.003 times the volume of the initial, Na2 CO3 solution). Finally, we select the parameters that we wish to track during the titration. The parameters of interest include the concentrations of the two solids and all the soluble, Zn-containing species, and possibly some other parameters (e.g., the IAPs of the two solids). Once these choices are made, we return to the main menu and run the program. The results of the simulations can be summarized in a series of graphs. First, Figure 11.23 shows the total concentrations of Zn and CO3 (including both soluble species and solids) at various mixing ratios (designated here as r). Even though the total mass of carbonate species in the system is constant throughout the process, T OT CO3 declines throughout the titration because the total volume of the solution is increasing. For the same reason, T OT Zn increases nonlinearly as r increases. Figure 11.24 shows the solution pH and the distribution of T OT Zn among the three possible phases in which it can exist; it also includes the T OT Zn curve from Figure 11.23 for reference. This figure shows that the titration consists of three distinct stages. At mixing ratios from r = 0.003 to r ⇡ 0.021 (Stage 1), virtually all of the Zn that is added precipitates as Zn(OH)2 (s). In this stage, no ZnCO3 (s) is present at all, and the concentration of dissolved Zn is negligible compared to the precipitated Zn(OH)2 (s). (The numerical data indicate that > 80% of the T OT Zn precipitates even at the lowest value of r investigated [0.003].) During this stage, the pH drops from an initial value of 11.1 to approximately 10.5. Stage 2 begins when r exceeds 0.021, at which point abrupt changes occur in the trends of several of the parameters: ZnCO3 (s) begins to precipitate, Zn(OH)2 (s) begins to dissolve, and the decline in pH becomes much more gradual. This stage 636
Figure 11.23 Concentrations of T OT Zn and T OT CO3 at various mixing ratios for titration of 0.02 M Na2 CO3 with 0.1 M ZnCl2 .
persists up to r ⇡ 0.20, when the last bits of Zn(OH)2 (s) that had precipitated at lower mixing ratios dissolve, and ZnCO3 (s) accounts for virtually all of the Zn in the system; i.e., a negligible portion of the T OT Zn is in solution.
Figure 11.24 Distribution of T OT Zn among the three phases (the solution and the two solids) at various mixing ratios in the example system.
As r is increased above ⇠0.20 (Stage 3), the precipitated ZnCO3 (s) starts dissolving and no other solid forms, so the total dissolved zinc concentration increases slightly more rapidly than the rate at which zinc is added with the titrant. In addition, the pH undergoes a dramatic decline at r slightly larger than 0.20, and then continues to decline (albeit more slowly) as r increases to 0.30. 637
Consistent with the observations about each solid, the solubility quotient for Zn(OH)2 (s) (i.e., Qs0,Zn(OH)2 (s) ) equals the corresponding solubility product (Ks0,Zn(OH)2 (s) ) in Stages 1 and 2 of the process and is less than Ks0,Zn(OH)2 (s) in Stage 3, whereas the solubility quotient for ZnCO3 (s) is less than its solubility product in Stage 1, but equal to it in Stages 2 and 3 (Figure 11.25).11
Figure 11.25 Comparison of the solubility quotient for the two solids in the example system.
The concentrations of the species that contribute 10 7 M or more to the total dissolved Zn concentration are shown in Figure 11.26. In both Stages 1 and 2, the dominant dissolved species is Zn(OH)2 , with Zn(OH)–3 and the Zn CO3 complexes contributing slightly. Note that whenever Zn(OH)2 (s) is present, the concentration of the dissolved, dihydroxo complex [Zn(OH)2 ] is very nearly steady at 10 4.42 M, and whenever ZnCO3 (s) is present, the concentration of the complex ZnCO3 is similarly steady at 10 6.04 M. These results stem from the fact that we can write reactions forming these two dissolved species directly from the corresponding solids, without production or consumption of any other species:
Ks2,Zn(OH)2 (s) = Ks1,ZnCO3 (s) =
Zn(OH)2 (s) )* Zn(OH)2
(11.51)
ZnCO3 (s) )* ZnCO3
(11.52)
{Zn(OH)2 } = {Zn(OH)2 } = 10 {Zn(OH)2 (s)} {ZnCO3 } = {ZnCO3 } = 10 {ZnCO3 (s)}
6.04
4.42
(11.53) (11.54)
11 In Figure 11.25, the IAP values reported by Visual Minteq for Zn(OH) (s) have been converted to 2 the conventional Qs0 values via the relationship Qs0 = IAP ⇥ Kw2 . This conversion makes the reaction for dissolution of Zn(OH)2 (s) similar to that of ZnCO3 (s) (both reactions have Zn2+ and free ligands as products) making it more convenient to plot and compare the data on a single graph.
638
Figure 11.26 Significant soluble Zn species in the example system.
Because these species are uncharged, their activity coefficients remain very close to 1.0, even if the ionic strength changes significantly. (Recall that Visual Minteq uses Equation (2.8) to compute activity coefficients for neutral species, so those activity coefficients are slightly different from 1.0.) The combination of a constant chemical activity and a nearly constant activity coefficient causes the concentrations of these species to be essentially constant whenever the solid is present. Finally, the distribution of carbonate in the system is shown in Figure 11.27. All the carbonate in the initial solution remains there, primarily as CO2– 3 , until ZnCO3 (s) starts to precipitate. Thereafter, soluble carbonate is converted to ZnCO3 (s) approximately linearly with increasing r until, at r ⇡ 0.20, virtually all of the carbonate has precipitated. Having reviewed the changes in individual system parameters during the titration, we can now interpret the results in a more integrated fashion and gain additional insight into the key reactions. In Stage 1, almost all of the zinc that enters the system is precipitated as Zn(OH)2 (s), so the dominant reaction of the added titrant is ZnCl2 + 2 OH )* Zn(OH)2 (s) + 2 Cl
(11.55)
This reaction consumes OH– and therefore lowers the pH of the solution, as indicated in Figure 11.24. The high pH of the original solution and the insolubility of Zn(OH)2 (s) combine to keep the soluble Zn concentration low (⇠ 3.8 ⇥ 10 5 M, or about 2.4 µg/L) during this stage, and low enough that ZnCO3 (s) remains undersaturated. However, as the pH declines, the Zn2+ activity and concentration required to maintain equilibrium with Zn(OH)2 (s) increase, leading to a corresponding increase in Qs0,ZnCO3 (s) . When r = 0.021, Qs0,ZnCO3 (s) equals the solubility product for ZnCO3 (s), and that solid begins to precipitate, initiating Stage 2. 639
Figure 11.27 Distribution of carbonate-containing species between the dissolved and solid phases in the example system.
Figure 11.24 indicates that, throughout Stage 2, the concentration of ZnCO3 (s) increases steadily, while that of Zn(OH)2 (s) declines, and a negligible portion of the total Zn is in solution. The increase in the concentration of ZnCO3 (s) is much larger than the decrease in the Zn(OH)2 (s) concentration, indicating that a substantial portion (in fact, virtually all) of the added Zn is precipitating as ZnCO3 (s). Because the pH during most of this stage is between 10.5 and 9.9, we can make the rough approximation that half of the carbonate that reacts is in the form of HCO–3 , and half in the form of CO2– 3 , so the stoichiometry of the precipitation reaction can be approximated as ZnCl2 + 0.5 HCO3 + 0.5 CO23 )* ZnCO3 (s) + 2 Cl + H2 O + 0.5 H+ (11.56) This reaction generates protons which, in the absence of other reactions, would cause the pH to decline. However, that decline is mitigated (i.e., the pH is buffered) both by protonation of some of the dissolved CO2– 3 and by conversion of a small amount of Zn(OH)2 (s) to ZnCO3 (s): Zn(OH)2 (s) + 1.5 H+ + 0.5 HCO3 + 0.5 CO23 )* ZnCO3 (s) + H2 O
(11.57)
The combination of Reactions (11.56) and (11.57) accounts for all the important changes in the solution during Stage 2 — the precipitation of ZnCO3 (s), the dissolution of Zn(OH)2 (s), the decrease in T OT CO3,diss , and the slight decline in pH. (Although Reaction (11.57) consumes more H+ than is released by Reaction (11.56) per mole of ZnCO3 (s) precipitating, Reaction (11.56) proceeds to a greater extent than Reaction (11.57) when an aliquot of ZnCl2 is added, so the net result is H+ release to solution.) 640
The fact that both solids can be present in Stage 2 while the pH changes might seem surprising in light of the discussion in prior sections, in which we found that Cd(OH)2 (s) and CdCO3 (s) either could not coexist at all or could coexist at only a single pH value. The difference between those cases and the current one is that, in the analysis of Cd precipitation, we imposed an additional constraint that either PCO2 (g) (and therefore the activity of H2 CO3 ) or T OT CO3,diss was constant. In the current analysis, no such constraint is imposed; the activity of H2 CO3 and T OT CO3,diss are free to take on whatever values are needed for both solids to be present at the equilibrium pH, and the system adjusts to meet those needs at each pH. Stage 2 ends when all of the Zn(OH)2 (s) has dissolved, most of the carbonate has been removed from solution by precipitation of ZnCO3 (s), and the pH has dropped to around 9.5, so that most of the carbonate remaining in solution is present as HCO–3 . All of these changes reduce the solution’s buffer intensity, so that precipitation of just a little more ZnCO3 (s) causes the pH to plummet. As a result, all of the Zn added at this point in the titration remains in solution. Early in Stage 3, the pH is in a range where much of the Zn2+ hydrolyzes, so a substantial fraction of the added Zn undergoes the following reaction, causing the pH to drop even further: ZnCl2 + H2 O )* ZnOH+ + H+
(11.58)
However, because of the low buffer intensity of the solution at this point, the pH quickly drops to values where hydrolysis is no longer favored. At that point, the added ZnCl2 simply splits into Zn2+ and Cl– ions, and the pH stabilizes. The preceding analysis demonstrates the wealth of information and understanding that can be acquired when the simple concepts of acid/base and metal/ligand chemistry are combined with the power of computer-based numerical solution techniques. The numerical output from the analysis could be generated in a matter of minutes, and the results include aspects that would have been very difficult to foresee by a strictly intuitive or conceptual analysis. The same results could have been obtained using a numerical or graphical analysis, but either of these approaches would have required an immense amount of time and effort. On the other hand, using a program to generate the output without having a good grasp of the underlying chemistry would leave us with a correct numerical solution, but no useful insights into why the system behaves the way it does. The analysis thus serves as an example of the synergism that can be achieved between computational and conceptual approaches to these types of problems.
11.10
PREDOMINANCE AREA DIAGRAMS CONSIDERING POSSIBLE PRECIPITATION OF SOLIDS
The final section of this chapter returns to the preparation of predominance area diagrams, extending the analysis in Chapter 10 to include the possibility that a solid will 641
precipitate. Often, this aspect of predominance area diagrams is what makes them valuable, because they can tell us the broad conditions under which a metal is expected to be present primarily in soluble versus particulate form. This distinction is often of greater interest than the determination of regions where particular dissolved species are dominant over other dissolved species. The inclusion of solids in predominance area diagrams raises a few issues that need not be considered if only soluble species are being considered. First, if only soluble species are being considered, the diagrams are independent of the total concentration of metal in the system.12 That is, for example, the ratio of the activities of PbOH+ and PbCl+ depends on {OH– } and {Cl– } but not on T OT Pb, just as the activity ratio {Ac }/{HAc} depends on pH but not on T OT Ac. However, when considering the possibility that a solid might form, the total concentration of Pb in the system becomes an important variable because, at a given pH, the solid can be present at high T OT Pb and absent at low T OT Pb, even if all other system conditions remain the same. Furthermore, because the activity of the solid is 1.0 regardless of how much solid is present, we need to be clear that, in predominance area diagrams, ‘predominance’ refers to activity, not concentration. Therefore, since such diagrams are almost always prepared for systems with the constituents present at total concentrations less than 1.0 M, the solid is considered to be dominant whenever it is present, and soluble species are considered to be dominant only if no solid is present. Predominance area diagrams that include consideration of solids are prepared by following essentially the same steps as described in Chapter 10 for the comparable diagrams in the absence of solids. That is, we begin by writing the reaction and evaluating the equilibrium expression for each possible pair of species. Keep in mind that these relationships are used only to establish the boundaries between regions where the two species being compared are the dominant species. Thus, whenever one of the species is a solid and the other is a solute, the two possibilities are that the solid is present (in which case we treat it as being dominant) or the solid is absent and the soluble species is dominant. To determine whether a solid is present, we first assume that it is not present, and then test that assumption by determining whether the solution composition is undersaturated or supersaturated with respect to the solid. When carrying out these calculations, it is common to make the approximation that the dominant soluble species is present at a concentration equal to T OT Me. Thus, for instance, if PbOH+ has been determined to be the dominant soluble Pb species in a certain region of a predominance area diagram, and we wish to determine whether Pb(OH)2 (s) would precipitate 12 This
statement is strictly correct only if all the dissolved species are monomers. If polymeric dissolved species (i.e., species containing two or more metal ions) are present, then the predominance area diagram will be sensitive to T OT Me, just as it is when solids must be considered.
642
(and therefore be dominant) in this region, we carry out the following calculations: Pb(OH)2 (s) + H+ )* PbOH+ + H2 O Q=
(11.59)
{PbOH+ }{H2 O} {PbOH+ } T OT Pb = ⇡ {Pb(OH)2 (s)}{H+ } {H+ } {H+ }
(11.60)
If we find that the approximate value of Q computed using Equation (11.60) exceeds the equilibrium constant for Reaction (11.59), we conclude that the solid is supersaturated. The approximation in Equation (11.60) is reasonable only in a region where PbOH+ is the dominant soluble species, but, as noted above, that is the only region in which this calculation is used. Furthermore, we should recognize that predominance area diagrams are commonly used only to identify the broad ranges over which different species are dominant. Under the circumstances, the error associated with this approximation has only a small effect on the diagram. For instance, if the dominant species accounts for 80% rather than 100% of T OTi, the line on the predominance area diagram would shift by less than 0.1 log unit. Coupling this result with the fact that many solubility products are known to an accuracy of only a few tenths of a log unit, approximations like the one in Equation (11.60) are almost always acceptable. Example 11.9 Incorporate the possible precipitation of Pb(OH)2 (s) (pKs0 = 19.85) into the PbOH-Cl predominance area diagram shown in Figure 10.19, for a system that contains 10 8 M T OT Pb. Solution The predominance area diagram in Figure 10.19 identifies regions where eight different species are dominant. To include Pb(OH)2 (s) in the diagram, we begin by writing the reaction for equilibrium of the solid with each of those eight species and the corresponding Ksi equations, and we add that information to Table 10.6. For instance, for the Pb(OH)2 (s)/Pb2+ line, we carry out the following calculations: Pb(OH)2 (s) + 2 H+ )* Pb2+ + 2 H2 O
⇤
Ks0 =
Ks0 = 108.15 Kw2
If Pb(OH)2 (s) is present and the activity of Pb2+ is 10 8 , {Pb2+ }{H2 O}2 = 108.15 {Pb(OH)2 (s)}{H+ }2 643
(11.61)
log
⇣ ⌘ 10 8.0 (1.0)2 8.15 = log 10 (1.0){H+ }2 8.0 + 2 pH = 8.15 pH = 8.08
Thus, the line separating areas of dominance of Pb2+ and Pb(OH)2 (s) in the system of interest is at pH 8.08. This result, and the corresponding information identifying conditions under which the seven other soluble species would have concentrations of 10 8 M when in equilibrium with Pb(OH)2 (s), are shown in Table 11.4. The three lines for equilibrium between Pb(OH)2 (s) and Pb OH species (including Pb2+ ) are shown in addition to all the lines from the prior predominance area diagram in Figure 11.28. The Pb2+ /Pb(OH)2 (s) line is at a higher pH than any portion of the region where Pb2+ is the dominant soluble species, meaning that the Pb2+ activity would have to be > 10 8 for the solid to precipitate in that region. Thus, Pb2+ remains the dominant species throughout the region, and the Pb(OH)2 (s)/Pb2+ line is irrelevant. By contrast, the Pb(OH)2 (s)/PbOH+ line falls in the middle of the region where PbOH+ is the dominant soluble species (at pH 8.55). The solid is undersaturated (so PbOH+ is dominant) at lower pH, but it is supersaturated and therefore dominant at higher pH. The situation for the Pb(OH)2 (s)/Pb(OH)3 comparison is similar to that for Pb(OH)2 (s)/PbOH+ , in that the line demarking saturation with the solid falls in the middle of the region where Pb(OH)–3 is the dominant soluble species. However, in this case, the solid is present on the low-pH side of the line (because the complex is more basic than the solid). When the above considerations are taken into account, the diagram can be modified as shown in Figure 11.29. The assessment of the relative activities of Pb(OH)2 (s) and Pb(OH)2 is slightly different from that for other solutes, because the activity of Pb(OH)2 in equilibrium with the solid is independent of pH. Whenever Pb(OH)2 (s) is present, Pb(OH)2 has an activity of 10 8.95 , meaning that a solution with a Pb(OH)2 activity of 10 8 would always be supersaturated with respect to the solid. As a result, Pb(OH)2 (s) is dominant throughout the region where Pb(OH)2 is the major dissolved species of Pb. To this point, we have identified the area of dominance of Pb(OH)2 (s) only over the region where Pb OH complexes were dominant in the diagram for the soluble species; to assess the relative dominance in the regions where Pb Cl complexes are the dominant solutes, we need to carry out a similar analysis considering Pb(OH)2 (s)/PbClx pairs. The lines separating the regions of dominance of the species in each such pair can be drawn based on the final four equations in Table 11.4. When this is done, the final predominance area diagram can be prepared, yielding Figure 11.30. 644
645 K(22) K(3) K(20) K(4) K(24) K(5) K(25) K(6) K(26) K(7)
Pb(OH)2 (s)/Pb(OH)2
Pb(OH)2 (s)/Pb(OH)–3
Pb(OH)2 (s)/PbCl+
Pb(OH)2 (s)/PbCl2
Pb(OH)2 (s)/PbCl–3
Pb(OH)2 (s)/PbCl2– 4
22
23
24
25
26
27
log K(x) (b) Activity
(a)
K(21) K(2)
Pb(OH)2 (s)/PbOH+
21
9.53
9.95
10.05
9.71
19.95
8.95
0.55
8.15
Value of log K
log (Cl ) = ( 17.53 + 2pH)/4
log (Cl ) = ( 17.95 + 2pH)/3
log (Cl ) = ( 18.05 + 2pH)/2
log (Cl ) = 17.71 + 2pH
pH = 11.95
No pH (Pb(OH)2 (s) dominant)(b)
pH = 8.55
pH = 8.08
Equation of Line Where Soluble Species Has Activity of 10 8 When Pb(OH)2 (s) is Present
indicates the log K value for line x, where x is indicated in the first column. of Pb(OH)2 in equilibrium with Pb(OH)2 (s) is always < 10 8 . See text for explanation.
K(20) K(1)
⇤K s0
Pb(OH)2 (s)/Pb2+
20
Eqn. for K (a)
Species Compared (Reactant/Product)
Line #
Table 11.4 Additional equilibrium relationships needed to include consideration of Pb(OH)2 (s) in the predominance area diagram for the Pb2+ Cl– -OH– system, for T OT Pb = 10 8 M.
Figure 11.28 Reproduction of Figure 10.19, with lines added showing conditions required for equilibrium of Pb(OH)2 (s) with 10 8 M Pb2+ , PbOH+ , or Pb(OH)–3 .
Figure 11.29 Predominance area diagram for the Pb-OH-Cl system with 10 8 M T OT Pb, identifying the region where Pb(OH)2 (s) is dominant over Pb+2 and the Pb OH complexes. The regions shown for dominance of Pb Cl complexes have not yet been adjusted to account for the possible presence of the solid.
646
Figure 11.30 Final predominance area diagram for the Pb-OH-Cl system with 10 T OT Pb, considering possible precipitation of Pb(OH)2 (s).
8M
Although the predominance area diagram was developed here in two stages (considering soluble species first, then the solid), in general there is no need for such an approach. The diagram can be prepared in a single process by writing out the reactions and equilibrium expressions relating all possible pairs of species in the system and drawing the corresponding lines on the diagram. Because the diagram can become confusingly crowded if all the lines are drawn from the beginning, it is often convenient to begin with just the horizontal and vertical lines (including those that account for the solid), and then add the other lines one at a time. In more complex systems than the one considered here, other possible solids (e.g., carbonates or sulfides) might be taken into account. These systems have the added complexity that the speciation of the precipitating anion also depends on pH, so the number of calculations required to develop the entire diagram can become substantial. However, the approach is the same as that outlined here.
11.11
SUMMARY
This and the preceding chapter have covered the multitude of reactions of metal ions with other constituents of aqueous solutions. The defining characteristic of these reactions is the interaction between the metal ion and unshared electron pairs that are typically associated with oxygen, nitrogen, or sulfur atoms in other chemicals 647
(including water). The virtually unlimited number of chemicals in environmental samples that contain O, N, or S donor atoms leads to the possibility that metals can exist in a large number of species in natural aquatic systems. Because the speciation of metals has an enormous effect on their mobility and bioavailability, understanding speciation is critical to an understanding of metal behavior in the environment. The interactions of metals with donor atoms to generate soluble reaction products were described in Chapter 10. In many cases, when the concentrations of metals and/or ligands increase, Me L complexes form polymers that can grow to form a distinct solid phase. A solid has an activity that is independent of its concentration in a suspension and, by convention, any pure solid has an activity of 1.0. Many of the ligands that can combine with metals to form solids are the same ones that form soluble complexes, so those ligands might impose limits on metal solubility under some circumstances and provide a means for increasing metal solubility in others. Curves of metal solubility as a function of pH are often bowl-shaped. That is, the metal is highly soluble at low pH due to the large equilibrium concentrations of the uncomplexed metal ion and acidic complexes, and highly soluble at high pH due to the high equilibrium concentrations of basic complexes. Between these extremes (at slightly alkaline pH for many of the most common metals used commercially), metal solubility passes through a minimum that is often low (< 0.1 µM). By taking advantage of these dramatic changes in solubility, it is often possible to remove metals from solution very efficiently. However, in such processes, it is important to recognize that missing the target pH in either direction can lead to dramatic increases in the equilibrium solubility. It is also important to remember that metals often do not reach equilibrium with solid phases in the time frame of minutes to hours that is typical of engineered treatment processes. If multiple solids containing a given type of metal ion are supersaturated in a solution, at least one of those solids will be present when the system reaches equilibrium. However, the others might or might not be present, depending on how the precipitation of the most stable solid alters the solution chemistry. Determination of the equilibrium composition in such cases requires that we make assumptions about which solids will be present, solve the equations that characterize the system based on those assumptions, and then modify the assumptions as needed. Software for solving chemical equilibrium problems is extremely handy in such cases, since it can rapidly carry out numerous iterations based on the different assumptions. When the software is used, it is important to distinguish between solids that are known to be present at equilibrium and those that might or might not be present, depending on the solution chemistry. The Gibbs phase rule relates the number of chemical components that are needed to describe a system and the number of phases that are present at equilibrium to the number of independent parameters that can be specified to characterize the system. 648
Because of the multiplicity of species that can form in systems containing metals and ligands, simplified diagrams that identify the dominant metal species but do not display the concentrations or activities of nondominant species are sometimes prepared. Such diagrams can incorporate consideration of solids as well as soluble species and can therefore display critical information about metal speciation over a very wide range of conditions.
11.12
PROBLEMS
Please re-read the note preceding the Chapter 8 Problems on p.457. The same guidelines apply to the following problems. 1. In some water softening operations, Mg2+ must be removed from solution along with Ca2+ . What pH is required to lower T OT Mg to 10 mg/L as CaCO3 by precipitation of Mg(OH)2 (s)? 2. What is the solubility of ferrihydrite (amorphous Fe(OH)3 (s)) in an ideal pH 7.0 solution containing 10 2 M T OT SO4 ? Give your answer in µg/L T OT Fe. 3. One gram of solid Cd(OH)2 (s) is dispersed in a solution at pH 9.0 containing 1.0 mg/L of total dissolved Cd. The solution was prepared by adding only Cd(NO3 )2 and NaOH to pure water. Do you expect some solid to dissolve, some dissolved Cd to precipitate, or no change in the solution composition? 4. A groundwater at pH 7.5 is in equilibrium with PCO2 = 10 1.5 atm and with siderite (FeCO3 (s)). Determine the total dissolved Fe concentration. 5. A natural water sample with the following solute concentrations has been placed in a tightly sealed, headspace-free glass bottle for storage. Determine whether the solution is stable or whether calcite is expected to precipitate. [SO24 ] = 10
I = 0.012 M Alkalinity = 1.5 ⇥ 10 +
[Na ] = 2 ⇥ 10 [Ca2+ ] = 10
3
3
3
equiv/L
M
3
M
[Cl ] = 0.5 ⇥ 10
3
M
pH = 8.9
M
6. One milligram of NiCO3 (s) is added to 1.0 L of a solution containing 10 3 M T OT CO3 . Over what pH range(s) will the solid completely dissolve? Assume ideal solution behavior. 7. What is the solubility of Ca3 (PO4 )2 (s) in mg/L, in a solution buffered at pH 7.2 using a phosphate buffer such that T OT PO4 = 10 2 M? 649
8. An aqueous solution at pH 7.5 and containing 10 6 M T OT Cu(II) is equilibrated with a gas phase having PH2 S = 10 8 atm. Will covellite (CuS(s)) form in this system? If so, how much? If not, how much higher would PH2 S have to be to cause the solid to precipitate? 9. Ten milligrams of gibbsite [a crystalline form of Al(OH)3 (s)] is added to 1.0 L of 10 3 M Na2 CO3 . How much aluminum will be dissolved at equilibrium, and what will the equilibrium pH be? 10. Determine the solubility of calcite, CaCO3 (s), in mg/L, in pH 12 solutions of ionic strength 0.01 and 0.1 M. What would the solubility be in a solution with I = 0.1 M at pH 8.0? Use the Davies equation to compute activity coefficients. If you solve the problem manually, ignore the acid/base reaction of Ca2+ . 11. Compare the solubility of mercuric oxide [HgO(s), montroydite] in two solutions at pH 8 and with ionic strength 0.1 M – one containing 10 3 M NaCl and the other containing 10 1 M NaCl. Use the Davies equation to compute activity coefficients. What is the dominant Hg species in each solution? 12. Weathering of the aluminosilicate mineral albite Na2 Al2 Si6 O16 (s) can generate the clay kaolinite Al2 Si2 O5 (OH)4 (s) via the following reaction, with log K = 0.68: 0.5 Albite + H+ + 4.5 H2 O )* 0.5 Kaolinite + 2 H4 SiO4 + Na+ Determine the pH and alkalinity of groundwater with PCO2 = 10 1.5 atm that equilibrates with albite if the groundwater contains negligible amounts of solutes other than carbonate species. (Hint: Like the problems in preceding chapters, this problem requires combining known information with mass and/or charge balances and equilibrium constants. And, as usual, there are many solution approaches that can be used successfully. One approach is as follows: (1) identify all solutes expected to be present at equilibrium; (2) write the CB for the solution; (3) use the given K and reaction stoichiometry to express (Na+ ) as a function of pH; (4) write other terms in the CB as a function of pH; and (5) find the pH at which the CB is satisfied.) 13. Prepare a log C-pH diagram for a system in equilibrium with atmospheric CO2 (g) and calcite at 15 C. Then, determine the equilibrium pH, alkalinity, and Ca2+ concentration for the following two solutions: (a) A solution for which atmospheric CO2 (g) and calcite are the only inputs. (b) The solution in part (a) after addition of 10 with the two nonaqueous phases. 650
2.7 M
HCl and re-equilibration
14. “pHistory of water”. (a) A cloud passes over a smokestack, where a raindrop absorbs 10 5 M H2 SO4 . Find the solution pH, assuming the droplet contains only water, the absorbed H2 SO4 , and carbonate species acquired by equilibration with the atmosphere. (b) The droplet passes over a feedlot and comes to equilibrium with PNH3 = 10 10.2 atm while remaining in equilibrium with atmospheric CO2 . Find the new pH and the total dissolved N. (c) Finally, the droplet falls to earth and percolates into the soil. There, it equilibrates with a rock containing amorphous silica, without any gas exchange. Prepare a log C-pH diagram for the solution (droplet) at this point, covering the pH range 3 to 12 and showing all species that are present at concentrations >10 12 M. Find the pH and the values of T OT CO3 , T OT N, and T OT Si at equilibrium. 15. The hypolimnetic (deep) water of a eutrophic lake contains 10 8 M total dissolved sulfide [T OT S(II)]. If the water is at pH 7.0 and in equilibrium with CdS(s), what is the concentration of total dissolved cadmium, considering Cd complexes with HS– and S2– ? (The hydrolysis of cadmium is negligible at this pH.) 16. A drinking water at pH 7.2 contains 8 ⇥ 10 4 M T OT Ca and has an alkalinity of 80 mg/L as CaCO3 , provided entirely by species of the carbonate group. It is desired to adjust the pH of this water to cause calcite to be slightly supersaturated for purposes of corrosion control. How much lime must be added to achieve a saturation index of 0.2? What would the pH of the water be? 17. Although Mg2+ is soluble under most conditions encountered in natural waters and wastewaters, it can form the mineral struvite (MgNH4 PO4 (s)) in solutions that have relatively high concentrations of ammonia and phosphate species, such as tanks where sewage sludge is anaerobically digested. Consider a solution that contains 4 ⇥ 10 3 M T OT Mg, 7 ⇥ 10 3 M T OT NH3 , and 9 ⇥ 10 4 M T OT PO4 and that is in contact with struvite as it exits a digester. Determine the pH values in the range 6 to 10 where additional struvite will precipitate and the values where some of the struvite on the pipe will dissolve. If you solve the problem manually, consider acid/base reactions but ignore the formation of complexes. If you use Visual Minteq, allow all complexes to be included in the calculations. The solubility constant for struvite and the corresponding reaction stoichiometry are given in Section 11.7. 18. A surface water being evaluated as a possible drinking water supply source has pH = 7.5 and T OT Ca = 185 mg/L. 651
(a) What is the alkalinity of the water, if it is in equilibrium with the atmosphere? (b) Is the water undersaturated or supersaturated with respect to calcite? (c) The water is to be dosed with 100 mg/L of lime, Ca(OH)2 . What will the new alkalinity and pH be if the lime dissolves completely, the solution re-equilibrates with the atmosphere, and no solids form? Will calcite be undersaturated or supersaturated at this point? 19. Sodium cyanide is slowly added to a solution containing a relatively small amount of mercuric ion (Hg2+ ). Analysis of the solution composition at various points in the titration indicates that all the mercury remains in solution when small amounts of the cyanide are added, some mercury precipitates when intermediate amounts are added, and the precipitate then re-dissolves when large amounts of cyanide are added. Explain these changes. 20. An industrial wastewater at pH 6.5 contains 10 mg/L T OT Cd and 100 mg/L dissolved inorganic carbon (DIC). NaOH is added to the wastewater to raise the pH to 12. (a) Considering Cd(OH)2 (s) and CdCO3 (s) as possible precipitates, will neither, one, or both of these solids begin to precipitate when the pH is first raised to 12? (b) Will neither, one, or both solids be present once equilibrium is attained, if the pH is maintained at 12? (c) What will the total dissolved concentration and the dominant dissolved species of cadmium be at equilibrium, again assuming the pH is held at 12? 21. Prepare a log C-pH diagram over the range pH 3-11 for an ideal solution in equilibrium with Al(OH)3 (am). Include lines for Al3+ and all Al OH complexes. On the same diagram, show the concentration of Al(OH)3 (aq) in equilibrium with gibbsite, a crystalline aluminum hydroxide. What conclusion do you draw? 22. Zinc orthophosphate is sometimes added to water to mitigate corrosion of pipes in water distribution systems. Prepare log C-pH diagrams showing the speciation of T OT Zn and T OT PO4 in a system made by adding 8 mg/L Zn3 (PO4 )2 · 4 H2 O(s) to pure water. Consider both Zn3 (PO4 )2 · 4 H2 O(s) and Zn(OH)2 (s) as solids that might be present at equilibrium. 23. Carbonate is sometimes added to treated drinking waters with low alkalinity to control corrosion of lead pipe and solder in the distribution system. If 652
carbonate is added to a finished water to increase T OT CO3 to 10 3 M and the pH to 8.0, what is the maximum concentration of dissolved lead that can be expected in the drinking water? Consider PbCO3 (s), Pb(OH)2 (s), and Pb3 (CO3 )2 (OH)2 (s) as possible solids in the system, and also consider the formation of PbOH+ and PbCO3 (aq) complexes. Which solid controls Pb solubility? 24.
(a) Use Visual Minteq to construct a log C-pH diagram for nickel in an industrial wastewater that contains 10 mg/L T OT Ni and 10 4 M T OT CO3 . Assume the system is closed to the atmosphere, and consider the pH range 4-11. At what pH will nickel solubility be minimized, and what will the nickel concentration be in mol/L and mg/L at this pH? (b) Repeat part (a) for an open system.
25. The owner of a limestone (calcite) quarry would like to use it to grow fish. She plans to fill the quarry with 100,000 m3 of water that is at pH 6.9 and has equilibrated with the atmosphere. For optimal fish productivity, the pH in the quarry should be between 7.5 and 8.0. (a) Over time, the water is expected to equilibrate with the limestone while remaining in equilibrium with the atmosphere. How much hydrochloric acid (HCl) or caustic (NaOH) should be added to the water so that its ultimate pH is within the desired range? Give your answer in kg of chemical added. (b) Depending on the location in the quarry, the water might equilibrate faster with the atmosphere or the limestone. If the amount of acid or base determined in part (a) is added to the water as it is pumped into the quarry, what would the pH be at the following locations: i. Near the bottom of the quarry, where the solution equilibrates with CaCO3 (s) but not the atmosphere; ii. Near the top of the quarry, where the solution equilibrates with the atmosphere but not CaCO3 (s); and iii. At mid-depth in the middle of the quarry, where the solution does not equilibrate with either the atmosphere or the CaCO3 (s). 26. A monument made of calcite has been exposed to the weathering action of atmospheric CO2 over a period of >2500 years. Starting around 1950, the rate of weathering drastically increased, primarily as a result of the elevated SO2 content of the atmosphere from the burning of fossil fuels. Assume that rainfall and PCO2 have been constant over the years at 40 cm/yr and 3.8 ⇥ 10 4 atm, respectively, and that the SO2 content of the atmosphere 653
has been 1 ppm (by volume) since 1950. (This value of PCO2 is the modern value and includes the increase since the beginning of the Industrial Age. It therefore overstates the acidity of the rain for most of the period of interest. However, the focus of this question is on the effect of the SO2 , and that effect is much greater than the effect of recent increases in the atmospheric CO2 concentration.) Calculate the maximum possible extent of weathering (dissolution of CaCO3 ) for the 2500 years preceding 1950 and for the period 1950 to 2010. Express the results as grams of CaCO3 (s) dissolved per cm2 cross-sectional area of the monument. Note that the actual surface area of the statue is irrelevant, because once the rain equilibrates with the solid, no more dissolution will occur, no matter how much surface area is available. However, the crosssection of the statue does matter, because that determines how much water hits it. Carry out the calculations based on two sets of assumptions. First, assume that the rain is in equilibrium with the atmosphere before striking the monument, but that no gas exchange occurs as it drips down the monument’s surface. Second, assume that the rain is in continuous equilibrium with the air as it drips. If you use Visual Minteq for these simulations, allow the program to account for nonideal solute behavior, but instruct the program to exclude consideration of CaSO3 and S2 O2– 5 . (Note: Visual Minteq might fail to converge in some cases where calcite is input as an infinite solid. If that occurs, try inputting calcite as a finite solid with an initial concentration of 1 mol/L. If the output indicates that some calcite remains at equilibrium, then the solution will be in equilibrium with that solid and therefore will have the same composition as if it were in equilibrium with an ‘infinite’ amount of solid.) 27. Consider a treated domestic wastewater containing 12 mg/L PO4 P and with an ionic strength of 0.007 M. You wish to reduce the P concentration to 0.2 mg/L by precipitation of AlPO4 (s). (a) Prepare a log C-pH diagram showing curves for the concentration of PO3– 4 in the untreated and treated solutions. On the same diagram, show the Al3+ activity that would cause the PO3– 4 concentration in each solution to be in equilibrium with AlPO4 (s). (b) Alum [Al2 (SO4 )3 · 14 H2 O] is added to the initial solution at a dose corresponding to 10 3.0 M T OT Al. Add a line to the diagram representing the Al3+ concentration as a function of pH immediately after the alum dissolves, before any solid forms. (Hydrolysis and complexation reactions are typically faster than precipitation reactions.) At what pH values in the range 4 pH 12 is AlPO4 (s) supersaturated? (c) How much alum (in mg/L) must be added to the original solution to 654
achieve the treatment goal, if the pH is well-buffered at 7.5? Keep in mind that one Al ion precipitates for each PO4 ion removed. For now, ignore possible precipitation of aluminum hydroxide and aluminum oxide. (d) Compute the Al3+ concentration that would be in equilibrium with gibbsite at pH 7.5 in the original solution. Discuss the consequences of this result for your answer to parts (b) and (c). 28. The solubility product for vivianite is reported as 10 37.76 in Table 11.1. What is the standard Gibbs energy of formation (G ) of vivianite? 29. A drinking water source contains 170 mg/L T OT Ca, has an alkalinity of 115 mg/L as CaCO3 , and is at pH 8.8. Answer the following questions, using Visual Minteq for the calculations. (a) What is Q/K for precipitation of calcite? (b) What would the composition of the solution be if it equilibrated with calcite? Why does the solution pH change? (c) To soften the water, its pH is increased to 10.8 by addition of lime. How much lime is needed, and by how much will T OT Cadiss be reduced, if solid/solution equilibrium is achieved? Why is the reduction in T OT Cadiss so limited? (d) What would the composition of the solution and the amount of calcite precipitated be if, instead of adding only lime, both soda ash (Na2 CO3 ) and lime were added to adjust the pH to 10.8 and simultaneously achieve stoichiometric softening conditions (T OT Ca = T OT CO3 )? (e) After all the CaCO3 (s) generated in part (d) is removed by settling and filtration, the softened water is recarbonated by bubbling the solution with pure CO2 (g). Plot T OT Cadiss as a function of pH during this process, and explain why it changes in the way it does. How much CO2 must dissolve to lower the pH to 8.5? How much CO2 would have to dissolve to reach the same pH if the calcite removal step were not completely efficient, and 25 mg/L calcite remains in the settled water? 30.
(a) Draw a log C-pH diagram showing total dissolved manganese from pH 5 to 13 in a system containing 10 4 M T OT Mn (total in both dissolved and solid phases) and 3 ⇥ 10 3 M T OT CO3 . Consider the formation of all the Mn-OH and Mn-CO3 complexes in the Visual Minteq database, and also the possible formation of pyrochroite [Mn(OH)2 (s)] and MnCO3 (am). On a separate graph, plot the concentrations of the two solids as a function of pH. (b) What PCO2 would be required for both solids to be present in an equilibrium system? 655
31.
(a) Determine the total dissolved Zn concentration in equilibrium with ZnCO3 (s) in a system with 10 2 M T OT Cl and 5 ⇥ 10 3 M T OT CO3 , at pH 8.0.
(b) Draw a log C-pH diagram for a system containing 3 ⇥ 10 5 M T OT Zn, 10 2 M T OT Cl, and 5 ⇥ 10 3 M T OT CO3 , in in which ZnCO3 (s) is the only possible precipitate. Show the concentrations of the following species or groups of species in solution: Zn2+ , T OT Zndiss , the sum of all Zn OH complexes, the sum of all Zn CO3 complexes, and the sum of all Zn Cl complexes. (c) Repeat part (b), but this time consider possible precipitation of only Zn(OH)2 (am).
(d) Re-draw the lines for (Zn2+ ), T OT Zn, and the sum of all Zn CO3 complexes from parts (b) and (c) on a separate graph. Discuss the significance of the intersection points and the implications of the graph in pH regions on the acid and base side of the intersection points. Confirm your discussion by plotting the concentrations of ZnCO3 (s) and Zn(OH)2 (am) in a system with the same total concentrations of Zn, CO3 , and Cl, in which both solids are considered as possible precipitates. 32. The reported composition of a groundwater is as follows. T OT Ca T OT Mg T OT SO4 ALK pH
467 mg/L as CaCO3 257 mg/L as CaCO3 470 mg/L 337 mg/L as CaCO3 (from CO3 species) 7.7
The only other ion expected to be present at a significant concentration is Na+ . It is desired to treat the water to reduce the concentrations of all the ions listed. Consultants have recommended adding BaCl2 to remove the SO2– 4 from solu9.98 tion by precipitation of BaSO4 (s) (Ks0 = 10 ), and then using conventional softening to reduce the Ca2+ and Mg2+ concentrations. However, BaCO3 (s) is quite insoluble (Ks0 = 10 8.57 ), so precipitation of that solid could consume much of the Ba2+ that is added. Determine how much, if any, BaSO4 (s) can be precipitated without precipitating any BaCO3 (s). Assume that the ionic strength remains the same as in the raw water. This question can be answered using either manual calculations or Visual Minteq. If you use the latter approach, you might have to add BaCl2 (try 10 3 M or more) to the input to get the program to run successfully. 33. Both turbidity and natural organic matter can be removed from drinking water by coagulation with alum (aluminum sulfate, [Al2 (SO4 )3 ]). When alum is 3–y added to water, Al3+ ions hydrolyze to form soluble monomeric Al(OH)y and 656
3x–y
polymeric Alx (OH)y species, [Al(OH)3 (am)] can form as well.
and
aluminum
hydroxide
solids
(a) Construct a log C-pH diagram for Al(III) species in equilibrium with Al(OH)3 (am). Consider the Al OH complexes in the Visual Minteq 3x–y database, including the polymeric complexes Alx (OH)y . Identify regions where Al(OH)3 (am) precipitates in a system containing 10 4 M T OT Al. (b) Use the diagram from part (a) to calculate the extent to which the solubility of Al(III) is increased if a lake water is subjected to acid precipitation and the pH decreases from 7.0 to 5.0, considering only the monomeric dissolved Al(III) species. 34. Although fluoride (F– ) forms relatively weak complexes with most metals, its complexes with Al3+ and Fe3+ are fairly strong. A log C-log (F ) diagram for a solution at pH 6.5 that contains 10 3 M T OT Al is shown below, ignoring polymeric species and the possibility that a solid might precipitate.
(a) Using only the information in the graph, determine log b3 for Al F complexes. (b) Is Al(OH)3 (am) supersaturated at any condition(s) shown on the plot? If so, what condition(s)? (c) A solution is well buffered at pH 6.5 and contains 10 3 M T OT Al. If you concluded in part (b) that Al(OH)3 (am) is supersaturated under some conditions shown on the graph, determine whether precipitation could be prevented by adding fluoride, and if so, find the value of T OT F that would prevent precipitation. If you concluded that Al(OH)3 (am) is not 657
supersaturated under any conditions shown, determine what T OT Al would have to be present to cause Al(OH)3 (am) to precipitate if the F– activity were fixed at 10 4.5 . 35. A wastewater at pH 7.1 containing 10 mg/L total dissolved lead and 10 3 M total dissolved carbonate is to be treated by raising the pH to precipitate either Pb(OH)2 (s) or PbCO3 (s). Although the initial solution might be supersaturated, no solids are initially present. (a) What is the minimum lead solubility at 25 C that can be achieved? (b) To what pH should the solution be adjusted to achieve this minimum concentration? (c) How much caustic (NaOH) must be added to reach the pH of minimum solubility? 36. Use the Gibbs Phase Rule to determine whether it would be possible (under some theoretical, but not necessarily practical, conditions) for AgCl(s), Ag2 CO3 (s), and AgOH(s) to all be present in an equilibrium system at standard T and P. 37. A system containing 10 2 M T OT Cl and 10 3 M T OT CO3 , initially at pH 9.0, is titrated with AgNO3 until 1.0 M Ag has been added. Answer the following questions, considering the possible precipitation of three solids: AgOH(s), AgCl(s), and Ag2 CO3 (s) (pKs0 = 11.09). Solve parts (a) and (b) using manual calculations, and part (c) using Visual Minteq. (Note: the questions can be answered by simulating the titration and determining the sequence of precipitation reactions, but it is also possible to reach the correct conclusion by manipulating the various solubility constants and the information about the initial solution composition without actually determining the Ag+ concentration under the conditions of interest. If you simulate the titration using Visual Minteq, you will have to add AgOH(s) to the program’s database. Either approach is acceptable.) (a) Which solid is expected to precipitate first, and what will the activity of Ag+ be when that occurs? How much AgNO3 has been added at this point? (b) How high must the activity of Ag+ be to cause a second solid to precipitate, and which solid will it be? How much AgNO3 has been added when this condition is reached? (c) If the titration is continued, will the third solid eventually precipitate? If so, what are the conditions when this occurs? If not, explain (words or equations) why you reached that conclusion. 658
38. A solution that is well-buffered at pH 8.5 contains 4 ⇥ 10 iron and 10 4 M T OT CO3 .
4M
total ferrous
(a) Do you expect either FeCO3 (s) or Fe(OH)2 (am), or both, to begin precipitating, based on the initial solution composition? Assume ideal solution behavior, and that Fe2+ and FeOH+ are the only significant contributors to T OT Fe in the solution. (b) Which solid(s) would be present if the system reached equilibrium, and what would the solution composition be? 39. Lime is sometimes added to wastewater to remove phosphorus from solution by forming a Ca PO4 solid. Since wastewaters typically contain T OT CO3 concentrations of a few millimolar, CaCO3 (s) might also form. (a) A waste contains 2 ⇥ 10 4 M T OT PO4 and 2 ⇥ 10 3 equiv/L alkalinity, with the alkalinity comprised entirely of carbonate and phosphate species. Prepare a logC -pH diagram showing the Ca2+ concentration that would be in equilibrium with each of two possible solid phases: calcite (CaCO3 (s)) and hydroxyapatite (Ca5 (OH)(PO4 )3 (s)). The reactions and corresponding equilibrium constants in the Visual Minteq database for dissolution of the two solids are: CaCO3 (s) )* Ca2+ + CO2– 3
K = 10
8.48
Ca5 (PO4 )3 OH(s) + H+ )* 5 Ca2+ + 3 PO3– 4 + H2 O
K = 10
44.333
Note: Visual Minteq is not a convenient tool for addressing this question fully, but it can be used to determine the distribution of carbonate and phosphate species as a function of pH, and those results can then be used in conjunction with a spreadsheet to draw the curves for log(Ca2+ ) vs. pH. (b) The solution in part (a) is initially at pH 8.1 and contains negligible Ca2+ . Lime is added in an effort to precipitate hydroxyapatite. As the lime is added, the Ca2+ activity and the pH both increase. Which of the two solids considered in part (a) will be the first to precipitate, and at what pH does that occur? (c) Lime addition is continued until 95% of the initial phosphorus precipitates. How much lime has been added, and what is the solution pH, when the treatment goal is reached? (d) Is calcite expected to be present when the treatment goal is reached? If so, how much? If not, how much lime must be added to cause calcite to form, and what is the solution pH at that point? 659
40. Because cyanide is a strong complexing agent, it is used to prevent metals from precipitating in some electroplating operations. (Note: to solve this problem using Visual Minteq, you will need to add HCN(g) and Hg(CN)2 (s) to the program’s database. See the appropriate sections of Chapters 9 and 11 for explanations of how to do that.) (a) Consider a solution at pH 10 in equilibrium with montroydite, HgO(s). Compute the amount of T OT CN that would be in solution if the overlying air had PHCN = 10 8 atm. What can you conclude from the result? (b) Consider that mercury can also form the solid Hg(CN)2 (s), with pKs0 = 39.28. What concentration of T OT CN would be present in a pH 10 solution in equilibrium with this solid and with PHCN = 10 8 atm? Ignore the possible formation of HgO(s) for this part of the problem. (c) A solution that is well-buffered at pH 10 and that contains no Hg has equilibrated with a gas phase containing HCN at a partial pressure of 10 8 atm. If 0.75 M Hg(CN)2 (s) is added to the solution, will any solid be present at equilibrium?
660
12
REDOX CHEMISTRY
661
Contents
12.1 12.2 12.3 12.4 12.5
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . DETERMING OXIDATION NUMBERS . . . . . . . . . . . . . . BALANCING REDOX REACTIONS . . . . . . . . . . . . . . . . REDOX HALF-REACTIONS . . . . . . . . . . . . . . . . . . . . THE ACTIVITY OF FREE ELECTRONS; EQUILIBRIUM CONSTANTS FOR REDOX HALF-REACTIONS . . . . . . . . . 12.6 DEFINITION OF e AND pe . . . . . . . . . . . . . . . . . . . . 12.7 COMPUTING pe FROM SPECIES ACTIVITIES: THE NERNST EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.8 COMBINING REDOX REACTIONS . . . . . . . . . . . . . . . . 12.9 REDOX SPECIATION AND LOG C-pe DIAGRAMS . . . . . . . 12.10SUMMARY OF SOME KEY SIMILARITIES BETWEEN ACID/BASE AND REDOX SYSTEMS . . . . . . . . . . . . . . . 12.11REDOX REACTIONS INVOLVING EXCHANGE OF BOTH ELECTRONS AND PROTONS . . . . . . . . . . . . . . . . . . . 12.12COMPUTING EQUILIBRIUM SPECIATION IN REDOX SYSTEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.13MODELING REDOX REACTIONS WITH VISUAL MINTEQ . . 12.14OXIDATION AND REDUCTION OF WATER . . . . . . . . . . . 12.15ENERGY CHANGES ACCOMPANYING REDOX REACTIONS . 12.16REDOX TITRATIONS AND THE GEOCHEMICAL REDOX SEQUENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.17pe-pH PREDOMINANCE AREA DIAGRAMS . . . . . . . . . . . 12.18REDOX REACTIONS AND ELECTROCHEMISTRY . . . . . . . 12.19SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.20PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
662
663 667 670 671 674 679 683 685 687 689 692 703 710 719 728 737 740 744 756 757
12.1
INTRODUCTION
This chapter describes oxidation/reduction, or more simply, redox reactions — chemical reactions in which electrons are transferred from one atom to another. The loss of electrons is called oxidation, and an atom that has lost electrons is said to be oxidized, while the acquisition of electrons is called reduction, and an atom that has gained electrons is said to be reduced. Redox reactions involve simultaneous oxidation of one atom and reduction of another. Since any reduced element is capable of releasing electrons that can combine with (i.e., reduce) another element, reduced compounds are sometimes referred to as reducing agents or reductants. Similarly, oxidized elements are capable of acquiring electrons from other elements; in the process, the (initially) oxidized element becomes reduced and acts as an oxidizing agent or oxidant. Reductants are also commonly referred to as electron donors, and oxidants as electron acceptors, especially in the microbiological and biochemical literature. The charge or oxidation number (also called oxidation state) of any atom is the difference between the number of protons in the atom’s nucleus and the number of electrons assigned to that atom. In a redox reaction, the oxidation number of the atom being oxidized increases (i.e., it becomes either more positive or less negative), and that of the atom being reduced decreases. Redox reactions are of overwhelming importance in environmental chemistry. Their significance stems primarily from two characteristics. First, changes in the behavior of elements when they participate in redox reactions can be enormous. To cite just a few examples, Cr(III)1 species are very insoluble cations under normal conditions and pose little risk to health, whereas Cr(VI) species are very soluble anions and are potent carcinogens. Similarly, S(VI) species (e.g., sulfate ion) are highly soluble, nonvolatile, and relatively innocuous, while S( II) species (H2 S, HS– , and S2– , for instance) form very insoluble metal precipitates, can exist in significant concentrations in a gas phase (as H2 S), and can be toxic at low concentrations. Finally, we are all familiar with the very different properties of metallic iron and rust [Fe(0) in the former case, compounds of Fe(II) or Fe(III) in the latter]. The cost associated with preventing and/or repairing the effects of the redox reactions that convert Fe(0) to Fe(II) or Fe(III) runs into billions of dollars annually. A second factor contributing to the importance of redox reactions is that often they are strongly inhibited kinetically, so that systems that are severely out of equilibrium (Q/K values different from 1.0 by many orders of magnitude) can be stable for long periods of time. This characteristic represents an important distinction between many redox reactions and most other reactions that we have considered up 1 The representation X(n) is commonly used to designate the oxidation state of an element X, where n is the oxidation number assigned to the element, often given as a Roman numeral. The representation is meant to include all the species in the system in which X is present with that oxidation number.
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to this point (e.g., acid/base, gas/liquid, metal complexation, and metal precipitation reactions). Because the Gibbs energy that is released when a reaction proceeds is directly related to how far that reaction is from equilibrium [recall from Chapter 4 that DGr = RT ln(Q/K)], redox reactions often release much more energy, when they do proceed, than other reactions. The ability to capture the energy released in redox reactions is at the core of virtually all metabolic activity. Specifically, organisms produce catalysts (enzymes) that overcome the kinetic inhibition of redox reactions and allow the reactions to approach equilibrium more rapidly. The energy released as a result of those reactions is captured and used for life processes; if the reactions proceeded at any significant rate in the absence of the catalyst, the disequilibrium would be reduced or eliminated outside the cell, and insufficient energy would be available for cell functioning.
The redox reactions leading to internal corrosion of water supply pipes can cause multiple problems. Aside from outright failure of the pipe, the products of corrosion can clog the pipe or enter the water, increasing consumers’ intake of metals (Fe, Cu, Pb) and leading to complaints of “red water.” These photographs show the formation of iron oxide ’tubercles’ in a steel pipe (left) and the penetration of a pit almost all the way through a copper pipe section (right). (Reprinted from Internal Corrosion of Water Distribution Systems, 2nd edition, by permission. Copyright c 1996, American Water Works Association.)
In this chapter, redox reactions of environmental importance are introduced, and the energy changes accompanying those reactions are quantified. The chapter relates to and synthesizes information from virtually all the preceding chapters of the text, 664
while simultaneously pointing out the unique features of redox reactions. In addition to a good deal of new terminology, which is defined at various stages throughout the chapter, the topics covered include the analysis of redox speciation and its display in log C-pe diagrams (analogous to log C-pH diagrams, but with the electron activity replacing the proton activity as the master variable), and an introduction to electrochemistry and the functioning of electrodes. Example 12.1 The sugar glucose (C6 H12 O6 ) and dissolved oxygen can be converted to carbonic acid via the following reaction, which is representative of reactions in which organisms oxidize carbohydrates to power their metabolism. C6 H12 O6 + 6 O2 (aq) )* 6 H2 CO3 A solution at 25 C contains 4.5 mg/L glucose (25 µmol/L), 8 mg/L O2 (aq) (250 µmol/L), and 10 5 mol/L H2 CO3 . (a) Determine the standard molar Gibbs energy (DGr ) and the equilibrium constant, K, of the reaction shown in the problem statement. The standard Gibbs energies of formation for dissolved glucose, O2 (aq), and H2 CO3 are 908.01, 16.40, and 623.04 kJ/mol, respectively. (b) Determine the reaction quotient, Q, the extent of disequilibrium, Q/K, and the Gibbs energy of the reaction in the given solution. (c) Compare the Gibbs energy released per mole of oxygen reacting in the given solution with the corresponding value for dissolution of atmospheric oxygen into a solution that contains 2 mg/L O2 (aq), if the concentration of O2 (aq) in equilibrium with the atmosphere is 10 mg/L. Solution (a) The standard Gibbs energy of the reaction is DGr = 6GH
2 CO3
GC
6 H12 O6
6GO
2 (aq)
= 2928.6 kJ/mol of reaction
Because the system is at 25 C, the equilibrium constant for the reaction can be computed from Equation (4.127): log K = =
DGr 5.71 kJ/mol 2928.6 kJ/mol = 512.89 5.71 kJ/mol
K = 10512.89 = 7.76 ⇥ 10512 665
(4.116)
(b) Assuming ideal behavior of all the solutes, the reaction quotient and the extent of disequilibrium for the reaction are Q=
(H2 CO3 )6 C6 H12 O6
Q 1.64 ⇥ 10 = K 10512.89
O2 (aq) 4
6
=
= 2.11 ⇥ 10
(10 5 )6 = 1.64 ⇥ 10 (25 ⇥ 10 6 )(250 ⇥ 10 6 )6
4
517
Thus, under the specified, reasonable conditions, the reaction is out of equilibrium by > 500 orders of magnitude! However, the reaction has a large activation energy, so it proceeds at a negligible rate in the absence of a catalyst. Organisms provide such catalysts (in the form of enzymes), allowing the reaction to proceed so that they can capture and utilize the energy stored in the chemical bonds of the sugar and oxygen molecules. The molar Gibbs energy of reaction at 25 C is related to the Q/K ratio by Equation (4.131), so ✓ ◆ kJ Q DGr = 5.71 log (4.114) mol K ✓ ◆ kJ = 5.71 log (2.11 ⇥ 10 517 ) mol = 2950
kJ mol
Because the molar energy of reaction is defined for one mole of stoichiometric reaction, the above calculation corresponds to reaction of one mole of glucose with six moles of oxygen. The Gibbs energy change per mole of oxygen reacting is therefore one-sixth of the value shown, or 491.7 kJ/mol. (c) The oxygen dissolution reaction is simply O2 (g) )* O2 (aq) Concentrations of 2 and 10 mg/L O2 (aq) correspond to 6.25⇥10 5 and 3.125⇥ 10 4 mol/L, respectively, and the activity of O2 (g) in the atmosphere equals its partial pressure in atmospheres (0.21). Therefore, the values of Q, K, and DGr for O2 dissolution in the system of interest are Q=
O2 (aq)
actual
O2 (g)
=
6.25 ⇥ 10 0.21 666
5
= 2.98 ⇥ 10
4
O2 (aq)
3.125 ⇥ 10 4 = 1.49 ⇥ 10 3 0.21 O2 (g) ✓ ◆ ✓ ◆ kJ Q kJ 2.98 ⇥ 10 DGr = 5.71 log = 5.71 log mol K mol 1.49 ⇥ 10 K=
eq
=
4 3
= 3.99
kJ mol
The Gibbs energy of reaction is negative for both the redox reaction and the oxygen dissolution reaction, meaning that both reactions will proceed spontaneously to convert reactants to products. However, the Gibbs energy released per mole of oxygen reacting is more than two orders of magnitude larger for the redox reaction, reinforcing the point that large amounts of energy are often available from such reactions.
12.2
DETERMING OXIDATION NUMBERS
Although determining the oxidation number of an atom is simple in concept, it is not always easy in practice, because ambiguity arises as to which atom a particular electron “belongs” to. Indeed, since the formation of chemical bonds involves the sharing of electrons between or among atoms, all electrons that are involved in bond formation should logically be assigned only partially to any single atom. While such a practice is common in some fields of chemistry, the convention for defining oxidation numbers in redox reactions is to assign each electron in a chemical compound entirely to a specific atom. One rule and several conventions for assigning oxidation numbers to elements of environmental significance are summarized in Table 12.1. Table 12.1 Conventions for assigning charge to various atoms
Rule (can never be violated): Charge must be conserved; i.e., the charge on a molecule must equal the sum of the charges on the constituent atoms. Convention: 1. H has an oxidation number of +1. 2. O has an oxidation number of
2.
3. N has an oxidation number of many organic compounds.
3 when bonded only to H or C, as it is in
4. S has an oxidation number of many organic compounds.
2 when bonded only to H or C, as it is in
Application: The above conventions are applied in the order given, but can be violated if necessary to avoid violating the charge conservation rule. 667
Example 12.2 Determine the average oxidation number of the following elements: – (a) C in (i) CO2 , (ii) CO2– 3 , (iii) HCO3 , and (iv) H2 CO3 .
(b) C in acetate ion, C2 H3 O–2 . (c) C in the amino acid glycine, C2 H5 O2 N. – (d) Cr in Cr(OH)3 , Cr2 O2– 7 , and HCrO4 .
(e) H and O in hydrogen peroxide, H2 O2 . (f) H in H2 (g). Solution (a) (i) By convention 2, in carbon dioxide, each oxygen atom contributes a charge of 2. Since the molecule is neutral, the carbon must contribute a charge of +4. Therefore, its oxidation state or oxidation number is +4. (ii) In a carbonate ion, each oxygen atom has a charge of 2, so the three oxygen atoms contribute a total charge of 6 to the ion. Since the overall charge on the ion is 2, the carbon must once again have a charge of +4. (iii), (iv) A similar calculation to that in part (ii) indicates that the carbon atom has a charge of +4 in both bicarbonate ion and carbonic acid. Thus, neither hydration/ dehydration nor any acid/base reaction among carbonate species is a redox reaction. This result can be extended to all reactions in which the only changes occurring are hydration/dehydration or protonation/deprotonation. (b) In acetate ions, the three H atoms contribute a charge of +1 each, and each of the two O atoms contributes a charge of 2. Together, the H and O therefore contribute a net charge of 1, which is the charge on the molecule. Thus, the average oxidation state of the C atoms is zero. (c) In glycine, the charge contributions of the five H, two O, and one N atoms are +5, 4, and 3, respectively, for a total of 2. The molecule is neutral, so the total charge contributed by the two C atoms is +2, and the average charge on the C atoms is +1. (d) Charge balances analogous to those in parts (a) through (c) indicate that the – oxidation number of Cr is +3 in Cr(OH)3 and +6 in both Cr2 O2– 7 and HCrO4 . – Note that when Cr(OH)3 is oxidized to either Cr2 O2– 7 or HCrO4 , the charge 668
on the molecule decreases even though the charge on the central metal ion increases.2 (e) The first two conventions cannot both be applied while following the rule of charge conservation, so Convention 1 is followed and Convention 2 is violated. The average oxidation number of H in H2 O2 is +1, and that of O is 1. (f) Convention 1 cannot be applied without violating the rule that charge must be conserved, so the convention is ignored and the rule is followed. The oxidation number of each H atom in H2 (g) is zero. According to Example 12.2b, the average charge on C atoms in acetic acid is zero. However, the molecular structure of acetic acid suggests that neither carbon atom in the molecule is, in fact, uncharged. Acetic acid can be reasonably represented as a combination of a methyl (methane-like) group and a carboxyl (carbonate-like) group, as shown in Figure 12.1.
Figure 12.1 Oxidation states of the carbon atoms in acetic acid.
The oxidation number of C is +4 in carbonic acid and 4 in methane. Thus, it is most reasonable to think of the carbon in acetic acid as having an average oxidation 2 It
is common to refer to the entire molecule as being oxidized or reduced, since in reality it is the molecule that is losing or gaining electrons.
669
number of zero due to the balancing of nonzero charges on the two individual carbon atoms. The same type of situation can cause the average oxidation number of an atom in a molecule to be nonintegral, even though charges (i.e., electrons) can exist only in whole-number units. At times, analysis of the charge on individual atoms in a compound can provide insight into its behavior; in other cases, the average oxidation state of all the atoms of a given type in a molecule is all that interests us. In any case, it is important to remember that the procedure described above yields a value for the average charge on a group of atoms and does not necessarily indicate the charge on any particular atom in a molecule.
12.3
BALANCING REDOX REACTIONS
Free electrons, like free protons, are highly unstable in aqueous solution. As a result, the concentration of free electrons in any solution is always vanishingly small, and all the electrons released by one element must combine with another; that is, no reaction releasing electrons to solution can take place in isolation, nor can free electrons be acquired from bulk solution. At times, we know the reactants and products of a redox reaction, but the stoichiometry is not obvious. In such a case, we can use the information that free aquo electrons cannot be a significant reactant or product to help balance the reaction, following the algorithm summarized in Table 12.2. Table 12.2 Algorithm for balancing redox reactions
1. Determine oxidation numbers of all atoms in the reaction. Identify those atoms that are oxidized and those that are reduced in the reaction. 2. Choose one of the oxidized or reduced species to have a stoichiometric coefficient of 1. 3. Assign other stoichiometric coefficients that are established unambiguously by the choice made in Step 2. 4. Given that the total number of electrons lost by atoms that are oxidized must equal the total number of electrons gained by those that are reduced (electrons must be conserved) and given the formula of each compound, determine stoichiometric coefficients for the other oxidized and reduced species in the reaction. 5. Add H2 O to either side of the reaction to balance the oxygen atoms. 6. Add H+ to either side of the reaction to balance charge. 7. Hydrogen should automatically balance at this point. Check the H balance to verify.
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Example 12.3 Balance a redox reaction in which ferrous ion, Fe2+ , is oxidized to ferric hydroxide solid, Fe(OH)3 (s), while dichromate ion, Cr2 O2– 7 , is simultaneously reduced to the chromium dihydroxo complex Cr(OH)+2 . This type of reaction is often used to convert Cr from its more toxic to its less objectionable oxidation state (+6 to +3). Solution The unbalanced reaction can be written as follows: Fe2+ +
Cr2 O27 )*
Fe(OH)3 (s) +
Cr(OH)+ 2
In the reaction, Fe(II) is oxidized to Fe(III), and Cr(VI) is reduced to Cr(III). Choosing the stoichiometric coefficient of Cr2 O2– 7 to be 1, it is clear that the coeffi+ cient of Cr(OH)2 must be 2, since this is the only way that the Cr atoms can balance. Each Cr atom undergoing the conversion from Cr(VI) to Cr(III) gains three electrons, so six moles of electrons are transferred per mole of stoichiometric reaction. Each Fe(II) atom releases only one electron when it is converted to Fe(III), so six moles of Fe2+ must be oxidized to provide enough electrons to reduce one mole of Cr2 O2– 7 . The coefficients on the four redox-active species are therefore as follows: 6 Fe2+ + 1 Cr2 O27 )* 6 Fe(OH)3 (s) + 2 Cr(OH)+ 2 The reaction has seven oxygen atoms on the reactant side and 22 on the product side. To balance the oxygen, we add 15 water molecules to the reactant side. 6 Fe2+ + 1 Cr2 O27 + 15 H2 O )* 6 Fe(OH)3 (s) + 2 Cr(OH)+ 2 Finally, we note that the total charge on the reactant side is +10, while that on the product side is +2. Adding eight H+ ions to the product side balances the charge and, as it must, also balances the H atoms (30 on each side). The final, balanced reaction is + 6 Fe2+ + 1 Cr2 O27 + 15 H2 O )* 6 Fe(OH)3 (s) + 2 Cr(OH)+ 2 + 8H
Note that H+ and H2 O appear in the reaction, but neither hydrogen nor oxygen is oxidized or reduced; their oxidation states are +1 and 2, respectively, in all compounds on both the reactant and product sides of the reaction.
12.4
REDOX HALF-REACTIONS
It is sometimes convenient to think of electron transfers as taking place in two steps — one in which electrons are released by the species being oxidized and another 671
in which they are consumed by the species being reduced. The individual reactions releasing or consuming electrons cannot take place in isolation and are therefore often called half-reactions. The redox half-reactions that can be combined to generate the overall reaction in the preceding example [oxidation of Fe2+ to Fe(OH)3 (s) and + reduction of Cr2 O2– 7 to Cr(OH)2 ] are Fe2+ + 3 H2 O )* Fe(OH)3 (s) + 3 H+ + e
(12.1)
Cr2 O27 (aq) + 10 H+ + 6e )* 2 Cr(OH)+ 2 + 3 H2 O
(12.2)
Redox half-reactions representing oxidation of any compound and reduction of any other can easily be combined (on paper) to form a virtually limitless variety of overall redox reactions. One half-reaction that is of overwhelming importance in environmental systems is the reduction of dissolved oxygen, because of the widespread availability of oxygen in the environment, the large amount of energy released by oxygen reduction, and the role that oxygen plays in the metabolism of many bacteria and all plants and animals. The maximum amount of dissolved oxygen that can potentially be consumed (i.e., reduced, forming H2 O) by oxidation of the organic and inorganic constituents of a solution has therefore been widely adopted as a useful water quality parameter. Unfortunately, direct measurement of oxygen consumption by such constituents is usually not a viable approach for assessing the total oxygen-consuming capacity of the solution, both because the reactions are often slow and because the potential O2 consumption might be far greater than the amount of O2 available. Analysis of all the individual compounds that can react with oxygen is also usually impractical, because the systems typically contain a plethora of oxidizable compounds, each present at a low concentration. To address this problem, the chemical oxygen demand (COD) test has been developed. In this test, a very strong, very acidic oxidizing agent (chromic acid, which dimerizes in concentrated solutions to form dichromate ion, Cr2 O2– 7 ) is added to the test solution, along with some catalysts and other reactants, and the mixture is heated. Under these conditions, most of the oxidizable chemicals in the original solution are efficiently oxidized. The difference between the chromic acid dose and the amount of chromic acid remaining at the end of the test indicates how much Cr(VI) was reduced to Cr(III), and therefore how many electrons were released by the substances that were oxidized. If the same oxidation process were carried out using oxygen as the oxidant, oxygen molecules would have to acquire the same number of electrons as the Cr(VI) did in the actual test. Therefore, one can use the results of the COD test [i.e., the concentration of Cr(VI) consumed] to compute how much oxygen would be required to carry out the same reactions. The half-reaction for reduction of oxygen is O2 (aq) + 4 H+ + 4e )* 2 H2 O 672
(12.3)
According to this reaction, each mole of O2 (aq) that gets reduced consumes four moles of electrons, whereas Reaction (12.2) indicates that reduction of one mole of Cr(VI) consumes only three. The molar concentration of oxygen that would be reduced in any redox reaction is therefore only three-fourths as much as the molar concentration of Cr(VI); correspondingly, the COD of a sample in moles of O2 per liter always equals 0.75 times the number of moles per liter of Cr(VI) reduced in the COD test; on a mass basis, this ratio corresponds to 0.42 mg COD per mg Cr(VI) consumed. Since oxygen is a weaker oxidant than Cr(VI), and since environmental conditions are never as favorable for the oxidation reactions as they are in the COD test, the actual oxygen consumption accompanying redox reactions in the environment is almost always less than the value computed based on the COD test. Nevertheless, the test provides a relatively quick, consistent estimate of the maximum O2 consumption that could be caused by constituents in the water, and it is therefore a useful indicator of one aspect of water quality. Historically, the efficiency of treatment processes at removing oxidizable compounds from wastewaters has frequently been based on a comparison of the COD values of the influent and effluent. Example 12.4 A wastewater contains 15 mg/L phenol (1.6 ⇥ 10
4 M).
(a) Write the redox half-reaction for oxidation of phenol to carbon dioxide, and 3+ combine this reaction with the half-reaction for Cr2 O2– to 7 reduction to Cr develop the overall reaction for oxidation of phenol in the COD test.
(b) How much Cr(VI) is reduced by the phenol in a COD test? How much COD does the phenol contribute to the solution, expressed as mg O2 /L? Solution (a) Following the steps in Table 12.2, balanced half-reactions for phenol oxidation and dichromate reduction are C6 H6 O + 11 H2 O )* 6 CO2 + 28 H+ + 28 e Cr2 O27 + 14 H+ + 6 e )* 2 Cr3+ + 7 H2 O When these reactions are combined to generate a net, overall reaction, all the electrons released by phenol oxidation must be consumed by dichromate reduction. Therefore, 28 moles of dichromate must be reduced for every six moles of phenol oxidized. Adding the reactions in this ratio (14 times the dichromate reaction plus 3 times the phenol reaction), we can write the overall reaction as 14 Cr2 O27 + 3 C6 H6 O + 112 H+ )* 28 Cr3+ + 18 CO2 + 65 H2 O 673
(b) The concentration of electrons acquired by Cr2 O2– 7 in the COD test is ! ! ✓ ◆ 14 mol Cr2 O27 6 mol e 4 mol C6 H6 O 1.6 ⇥ 10 L 3 mol C6 H6 O mol Cr2 O27 = 4.48 ⇥ 10
3 mol
e
L
Because the same number of electrons is required to oxidize the phenol regardless of which oxidant is used, the concentration of oxygen that would be consumed if it were the oxidant is ✓ ◆✓ ◆ 1 mol O2 mol O2 3 mol e 4.48 ⇥ 10 = 1.12 ⇥ 10 3 L 4 mol e L ✓ ◆✓ ◆ mol O2 mg O2 mg O2 1.12 ⇥ 10 3 32, 000 = 35.7 L mol O2 L Thus, the phenol in the wastewater contributes a COD of 35.7 mg/L.
12.5
THE ACTIVITY OF FREE ELECTRONS; EQUILIBRIUM CONSTANTS FOR REDOX HALF-REACTIONS
We can write a generic redox half-reaction and the corresponding equilibrium constant as follows: Ox + ne e )* Red K(12.4) =
{Red} {Ox}{e }ne
(12.4) (12.5)
where “Ox” and “Red” represent an oxidized species and the conjugate reduced species, respectively. If multiple species appear on either side of the reaction, the same equation applies for the equilibrium constant, with the understanding that {Red} represents the product of the activities of all species on the product side of the reaction, and {Ox} represents the product of the activities of all species on the reactant side, excluding the electrons. To evaluate K(12.4) by plugging values into the right side of Equation (12.5), we have to evaluate the activity of free electrons, {e }, in the solution. Although defining that term based on the convention used for most other dissolved species (i.e., a standard state concentration of 1.0 mol/L and a reference state of infinite dilution) 674
Redox reactions between metallic (zero-valent) iron and some chlorinated organic compounds can dechlorinate the organics. This reaction has been utilized at a number of field sites where iron has been placed in the path of contaminated groundwater in an attempt to dechlorinate trace organics in the water. This photograph shows an experimental site in Utah where such a process is being tested; the iron is in the closer pit. (D.L. Naftz, USGS.)
presents no conceptual problem, it does present a practical one: as noted previously, free electrons are extremely unstable in aqueous solutions, so the concentration of e in almost any real solution is immeasurably small. This difficulty is overcome by defining the standard state for dissolved electrons (i.e., the conditions under which the activity of electrons is 1.0 by definition) differently from that for most other dissolved species. Specifically, the standard state for dissolved electrons is chosen to be a solution in equilibrium with H+ and H2 (g), both present at an activity of 1.0. An activity of 1.0 for H+ ions corresponds to pH = 0 and, assuming that H2 (g) behaves according to the ideal gas law, an activity of 1.0 for H2 (g) corresponds to PH2 (g) = 1.0 atm, at 25 C. These species are related by the redox half-reaction: 1 H+ + e )* H2 (g) (12.6) 2 675
Because, by definition, {e } ⌘ 1.0 if Reaction (12.6) is at equilibrium and if both {H2 (g)} and {H+ } are 1.0, the equilibrium constant for Reaction (12.6) must equal 1.0: 1 1 {H2 (g)} 2 (1.0) 2 K(12.6) = = = 1.0 (12.7) {H+ }{e } (1.0)(1.0) equilibrium
Knowing the value of K(12.6) , we can use Equation (12.7) to compute the activity of e in any other equilibrium system containing H+ and H2 (g). For instance, in a pH 7.0 solution in equilibrium with a hydrogen partial pressure of 10 5 atm, the value of {e } would be 1
{e } =
{H2 (g)} 2 (10 5 )0.5 = = 10+4.5 K(12.6) {H+ } (1.0)(10 7.0 )
(12.8)
Using the “p” shorthand, the activity of electrons in this system can be represented as pe = log {e } = 4.5. Once the value of {e } in a solution is established based on a calculation like that in Equation (12.8), the equilibrium constant for any other redox half-reaction that is at equilibrium in the solution can be determined by analysis of the activities of the species participating in that reaction. For instance, assume that the solution described above (with pe = 4.5) also contains Fe2+ and Fe3+ . These species can be interconverted via the half-reaction Fe3+ + e )* Fe2+
(12.9)
Therefore, if we believed that Reaction (12.9) had reached equilibrium, we could determine K for the reaction by analyzing the activities of Fe2+ and Fe3+ and inserting those values in the equilibrium constant expression: K(12.9) =
{Fe2+ } {Fe2+ } = {Fe3+ }{e } {Fe3+ }{10 4.5 }
(12.10)
Based on this type of reasoning (but using somewhat different experimental methods), the equilibrium constant K(12.9) has been determined to be 10+13.03 . This type of analysis has been conducted for many redox couples, and the equilibrium constants for those couples have been tabulated in numerous collections. The equilibrium constants for several half-reactions of importance in environmental systems are provided in Table 12.3. The equilibrium constants given in Table 12.3 can be combined with others (e.g., Ka , H, or Ki values) to yield K values for redox half-reactions between other, related species. For instance, if we wished to know K for the half-reaction in which SO2– 4 is 2– reduced to H2 S(aq), we could combine the given reaction for reduction of SO4 to HS– with Ka1 for H2 S(aq) as follows: 676
Table 12.3 Equilibrium constants for some environmentally important redox half-reactions* Reaction NO–3 + 2 H+ + 2e
log K 28.57
)* NO–2 + H2 O
NO–3 + 10 H+ + 8e
)* NH+4 + 3 H2 O
119.077 14.885
2 NO–3 + 12 H+ + 10e )* N2 (aq) + 6 H2 O – + SO2– 4 + 9 H + 8e )* HS + 4 H2 O 2– + SO2– 4 + 2 H + 2e )* SO3 + H2 O – + SeO2– 4 + 3 H + 2e )* HSeO3 + H2 O + * AsO3– 4 + 5 H + 2e ) H3 AsO3 + H2 O 2– + CrO4 + 6 H + 3e )* Cr(OH)+2 + 2 H2 O OCN– + 2 H+ + 2e )* CN– + H2 O 2 H+ + 2e )* H2 (g) 2 H+ + 2e )* H2 (aq) O2 (g) + 4 H+ + 4e )* 2 H2 O O2 (aq) + 4 H+ + 4e )* 2 H2 O O2 (aq) + 2 H+ + 2e )* H2 O2 (aq) H2 O2 (aq) + 2 H+ + 2e )* 2 H2 O O3 (g) + 2 H+ + 2e )* O2 (g) + H2 O Cl2 (aq) + 2e )* 2 Cl– ClO–3 + 6 H+ + 6e )* Cl– + 3 H2 O OCl– + 2 H+ + 2e )* Cl– + H2 O ClO2 + 4 H+ + 5e )* Cl– + 2 H2 O ClO–2 + 4 H+ + 4e )* Cl– + 2 H2 O HOBr + H+ + 2e )* Br– + H2 O 2 HOBr + 2 H+ + 2e )* Br2 (aq) + 2 H2 O BrO–3 + 6 H+ + 6e )* Br– + 3 H2 O 3+ * Al + 3e
)
Zn2+ + 2e
)* Zn(s)
2+
pe pe (W ) EH , mV 14.285 7.285 845
Al(s)
6.135
881
207.08 20.708 12.308
1225
33.66
4.208
3.668
249
3.65
1.82
8.82
215
36.308 18.154
7.654
1074
39.987 19.994
2.494
1183
67.376 22.459
8.459
1329
2.44
9.44
144
0
7.00
0
3.10
1.55
8.55
92
83.12
20.78
13.78
1229
86.00
21.50
14.50
1272
26.34
13.17
6.17
779
59.59
29.795 22.795
1763
70.12
35.06
28.06
2074
47.20
23.6
23.60
1396
147.02 24.503 17.503
1450
57.73
28.865 21.865
1708
126.67 25.334 19.734
1499
109.06 27.265 20.265
1613
45.36
22.68
19.18
1342
53.60
26.80
19.80
1585
146.10
24.35
17.35
1440
4.88 0.00
85.71
28.57
28.57
1690
25.789 12.895 12.895
763
Ni + 2e
)* Ni(s)
7.98
3.99
3.99
236
Pb2+ + 2e
)* Pb(s)
4.246
2.123
2.123
126
4+
)* Pb
2+
+
Pb + 2e
2+
57.28
28.64
28.64
1690
Cu + e )* Cu
2.69
2.69
2.69
159
Cu2+ + 2e
11.48
5.74
5.74
340
)* Cu(s)
677
Table 12.3 – Continued from previous page Reaction Fe3+ + e )* Fe2+
log K pe pe (W ) EH , mV 13.032 13.032 13.032 771
Fe2+ + 2e
)* Fe(s)
Hg2+ 2 + 2e
)* 2 Hg(l)
14.90 26.91
+
Ag + e )* Ag(s) +
2 Hg(OH)2 + 4 H + 2e +
MnO2 (s) + 4 H + 2e
)* Hg2+ 2 + 4 H2 O 2+ )* Mn + 2 H2 O
Mn3+ + e )* Mn2+ MnO–4 + 8 H+ + 5e )* 3+ * 2+ Co + e )
2+
Mn + 4 H2 O
Co
+ 4+ UO2+ 2 + 4 H + 2e )* U + 2 H2 O + PuO2+ 2 + e )* PuO2 PuO+2 + 4 H+ + e )* Pu4+ + 2 H2 O NpO+2 + 4 H+ + e )* Np4+ + 2 H2 O
7.45
7.45
441
13.455 13.455
796
13.507 13.507 13.507
799
43.185 21.593
7.593
1277
41.60
20.8
6.80
1230
25.35
25.35
25.35
1500
127.82 25.564 14.364
1512
32.40
32.40
1917
9.216
4.608
15.86
15.86
16.63
16.63
11.37
984
10.07
10.07
17.93
596
32.40 9.392 15.86
273 938
*For reactions included in the Visual Minteq database, values are from that database. Other values are computed from the Gibbs energy of reaction or from various other reference sources. The meaning of the terms in the last three columns is discussed later in the chapter.
Reaction
log K
– + SO2– 4 + 8e + 9 H )* HS + 4 H2 O
33.66
HS– + H+ )* H2 S(aq)
6.99
+ SO2– 4 + 8e + 10 H )* H2 S(aq) + 4 H2 O
40.65
It should be apparent that the definition of electron activity bears many similarities to that for H+ activity. For instance, as is true for dissolved electrons, the concentration of free dissolved protons is always negligible. Nevertheless, we choose to define the activity of H+ in a way that makes its value finite. Specifically, we assign {H+ } a value of 1.0 in an ideal solution in which {H2 O} = 1.0 and in which the sum of the concentrations of all species of the type H2n + 1 O+n is 1.0 M. Based on this (arbitrary) definition, we can compute unambiguous values of {H+ } from experimental data in any other system. An important qualitative difference between H+ and e is that H+ is relatively stable when bound to one or more water molecules, whereas e is not (i.e., H3 O+ is a weakly stable species, while H2 O– is not). However, this distinction is not important in terms of how {H+ } and {e } are defined and used. 678
The electron activity determines the relative activities, at equilibrium, of the two species that make up a redox conjugate pair, just as the activity of H+ determines the relative activities of the protonated (acidic) and deprotonated (basic) species that make up an acid/base conjugate pair. This point can be illustrated by manipulating Equation (12.10) as follows: {e } =
{Fe2+ } K(12.9) {Fe3+ } 1
(12.11)
Because K(12.9) is a constant (at a given T and P), this equation demonstrates that {e } is a direct measure of the relative activities of Fe2+ and Fe3+ at equilibrium. Equation (12.11) indicates that any ratio of oxidized to reduced iron may be present in an equilibrium solution, and that each such ratio is associated with a specific value of {e }; the larger the value of {e }, the larger the equilibrium ratio of {Fe2+ } to {Fe3+ }. Similarly, any activity ratio of an acid and its conjugate base may exist in an equilibrium solution, and a given ratio corresponds to a specific value of {H+ }. The range of typical pe and pH values in a few natural environments is shown in Figure 12.2. Using this figure as a guideline, we could predict the dominant form of any element in a given environment at equilibrium. In practice, though, the impact of pe on speciation is often far less dramatic than such predictions imply, because of the tendency for many redox reactions to remain far from equilibrium. In fact, the speciation of many redox pairs does not change at all in response to the addition of an oxidant or reductant, even if the computed change in equilibrium speciation is large. For this reason, one must be very cautious when interpreting or applying equilibrium calculations for such systems.
12.6
DEFINITION OF e AND pe
Just as the generic parameter Ka has been defined to characterize a certain type of acid/base reaction, some conventional terminology has developed for certain types of redox reactions. Specifically, e is defined as the equilibrium constant for a redox half-reaction in which the reduced species releases one electron to form the corresponding oxidized species: 1 1 Red )* Ox + e ne ne e ⌘ K(12.12) =
{Ox}1/ne {e } {Red}1/ne
(12.12) (12.13)
An equivalent definition that is sometimes used and whose validity is obvious from Equation (12.13) is that e for a half-reaction is the equilibrium value of {e } 679
Figure 12.2 Typical pe and pH ranges in natural aquatic systems. Adapted from Garrels and Christ (1965).3
when all the reactants and products of the reaction other than e are in their standard states (and therefore have activities of 1.0). For this reason, e is sometimes called the standard electron activity for the reaction.4 By definition, pe is the negative logarithm of e : ! {Ox}1/ne pe = log (e ) = log {e } (12.14a) {Red}1/ne ! {Red}1/ne 1 pe = log (12.14b) {Ox}1/ne {e } The term in parentheses in Equation (12.14b) is the equilibrium constant for the reduction half-reaction and is the reverse of Reaction (12.12), so we can write: 1 1 Ox + e )* Red ne ne
(12.15)
3 Garrels, R., and Christ, C. Solutions, Minerals, and Equilibria, Freeman Cooper, San Francisco (1965). 4 Keep in mind that the reactant and product sides of the reaction might include other species that do not undergo changes in redox state. For instance, H2 O and/or H+ are often required to balance redox half-reactions. In such cases, the numerator of the fraction in Equation (12.13) includes the activities of all the reaction products (other than e ), and the denominator includes those of all the reactants.
680
log K(12.15) = log
{Red}1/ne 1 {Ox}1/ne {e }
!
= pe
(12.16)
Correspondingly, pe can also be defined as the base-10 logarithm of the equilibrium constant for the reduction half-reaction written for a one-electron transfer [Reaction (12.15)]. The preceding definitions of e and pe are universally accepted and are completely consistent with one another. However, some confusion can arise based on the fact that e is most easily defined by reference to the reaction written as an oxidation [Reaction (12.12)], whereas pe is most easily defined by reference to the reaction written as a reduction [Reaction (12.15)]. The second definition, establishing that pe is log K for the one-electron reduction reaction, is the one that is most widely used in environmental engineering and science, and it is the one that we will rely on here. Correspondingly, although the information in a redox reaction can be shown equally well whether the reaction is written as an oxidation or a reduction, the convention is to write the reactions as reductions. The larger the value of e for a reaction (and the lower the value of pe ), the more the oxidation reaction is favored, or equivalently, the more stable the oxidized species is compared to the reduced species. The oxidized species of redox couples with high e and low pe therefore have only a slight tendency to acquire electrons (i.e., they are weak oxidants), and the reduced species have a strong tendency to release them (i.e., they are strong reductants). The reverse is true for species forming a redox couple with a low e and high pe . These trends and properties have close analogs in acid/base chemistry, with the deprotonated (basic) species of acid/base couples corresponding to the oxidized species of redox couples. For example, if an acid/base couple has a high Ka and low pKa , the basic species is highly stable and has only a slight tendency to acquire a proton (i.e., it is a weak base), whereas the acidic species is a strong acid with a strong tendency to release protons. More such analogies are introduced shortly. Example 12.5 The half-reactions and pe values for the Pb4+ /Pb2+ , Cu2+ /Cu+ , and Co3+ /Co2+ redox couples are as follows: Pb4+ + 2 e )* Pb2+ Cu2+ + e )* Cu+
peo = 28.64
Co3+ + e )* Co2+
peo = 32.40
peo = 2.69
(a) What are the equilibrium constants for the three reactions? (b) Of the six metal species shown, which is the strongest oxidant, and which is the strongest reductant? 681
(c) If a solution that contains all the species shown has attained redox equilibrium at low pH (so that none of the metals undergoes significant hydrolysis), and if Pb2+ and Pb4+ are present at equal activities, which form of dissolved copper and which form of dissolved cobalt would be dominant? Solution (a) The value of log K for a reduction half-reaction is ne (pe ), where ne is the number of electrons transferred. Therefore, the log K values of the three reactions are 57.28, 2.69, and 32.40, respectively. (b) The oxidants in the reactions are the more positively charged species of each pair (Pb4+ , Cu2+ , and Co3+ ). The preceding discussion indicated that oxidant strength increases with decreasing e and increasing pe , so the strongest oxidant listed is Co3+ . Similarly, the strongest reductant is the reduced species of the couple with the lowest pe value, i.e., Cu+ . (c) Knowing that Pb2+ and Pb4+ are present with equal activities, we can apply Equation (12.5) to find the pe of the solution: KPb4+ /Pb2+ = {e } =
{Red} {Pb2+ } 1 = = 4+ 2 {Ox}{e }ne {e }2 {Pb }{e }
1
1/2 KPb4+ /Pb2+
=
1
(1057.28 )1/2
= 10
28.64
pe = 28.64(= pe ) Then, knowing the pe of the solution, we can determine the dominant species of the other two redox couples, as follows: KCu2+ /Cu+ =
{Cu+ } {Cu2+ }{e }
{Cu+ } = KCu2+ /Cu+ {e } = 102.72 (10 {Cu2+ } KCo3+ /Co2+ =
28.64
) = 10
25.92
⌧1
{Co2+ } {Co3+ }{e }
{Co2+ } = KCo3+ /Co2+ {e } = 1033.1 (10 {Co3+ }
28.64
) = 102.46 > 1
We conclude that the oxidized form of copper (Cu2+ ) and the reduced form of cobalt (Co2+ ) are the dominant species in the Cu2+ /Cu+ and Co3+ /Co2+ redox couples.
682
12.7
COMPUTING pe FROM SPECIES ACTIVITIES: THE NERNST EQUATION
By manipulating Equation (12.14b), we can calculate pe in any equilibrium solution of known composition if the pe values of redox-active couples in the system are known. Specifically, we can rearrange the equation as follows: ! {Red}1/ne 1 pe = log (12.14b) {Ox}1/ne {e } pe = log pe = pe
{Red}1/ne + pe {Ox}1/ne
(12.17)
1 {Red} log ne {Ox}
(12.18)
Equation (12.18) is known as the Nernst equation and is one of the fundamental equations of redox analysis. The Nernst equation can be used to determine the pe in a solution at equilibrium based on the speciation of a redox couple in the solution, or to determine the equilibrium speciation of a redox couple in a solution with a known pe. Example 12.6 A solution at pH 7.0 contains Mn2+ at an activity of 10 5 , as well as some MnO2 (s). The redox half-reaction for the MnO2 (s)/Mn2+ reaction is shown below. What is the pe of the solution, assuming the system is at equilibrium? MnO2 (s) + 4 H+ + 2e )* Mn2+ + 2 H2 O
pe = 20.8
Solution Using the Nernst equation, and noting that the activity of MnO2 (s) is 1.0 regardless of its concentration, we find pe = pe pe = 20.8
{Mn2+ }{H2 O}2 1 log 2 {MnO2 (s)}{H+ }4 1 (10 5.0 )(1.0)2 log = +9.3 2 (1.0)(10 7.0 )4
Example 12.7 The U.S. Environmental Protection Agency (EPA) maximum contaminant level (MCL) for arsenic in drinking water is 10 µg/L. Arsenic occurs naturally primarily in two oxidation states: As(III) and As(V). It is acidic in both oxidation states, but 683
much more so as As(V), in which case it is a chemical analog of phosphoric acid [i.e., P(V)]. The pKa values of arsenous acid, H3 AsO3 , and arsenic acid, H3 AsO4 , are shown below and indicate that H3 AsO3 is almost fully protonated at pH 7, whereas H3 AsO4 is extensively deprotonated. Partly because of the difference in ionic charge on the dominant species, it is much easier to remove As(V) than As(III) from solution by adsorption and ion exchange water treatment processes. pKa1
pKa2
H3 AsO3
9.17
14.10
H3 AsO4
2.20
7.01
pKa3 11.80
Workers at a water treatment plant want to ensure that >99% of the T OT As in the water is in the oxidized form prior to feeding it to an ion exchange process. The pH of the water is 6.7. If the As(V)/As(III) redox couple equilibrates rapidly in response to changes in pe in the solution, what is the acceptable pe range to meet the treatment goal? Solution The As(V)/As(III) redox reaction is given in Table 12.3 as AsO34 + 2e + 5 H+ )* H3 AsO3 + H2 O
log K = 39.99, pe = 19.99
This reaction and the corresponding log K relate the activities of AsO3– 4 and H3 AsO3 to one another. Furthermore, based on the pKa values and the solution pH, we infer that essentially all the As(III) in the solution is present as H3 AsO3 , but the As(V) is present predominantly as H2 AsO–4 , with only a small fraction present as AsO3– 4 (a3 = 2.69 ⇥ 10 6 ). Combining this information with the Nernst equation, we find ! (H3 AsO3 )(H2 O) 1 {Red} 1 pe = pe log = 19.99 log ne {Ox} 2 (AsO34 )(H+ )5 ✓ ◆ 1 T OT As(III) 1.0 ⇡ 19.99 log 2 (2.69 ⇥ 10 6 )T OT As(V) (10 6.7 )5 ✓ ◆ 1 39 T OT As(III) = 19.99 log 1.18 ⇥ 10 2 T OT As(V) Substituting the criterion that T OT As(V) is 99 times as large as T OT As(III) yields
pe = 19.99 = 1.45
✓ ◆ 1 1 log 1.18 ⇥ 1039 2 99 684
Thus, 99% of the arsenic will be in the more oxidized state if pe = 1.45. Increasing the pe shifts the speciation from As(III) to As(V), so the treatment goal will be met if the pe is greater than or equal to 1.45.
12.8
COMBINING REDOX REACTIONS
The definition of pe is particularly useful for quickly computing the value of the equilibrium constant for an overall redox reaction composed of any two halfreactions. For instance, say we wanted to write the balanced overall reaction and determine the equilibrium constant for oxidation of H2 S to SO2– 4 by reduction of – HOCl to Cl . This reaction is sometimes used to convert S( II) species to S(VI) species as sewage enters a wastewater treatment plant, both to reduce odors and to prevent toxic S( II) species from entering the biologically active reactor. Since the half-reaction corresponding to pe is for a one-electron reduction, adding the “pe reaction” for the species being reduced (HOCl) to the reverse of the pe reaction for the species being oxidized yields a balanced overall reaction of interest, from which the free electron species have been eliminated. Log K for this overall reaction is the sum of pe for the redox couple being reduced and pe for the couple being oxidized. For example, for the reaction of HOCl with H2 S, the overall reaction and corresponding log K are 1 1 1 5 H S + H2 O )* SO24 + H+ + e 8 2 2 8 4
log K = pe
SO24 /H2 S
1 1 1 1 HOCl + H+ + e )* Cl + H2 O 2 2 2 2 1 1 1 1 3 H2 S + HOCl )* Cl + SO24 + H+ 8 2 2 8 4
= 5.08
log K = peHOCl/Cl = 25.10 log K = peHOCl/Cl
pe
SO24 /H2 S
= 20.02
Although the electron does not appear as either a reactant or a product, the overall redox reaction obtained from this procedure always characterizes a one-electron transfer. Because of the way the reaction is generated, the value of log K for the overall reaction is often designated pe ; to minimize confusion with pe values for redox half-reactions, in this text, the pe value for overall reactions will be designated as DpeOx’t/Red’t , i.e.,5 DpeOx’t/Red’t = log Koverall = peOx’t 5 In
peRed’t
(12.19)
the literature, this quantity is frequently represented as peOx’t/Red’t . The ‘D’ is added here to avoid any confusion with the peOx/Red , which is used to characterize a single half-reaction, and also because remembering that the term represents a difference in two pe values is useful when evaluating the energy changes associated with the reaction. The meaning of the term is almost always obvious from context, because the subscript Ox/Red refers to oxidized and reduced species of a conjugate redox pair when referring to a half-reaction and to an oxidant and reductant from two different redox pairs when referring to an overall reaction.
685
where Ox’t and Red’t are the oxidant and reductant, respectively. By analogy with the designation of e as the standard electron activity, DpeOx’t/Red’t is sometimes called the standard pe of the reaction. Applying the Nernst equation to both the oxidation and reduction half-reactions in an overall reaction, we can find the difference in the pe values associated with the two half-reactions (DpeOx’t/Red’t ) under nonstandard conditions as DpeOx’t/Red’t ⌘ peOx’t = peOx’t
peRed’t
(12.20)
1 {Red} log ne {Ox}
= DpeOx’t/Red’t = DpeOx’t/Red’t
Ox’t
peRed’t
1 {Red} log ne {Ox}
Red’t
1 {Red}Ox’t {Ox}Red’t log ne {Ox}Ox’t {Red }Red’t 1 log Q ne
(12.21)
where Q is the conventional reaction quotient (i.e., the ratio of the activities of the products to those of the reactants) for the overall reaction, and ne is the number of electrons that are transferred in that reaction (even though, once again, the electrons do not appear explicitly as reactants or products). As is shown later in the chapter, Equation (12.21) is very useful when evaluating the energetics of overall redox reactions. Example 12.8 (a) Using the half-reactions shown in Table 12.3, write balanced reactions and determine DpeOx’t/Red’t values for the following overall reactions: (i) Oxidation of NH3 (aq) to NO–3 by reduction of O2 (g) to H2 O. (ii) Oxidation of cyanide (CN– ) to cyanate (OCN– ) by reduction of Cl2 (aq) to Cl– . (b) What is the equilibrium constant for the following reaction? NH3 (aq) + 2 O2 (aq) )* NO3 + H2 O + H+
Solution (a) Each overall reaction can be obtained by adding the half-reaction for the species being reduced to the reverse of the half-reaction of the species being oxidized.
686
(i) O2 /NH3 1 1 O2 (g) + H+ + e )* H2 O 4 2 1 3 1 9 NH3 (aq) + H2 O )* NO3 + H+ + e 8 8 8 8 1 1 1 1 1 NH3 (aq) + O2 (g) )* NO3 + H2 O + H+ 8 4 8 8 8
log K = pe = 20.78 log K = pe = 13.73 log K = DpeO
2 /NH3
= 7.05
(ii) Cl2 /CN– 1 1 Cl (aq) + e )* Cl 2 2 2 1 1 1 CN + OH )* CNO + H2 O + e 2 2 2 1 1 1 1 Cl (g) + CN + OH )* Cl OCN + H2 O 2 2 2 2 2
log K = pe = 23.60 log K = pe = 16.44 log K = DpeCl
2 /CN
= 40.04
(b) The reaction of interest is the same as that in part (a)(i), except that in this part of the question all the stoichiometric coefficients in (a)(i) are multiplied by 8, because we want to determine log K for an eight-electron transfer. Multiplying the reaction in part (a)(i) by 8, we obtain
NH3 (aq) + 2 O2 (g) )* NO3 + H2 O + H+ log K = 8DpeO
2 (g)/NH3
= 56.40
12.9
REDOX SPECIATION AND LOG C-pe DIAGRAMS
12.9.1
Redox Reactions with No H+ Exchange
Equilibrium speciation of redox couples can be represented on log C-pe diagrams that, not surprisingly, bear many similarities to log C-pH diagrams. Consider, for example, a system containing 10 4 M T OT Fe at a pH that is low enough that formation of hydroxo complexes can be ignored and in which both Fe2+ and Fe3+ behave ideally. By definition, e equals log K for the oxidation reaction, so we can write e =
(Fe3+ )(e ) (Fe2+ )
(12.22)
Thus, the fraction of T OT Fe that is in the +2 oxidation state in the given system is
687
aFe2+ =
=
(Fe2+ ) (Fe2+ ) = = T OT Fe (Fe2+ ) + (Fe3+ )
1 e 1+ (e )
aFe3+ = 1
=
(e ) (e ) + e
aFe2+ =
e (e ) + e
1 (Fe3+ ) 1+ (Fe2+ ) (12.23)
(12.24)
Equations (12.23) and (12.24) are essentially identical to the equations for a0 and a1 for a monoprotic acid, with e in the Fe3+ /Fe2+ system taking the place of H+ in an HA/A– system, and e corresponding to Ka . With these results and the fact that pe for the Fe3+ /Fe2+ couple is 13.03 (from Table 12.3), we can prepare the log Cpe diagram for a system with 10 4 M T OT Fe; the diagram is shown in Figure 12.3. As in a log C-pH diagram, the lines for the two species have slopes that differ by 1 (the number of electrons exchanged when one species is converted to the other); extrapolation of the linear parts of the curves shows that they would intersect at the point pe = pe and log C = log (T OT Fe); and the curvature near this point causes the curves to actually intersect at pe = pe and log C = log (T OT Fe) 0.3.
Figure 12.3 LogC-pe diagram showing the concentrations of Fe2+ and Fe3+ in an ideal solution containing 10 4 M T OT Fe, at a pH low enough that Fe2+ and Fe3+ are the dominant species in the +2 and +3 oxidation states, respectively.
688
12.10
SUMMARY OF SOME KEY SIMILARITIES BETWEEN ACID/BASE AND REDOX SYSTEMS
Numerous similarities between acid/base and redox systems have been alluded to in the preceding sections. For example, just as the equilibrium speciation of an acid/base couple in a given solution depends on both the pH and pKa , that of a redox couple depends on the pe (which characterizes the redox state of the solution) and pe (which characterizes the inherent tendency for that couple to retain or release electrons). The higher the pe of the solution and the lower the pe of the couple, the more favored the oxidized species is over the reduced species. Solutions with high pe values are referred to as being strongly oxidized or oxidizing (analogous to being strongly alkaline), and solutions characterized by a low pe are referred to as being reduced or reducing. If we wish to alter the pe of a solution, we can do so by adding an oxidized or reduced species (which will increase or lower the solution pe, respectively), just as we add a base or an acid to raise or lower the solution pH. These analogies and several others between redox and acid/base equilibria are summarized in Table 12.4. The major difference between these two groups of reactions is simply in the conventions used for writing the reactions: as noted previously, the convention for writing redox reactions is to show them as reductions, with the electron as a reactant, whereas the convention for acid/base reactions is to write them as dissociations, with the proton as a product. A visual representation of the pH scale is shown in Figure 12.4a, and the analogous pe scale in Figure 12.4b. At various values along each scale, dominance switches from the acidic or reduced form of a given couple to the conjugate basic or oxidized form. For reasons that are explained shortly, the crossover pe value for many conjugate redox pairs is at pe only if the solution pH is 0.0. At pH 7, the pe at which dominance switches [designated pe (W)] is often of interest; this value is shown in part (c) of the diagram and is discussed shortly. The information on the pH scale in Figure 12.4a can be interpreted as showing that HAc and OCl– cannot both be dominant in the same solution, because HAc is dominant only at pH < 4.76 while OCl– is dominant only at pH > 7.53. If we make the approximation that the interconversion between the acidic and basic species of a conjugate pair occurs entirely at pH = pKa , then we could say than HAc and OCl– cannot coexist in any equilibrium solution. By the same logic, HS– and HOCl cannot both be dominant redox species in an equilibrium solution at pH 0 (Figure 12.4b), because HS– is never dominant over – SO2– 4 at the pe values that are required for HOCl to be dominant over Cl . Making the – same approximation as above, then, we might say that HS and HOCl cannot coexist at equilibrium at pH 0; if both species are present in a solution, the solution cannot be at equilibrium, and a thermodynamic driving force will exist favoring oxidation of the HS– and transfer of the electrons to HOCl until one or the other of these reagents is (almost) entirely consumed. 689
690
Relationship between pe or pKa and log K
Ways of defining e and Ka (Two equivalent ways of expressing each definition are shown; the most common expression is shown in bold)
{Red} + ne pe {Ox}
1 1 (Acid) )* (Base) + H+ nH nH pKa = log K for the reaction: 1 1 (Base) + H+ )* (Acid) nH nH 1 pKa = log Kdeprotonation nH
pe = log K for the reaction: 1 1 Ox + e )* Red ne ne 1 pe = log Kreduction ne
K a ⌘ K for the reaction:
{Base} + nH pH {Acid}
{Base}{H+ }nH {Acid} log K = log
Kdissoc’n =
Acid )* Base + nH H+
Acid
Base
Acid/Base Reactions H+
1 1 Red )* Ox + e ne ne
e ⌘ K for the reaction
log K = log
General reaction, written in conventional form
log K
Ox + ne e )* Red
Molecules having e or H+ {Red} {Ox}{e }ne
Reduced species
Molecules lacking e or H+
Kreduction =
Oxidized species
Species being transferred
Equilibrium constant for reaction written in conventional form
Redox Reactions e
Table 12.4 Analogies between acid/base and redox reactions.
691
High pe: oxidizing Low pe: reducing
Solution Condition
High pH: alkaline Low pH: acid
At pH > pKa : {Base} > {Acid} At pH < pKa : {Acid} > {Base}
Acid/Base Reactions {Acid} = {Base}, when pH = pKa
* Relationships shown on this line apply only if the stoichiometric coefficients on Ox and Red (or Acid and Base) are equal, and if all non-redox-active species that participate in the reaction are present at unit activity.
At pe > pe : {Ox} > {Red} At pe < pe : {Red} > {Ox}
Condition away from crossover point*
Conditions at “crossover” point*
Redox Reactions {Ox} = {Red}, when pe = pe
Table 12.4 – Continued from previous page
Figure 12.4 Conditions required for equal activity of the two species of a conjugate pair in various systems. To the left of the vertical bar (|), the acidic or reduced form is dominant, and to the right, the basic or oxidized form is dominant. (a) Acid/base couples shown on a pH scale; (b) Redox couples shown on a pe scale, for pH = 0; (c) Redox couples on a pe scale, for pH = 7.0. The vertical bars are at pH = pKa in part (a), pe = pe in part (b), and pe = pe (W) in part (c).
Generalizing the above result, at pH 0, any oxidized species is predicted to oxidize any reduced species that is part of a redox couple with a lower pe . On the other hand, oxidized species are unreactive with any reduced species that is part of a couple with a higher pe . Analogous statements apply at pH 7, except that the critical pe at that pH is pe (W). An interesting conclusion from this analysis is that neither HOCl nor OCl– is ever stable in an aqueous solution at pH 0 or 7; if such a solution is prepared and the redox reactions reach equilibrium, essentially all the HOCl and OCl– will oxidize the oxygen ion in water to O2 (aq), while the Cl(+I) ion in the hypochlorite species will be reduced to Cl– . However, as is the case with many redox reactions, this reaction is kinetically inhibited at normal temperatures and pressures.
12.11
REDOX REACTIONS INVOLVING EXCHANGE OF BOTH ELECTRONS AND PROTONS
12.11.1
Definition of pepH and pe (W)
Often, the two species that form a redox conjugate pair have different acid strengths (i.e., Ka values); the most common situation is for oxidation to increase the acidity 692
of the molecule, causing protons to be released along with electrons in the oxidation half-reaction. In such cases, the redox speciation depends not only on solution pe, but also on pH. For instance, consider a generic half-reaction in which nH protons and ne electrons combine with the oxidized species to form the reduced species
Ox + nH H+ + ne e )* Red
(12.25)
By definition, pe for Reaction (12.25) equals 1/ne times the logarithm of the equilibrium constant for the reaction. Writing out this relationship and then rearranging it slightly, we obtain pe =
1 1 {Red} nH ne log K(12.24) = log + pH + pe ne ne {Ox} ne ne
1 {Red} log = pe ne {Ox}
pe
nH pH ne
(12.26a)
(12.26b)
Equation (12.26b) indicates that the larger the value of pe, the more the oxidized species is favored over the reduced species, consistent with the trends discussed previously. However, the activities of the oxidized and reduced species are not necessarily equal at pe = pe , because the right side of the equation depends on solution pH as well as pe. Rather, the criterion for the reduced and oxidized species to be present at equal activity is {Red} = {Ox}
if
pe = pe
nH pH ne
(12.27)
Equation (12.27) applies at any pH.6 For a solution at known pH, it is convenient to designate the expression on the right side of this equation as pepH , i.e., pepH ⌘ pe
nH pH ne
(12.28)
Substituting this expression into Equation (12.26b), we obtain 1 {Red} log = pepH ne {Ox}
pe
(12.29)
pepH can be thought of as a conditional pe that is specific to the given pH. The usefulness of pepH is that it tells us directly the pe at which dominance changes from 6 Equation (12.27) includes the implicit assumption that the stoichiometric coefficients on Ox and Red are identical. That assumption is usually, but not always, valid (see, for example, the HOBr/Br2 (aq) and Hg(OH)2 /Hg2+ 2 half-reactions in Table 12.3). In cases where the assumption is violated, a derivation like that shown here can be carried out, but the details are different, and the pe at which {Red} = {Ox} depends on both pH and T OTi.
693
Redox reactions are widely used for remediation of contaminated groundwater and soil. These images show (a) an ozone generator being used to inject ozone into the ground through Teflon R tubing; (b) a trench and entry points for injecting potassium permanganate; and (c) a tank of ferrous sulfate and injection pumps being transported to an injection site. (Courtesy of Pacific Groundwater Group.)
the oxidized to the reduced species if the system is at the specified pH. It thereby provides a direct measure of the relative oxidizing tendencies of different species at the given pH. By far the most common pH value used when reporting pepH values is 7.0; pe7.0 is frequently represented as pe (W), with the W signifying (neutral) water. Thus, nH pe (W) ⌘ pe7.0 = pe (7.0) (12.30) ne For example, for the SO24 /SO23 couple, nH = ne = 2, and pe = 13.58, so ✓ ◆ 2 pe (W) = 13.58 (7.0) = 6.58 2 694
From this value, we know that in a solution at pH 7, dominance switches between 2– SO2– 3 and SO4 at pe = 6.58. Table 12.3 includes a column showing pe (W) values for all the half-reactions listed. In general, the terms substituted for {Red} and {Ox} in the Nernst equation must include the activities of all the species on the reduced and oxidized sides of the reaction, respectively, other than e . However, when a system is at pH 7 and the standard pe is expressed as pe (W), the system pe is sometimes represented as pe(W), and the Nernst equation can be written as pe (W) = pe (W)
1 {Red} log ne {Ox}
(12.31)
When this equation is used, the term {H+ }nH is not included in the logarithmic argument on the right side even if H+ appears in the reaction, because its effect has already been accounted for in the conversion of pe to pe (W). Note that the calculated value of pe (W) is the same as that of pe; including the ‘W’ in the term on the left side of this equation serves simply as a reminder that the solution is at neutral pH. As noted previously, the pe values where dominance shifts between oxidized and reduced species of various redox couples in solutions at pH 7.0 are shown on a pe (W) scale in Figure 12.4c. By comparing pe (W) values of various redox couples, we can infer the relative stabilities of the oxidized and reduced species in those couples under conditions that are typical of natural systems and many engineered water or wastewater treatment systems. Specifically, the higher the value of pe (W), the more strongly oxidizing the redox couple is at pH 7.0, and the more stable the reduced species is compared to the oxidized species. For this reason, pe (W) values are often shown instead of or in addition to pe values in tabulations intended for environmental applications. In such tabulations, the reactions are often written with the dominant acid/base forms of all species as reactants and products, so that pe (W) corresponds to the pe where the majority of the redox-active element changes from its oxidized to its reduced state. This point is illustrated in the following example. Example 12.9 The log C-pe diagram for the Fe(III)/Fe(II) system was shown in Figure 12.3 for a solution whose pH was low enough that hydrolysis could be ignored. (a) Compute pe (W) for (i) the Fe3+ /Fe2+ couple, and (ii) the redox couple comprised of the dominant Fe(III) and Fe(II) species at pH 7, which are Fe(OH)+2 and Fe2+ , respectively. (b) Develop the log C-pe diagram for a pH 7 solution containing 10 4 M T OT Fe and relate it to the two pe (W) values determined in part (a). Assume ideal solute behavior, and ignore possible precipitation of any solids. Hydrolysis constants for Fe2+ and Fe3+ are as follows: 695
2+
Fe Fe3+
log⇤ KOH,1 9.50 2.19
log⇤ KOH,2 11.07 3.48
log⇤ KOH,3 10.43 7.93
log⇤ KOH,4 8.00
Solution (a) (i) The redox reaction for the Fe3+ /Fe2+ couple does not involve transfer of protons (nH = 0), so, according to Equation (12.30), pe (W) is the same as pe ; i.e., it is 13.03. (ii) The reaction between the Fe(III) and Fe(II) species that are dominant at pH 7.0 and the corresponding equilibrium constant can be derived by summing the Fe3+ /Fe2+ redox reaction and the reverse of the hydrolysis reactions that convert Fe3+ to Fe(OH)+2 : Fe3+ + e )* Fe2+
log K = peFe3+ /Fe2+ = 13.03
FeOH2+ + H+ )* Fe3+ + H2 O
log K =
log ⇤ KOH,1,Fe3+ = +2.19
Fe(OH)+2 + H+ )* FeOH2+ + H2 O
log K =
log ⇤ KOH,2,Fe3+ = +3.48
Fe(OH)+2 + 2 H+ + e )* Fe2+ + 2 H2 O
log K = 18.70
Because the composite reaction is written as a 1-electron reduction, pe = log K, i.e., pe = 18.70. pe (W) can then be calculated using Equation (12.30) as 2 pe (W) = 18.70 (7.0) = 4.70 1 (b) The log C-pe diagram for a solution at pH 7.0 containing 10 4 M T OT Fe can be derived using an approach similar to that for developing log C-pH diagrams. We begin by writing the mass balance on Fe, taking into account all the species in both oxidation states. T OT Fe = (Fe2+ ) + (FeOH+ ) + (Fe(OH)2 ) + (Fe(OH)3 ) + (Fe3+ ) + (FeOH2+ ) + (Fe(OH)+ 2 ) + (Fe(OH)3 ) + (Fe(OH)4 ) " # o + (Fe(OH) ) (FeOH ) (Fe(OH) ) 3 2 = (Fe2+ ) 1 + + + (Fe2+ ) (Fe2+ ) (Fe2+ )
(12.32a)
(Fe(OH)o3 ) (Fe(OH)4 ) (FeOH2+ ) (Fe(OH)+ 2) + (Fe ) 1 + + + + (Fe3+ ) (Fe3+ ) (Fe3+ ) (Fe3+ ) (12.32b) 3+
The calculations can be greatly simplified by recognizing that the acid/base relationships among the species in each oxidation state depend only on pH, 696
and not pe. Therefore, because the pH is known and we are assuming ideal behavior of the solutes, the concentration ratio of each Fe(II) OH complex to Fe2+ , and of each Fe(III) OH complex to Fe3+ , is constant, independent of pe. For example, ⇤K (FeOH+ ) 10 1 = = 2+ + (H ) 10 (Fe )
9.50 7.0
log (FeOH+ ) = log (Fe2+ )
= 10
2.50
(12.33a)
2.50
(12.33b)
The implication is that the curve for FeOH+ on the log C-pe diagram will be parallel to and 2.50 units below that for Fe2+ , independent of pe. Similar conclusions apply to all the other Fe(II) OH complexes and to all the Fe(III) acid/base species, with the offsets between the curves dependent on the ⇤ Ki values. When these relationships and the given value of T OT Fe are substituted into Equation (12.32b), that equation simplifies to ⇥ ⇤ 1 + 10 2.50 + 10 7.43 + 10 11.00 ⇥ ⇤ + Fe3+ 1 + 104.81 + 108.33 + 107.40 + 106.40
10
4
= Fe2+
10
4
=1.003 Fe2+ + 2.41 ⇥ 108 Fe3+
(12.34)
Finally, we can use the eFe3+ /Fe2+ relationship to substitute for (Fe3+ ) in this equation, yielding 10
4
= 1.003 Fe2+ + 2.41 ⇥ 108
Fe2+ =
10 4 2.25 ⇥ 10 1.003 + (e )
5
Fe2+ 10 (e )
13.03
(12.35)
By substituting various values for (e ) into Equation (12.35), we can determine (Fe2+ ) as a function of pe and plot the corresponding curve on the diagram. We can then use peFe3+ /Fe2+ and the (now) known value of (Fe2+ ) at each pe to find
(Fe3+ ) at that pe, and finally use the various stability constant expressions to determine the concentrations of the Fe(II) OH and Fe(III) OH complexes. The complete log C-pe diagram is shown in Figure 12.5 and confirms that all the curves representing iron species in a given oxidation state are parallel. Consistent with the calculations in part (a), (Fe3+ ) = (Fe2+ ) at peFe3+ /Fe2+ (W) (= 13.03), and (Fe3+ ) = (Fe(OH)+ 2 ) at peFe(OH)+ /Fe2+ (W) (= 4.70). In most 2 cases, the latter crossover point would be of more interest, because it is the pe 697
Figure 12.5 log C-pe diagram for a solution containing 10
4M
T OT Fe at pH 7.0.
at which dominance switches between T OT Fe(II) and T OT Fe(III). Note how grossly wrong we would have been about the concentrations of T OT [Fe(II)] and T OT [Fe(III)] at, say, pe = 10 if we had considered only the reactions between the unhydrolyzed species: at that pe, Fe2+ is more than 1000 times as concentrated as Fe3+ , but there is still 200,000 times as much oxidized Fe [T OT Fe(III)] as reduced Fe [T OT Fe(II)] in the solution. As the preceding example demonstrates, pe for the reaction between the dominant redox-active species in a system can sometimes be more informative than that for the reaction between the simplest species in each oxidation state, because the former value allows us to assess directly the pe at which most of the component shifts between its oxidized and reduced states. This advantage applies not just to Me OH complexes, but to complexes with any ligand, as illustrated in the following example. Example 12.10 Both Fe(II) and Fe(III) form extremely strong complexes with cyanide ion (CN– ). As a result, the dominant Fe species in solutions containing cyanide are often Fe(CN)4– 6 (ferrocyanide) and/or Fe(CN)3– 6 (ferricyanide), respectively. The stability constants 52.63 for for formation of these complexes (b6 values) are 1045.61 for Fe(CN)3– 6 and 10 3– 4– 3– Fe(CN)6 . In a solution in which Fe(CN)6 and Fe(CN)6 are expected to be the dominant soluble Fe species, at what pe would most of the Fe(II) be oxidized to Fe(III)?
Solution In the given solution, the shift of dominance between Fe(II) and Fe(III) occurs 3– when the dominant species changes from Fe(CN)4– 6 to Fe(CN)6 . We can determine 698
pe for the Fe(CN)36 /Fe(CN)46 redox couple by combining the Fe3+ /Fe2+ half reaction with the two complexation reactions, as follows: log K
Reaction Fe3+ + e )* Fe2+
13.03
Fe3+ + 6 CN– )* Fe(CN)4– 6
45.61
– * 3+ Fe(CN)3– 6 ) Fe + 6 CN
52.63
4– * Fe(CN)3– 6 + e ) Fe(CN)6
6.01
Because the final reaction shown is a one-electron reduction, log K is pe . Writing out the log K expression, we find ⇣ ⌘ Fe(CN)46 ⇣ ⌘ = 106.01 (e ) 3 Fe(CN)6 Therefore, in the given solution, most of the Fe(CN)4– 6 would be oxidized to 3– – 6.01 Fe(CN)6 when (e ) was less than 10 , corresponding to pe > 6.01, and that is the pe at which the dominant oxidation state of Fe would shift from II to III. Note that, because nH = 0 for this reaction, the shift in dominance occurs at the same pe regardless of the solution pH.
12.11.2
Redox Reactions Involving Exchange of Multiple Electrons
Although some redox transformations, like that between Fe(II) and Fe(III) species, involve transfer of a single electron, the more common situation is for electrons to be transferred in pairs, so that differences of four, six, or even eight electrons between the most oxidized and most reduced forms of an element are fairly common. This tendency differentiates redox reactions from most acid/base reactions: while acids can be multiprotic, it is generally the case that the protons are released one at a time, and it is rare for the most protonated and the most deprotonated species of environmental significance to differ by more than four protons. This difference between acid/base and many redox reaction has important implications for log C-pe diagrams, as becomes evident if we prepare such a diagram for the S( II)/S(VI) redox couple under conditions similar to those for Figure 12.3 (T OTi of 10 4 M, and pH at which only one species in each redox state need be considered). For this exercise, we will prepare the diagram for a solution at pH 4.5, at which H2 S and SO2– 4 are far more – – concentrated than HS and HSO4 , respectively (pKa is 7.02 for H2 S and 1.99 for HSO–4 ). 699
A balanced redox half-reaction between the two species of interest is SO24 + 10 H+ + 8e )* H2 S + 4 H2 O
(12.36)
Log K and pe for the SO24 /HS redox couple are given in Table 12.3, and these values can be combined with Ka1 for H2 S to derive the corresponding values for Reaction (12.36). For instance, log K for the reaction can be computed as + SO2– 4 + 9 H + 8e
)* HS– + 4 H2 O
HS– + H+ )* H2 S + SO2– 4 + 10 H + 8e
log K = 33.66 log K = pKa1 = 7.02
)* H2 S + 4 H2 O
log K = 40.68
(12.37)
Because Equation (12.37) is an eight-electron reduction, peo is (log K)/8, or 5.085. peopH 4.5 can then be found from Equation (12.28): peopH 4.5 = 5.085
10 (4.5) = 0.54 8
(12.38)
This result indicates that, in a solution at pH 4.5, the activity of SO2– 4 equals that 2– of H2 S at pe = 0.54, with H2 S dominant at lower pe and SO4 dominant at higher pe. Furthermore, because eight electrons are transferred when one mole of H2 S is converted to one mole of SO2– 4 , the slopes of the lines for those two species on a log C-pe diagram are different by 8. With this information, we can draw the log C-pe diagram, which is shown in Figure 12.6.
Figure 12.6 log C-pe diagram showing the concentrations of H2 S and SO2– 4 in a system containing 10 4 M T OT S at pH 4.5.
700
At any pe, the curve for the nondominant species is very steep, which is a direct consequence of the fact that the redox reaction involves transfer of multiple electrons. In fact, the slope for the nondominant species is so steep that the concentration of that species becomes insignificant at a pe just a few tenths of a unit away from the crossover point (i.e., pe4.5 ). As a result, if significant amounts both SO2– 4 and H2 S are present in a solution that has a pe different from pe4.5 by more than ⇠ 0.5, the SO4 /H2 S redox reaction will be out of equilibrium by many orders of magnitude, meaning that a large amount of energy will be released when the reaction proceeds; similar trends apply to all redox couples that are linked by exchange of multiple electrons. Example 12.11 Construct a log C-pe diagram showing the speciation of nitrogen in an ideal, pH 8.0 solution that is in equilibrium with atmospheric N2 (g) [PN2 = 0.79 atm; HN2 = 1513 atm/(mol/L)]. Consider the possibility that dissolved N can be present as N(0) [in the form of N2 (aq)], N( III) (as NH3 and NH+4 ), or N(V) (as NO–3 ). Solution Because the solution is in equilibrium with atmospheric N2 (g), the activity and concentration of N2 (aq) are fixed by Henry’s law, independent of pe. (N2 (aq)) =
N2 (g) 0.79 [atm] = = 5.2 ⇥ 10 HN2 1513 [atm/(mol/L)]
4
= 10
3.28
The activity of N2 (aq) is therefore represented on the log C-pe diagram by a horizontal line at log C = 3.28. The reactions in Table 12.3 relating the three N oxidation states of interest are 2 NO3 + 12 H+ + 10e )* N2 (aq) + 6 H2 O
log K = 207.08
NO3 + 10 H+ + 8e )* NH+ 4 + 3 H2 O
log K = 119.007
Substituting the known values of pH and the activity of N2 (aq) into the equilibrium constant expressions for these reactions yields equations for the curves representing NO–3 and NH+4 on the log C-pe diagram. 207.08 = log
N2 (aq) = log (N2 (aq)) (NO3 )2 (H+ )12 (e )10
207.08 = 3.28
2 log NO3 + 12 pH + 10 pe
2 log NO3 + 12(8.0) + 10pe log NO3 = 57.18 + 5pe
119.077 = log
3 ⇠⇠ (NH+ (H⇠ 4 )⇠ 2 O) = log NH+ 4 (NO3 )(H+ )10 (e )8
701
log NO3 + 10 pH + 8 pe
119.077 = log NH+ 4
[ 57.18 + 5pe] + 10(8.0) + 8pe log NH+ 4 = 18.103
3pe
Finally, the activity of NH3 (aq) can be determined based on the fact that, at pH 8.0, it is always 1.24 log units less than the activity of NH+4 (because pKa is 9.24). Thus, log NH3 (aq) = log NH+ 4 = 18.103
1.24 3 pe
1.24 = 19.343
3 pe
Having derived expressions for the activities of N2 (aq), NO–3 , NH+4 , and NH3 (aq), we can prepare the log C-pe diagram, which is shown in Figure 12.7.
Figure 12.7 log C-pe diagram for the N( III)/N(0)/N(V) system in equilibrium with atmospheric N2 (g) at pH 8.0. Note that, in the region where N2 (aq) is the dominant species, T OT N is approximately 2 ⇥ N2 (aq) .
The concentration of each N-containing species is represented by straight line on the diagram, and the total dissolved nitrogen follows a bowl-shaped pattern. In these ways, the diagram bears an obvious resemblance to log C-pH diagrams for solutions in equilibrium with a metal-bearing solid. Recall that, in the case of the metal, the bowl-shaped curve for T OT Me reflected the fact that one species (the soluble species with the same composition as the solid) was present at a fixed concentration, independent of pH, while others were more acidic (and hence had concentrations that increased with decreasing pH) or more basic (and had concentrations that increased with increasing pH). Similarly, in the current example system involving redox reactions, one species [N2 (aq)] has a fixed concentration, independent of pe (because of the gas/liquid equilibrium); others (NH3 and NH+4 ) are more reduced than the species 702
with fixed concentration; and another (NO–3 ) is more oxidized. The lack of curvature in any of the lines reflects the assumption that an essentially infinite reservoir of the soluble species is available (as a solid, in the former systems, and as a gas here) that can dissolve as necessary to meet the requirements of the various equilibrium relationships.
12.12
COMPUTING EQUILIBRIUM SPECIATION IN REDOX SYSTEMS
12.12.1
Governing Equations and Preparation of logC - pe Diagrams
As is true for log C-pH diagrams, log C-pe diagrams contain all the system information about T OTi and the equilibrium constants relating the species being displayed. In the analysis of acid/base problems, once the diagram was prepared, we could determine the equilibrium composition of a system by identifying the pH where the T OT H equation was satisfied. Similarly, when analyzing redox systems, we can solve for the equilibrium composition by identifying the pe where the mass balance on e (i.e., the T OT e equation) is satisfied. This process is described next. We begin by developing the log C-pe diagram for a solution in which 5 ⇥ 10 4 M T OT Cl is distributed among three species: Cl– , HOCl, and OCl– . pe for the HOCl/Cl couple is 25.1, and pKa for HOCl is 7.53. Because the HOCl/OCl speciation is pH-dependent, we anticipate that the log C-pe diagram will look different at different pH values. Therefore, our approach will be to derive the equations needed to draw the diagram leaving pH as an adjustable parameter, and then insert different pH values into the equations to explore the effect of pH. We will then add information to the diagram to account for the presence of 1 ⇥ 10 4 M T OT Fe in the same solution. Finally, we will determine the equilibrium pe and solution composition for solutions which have been prepared by adding the specified concentrations of T OT Cl and T OT Fe in a couple different ways. In the example system, the activities of the three Cl-containing species are related by three equations: a redox equilibrium constant (e.g., pe for either the HOCl/Cl or the OCl /Cl couple), an acid/base equilibrium constant (Ka for HOCl/Cl ), and the mass balance on Cl. The mass balance on Cl (assuming, as usual, ideal behavior of the solutes) is7 7 As in previous chapters, when characterizing a conventional solute that behaves ideally, parentheses
( ) are used to indicate both the activity of the solute and its concentration in mol/L. For H2 O (the solvent) and e (an unconventional solute, in terms of its activity-concentration relationship), we also use parentheses to indicate the species’ activity, but square brackets [ ] when referring to its molar concentration.
703
T OT Cl = (HOCl) + (OCl ) + (Cl )
(12.39)
We can use the acidity constant expression to relate the activities of OCl– and HOCl by (OCl ) = 10pH
pKa
(12.40)
(HOCl)
Similarly, the redox equilibrium expression can be used to write (Cl– ) in terms of one of the Cl(+I) species. For instance, the HOCl/Cl half-reaction and equilibrium constant are 1 1 1 1 HOCl + H+ + e )* Cl + H2 O (12.41) 2 2 2 2
K(12.41) = 10pe =
(Cl )1/2 (H2 O)1/2 (HOCl)1/2 (H+ )1/2 (e )
(Cl ) = 102pe (HOCl)(H+ )(e )2 Substituting 10 becomes
pH
and 10
2pe
(12.42a) (12.42b)
for (H+ ) and (e )2 , respectively, Equation (12.42b)
(Cl ) = 102pe
pH 2pe
(12.43)
(HOCl)
Finally, the expressions for (OCl– ) and (Cl– ) can be substituted into the mass balance on Cl to yield an expression for (HOCl) as a function of pe for any given T OT Cl and pH. T OT Cl = (HOCl) + 10pH HOCl =
1 + 10pH
pKa
T OT Cl + 102pe
pKa
(HOCl) + 102pe pH 2pe
pH 2pe
(HOCl)
(12.44a) (12.44b)
Equation (12.44b) provides the information needed to draw the curve for HOCl on a log C-pe diagram for any given pH. Once the value of (HOCl) is determined, the corresponding values of (OCl– ) and (Cl– ) can be computed from the equilibrium expressions, so that lines for all three species can be drawn, completing the diagram. The diagrams for three pH values are shown in Figure 12.8. Several general observations apply to all three diagrams. First, as we would expect, in all cases the two oxidized species (HOCl and OCl– ) dominate at high pe, and the reduced species (Cl– ) dominates at low pe. However, because interconversion of HOCl and Cl– involves transfer not only of elections but also of a proton, the crossover pe between these two species depends on the solution pH.
704
Figure 12.8 log C-pe diagram for a system containing 5 ⇥ 10
705
4M
T OT Cl at pH 4, 7, and 10.
Second, the lines for HOCl and OCl– are always parallel, reflecting the fact that interconversion of HOCl and OCl– involves transfer of a proton but not of electrons. Because each diagram characterizes a system at a single pH, the ratio (HOCl)/(OCl ) is constant across the whole diagram. Thus, at pH 4.0, (HOCl)/(OCl– ) is 103.53 , so the line for HOCl is always 3.53 log units above that for OCl– . Correspondingly, at pH 7.0, the HOCl line is 0.53 log unit above the OCl– line, and at pH 10.0, it is 2.47 log units below the OCl– line. Third, the slope of the Cl– line is always 2 less than that of the HOCl and OCl– lines, consistent with the fact that two electrons are exchanged in the reaction between one molecule of Cl– and one molecule of either HOCl or OCl– . And, finally, the redox reaction of interest [Reaction (12.41)] has the form of Reaction (12.25), so peopH for the HOCl/Cl couple can be computed from Equation (12.28), using ne = 1 and nH = 12 . The computed values of peopH are 23.1, 21.6, and 20.1 at pH 4.0, 7.0, and 10.0, respectively; as the diagrams show, (HOCl) = (Cl ) at peopH in each system. Next, assume that 10 4 M T OT Fe is present in the same, pH 7.0 solution. The log C-pe diagram for this T OT Fe and pH was developed in Example 12.9. Combining that diagram with Figure 12.8 yields Figure 12.9.
Figure 12.9 log C-pe diagram for a solution containing 5 ⇥ 10 T OT Fe, at pH 7.0.
4M
T OT Cl and 10
4M
This figure characterizes the distribution of Fe and Cl species in the given solution over a wide range of pe values. By using the diagram in conjunction with information about exactly how the solution was prepared (i.e., what chemicals were added to provide the given T OT Fe and T OT Cl), we can determine the pe that will actually develop and the corresponding solution composition. 706
12.12.2
Choosing Components and Preparing the Tableau
As is the case for acid/base systems, all of the information about the system can be concisely presented in a system tableau. To prepare the tableau, we first need to choose the system components. Because e– appears in all the redox reactions, it seems wise to choose e as a component. In that case, we can include either Fe3+ or Fe2+ as an additional component, but not both. This restriction arises because any of these three species could be made by a combination of the other two. For instance, Fe2+ can be made by the reaction Fe3+ + e )* Fe2+
(12.45)
A similar constraint applies to the representation of Cl in the tableau: if e is chosen as a component, we can choose either Cl– or one of the Cl(+I) compounds (HOCl or OCl– ) as an additional component, but we cannot include compounds from both oxidation states along with e as components. Excluding Fe3+ or Fe2+ from the component set poses no theoretical or fundamental problems, but it does complicate the description of the system. For example, if Fe3+ and e are chosen as components and Fe2+ is not, the reactions for forming all Fe(II) species have to be written in terms of Fe3+ and e . For example, assuming that H2 O and H+ are also included as components, the reaction for formation of Fe(OH)–3 would be Fe3+ + e + 3 H2 O
3H+ )* Fe(OH)3
(12.46)
The equilibrium constant for this reaction can be derived by combining Reaction (12.45) with the reactions for Fe(II) hydrolysis, but carrying out that type of calculation for all species containing Fe(II) would be tedious at best. Fortunately, the calculation of these values and their inclusion in the tableau are often unnecessary, because that information is already embedded in the log C-pe diagram for the system (via the slopes and intersection points of the various curves). Once such a diagram has been prepared, the system composition at equilibrium can be identified by locating the pe where the T OT e equation is satisfied. Although the stoichiometric coefficients that appear in the tableau are needed to write and solve this equation, the log K values are not. For example, assume that the system of interest was prepared as a mixture of 3 ⇥ 10 4 M HOCl plus 1 ⇥ 10 4 M FeCl2 . A tableau for this system, using Fe3+ and OCl– as components, is shown in Table 12.5. For completeness, the log K values for all the species are included, but, as noted in the preceding discussion, those values are not needed if the log C-pe diagram for the system is available. Although acids or bases would have to be added to adjust the initial pH to 7.0 and keep it at that value as the solution equilibrates, we will assume that those chemicals do not interact with other species in the system and do not undergo redox reactions, so they can be ignored in this analysis. Also, because complexes of both Fe(II) and Fe(III) with Cl– are weak, those complexes have been left out of the tableau. 707
Table 12.5 System tableau for a solution of 3 ⇥ 10
4M
Stoichiometric Coefficient H2 O H+ e– OCl– Fe3+ Species H2 O H+ e– OCl– Fe3+ OH– FeOH2+ Fe(OH)+2 Fe(OH)3 Fe(OH)–4 HOCl Cl– Fe2+ FeOH+ Fe(OH)2 Fe(OH)–3
1 0 0 0 0 1 1 2 3 4 0 1 0 1 2 3
0 1 0 0 0 1 1 2 3 4 1 2 0 1 2 3
0 0 1 0 0 0 0 0 0 0 0 2 1 1 1 1
0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0
0 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1
Inputs FeCl2 HOCl
2 0
4 1
5 0
2 1
1 0
HOCl plus 1 ⇥ 10
log K
4M
FeCl2 .
Conc’n
0.00 0.00 0.00 0.00 0.00 14.00 2.02 5.75 15.00 22.70 7.53 50.20 13.03 3.63 7.46 17.96 1 ⇥ 10 3 ⇥ 10
4 4
Like the species expected to be present at equilibrium, the inputs must be specified in terms of components. The input of FeCl2 is therefore represented in the bottom portion of the tableau as a mixture of components based on the reaction Fe3+ + 2 OCl + 4 H+ + 5 e
2 H2 O )* FeCl2
(12.47)
The T OT e equation is a formal, mathematical expression of the idea that the total concentration of electrons in the species present in the equilibrated solution must be the same as that in the input chemicals. The concentration of electrons associated with each species equals the concentration of that species times the number of electrons required to make it from the components (i.e., the coefficient shown in the e column for that species), so we can write for the example system: 708
T OT ein =T OT eeq ⇥ ⇢ ⇤ 5(1 ⇥ 10 4 ) + 0(3 ⇥ 10 4 ) = e⇢ + 2 (Cl ) + 1 (Fe2+ ) + 1 (FeOH+ ) ⇢
+ 1 (Fe(OH)2 ) + 1 Fe(OH)3
5 ⇥ 10
4
=2 (Cl ) + T OT Fe(II)
(12.48)
In Equation (12.48), [e ] has been assumed to be negligible because T OT e is a sum of concentrations (not activities), and the concentration of free electrons is negligible in all solutions. As a result, this assumption applies to all T OT e equations. ⇥ ⇤ Scanning the log C-pe diagram and noting that a curve representing log 2(Cl ) is 0.3 log unit above the curve for log (Cl ), we see that Equation (12.48) is satisfied at pe > 20, where (Cl ) (Fe2+ ). Under these conditions, the equation simplifies to (Cl ) = 2.5 ⇥ 10 4 . An expanded version of this region of the plot (Figure 12.10) indicates that the equilibrium pe is 21.54.
Figure 12.10 Expansion of the region of Figure 12.9 where the T OT e equation is satisfied for a mixture of 1 ⇥ 10 4 M FeCl2 and 3 ⇥ 10 4 M HOCl. The curves for Fe(III) species are not shown, because those species do not appear in the T OT e equation. The point where the equation is satisfied is shown by the circle.
As is true for log C-pH diagrams, a single log C-pe diagram can be used to analyze multiple systems that have the same total concentrations of the key reactive groups (in this case, the redox groups). For example, if a pH 7.0 solution were prepared with 10 4 M FeCl2 , 2 ⇥ 10 5 M HOCl, and 2.8 ⇥ 10 4 M NaCl, it would have the same T OT Fe and T OT Cl as the one considered above, so Figure 12.9 would still apply, but with a different T OT e equation. For such a solution, the bottom (inputs) portion of the tableau and the corresponding T OT e equation would be as follows. 709
Stoichiometric Coefficient H2 O H+ e OCl– Fe3+ Na+
log K
Conc’n
Inputs FeCl2
2
4
5
2
1
0
NaCl
0
2
2
1
0
1
HOCl
0
1
0
1
0
0
1.0 ⇥ 10
4
2.0 ⇥ 10
5
2.8 ⇥ 10
4
T OT ein =T OT eeq ⇥ ⇢ ⇤ 5(1 ⇥ 10 4 ) + 2(2.8 ⇥ 10 4 ) + 0(2 ⇥ 10 5 ) = e⇢ + 2(Cl ) + 1(Fe2+ ) + 1(FeOH+ ) ⇢
+ 1 (Fe(OH)2 ) + 1 Fe(OH)3
1.06 ⇥ 10
3
=2(Cl ) + T OT Fe(II)
(12.49)
This T OT e equation is satisfied in a region of the log C-pe diagram where the curve representing 2 (Cl– ) is almost horizontal, making it difficult to identify the equilibrium pe. As we did when we encountered this situation with log C-pH diagrams, we can solve the problem by subtracting a relatively large term from each side of the equation. In the current case, the easiest thing to do is to subtract 2 ⇥ T OT Cl from each side. Representing this quantity by a numerical value on the left and a sum of species concentrations on the right, the subtraction yields 1.06 ⇥ 10
3
2(5 ⇥ 10 4 ) = 2(Cl ) + T OT Fe(II) 6.0 ⇥ 10
6.0 ⇥ 10
5
5
= T OT Fe(II)
2(HOCl)
+ 2T OT OCl = T OT Fe(II)
2 (Cl ) + (HOCl) + (OCl ) 2(OCl ) (12.50)
When we scan the log C-pe diagram, it is clear that 2(T OT OCl) is negligible compared to 6.0 ⇥ 10 5 in the pe region where the equation is satisfied, so the equation simplifies to T OT Fe(II) = 6.0 ⇥ 10 5 = 10 4.22 ; the equilibrium pe can be read off the graph as 4.60.
12.13
MODELING REDOX REACTIONS WITH VISUAL MINTEQ
12.13.1
Using Pseudospecies to Allow Both Species in the Redox Couple to be Components
In principle, the approaches described in previous chapters for modeling equilibrium speciation with Visual Minteq can also be applied directly to redox reactions. That 710
is, if we input the stoichiometry and equilibrium constant for each redox reaction to be considered, we would expect Visual Minteq to solve for the equilibrium redox speciation, just like it does for reactions that characterize protonation/deprotonation, exchange of volatile species between solution and a gas phase, metal complexation, or solid precipitation. However, the fact that the oxidized and reduced species of a redox couple and e cannot all be designated as components leads to some complications that are most easily resolved by using a slightly different data input approach. This approach involves defining a new, imaginary pseudospecies for each redox half-reaction to be considered in the analysis. The pseudospecies is treated as an additional product of the reduction reaction and is assigned a fixed activity of 1.0. Within the program, this species is represented as Red/Ox (where Red and Ox, as usual, refer to the reduced and oxidized species of the redox couple); here, we represent it as Red/Ox. Thus, for example, in Visual Minteq, the Fe3+ /Fe2+ redox reaction is considered to be Fe3+ + e )* Fe2+ + Fe2+ /Fe3+
(12.51)
Including the pseudospecies as a product of the redox reaction makes it impossible to form Fe3+ , Fe2+ , or e strictly from a combination of the other two, and therefore makes it acceptable to treat all three as components. Furthermore, the stoichiometry for the reaction forming Fe2+ /Fe3+ is essentially identical to that for the real redox half-reaction that we want to specify and, taking into account the fact that the activity of the pseudospecies is always 1.0, its log K is ⇠⇠
log KFe2+ /Fe3+
⇠ 3+ ) ⇠/Fe (Fe2+ )⇠ (Fe ⇠2+ = = log K(12.45) (Fe3+ )(e )
(12.52)
Equation (12.52) indicates that, if the pseudospecies is assigned a log K value equal to that for the actual redox reaction of interest [Reaction (12.45)], then whenever the equilibrium constant for the pseudospecies is satisfied (as it will be when Visual Minteq solves the problem), the equilibrium constant for the actual redox reaction will automatically be satisfied as well. Thus, when we specify the properties of the pseudospecies, we are effectively inputting the stoichiometry and equilibrium constants of the redox half-reactions we want the program to consider. This approach also makes it easy to instruct the program not to consider certain half-reactions (if, for example, they are thought to be kinetically inhibited), simply by not including the corresponding pseudospecies in the input list. To illustrate the mechanics of modeling redox reactions in Visual Minteq, we next develop the log C-pe diagram and solve for the equilibrium composition of the same example solutions as we investigated using a manual/graphical approach in the preceding sections, i.e., ideal solutions at pH 7.0 containing 10 4 M T OT Fe and 5 ⇥ 10 4 M T OT Cl, in which the T OT Fe and T OT Cl were added in two different ways. 711
12.13.2
Preparing log C-pe Diagrams with Visual Minteq
To prepare the log C-pe diagram, we begin by choosing Fe2+ , Fe3+ , Cl– , and OCl– as components, and we specify that the pH is fixed at 7.0 and the ionic strength at 0.0. Because the program will determine the equilibrium distribution of T OT Fe and T OT Cl at each pe, the way in which we specify their inputs for this part of the exercise is irrelevant. That is, we could indicate that T OT Fe was input entirely as Fe2+ , entirely as Fe3+ , or as any combination of those components. Next, from the “Redox” pulldown menu, we select “Specify redox couples” and choose Fe2+ /Fe3+ from the list. Doing so automatically causes the log K and DHr (i.e., DH r ) fields to be filled in, and the stoichiometry of the reaction to appear in the dialog box; all this information is stored in the default database. Clicking the “Add” button tells the program that this redox couple should be included in the calculations. The program then automatically adds e (shown in the program as E ) to the list of components and to the system tableau. If a redox reaction of interest is not included the pulldown list, the reaction can be added to the database. The redox reaction database is accessed via the “Database management/ Redox couples” links, after which the name of the new pseudospecies, its stoichiometry, and its log K value (equal to ne pe if the reaction is input as a reduction) can be input following steps that are virtually identical to those for adding a new solid species (see Section 11.7).8,9 When the first redox reaction is added, the program alerts the user that the default option is for the equilibrium pe to be fixed at 8.0. We will change that option shortly. First, though, we input the Cl /OCl redox reaction by following the same steps as we used to input the Fe2+ /Fe3+ reaction. Once this reaction is entered, the redox reactions that have been input can be reviewed on the “List of fixed species” page, which is accessed either from the “Redox” or the page showing the current “List of components”. 8 Note that it is necessary and allowable to enter only one reaction between any two oxidation states. That is, Visual Minteq has all the information for formation of Fe(II) OH and Fe(III) OH complexes in its database and will include that information in the analysis, so it would be inappropriate to input, for example, the FeOH+ /Fe(OH)4 reaction in addition to the Fe2+ /Fe3+ reaction. 9 The coding in Visual Minteq precludes listing a compound as both a solid and part of a redox couple. To circumvent this complication, only reactions that relate pairs of dissolved components are listed in the “Redox” menu, and any solids that contain redox-active elements are input via the “Solids and excluded species” menu. In some cases, an element is stable in one oxidation state as a solid, but is so insoluble that no dissolved species in that oxidation state are known. As a result, it is not possible to write a redox reaction that comprises only soluble species and includes that oxidation state. In such cases, the solid is input via the “Solid phases” menu, but a redox reaction is used to specify its stoichiometry. When this is done, electrons and the redox conjugate partner of the solid are automatically added to the component list, and the redox reaction involving the solid is included in the analysis. For example, to consider redox reactions of pure metallic Cu, we select “Cu metal” from the Visual Minteq “Solid phases” list. The stoichiometry for forming Cu metal appears on the page, indicating that it is made by combining one Cu+ ion with one e– . Visual Minteq adds these components to the problem, and the Cu+ /Cu metal redox reaction is then included in the analysis.
712
We then use the “Multi-problem/Sweep” menu to instruct the program to carry out a series of simulations over a range of pe values, exactly as we would if we were exploring a range of pH values. Images of a few of the input screens for the current example system are shown in Figure 12.11. When the program is run and the output is plotted, the log C-pe diagram that is generated is identical to that shown earlier as Figure 12.9.10
12.13.3
Solving for the Equilibrium pe and Solution Composition Using Visual Minteq
To utilize Visual Minteq to determine the equilibrium solution composition in systems where redox reactions can occur, we use most of the features described previously for preparing log C-pe diagrams. However, extra care is often needed to assure that the inputs are consistent with the makeup of the solution we wish to model. For instance, the first solution that we considered in the manual analysis contained 10 4 M FeCl2 plus 3 ⇥ 10 4 M HOCl and had a fixed pH of 7.0. Because Fe2+ , Cl– , and OCl– are all components, we can specify their input concentrations exactly as they were added. We could also indicate that the input T OT H is 5 ⇥ 10 4 M, but that value will be overridden to maintain the pH at 7.0 anyway, so there is no point in specifying it. Next, we invoke the option to have the program determine the equilibrium pe based on the chemical inputs. This choice is entered via the “Parameters/Specify pe and Eh” link on the main menu, which opens a page showing the available options for computation of pe. (The options that refer to Eh are explained later in this chapter.) When the button for computing and making a guess for pe is clicked, a field appears where the user can enter an initial guess for the equilibrium pe, as shown in Figure 12.12.11 Because chemical speciation is extremely sensitive to pe in many redox systems, Visual Minteq (and most other chemical equilibrium software) encounters numerical problems much more frequently when modeling redox reactions than when modeling other types of reactions. Making a good guess for the equilibrium pe reduces the likelihood of such problems. To guess the pe in the current solution of interest, we can note that OCl species are strong oxidants, and that the initial T OT Cl(+I) is 10 As shown in Figure 12.11a, the Visual Minteq database includes equilibrium constants for the formation of FeCl+ and FeCl2 complexes. However, these species account for a negligible portion of T OT Fe in the equilibrated solutions, and they were not considered in the manual analysis, so they have been left out of the log C-pe diagram. 11 The default option is for the program to assign a fixed pe, which is the appropriate option if one is carrying out simulations to prepare a log C-pe diagram. For this reason, the “Specify pe and Eh” link was not mentioned when the steps for preparing such diagrams were described. However, if a different option for computing pe has been selected and one then wants to prepare a log C-pe diagram, it is necessary to return to the “Specify pe and Eh” page and re-select “Fixed pe” (even if the instruction to investigate a range of fixed pe’s has been selected on the “Multi-problem/Sweep” page).
713
Figure 12.11 Screenshots summarizing some inputs to the example problem. (a) The input tableau; (b) Summary of the input species (including pseudospecies) with fixed activities; (c) The Multi-problem/Sweep page.
714
Figure 12.12 Screenshot showing the selection of the approach for determining the final pe.
three times that of T OT Fe(II). Each mole of T OT Cl(+I) can oxidize two moles of T OT Fe(II), so the available Cl(+I) is more than sufficient to oxidize all of the Fe(II). We might therefore guess that, once the solution reaches equilibrium, some Cl(+I) and very little Fe(II) will remain. Figure 12.8 indicates that solutions containing any substantial concentration of Cl(+I) have pe values of about 20 or higher, so a guess of pe = 20 might be appropriate; this value has been entered in the screen image shown in Figure 12.12. The Visual Minteq output using the specified inputs is summarized in Figure 12.13. As we would expect, it identifies the same equilibrium conditions (e.g., pe = 21.54) as we found when we analyzed the problem manually. The Visual Minteq output for the second scenario analyzed previously (10 4 M FeCl2 , 2⇥10 5 M HOCl, and 2.8⇥10 4 M NaCl) also matches the result we obtained manually, with a reported equilibrium pe of 4.605; the simulation of this system is left as an exercise. While the similarities between determining the equilibrium pH in acid/base problems and the equilibrium pe in redox problems are evident, it is important to recognize one significant difference between the two types of analyses. Whereas any single acid/base reaction can proceed in isolation in an aqueous solution, redox reactions are commonly written as half-reactions that cannot occur in isolation. For example, we can imagine a single acid (e.g., HAc) being added to solution, undergoing partial dissociation, and reaching acid/base equilibrium. For such a scenario, if we knew the total amount added and the pKa , we could determine the pH. In doing so, we implicitly assume that the H+ released from the acid binds to water molecules. By contrast, if a single reduced species is added to solution (e.g., Fe2+ , added as part of FeCl2 ), it cannot dissociate (into Fe3+ and e ) and reach redox equilibrium unless another species is present that can accept the released electrons. For this reason, unlike simulations of acid/base reactions, all redox simulations in which the goal is 715
Figure 12.13 Visual Minteq screen showing a summary of the output from the example simulation, for inputs of 10 4 M FeCl2 and 3 ⇥ 10 4 M HOCl at a fixed pH of 7.0.
to determine the equilibrium pe (whether using Visual Minteq or solving the problem manually) must include at least two redox half-reactions. This restriction does not apply when preparing log C-pe diagrams, because in that case we specify the pe independently, implicitly assuming that some other redox reactions are occurring to control the pe at the given value, even though we do not specify exactly what those reactions are. Example 12.12 When brass [an alloy of Cu(0) and Zn(0)] is exposed to certain water qualities, the copper in the brass can corrode to Cu(I) and form the mineral nantokite, CuCl(s). However, under some circumstances, Cu(I) can undergo a so-called disproportionation reaction, in which it is simultaneously oxidized and reduced via a reaction that can be represented generically as follows: 2 Cu(I) )* Cu(II) + Cu(0)
716
(12.53)
Imagine that a drinking water comes into contact and equilibrates with a brass fixture that has partially corroded, so that it reaches equilibrium with both CuCl(s) and Cu(s). The water entering the fixture is at pH 8.1 and contains essentially no copper, 55 mg/L Cl– , 93 mg/L Na+ , and enough T OT CO3 to provide 125 mg/L ALK as CaCO3 . These species account for essentially all the ionic strength of the solution. Use Visual Minteq to determine the equilibrium composition of the solution, assuming the pH remains at 8.1. Consider the possible formation of complexes between Cu+ or Cu2+ and OH– or Cl– , but assume that no solids other than copper metal and nantokite can form. Compare the predicted concentration of total dissolved copper with the EPA Action Level of 1.3 mg/L (2.05 ⇥ 10 5 M). (If the Action Level is exceeded in a sufficiently large fraction of the households receiving the water, the utility would be required to take action to reduce the corrosivity of the water.) Solution Because the pH of the initial solution is 8.1, we can assume that essentially all the T OT CO3 is present as HCO–3 , and that this HCO–3 accounts for all the alkalinity. T OT CO3 can therefore be estimated as12 ⇤ 125 mg/L as CaCO3 meq ⇥ = 2.50 ⇡ HCO3 50 mg CaCO3 /meq L ⇥ ⇤ T OT CO3 ⇡ HCO3 = 2.5 ⇥ 10 3 M ALK =
T OT Cl, T OT Na, T OT CO3 , and the fixed pH can then be input into Visual Minteq in the usual way. These inputs will assure that the acid/base and metal complexation reactions will be taken into account. In addition, we need to input information that will cause the program to account for the following redox and precipitation/dissolution reactions: Cu+ + e )* Cu(s) Cu2+ + e )* Cu+ +
Cu + Cl )* CuCl(s)
(12.54) (12.55) (12.56)
To account for the redox reactions and conditions, we “Add” the Cu+ /Cu2+ couple from the “Redox” menu and use the “Parameters/Specify pe and Eh” menu to indicate that the program should calculate the equilibrium pe. Lacking insight into what we expect the pe to be, we can leave the guess at the default value of 0. Finally, because the problem statement specifies that the solution reaches equilibrium with both nantokite and metallic copper, we specify that both of these solids are present as “infinite solid phases” via the “Solid phases and excluded species” menu. 12 Alternatively,
we could enter the alkalinity directly via the “Parameters/Specify alkalinity” menu and allow the program to compute T OT CO3
717
When the program is run, the output screen shown in Figure 12.14a appears, indicating that the equilibrium pe is 3.385; the equilibrium values of T OT Cl, T OT Cu(I), and T OT Cu(II) can be found on the “Equilibrated mass distribution” page (Figure 12.14b). Note that less than 0.2% of the dissolved T OT Cu(I) and T OT Cu(II) are present as the free ions. More than 99% of the Cu(I) is present as Cu(I) Cl complexes (of which CuCl–2 is dominant), while the Cu(II) is present primarily as hydroxo complexes (with Cu3 (OH)2+ 4 dominant).
Figure 12.14 Two screens summarizing the Visual Minteq output for the example problem.
718
To understand the steps that lead to the ultimate equilibrium condition, consider first the following mass balance, showing the change in the dissolved chloride concentration (T OT Cldiss ) as the solution equilibrates with the solids: DT OT Cldiss = T OT Cldiss,fin
T OT Cldiss,init
2
= 5.89 ⇥ 10 M
1.55 ⇥ 10 3 M = 5.74 ⇥ 10
2
M
The only possible source of the Cl entering solution is dissolution of nantokite, so we infer that 5.74 ⇥ 10 2 M nantokite dissolves as the water equilibrates. This dissolution reaction must have also supplied 5.74 ⇥ 10 2 M T OT Cu(I) to solution. However, the initial solution contained no Cu(I), and the equilibrated solution contains only 2.89⇥10 3 M, so some of the Cu(I) must have been removed from solution after it entered. The concentration of Cu(I) removed in this process ( DCu(I)) is D(Cu(I)) = 5.74 ⇥ 10
2
2.89 ⇥ 10
3
= 5.45 ⇥ 10
2
The only reaction under consideration that can account for this loss of Cu(I) is disproportionation, which generates one mole each of Cu(s) and Cu(II) for every two moles of Cu(I) reacting. Consistent with this analysis, the program reports that T OT Cu(II) in the equilibrated solution is one-half of D(Cu(I)), or 2.725 ⇥ 10 2 M. As has been noted a couple times, the range of possible equilibrium activities for dissolved species is enormous when redox reactions are important. This scenario is especially common if several electrons are transferred between the reduced and oxidized species because, for such couples, the concentrations of the nondominant species can decline by many orders of magnitude for each unit change in pe. Visual Minteq calculations can sometimes fail to converge when such systems are simulated, especially if the initial guess of the equilibrium pe is poor. If this occurs, a trial and error approach using different guesses for pe can often overcome the problem. The efficiency of this approach might be enhanced by using Visual Minteq to prepare a log C-pe diagram for the system to help identify the approximate pe at equilibrium. (Visual Minteq is almost always able to correctly compute speciation at fixed pe values, and hence has no problem generating the data needed to prepare such a diagram.) If the program fails to converge even when the initial guess for pe is thought to be close to the equilibrium value, manual solution of the T OT e in conjunction with the log C-pe diagram will always yield a correct result.
12.14
OXIDATION AND REDUCTION OF WATER
12.14.1
The LOG C-pe Diagram for the H2 /H2 O/O2 Redox System
Redox-active species such as HOCl and HS– may or may not be present in any given water sample. However, one redox-active species that is always present is water itself, 719
which can undergo either oxidation or reduction according to the following reactions:
Oxidation : Reduction :
2 H2 O )* O2 (aq) + 4 H+ + 4e 2 H2 O + 2e )* H2 (aq) + 2 OH
(12.57) (12.58)
Reactions (12.57) and (12.58) are often combined with Henry’s law reactions for O2 and H2 , respectively, so that the reactions characterize equilibrium between water and gaseous, rather than dissolved, H2 or O2 . The various forms of these reactions and the corresponding equilibrium constants and pe values at 25 C are shown in Table 12.6. Table 12.6 Representations of the Redox Reactions of Water
Two Representations of the Water Oxidation Reaction 2 H2 O )* O2 (aq) + 4 H+ + 4e O2 (aq) )* O2 (g) 2 H2 O )* O2 (g) + 4 H+ + 4e
K(12.57) = 10
86.0
pe = 21.50
(12.57)
83.12
pe = 20.78
(12.59)
pe = 15.55
(12.58)
pe = 1.55
(12.60)
HO2 = 102.88 K(12.59) = 10
Three Representations of the Water Reduction Reaction 2 H2 O + 2e
)* H2 (aq) + 2 OH–
K(12.58) = 10
31.1
2 H+ + 2 OH– )* 2 H2 O
K = Kw 2 = 1028.0
2 H+ + 2e
K(12.60) = 10
)* H2 (aq)
2 H+ + 2e
)* H2 (aq)
H2 (aq) )* H2 (g) 2 H+ + 2e
)* H2 (g)
K(12.60) = 10
3.1
3.1
HH2 = 103.1 K(12.61) = 100.0 = 1.0
pe = 0.0
(12.61)
Manipulation of the equilibrium constant expressions for the preceding reactions yields the following equations for the activities of O2 and H2 in redox equilibrium with water: log{O2 (aq)} = 86.00 + 4pH + 4pe
(12.62)
log{O2 (g)} = 83.12 + 4pH + 4pe
(12.63)
720
log{H2 (aq)} = 3.10 log{H2 (g)} = 2pH
2pH 2pe
2pe
(12.64) (12.65)
These equations indicate that any aqueous solution that is at equilibrium with respect to the redox reactions of water contains some dissolved oxygen and hydrogen, and that the activities of those solutes are controlled by solution pH and pe. Furthermore, via the Henry’s law relationships, each equilibrium activity of dissolved H2 or O2 can be in equilibrium with only one partial pressure of H2 (g) or O2 (g), respectively. Thus, if the O2 /H2 O/H2 redox group has reached equilibrium, specifying the solution pe and pH fully estabishes the activities of H2 (aq), O2 (aq), H2 (g), and O2 (g). A log ai -pe diagram showing the activities of O2 (g) and H2 (g) at pH’s of 4, 7, and 10 is provided in Figure 12.15, and the activities of both dissolved and gaseous H2 and O2 in equilibrium with water at pH 7.0 are shown in Figure 12.16. The lines representing the activities of H2 and O2 in Figures 12.15 and 12.16 have slopes of 2 and +4, respectively, and they have no curvature. This latter feature is reminiscent of the H+ and OH– lines on log C-pH diagrams, and, in both cases, it is a consequence of the assumption that the activity of H2 O remains at 1.0 regardless of how much dissociation occurs.
Figure 12.15 Log a-pe diagram for the H2 (g)/H2 O/O2 (g) system at various pH values. Note that the y axis is the logarithm of the activity of the species shown, not the logarithm of its concentration. Thus, for the gases, the value on the ordinate is the logarithm of the partial pressure.
In most solutions of interest, upper limits are placed on the equilibrium values of {H2 (aq)} and {O2 (aq)} by Henry’s law in combination with the composition and total pressure of a gas phase. For example, the partial pressure of gaseous hydrogen in the normal atmosphere is 5 ⇥ 10 7 atm; combining this value with the Henry’s 721
Figure 12.16 Activities of dissolved and gaseous H2 and O2 in equilibrium with water for a system at pH 7.0 and 25 C.
constant for H2 at 25 C, we find that the activity of dissolved H2 (aq) in equilibrium with the atmosphere is 10 9.40 . Then, using either Equation 12.64 or 12.65, we find that, for a solution at pH 7.0, atmospheric H2 (g) is in redox equilibrium with H2 O only at pe = 3.8. If a solution at pH 7 is in equilibrium with H2 (g) in the atmosphere, but the pe of the solution is greater than 3.8, a driving force will exist favoring oxidation of the H2 (aq). Similarly, if pe is less than 3.8, the driving force will favor reduction of H+ to form more H2 (aq). Corresponding statements apply to oxygen, except in that case equilibrium with both water and the atmosphere at pH 7 requires that the pe be 13.6 and the O2 (aq) activity be 10 3.60 (i.e., 8.0 mg/L). The preceding discussion suggests that most aqueous solutions that are in contact with the atmosphere are likely to be out of equilibrium with respect to the H2 (aq)/H2 O redox reaction, the H2 (aq)/H2 (g) gas transfer reaction, or both; the same statement applies to the O2 (aq)/H2 O and O2 (aq)/O2 (g) reactions. Nevertheless, because equilibration of the H2 (aq)/H2 O/O2 (aq) redox system is very slow in the absence of a catalyst, neither hydrogen nor oxygen is consumed or generated at a detectable rate by these redox reactions under normal circumstances, regardless of the extent of disequilibrium. If a catalyst is present that does allow the redox reactions to equilibrate, they can change the dissolved H2 (aq) and/or O2 (aq) concentrations and thereby drive the gas/liquid reaction out of equilibrium, inducing transfer of H2 (g) or O2 (g) across the gas/liquid interface. Such gas transfer typically occurs by passive diffusion and so is not visible to the naked eye. However, if the solution pe is driven to extreme values where the equilibrium partial pressure of hydrogen or oxygen is greater than the 722
ambient total pressure, a pure H2 (g) or O2 (g) phase can form in the solution, causing visible bubbles of these gases to appear. Demonstrations of this phenomenon, in which H2 (g) is generated at the surface of a piece of metal immersed in an acidic solution, are common in introductory chemistry classes. In these demonstrations, the metal serves as both a source of electrons and a catalyst. Example 12.13 In what pe range will gas bubbles of pure hydrogen be generated in a solution at pH 2.0 if Ptot = 1.0 atm? Solution For hydrogen bubbles to form, the partial pressure of H2 (g) must be > Ptot , or, in this case, 1.0 atm. This pressure corresponds to an activity of 1.0 for H2 (g). According to Equation (12.65), the pe value in a solution at pH 2.0 and in which the activity of H2 (g) is 1.0 is 1 1 log{H2 (g)} = 2.0 log 1.0 = 2.0 2 2 Thus, H2 (g) bubbles should appear if the pe of the solution is less than or equal to 2.0. pe = pH
12.14.2
The Impossibility of Redox Equilibrium Between H2 O and the Atmosphere
As has been noted, the partial pressure of H2 in the normal atmosphere is 5 ⇥ 10 7 atm. Therefore, according to Equation (12.65), the pe and pH in a solution in equilibrium with the atmosphere must be related by 1 log (5 ⇥ 10 7 ) 2 pe = 3.15 pH pe =
pH
(12.66a) (12.66b)
Similarly, according to Equation (12.63), for a solution to be in equilibrium with atmospheric O2 (i.e., with an oxygen partial pressure of 0.21 atm), the pe and pH must be related by 1 pe = (83.12 + log 0.21 4 pe = 20.60 pH
4pH)
(12.67a) (12.67b)
Clearly, Equations (12.66b) and (12.67b) cannot be satisfied simultaneously. We therefore conclude that, at their normal atmospheric concentrations, oxygen and hy723
drogen cannot simultaneously be in redox equilibrium with H2 O in any aqueous solution. To determine the extent of disequilibrium of the H2 /H2 O/O2 system under typical environmental conditions, we can generate the following overall reaction for redox dissociation of water into H2 (g) and O2 (g) (analogous to acid dissociation of H2 O into H+ and OH– ): 2 H2 O )* O2 (g) + 4 H+ + 4e 4 H+ + 4e
K(12.59) = 10
83.12
2 K(12.68) = 100.0
)* 2 H2 (g)
2 H2 O )* O2 (g) + 2 H2 (g)
K = 10
83.1
(12.59) (12.68) (12.69)
Reaction (12.69) indicates that for liquid water to be in equilibrium with gaseous oxygen and hydrogen, the product {O2 (g)}{H2 (g)}2 must equal 10 83.12 . In the atmosphere, the actual value of that product is (0.21)(5 ⇥ 10 7 )2 , or 5 ⇥ 10 14 . Remarkably, this value is more than 69 orders of magnitude larger than the equilibrium value. The same result would be obtained if we carried out the calculations using dissolved H2 (aq) and O2 (aq) concentrations, if these dissolved species were in equilibrium with the atmosphere. Thus, under ambient environmental conditions, there is an enormous driving force favoring the reaction of H2 and O2 to form H2 O, regardless of whether the H2 and O2 are gases in the atmosphere or dissolved in an aqueous solution. These calculations suggest that, if a solution containing H2 (aq) and O2 (aq) were isolated from the atmosphere and the redox reaction proceeded to equilibrium, essentially all the H2 would be consumed (the concentration of O2 is much larger than that of H2 , so the solution would run out of H2 before a significant fraction of the O2 was consumed). However, despite the large driving force for such a reaction, the two gases coexist in the nonequilibrated condition because, normally, the rate of the redox reaction is extremely slow, and inputs of H2 (g) and O2 (g) from the atmosphere replenish the dissolved species as fast as they are consumed by the reaction. As a result, the reaction remains in a continuous state of extreme disequilibrium.
12.14.3
Considering Redox Activity of Water When Analyzing pe of a Solution
To illustrate the potential effect of the H2 /H2 O/O2 redox system on other redox couples in solution, we next reanalyze the example solution considered previously containing 10 4 M FeCl2 plus 3 ⇥ 10 4 M HOCl at pH 7.0, but we now assume that the H2 /H2 O/O2 redox group reaches equilibrium. For the purposes of this example, we assume that the solution is initially in equilibrium with atmospheric H2 (g) and O2 (g) with respect to Henry’s law, but that no gas exchange occurs thereafter. The initial concentrations of H2 (aq) and O2 (aq) are therefore 724
[H2 (aq)] =
PH2 5.0 ⇥ 10 7 atm = atm = 4.0 ⇥ 10 HH2 103.10 mol/L
[O2 (aq)] =
PO2 0.21 atm = atm = 2.77 ⇥ 10 HO2 102.88 mol/L
10 mol
L
4 mol
L
The log C-pe diagram for the system is identical to Figure 12.9, except lines are added to represent the activities of H2 O, H2 (aq), and O2 (aq) at pH 7. These lines can be drawn based on reactions (12.57) and (12.58) and are the same as those in Figure 12.16. The modified diagram is shown in Figure 12.17.
Figure 12.17 Figure 12.9 modified to include equilibrium among H2 O, O2 (aq), and H2 (aq).
The T OT e equation for the system ignoring H2 /H2 O/O2 redox reactions was developed earlier [Equation (12.48)] based on the tableau shown in Table 12.5. Modifying that tableau to include H2 (aq) and O2 (aq) as species that are present initially at concentrations controlled by the atmospheric gases and that subsequently reach redox equilibrium with the solution yields Table 12.7. Applying the assumption of ideal solute behavior, the expressions for T OT e in the initial and equilibrated solution are T OT ein = 2(4⇥10 10 ) 4(2.77 ⇥ 10 4 ) + 5(10 4 ) + 0(3 ⇥ 10 4 ) = 6.08 ⇥ 10 4 ⇥ ⇢ ⇤ T OT eeq = e⇢ + 2 (H2 (aq))eq 4 (O2 (aq))eq + 2(Cl )eq + 1(Fe2+ ) + 1(FeOH+ ) ⇢
+ 1 (Fe(OH)2 ) + 1 Fe(OH)3 725
Table 12.7 A system tableau for a closed solution containing 10 4 M FeCl2 and 3 ⇥ 10 HOCl at pH 7.0, initially in equilibrium with atmospheric H2 (g) and O2 (g).
Stoichiometric Coefficient H2 O H+ e– OCl– Fe3+ H2 O H+ e– OCl– Fe3+ OH– H2 (aq) O2 (aq) FeOH2+ Fe(OH)+2 Fe(OH)3 Fe(OH)–4 HOCl Cl– Fe2+ FeOH+ Fe(OH)2 Fe(OH)–3
1 0 0 0 0 1 0 2 1 2 3 4 0 1 0 1 2 3
0 1 0 0 0 1 2 4 1 2 3 4 1 2 0 1 2 3
0 0 1 0 0 0 2 4 0 0 0 0 0 2 1 1 1 1
0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0
0 0 0 0 1 0 0 0 1 1 1 1 0 0 1 1 1 1
H2 (aq) O2 (aq) FeCl2 HOCl
0 2 2 0
2 4 4 1
2 4 5 0
0 0 2 1
0 0 1 0
log K
5M
Conc’n
0.00 0.00 0.00 0.00 0.00 14.00 3.10 86.00 2.02 5.75 15.00 22.70 7.53 50.20 13.03 3.63 7.46 17.96 4.00 ⇥ 10 2.77 ⇥ 10 1.00 ⇥ 10 3.00 ⇥ 10
10 4 4 4
Equating these two expressions and rearranging so that both sides of the equation are strictly sums, we find 6.08 ⇥ 10
4
+ 2 (H2 (aq))eq + 2(Cl )eq + T OT Fe(II) = 4 (O2 (aq))eq
The portion of Figure 12.17 containing the point where the T OT e equation is satisfied is shown with expanded axes in Figure 12.18. The equilibrium pe is 12.93. At this pe, virtually all the Cl is in the reduced form (the computed value of T OT Cl(+I) is < 10 20 ), and virtually all the Fe is oxidized (T OT Fe(II) < 10 14 ). While the iron 726
speciation is roughly the same as in the prior analysis (in which the H2 /H2 O/O2 equilibria were not considered), the speciation of Cl is different: essentially all the OCl is reduced in the system where the H2 /H2 O/O2 equilibria are considered, because it oxidizes not only the Fe(II) but also H2 O, to form O2 (aq). As a result, oxygen is generated in the solution, so that at equilibrium, the concentration of O2 (aq) is greater than that in the initial solution (3.98 ⇥ 10 4 M vs. 2.77 ⇥ 10 4 M). If this solution were allowed to equilibrate with the atmosphere, an amount of O2 (aq) equal to the amount that was generated by the redox reaction would transfer into the gas phase, thereby returning the O2 (aq) concentration to its original value. The equilibrium speciation in the systems with and without consideration of the H2 /H2 O/O2 redox reactions is summarized in the Table 12.8.
Figure 12.18 Expanded view of a portion of Figure 12.13, showing the RHS and LHS of the T OT e equation. The concentrations of all Fe(II) species are ⌧ 10 5 M.
The log C-pe diagram for the system indicates that Fe(II) is easier to oxidize than H2 O at pH 7.0 (dominance shifts from Fe(II) to Fe(III) at pe ⇠ 5.7, but the activity of O2 (aq) does not become larger than that of H2 O until the pe is >13). Put another way, in this system, Fe(II) gives up its electrons more readily than does H2 O. As a first approximation, we can model the reactions as though all the Fe(II) is oxidized to Fe(III) before any H2 O is oxidized to O2 , much like the approximation that stronger bases become fully protonated before weaker ones are protonated at all. Thus, the electrons released by Cl(+I) species can be thought of as first reacting with Fe(II), consuming 1 ⇥ 10 4 M electrons and reducing 0.5 ⇥ 10 5 M OCl. After that, the remaining 2.5 ⇥ 10 4 M Cl(+I) is reduced, consuming 5 ⇥ 10 4 M electrons from the oxidation of H2 O and generating 1.25 ⇥ 10 4 M O2 (aq). 727
Table 12.8 Equilibrium speciation in a system prepared by adding 10 5 ⇥ 10 4 M Cl(+l) to water at a fixed pH of 7.0.
pe HOCl OCl– Cl– Fe2+ T OT Fe(II) Fe3+ T OT Fe(III) O2 (aq) H2 (aq)
12.15
H2 /H2 O/O2 Does Not Equilibrate
H2 /H2 O/O2 Equilibrates
21.54 1.93 ⇥ 10 4 5.70 ⇥ 10 5 2.50 ⇥ 10 4 1.72 ⇥ 10 21 1.72 ⇥ 10 21 5.58 ⇥ 10 13 1.00 ⇥ 10 4 2.77 ⇥ 10 4 4.0 ⇥ 10 10
12.93 2.29 ⇥ 10 6.74 ⇥ 10 5.00 ⇥ 10 7.06 ⇥ 10 7.09 ⇥ 10 5.58 ⇥ 10 1.00 ⇥ 10 3.98 ⇥ 10 1.10 ⇥ 10
4M
S( II) and
21 22 4 13 13 13 4 4 43
ENERGY CHANGES ACCOMPANYING REDOX REACTIONS
The fact that redox reactions might proceed at imperceptible rates even under conditions that are far from equilibrium has been noted a few times in this chapter. Nevertheless, it is often possible to induce the reactions to proceed rapidly by changing the ambient conditions. Thus, for instance, coal that is stable indefinitely at room temperature burns readily when ignited. Also, by changing the redox environment, we can often induce a change in the opposite direction from the one that would otherwise occur. For instance, the natural tendency of most zero-valent metals is to oxidize (corrode) in the normal atmosphere. However by appropriate manipulation of the pe of the system (either chemically or by perturbing the system electrically), we can reduce oxidized metals to their zero-valent state, thereby preparing materials of much greater value. In this section, we explore more thoroughly the thermodynamics of such reactions.
12.15.1
The Molar Gibbs Energy of Electrons
As shown in Chapter 4, the standard Gibbs energy change of any reaction is related to the equilibrium constant for that reaction by DGr = RT ln K = 2.303 RT log K 728
(12.70)
By applying Equation (12.70) to the H2 (g)/H+ half-reaction [Reaction (12.6)], we can compute the standard Gibbs energy of electrons, as follows: H+ + e )*
1 H (g) 2 2
DGr =
1 G 2 H2 (g)
GH+
1 (0) 2
Ge
RT ln K(12.6) =
K(12.6) ⌘ 1.0
0
0 =Ge
Ge
(12.6) (12.71a) (12.71b) (12.71c)
The molar Gibbs energy of e– in any solution can then be determined from Equation (4.66) as Ge = ⇢ Ge⇢ + RT ln{e } = 2.303 RT pe (12.72) Substituting for RT , we obtain the following simple expression for the molar Gibbs energy of electrons at 25 C.13 ✓ ◆ kJ Ge = 5.71 pe (12.73) mol
12.15.2
An Alternative Way of Expressing Electron Energy and Activity: The EH Scale
Like any value of energy, the molar Gibbs energy of electrons (Ge ) can be expressed in a variety of ways. Because of the link between electrons and electrical energy, one common way to express Ge is in terms of electrical potentials. The electrical potential (y, typically expressed in volts) at any location was defined in Chapter 4 as the electrical potential energy of an object (PEelec ) at that location per unit charge on the object (s , typically in coulombs). When dealing with redox systems, we are more often interested in the electrical potential energy per mole of a species, with conventional units of kJ/mol. Representing this term as PEi,elec , we can write PEi,elec = zi Fy 13 According
(12.74)
to the approach used here, the activity of free electrons in solution is defined based on Reaction (12.6) and Equation (12.7), and Ge = 0 is derived as a result of the relationship between the standard free energy of reaction and the equilibrium constant. These results are often presented in the reverse order. That is, the standard free energy of formation of electrons in solution is defined to be zero, and the equilibrium constant of 1.0 for Reaction (12.6) is derived. The approaches are effectively identical, since the assignment of a value to one of the terms (Ge or K(12.6) ) leads directly to the correct value for the other.
729
where PEi,elec is the molar electrical energy of i in kJ/mol; zi is the molar charge of i, with units of equivalents of charge per mole; y is the electrical potential, in volts; and F is the Faraday constant, equal to 96.48 (kJ/equiv)/V. Although Equation (12.74) is applicable most directly to systems where the electrical potential is nonzero, it also can be used to define a “virtual” electrical potential in cases where no such potential exists. That is, one way of thinking about Equation (12.74) is that it characterizes what the electrical potential y would have to be at a location for a species with a charge zi at that location to have electrical energy per mole equal to PEi,elec .14 This idea of a virtual potential is commonly used to describe the molar energy of electrons, Ge , in a real solution in the absence of any real electrical field. If we let i = e in Equation (12.74), then PEi,elec is just another name for Ge . The value of y on the right side of the equation then indicates the electrical potential to which the solution would have to be subjected to cause the electrons to have an amount of electrical energy per mole equal to the Gibbs energy per mole that they really do have. When the equation is used in this way, y is conventionally replaced by the designation EH ,15 thereby distinguishing between the virtual potential that we are computing and any real electrical potential that the solution might be exposed to. When this substitution is made along with the assignment of z = 1 for electrons, the equation becomes Ge = ze FEH = FEH (12.75) Combining Equations (12.72) and (12.75), we find EH =
Ge 2.303 RT =+ pe F F
(12.76)
and, for a system at 25 C EH = (0.0592 V) pe
(12.77)
Equation (12.76) demonstrates that EH is a surrogate indicator for the molar energy of electrons, expressed in conventional electrical units rather than those typically used for aqueous solutes; once the temperature is specified [as in Equation (12.77)], it is also a surrogate for pe. EH is referred to as the oxidation potential or redox potential of the solution. When pe for a half-reaction is inserted on the right side of Equation (12.76) or (12.77), the corresponding value of EH is designated EH and is 14 The electrical potential is zero in any bulk solution unless an electrical field is imposed from outside the solution phase. Later in this chapter and in Chapter 13, we will consider a few systems where such an electrical potential is imposed. However, for the vast majority of systems of interest, the actual electrical potential in the solution (y) is 0. 15 The designation E was chosen to indicate that that the value is an electrical (E) potential and that H the baseline for measuring this potential is the assignment of EH = 0 in a system containing H+ and H2 (g) in their standard states (thus causing the EH scale to conform to that for Ge ).
730
called the standard oxidation potential, standard redox potential, or simply the standard potential of the half-reaction. Equations (12.73) and (12.77) indicate that when the pe of a solution at 25 C increases by one unit (i.e., when the electron activity decreases by a factor of 10), the Gibbs energy of the electrons in the solution decreases by 5.71 kJ/mol, and the redox potential of the solution increases by 59 mV. These simple relationships among the three parameters emphasize the fact that they are really just different ways of describing the same quantity – the chemical potential energy of electrons in the system. Based on these relationships, we can write the Nernst equation [Equation (12.18)] in terms of EH as follows: EH = EH
2.303 RT {Red} log ne F {Ox}
(12.78a)
= EH
0.0592V {Red} log ne {Ox}
(12.78b)
where the latter equation applies at 25 C. Keep in mind that, in these equations, {Red} and {Ox} refer, respectively, to the activity products of all species on the reduced and oxidized sides of the equation, other than e– . The relationship between the electron energy level (expressed as pe, Ge , or EH ) and the equilibrium redox speciation is shown schematically in Figure 12.19 for three redox couples [HOCl/Cl– , + SO2– 4 /H2 S(aq), and H /H2 (g)].
Figure 12.19 Schematic of the relationship between electron energy levels (quantified in three different ways) and the speciation of various redox couples, based on the Nernst equation. The arrows indicate the effect of changing the speciation of each couple from the standard state to a particular nonstandard state.
731
12.15.3
The Molar Gibbs Energy Change Accompanying Redox Reactions
We saw earlier [Equation (12.19)] that log K for a redox reaction involving transfer of one electron equals the difference between the pe values of the reduction and oxidation half-reactions. Correspondingly, log K for an ne -electron transfer is just ne times this difference: log K = ne DpeOx’t/Red’t (12.79) We can combine this relationship with that between the standard Gibbs energy change of reaction and the equilibrium constant [Equation (12.70)] to write the standard molar Gibbs energy of any redox reaction as DGr = 2.303ne RT DpeOx’t/Red’t and, at the convenional standard-state temperature of 25 C, ✓ ◆ kJ DGr = 5.71 ne DpeOx’t/Red’t mol
(12.80)
(12.81)
Equations (12.80) and (12.81) can be converted to the following expressions that apply to any conditions (i.e., not just standard-state conditions): DGr = 2.303 ne RT DpeOx’t/Red’t or, at 25 C, DGr =
✓ ◆ kJ 5.71 ne DpeOx’t/Red’t mol
(12.82)
(12.83)
Equations (12.82) and (12.83) indicate that the molar Gibbs energy of a reaction is proportional to the difference in the pe values associated with the two half-reactions. This result makes intuitive sense: because pe is a direct measure of the molar Gibbs energy of electrons in a redox couple, the difference in pe between two couples indicates the difference in electron energy levels between them. If a path is available for electron transfer between the couples, we expect electrons to transfer from the higher to the lower Ge environment (lower to the higher pe), releasing a corresponding amount of Gibbs energy as they do. Such a transfer of electrons corresponds to oxidation of the couple at lower pe and reduction of the couple at higher pe. This process will continue until the pe values of the two couples are equalized, at which point the driving force for the transfer will have dissipated. Though these reactions might not proceed rapidly, they certainly will occur as the system approaches equilibrium. The only difficulty with the preceding description is the idea that different redox couples in the same solution can have different pe values. Certainly a well-mixed solution has only one pH, one temperature, one Cl– activity, etc.; so how can it have more than one pe value? The answer is that systems can be characterized by a uniform 732
value of some variable only if they are in equilibrium with respect to that variable. For instance, a solution at thermal equilibrium has a single temperature, but not all solutions are at thermal equilibrium. If temperature varies from one location to another in a solution, then the question “What is the temperature of the solution?” has no answer. One might compute the theoretical temperature of the solution once it equilibrates or give a long answer describing the temperature as a function of location, but in either case the original question would be meaningless. The same situation applies to a redox system that is not at equilibrium, except that the variation is from one redox couple to the next, rather than from one location to another. The pe of a given redox couple can be defined as the pe that would cause the half-reaction between the species to be at equilibrium for the given system composition (activities of the oxidized species, reduced species, H+ , and any other chemicals that participate in the reaction). For instance, if the activities of HOCl and Cl– were 10 4 and 10 3 , respectively, in a pH 7 solution, we could compute the pe of the HOCl/Cl– redox couple from the Nernst equation, as follows: 1 {Cl } log ne {HOCl}{H+ }
peHOCl/Cl– = peHOCl/Cl–
1 10 3 log = 21.10 2 (10 4 )(10 7 )
= 25.10
3 in the same solution, the pe If the activities of both H2 S(aq) and SO2– 4 were 10 of that couple could be computed using the same approach:
peSO2– /H 4
2 S,pH 7.0
= peSO2– /H 4
= 5.085
2S
{H2 S} 1 log 2 8 {SO4 }{H+ }10
1 10 3 log 8 (10 3 )(10
7.0 )10
= 3.67
These calculations indicate that the pe of the solution would have to be 21.10 for the HOCl/Cl– couple to be in equilibrium and 3.67 for the SO2– 4 /H2 S couple to be in equilibrium. (Note that these are the values shown in Figure 12.19.) The solution cannot simultaneously have both of these pe values, so it must not be at equilibrium. Assuming that the system is at 25 C, we can use Equation (12.73) to compute Ge of the exchangeable electrons in the HOCl/Cl– and SO2– 4 /H2 S couples to be 120.48 and 20.96 kJ/mol, respectively. Thus, the electrons associated with the SO2– 4 /H2 S couple have substantially more energy than those associated with the HOCl/Cl– couple, and Gibbs energy could be released if electrons transferred from the former to the latter. This, then, is the direction in which we expect the reaction to proceed. We can even infer the quantitative result from this analysis: 733
DGr = ( 120.48
20.96) kJ/mol = 141.44 kJ/mol
Therefore, 141.44 kJ of Gibbs energy will be released for each mole of electrons transferred. The overall process is shown schematically in Figure 12.20.
Figure 12.20 Revision of Figure 12.19 showing the molar Gibbs energy of reaction for oxidation of H2 S(aq) by HOCl under the conditions shown. The dashed arrows indicate the direction of change as the reaction proceeds (see text).
The same result can be obtained by combining the half-reactions and applying the relationship DGr = RT ln(Q/K) to the overall reaction. 1 1 1 5 H S(aq) + H2 O )* SO24 + H+ + e 8 2 2 8 4 1 1 + 1 1 HOCl + H + e )* Cl + H2 O 2 2 2 2 1 1 1 1 3 H S(aq) + HOCl )* Cl + SO24 + H+ 8 2 2 2 8 4
Q=
(SO24 )1/8 (Cl )1/2 (H+ )3/4 (H2 S(aq))1/8 (HOCl)1/2
=
log K = pe = 5.08
(12.84)
log K = pe = 25.10
(12.85)
log K = 20.02
(10 3 )1/8 (10 3 )1/2 (10 7 )3/4 (10 3 )1/8 (10 4 )1/2
= 10
✓ ◆ ✓ ◆ kJ Q kJ 10 4.75 kJ DGr = 5.71 log = 5.71 log 20.02 = 141.44 mol K mol 10 mol 734
(12.86)
4.75
(12.87)
Recall that DGr = 0 for any reaction at equilibrium, and that DGr for any overall reaction equals the sum of the DGr values for the reactions that are added to obtain the overall reaction. Thus, the Gibbs energy of reaction for oxidation of HS– by OCl– in the same solution where we considered oxidation of H2 S by HOCl can be computed as 1 1 1 3 1 H2 S(aq) + HOCl )* Cl + SO24 + H+ 8 2 2 8 4 1 1 + 1 HS + H )* H2 S(aq) 8 8 8 1 1 + 1 OCl + H )* HOCl 2 2 2 1 1 1 1 1 HS + OCl )* Cl + SO24 + H+ 8 2 2 8 8
DGr = 141.44
kJ mol
kJ mol kJ DGr = 0 mol DGr = 0
DGr = 141.44
kJ mol
(12.88)
Because DGr = 0 for the two acid/base reactions (based on the assumption that they are always at equilibrium), DGr for the HS– /OCl– reaction is identical to that for the H2 S(aq)/HOCl reaction. Generalizing this result, we conclude that DGr for a redox reaction is the same regardless of which of the acid/base species participates. HOCl is often reported to react more readily than OCl– with reduced species, and this observation is sometimes explained by suggesting that the driving force for the reaction with HOCl is greater than with OCl– . Such arguments are usually based on comparison of DpeOx’t/Red’t values for the reaction of the reduced species with the two different oxidants. However, such arguments are fallacious, because the driving force is actually related to DpeOx’t/Red’t , not DpeOx’t/Red’t , and DpeOx’t/Red’t is the same regardless of which oxidant participates. Reactions with HOCl do indeed often proceed more rapidly than those with OCl– , but the reason has to do with the details of the reaction pathways that cause the reactions to have different activation energies, not differences in the thermodynamic driving force. When the relationships introduced in the preceding sections are combined with expressions for K and Q, a truly dizzying array of equations can be developed, each of them useful under slightly different conditions. Several of these relationships are collected in Table 12.9. In the table, Row 1 describes the molar Gibbs energy of electrons in a system at a given pe or EH , and Row 2 shows expressions for computing the pe or EH from the solution composition (assuming redox equilibrium). Keep in mind that no absolute value of Ge , pe, or EH can be defined without reference to some arbitrary baseline; the baseline for all three expressions in the first two rows of the table is the H2 (g)/H+ half-reaction at standard state. Rows 3 through 5 describe the energy change that occurs when electrons shift from one host (i.e., the reduced species in one redox couple) to another, and Rows 6 735
Table 12.9 Equations relating parameters that describe electron energy levels for a halfreaction and energy changes accompanying an n-electron reduction reaction(a)
1 Energy of exchangeable electrons in a system
Ge = ⇢ Ge⇢ + RT ln{e } = 2.303RT pe = FEH
2 Nernst equation: energy of exchangeable electrons associated with a given halfreaction
pe = pe
1 {Red} log ne {Ox}
EH = EH
0.0592 V {Red} log ne {Ox}
3 Gibbs energy of reaction in terms of energy of electrons associated with the two halfreactions(b)
DGr = ne DGe
4 Gibbs energy of reaction in terms of extent of disequilibrium
DGr = DGr +
5 Change in electron energy in a reaction in terms of extent of disequilibrium(b)
Dpe = Dpe
DGr = 2.303ne RT Dpe DGr = ne FDEH
DEH = DEH
2.303 RT log Q ne 1 log Q ne 2.303 RT log Q ne F
DGr 2.303 RT log K = ne Dpe ne F log K = DE 2.303 RT H
6 Equilibrium constant in terms of energy change at standard state(b)
log K =
7 Conditions for conversion of reactants to products(b)
DGr < 0 Dpe > 0 DEH > 0 DGr = Dpe = DEH = 0
8 Conditions at equilibrium(b)
(a) At 25 C, 2.303 RT = 5.71 kJ/mol and 2.303 RT /F = 59 mV. (b) {Red} and {Ox} refer, respectively, to the activity products of all terms on the reduced and oxidized sides of the reaction, other than e– . (c) Dpe = peOx’t peRed’t , where peOx’t and peRed’t are the pe values of the half-reactions for the oxidant and reductant, respectively. Similarly, DEH,Ox’t/Red’t = EH,Ox’t EH,Red’t .
736
through 8 characterize the likelihood that such shifts will occur. Row 6 expresses the equilibrium constant for a redox reaction in terms of the change in electron energy under standard state conditions; Row 7 indicates the conditions that correspond to a downhill (favorable) gradient in electron energy; and Row 8 indicates that when this gradient disappears, the system is at equilibrium and will react no more.
12.16
REDOX TITRATIONS AND THE GEOCHEMICAL REDOX SEQUENCE
Just as the pH of a solution responds to additions of acids or bases, the pe responds to additions of reducing or oxidizing agents. Thus, for example, if a solution is titrated with a strong reductant, electrons transfer from that reductant to the oxidized species that has the greatest affinity for them under the extant conditions (assuming the solution equilibrates after each addition of titrant). To a good approximation, each oxidized species in the system is converted completely to its reduced form before any weaker oxidants react. For instance, consider the changes in pe when a solution that contains 10 3 M T OT Cl and 5⇥10 4 M T OT N, at an initial pe of 25 (highly oxidizing conditions) and at pH 8.0, is titrated with NaHS. We will assume that the redox reactions among Cl, N, and S equilibrate rapidly, but that reactions of the H2 /H2 O/O2 group are kinetically inhibited, and that the pH is maintained at 8.0 throughout the titration. The species that we will consider include HOCl, OCl– , and Cl– ; NO–3 , NO–2 , NH3 (aq), and NH+4 ; and H2 S, HS– , and SO2– 4 . A log C-pe diagram for the initial solution (before any titrant has been added) is shown in Figure 12.21 and indicates that HOCl, OCl– , and NO–3 are the only species present at significant concentrations. – pe8.0 for the SO2– 4.79. Therefore, when a small amount of 4 /HS couple is NaHS is added to the solution at pe 25, a strong driving force exists for the S( II) to be oxidized to SO2– 4 . The electrons released from this oxidation reaction will transfer to the strongest oxidizing agent available, which can be identified in Figure 12.21 as the oxidized species of the redox couple with the highest pe8.0 ; i.e., either HOCl or OCl– . By transferring to these species, the electrons will experience the largest possible Dpe, and the transfer will lead to the largest possible release of Gibbs energy.16 The large separation between the pe8.0 values for the S(VI)/S( II) and Cl(+I)/ Cl( I) couples implies that it is impossible for significant concentrations of S( II) and Cl(+I) species to coexist at equilibrium. Because we are assuming that equilibrium is reached rapidly after each dose of NaHS is added, we conclude that the added S( II) will react with Cl(+I) species until one or the other is essentially completely consumed. In the current case, where we are adding just a small amount of S( II) to 16 Recall that the driving force is the same for reaction with all species that are in equilibrium with one
another. Therefore, in terms of the system thermodynamics, whether the Cl(+I) species that acquires the electrons is HOCl or OCl– is irrelevant.
737
Figure 12.21 LogC-pe diagram for a system containing 10 3 M T OT Cl and 5 ⇥ 10 4 M T OT N at pH 8.0, assuming that N can exist in the 3, +3, and +5 oxidation states.
a solution with a significant concentration Cl(+I), the S( II) will be fully consumed, and the solution will retain a significant Cl(+I) concentration, but it will also contain a small amount of Cl( I). Figure 12.21 suggests that the equilibrium pe in the solution under those conditions will be slightly greater than 21. As more NaHS is added and reacts with the Cl(+I), the same scenario will repeat itself, with the concentration of Cl( I) steadily increasing and that of Cl(+I) steadily decreasing. As long as significant concentrations of both those oxidation states are present, the pe will remain near 21. Only after enough NaHS has been added to reduce essentially all the Cl(+I) will the next strongest oxidant (in this case, NO–3 ) start to be consumed. As soon as that process begins, significant amounts of both NO–3 and NO–2 will be present in solution. Therefore, when the system undergoes the transition from reduction of Cl(+I) to reduction of N(V), the pe will decline from near 21 to near 7. It will then remain around 7 until all the N(V) has been converted to N(III), after which it will undergo another relatively rapid decline to near 4. The preceding discussion is essentially identical to the description of changes in pH when a solution containing several weak bases is titrated with acid. In that case, the pH drops continuously as the titrant is added, but the rate of decline is not steady. Rather, the decline is slow near points where the identity of the dominant species is changing and rapid between those points. By analogy, then, we can say that the pe of a solution is buffered near pe values where the dominant species of a redox couple changes, if that couple is present at a significant concentration. As we have seen, the pe value where dominance changes equals pepH for redox couples whose half-reactions involve only the oxidized species, reduced species, H+ , and H2 O. Correspondingly, we can intentionally stabilize the solution pe near a desired 738
value by adding appropriate amounts of redox species whose pepH is near that value.17 As is the case for acid/base titrations, we can use redox titrations to infer the identities and concentrations of unknown species in a solution. For instance, in the example solution just discussed, the dose of NaHS required to consume all the Cl(+I) can be computed by an electron balance: each Cl(+I) molecule undergoing reduction consumes two electrons, and each S( II) molecule undergoing oxidation provides eight electrons, so the NaHS requirement to reduce all the Cl(+I) is one-fourth of T OT Cl(+I), or 2.5 ⇥ 10 4 M. Thus, if we didn’t know the composition of the initial solution, we could use the results of the titration to infer that Cl(+I) was present (because the pe was buffered near 21) and to estimate its concentration based on the concentration of NaHS required to drive the pe from ⇠25 to ⇠15. An interesting application of redox titrations occurs naturally near the bottom of water bodies with intense bioactivity. In this case, the titrations are mediated by organisms that oxidize organic matter, reducing the strongest oxidant available to them. The preference for the strongest oxidant can be viewed as a result of natural selection: the strongest oxidant is the one that is farthest from equilibrium with the organic matter, so the reaction between it and organic matter releases the most Gibbs free energy. Organisms that can mediate these reactions and capture the energy are at a competitive advantage over those that, for whatever reason, can only use weaker oxidants. The former groups of organisms proliferate until essentially all of that oxidizing agent is consumed. Thereafter, either those organisms or, more often, a different microbial community starts utilizing the next most powerful oxidant to gain energy from the remaining organic matter. When this sequence occurs in lake sediments, the major oxidants, in order of preference, are typically O2 (aq), NO–3 , Mn(IV), Fe(III), SO2– 4 , and, finally, organic molecules that are more oxidized than the organic matter being used as food. The sequence described above can occur over time in a batch system, but in a natural system, it is more often manifested as a steady-state gradient in space, especially in sediments. This gradient is established by diffusion of organic matter downward from the water-sediment interface, along with the dissolved oxidants (O2 (aq), NO–3 , – and SO2– 4 ). In the upper sediment layers, the oxygen is consumed, while the NO3 and 2– SO4 remain in solution. At some depth, the oxygen is almost completely consumed, and the local organisms use NO–3 as the electron acceptor for their energy-producing metabolic reactions. Once all the NO–3 is depleted (at a lower depth), other groups of organisms use the available Fe(III) and Mn(III) or Mn(IV) as electron acceptors. These oxidized metals are often present as solids in the sediments, but when they are reduced to their +II state, they typically dissolve and diffuse away from the solid; those that diffuse upward (to more oxidizing regions) might re-precipitate. Once the oxidized Fe and Mn are depleted, organisms use SO2– 4 as the oxidant. Finally, when even that 17 If the half-reaction includes other species (e.g., HCO– ), the crossover point can be slightly different 3 from pepH , but not dramatically so.
739
source of oxidizing power is consumed, some organisms can obtain relatively small amounts of energy by mediating redox reactions in which complex organic molecules are converted to carbonate species [C(+IV)] and methane [C( IV)]. In these reactions, referred to as fermentation reactions, electrons are simply transferred among different carbon atoms in the organic molecules, without the direct involvement of any other inorganic chemicals.
12.17
pe-pH PREDOMINANCE AREA DIAGRAMS
Because pe can vary over a wide range and nondominant redox species are often present at insignificant concentrations, pe-pH predominance area diagrams offer a convenient way to represent the major species in redox-active systems. In such diagrams, the dominant form of each element (considering both acid/base and redox equilibria) is identified for all possible combinations of pH and pe, but the concentrations of nondominant species are not shown. pe-pH predominance area diagrams can be prepared following essentially identical procedures to those described in Chapters 10 and 11 for other types of predominance area diagrams. Specifically, we write reactions and derive equilibrium constants relating each pair of species in the system. The conditions under which the two species have equal activity can then be expressed by an equation of the form pe = m pH + b. When this line is plotted in pe-pH space, the two species are dominant on opposite sides of the line. When lines for additional pairs of species are added to the diagram, regions where a single species is dominant over all the others become apparent; once these regions have been determined over the whole range of pe and pH of interest, the diagram is complete. To demonstrate this procedure, we will prepare a pe-pH predominance area diagram for nitrogen, considering the species NH3 , NH+4 , NO–2 , and NO–3 . The oxidation state of N is 3 in NH3 and NH+4 , +3 in NO–2 , and +5 in NO–3 . NH+4 and NH3 can be interconverted by acquisition or release of one proton without exchange of electrons, so the line separating areas of dominance of these two species is trivial to derive: It is defined by the equation pH = pKa,NH+ (= 9.24), with NH+4 dominant at lower pH 4 and NH3 dominant at lower pH. To determine the equations of the lines characterizing all the other species pairs, we write the reaction between each pair in the form of Reaction (12.25) and then apply Equation (12.27) to determine the conditions under which the two redox species have equal activity:18 18 Reaction (12.25) implies that the oxidized and reduced species have equal stoichiometric coefficients, and Equation (12.27) relies on that assumption. If the stoichiometric coefficients on Red and Ox are not identical [e.g., for comparison of N2 (aq) and NH3 (aq)], then {Ox} and {Red} have different exponents in Equation (12.27), and the line along which their activities are equal depends on the total concentration of the constituent in the system. If other species appear in the reaction, their activities must be specified and included in Equation (12.27).
740
Ox + nH H+ + ne e )* Red {Red} = {Ox}
if:
pe = pe
(12.25) nH pH ne
(12.27)
The corresponding equations for all possible species pairs in the system of interest are collected in Table 12.10. Table 12.10 Equilibrium relationships relevant for developing a pe-pH predominance area diagram for a system containing N( III), N(+III), and N(V)
Line #
Species compared (Reactant/Product)
Log K
Equation of line where species have equal activities
1
NH+4 /NH3
9.24
pH = 9.24
2
NO–2 /NH+4
90.51
pe = 15.08
3
NO–3 /NH+4
119.077
pe = 14.88
4
NO–2 /NH3
81.27
pe = 13.54
5
NO–3 /NH3
109.84
pe = 13.73
6
NO–3 /NO–2
28.57
pe = 14.29
4 3 5 4 7 6 9 8
pH pH pH pH
pH
(a)
log K is for a reaction relating the two species shown in the second column, written with the minimum integer values for the stoichiometric coefficients of both species (in this case, a stoichiometric coefficient of 1 for both species in all reactions).
The easiest line to draw on the diagram is the vertical line at pH 9.24 that separates the region where (NH+ 4 ) > (NH3 ) from the region where the reverse is true. Based on this comparison, we know that the possible dominant species at pH < 9.24 are NH+4 , NO–2 , and NO–3 , and those at pH > 9.24 are NH3 , NO–2 , and NO–3 . Considering the low-pH region first, we can prepare the graph shown in Figure 12.22, which includes Lines #1-3 from Table 12.10. In the region below both the NO–2 /NH+4 and NO–3 /NH+4 lines, NH+4 is dominant over both of the more oxidized species, so there is no question that it is the N species with the highest activity. Above both of those lines, both of the oxidized species have higher activity than NH+4 , so we know that one of those two species will be dominant, but we cannot say which one without further analysis. In the region between the two lines and at pH > 2.4, the 741
activity of NO–2 is greater than that of NH+4 , and the activity of NH+4 is greater than that of NO–3 , so it is certain that NO–2 is dominant. The reverse is true in the thin – slice between the two lines at pH < 2.4 [i.e., (NO3 ) > (NH+ 4 ) > (NO2 )], so NO3 is dominant in that region.
Figure 12.22 Preliminary pe-pH predominance area diagram considering NH+4 , NO–2 , and NO–3 at pH<9.24.
To assess where NO–2 and NO–3 are each dominant in the region above the two lines, we can add Line #6 from Table 11.9. To do so, we can either rely on the equation in the table, or simply note that the activities of NO–2 and NO–3 must be equal at the intersection of the NH+4 /NO–2 and NH+4 /NO–3 lines, and that the slope of the NO–2 /NO–3 line is 7/6, based on the equation for Line #6 in Table 12.10. Using either approach, we can draw the line and distinguish the regions of dominance of those two species, as shown in Figure 12.23. Once we do that, we have completed the diagram for solutions with pH < 9.24. We can follow essentially the same procedures to draw the remaining portion of the diagram, for the conditions at pH > 9.24. The NO–2 /NO–3 line is still relevant in this region, and that line is independent of the relative activities of NH+4 and NH3 , so we can just extend the existing line to the higher pH region. As we did for the NO–2 /NO–3 line, we can draw the NO–2 /NH3 line based on either the equation in Table 12.10 or the knowledge that this line must pass through the intersection of the NO–2 /NH+4 and NH+4 /NH3 lines, and must have a slope of 7/6. Once we have drawn that line, we see that the regions of dominance of NH3 and NO–3 do not intersect anywhere in pepH space, so we need not draw the corresponding line. The final diagram, identifying the dominant species over the entire pe and pH range, is shown in Figure 12.24.
742
Figure 12.23 pe-pH predominance area diagram considering NH+4 , NO–2 , and NO–3 at pH < 9.24.
Figure 12.24 pe-pH predominance area diagram considering NH+4 , NH3 , NO–2 , and NO–3 , at any values of pe and pH.
The diagram accomplishes the goal of indicating the conditions under which different species are expected to be dominant, allowing us to predict or interpret changes in speciation associated with major changes in system conditions. As in the diagrams developed in Chapters 10 and 11, three line segments with different slopes must emanate from every intersection point. If any of the species under consideration were solids, their areas of predominance could be identified by following the procedures discussed in Chapter 11. In such a case, each solid would be considered dominant whenever it was predicted to be present, and its region of dominance would depend on the total concentration of the element of interest. 743
12.18
REDOX REACTIONS AND ELECTROCHEMISTRY
12.18.1
Anodes, Cathodes, and Electrochemical Reactions
Throughout this chapter, the analogies between electron transfer in redox reactions and proton transfer in acid/base reactions have been emphasized. We now consider the most important difference between these reactions, and the characteristic that makes redox reactions unique among chemical reactions: the ability of electrons to travel easily through wires or other electrical conductors. This characteristic, in conjunction with the slowness of many redox reactions via direct molecular collisions, allows a degree of external control over redox reactivity that is not available for most other types of reactions. Reactions that proceed very slowly have high activation energies. In simplistic terms, a high activation energy implies that there is no ‘convenient” path for the electrons to take as they move from the species being oxidized to the one being reduced, and a great deal of energy must be provided either to free the electrons from the reductant or to get the electrons past some barrier interfering with their attachment to the oxidant. The kinetic inhibition can be surmounted either by providing energy to the system to overcome the activation energy barrier along the existing path or by providing an alternative reaction pathway with a lower activation energy barrier. Examples of these two alternatives include providing a spark to start coal burning and using enzymes to catalyze metabolic reactions, respectively. The surfaces of pure metals are good catalysts for many redox reactions occurring in solution. For instance, dissolved molecular hydrogen and oxygen react with one another to form water at a negligible rate if the only pathway for electron transfer is direct collision of the molecules, but the reaction proceeds immensely more quickly in the presence of metallic platinum. The catalysis is attributed to the tendency of hydrogen and oxygen molecules to bind to the metal‘s surface (a process known as adsorption) and to exchange electrons readily with the metal. This fact, in conjunction with the availability of delocalized, highly mobile electrons in the metal, allows electron transfer to proceed more rapidly via a hydrogen-to-metal-to-oxygen pathway than via the direct hydrogen-to-oxygen pathway (Figure 12.25). When the reaction is catalyzed by platinum, the H2 oxidation and O2 reduction half-reactions are still linked in the sense that they can only proceed in tandem, with electrons being released at the same rate that they are consumed.19 However, unlike in the direct-collision pathway, the electron transfers can take place at two different locations. That is, because the electrons are so mobile in the metal, it is not necessary for the H2 and O2 to bind to the platinum adjacent to one another. The location 19 In
theory, electrons could simply accumulate in the metal, allowing the oxidation reaction to proceed independently of the reduction reaction. However, in reality, the number of electrons that can accumulate in a metal beyond those that are part of the metallic structure is extremely small, so the rates of the two half-cell reactions are in fact closely linked.
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Figure 12.25 Schematic contrasting the rates of electron transfer from H2 (aq) to O2 (aq) in the absence and presence of metallic Pt as a catalyst.
on the metal where the oxidation occurs is called the anode, the location where the reduction occurs is called the cathode, and both the anode and cathode are referred to as electrodes. Anodes and cathodes might be distributed more or less randomly around the surface of the metal, or they might be localized.
12.18.2
Quantifying the Driving Force and Energy Change in Electrochemical Reactions
Now consider a slightly different scenario, in which a solution is separated into two portions, and a Pt electrode is immersed in each. The electrodes can be connected electrically via a switch, but initially the switch is in the open (disconnected) position. The two solutions are then equilibrated with different gases: one gas contains H2 at its atmospheric partial pressure, but no O2 , and the other contains O2 at its atmospheric partial pressure, but no H2 . Also, assume that both solutions contain identical concentrations of salts that are not redox-active (e.g., each might contain some NaCl), that both are well buffered at pH 7.0 and 25 C, and that the solutions are connected through a semipermeable barrier that allows small amounts of ions to diffuse from one solution to the other, but does not allow extensive mixing of the two solutions. The scenario is depicted schematically in Figure 12.26. When the gases dissolve in the solutions, some H2 adsorbs to the anode, and some O2 to the cathode. The two redox half-reactions then approach equilibrium, with electrons entering the anode as hydrogen molecules are oxidized in one cell, and electrons leaving the cathode to reduce adsorbed oxygen molecules in the other. These products of the redox half-reactions are then released from the electrode surfaces, causing H+ and OH– to accumulate in the anodic and cathodic solutions, respectively. The reactions thus generate a charge imbalance in each solution, but the imbalance is rapidly eliminated by diffusion of cations (Na+ and H+ ) from the anode chamber into the cathode chamber, and of anions (Cl– and OH– ) in the opposite direction. 745
Figure 12.26 Schematic of an electrochemical cell. The anodic cell is in equilibrium with atmospheric H2 (g) but no O2 (g), and the cathodic cell is in equilibrium with atmospheric O2 (g) but no H2 (g). The pe values shown apply when the circuit is open, and the flows of electrons and salt ions apply when the circuit is closed. The molar Gibbs energy of the electrons is shown in kilojoules per mole.
Applying the Nernst equation to each cell in conjunction with the known pH and partial pressures of the two gases, we can compute that once equilibrium is attained, the pe will be +13.6 and 3.85 in the solutions in contact with the oxygen- and hydrogen-containing gases, respectively. Although the pe of each solution changes substantially as the solution equilibrates with the electrode, the amount of reaction that occurs (in terms of the mass of oxygen or hydrogen consumed, or the number of electrons transferred into or out of the electrodes) is exceedingly small. Correspondingly, the charge imbalance that is generated and the amount of diffusion needed to counter it are also small. Consider now the thermodynamic status of electrons in the electrodes. As in any phase, the activity of the electrons (or any chemical species) can be characterized based on their concentration and their environment. The concentration of free electrons in each metal is huge compared to the number that transfer in or out, so this concentration is effectively constant as the redox reactions proceed. However, the environment in the electrodes, and specifically their electrical potential, changes substantially when even a small number of electrons enters or leaves. Because electrons can move freely in the metal electrodes, the electrical potential is nearly uniform throughout each electrode. Thus, once electrons transfer into 746
the anode, it will be characterized by an almost uniform negative potential, yanode , whereas the cathode will develop a positive potential, ycathode . As each half-reaction proceeds, the magnitude of the corresponding y increases. The reactions continue until the electrons in the electrode have the same potential as those in the adjacent solution; at that point, no potential energy would be released by transfer of additional electrons, and the reaction has reached equilibrium. We typically think of the potential in the electrode as an electrical phenomenon and that in the solution as a chemical phenomenon, but this distinction is irrelevant in terms of electron flow; the electrons always move in the direction of decreasing total electrochemical energy, regardless of how that energy is stored.20 Recall that EH can be defined as the electrical potential to which electrons would have to be exposed in order to have the same electrochemical potential energy as the electrons in a solution of interest. The scenario in the two cells is, in essence, the experimental realization of that conceptual definition. That is, when the half-reactions have reached equilibrium, ycathode = EH, oxygenated = (0.0592 V)peoxygenated = +0.802V solution
solution
yanode = EH, hydrogenated = (0.0592 V)pehydrogenated = 0.227V solution
solution
The electrical potentials of the electrodes propagate into solution a short distance, orienting the water molecules near the electrode surfaces and attracting or repelling ions in much the same way that dissolved ions affect the water molecules and other solutes that surround them. Now consider what happens if the switch between the two electrodes is closed. Because the electrons in the anode are at a higher electrical potential than those in the cathode, they travel through the wire from the former to the latter. This process serves as a sink for electrons that would otherwise accumulate in the anode, while replenishing the electron supply in the cathode. Although the EH values of the solutions remain at +0.802 and 0.227 V (being the hypothetical electrical potential to which electrons would have to be exposed to be in equilibrium with the solutions), ycathode and yanode become lower than +0.802 V and higher than 0.227 V, respectively. Thus, the electrochemical potential in the entire system decreases continuously in the sequence: hydrogenated solution>anode>cathode>oxygenated solution. This gradient in potentials drives electrons steadily from the hydrogenated solution through the anode, the wire, and the cathode, to the oxygenated solution. The electron flow and reactions continue until the electron activity is identical throughout the system; 20 This
situation is analogous to water flow in a pipe, which occurs in the direction of decreasing total energy (often referred to as ‘hydraulic head’ in that situation). The total hydraulic head includes contributions from pressure, gravity, and fluid velocity, but the flow responds to the sum of those terms, regardless of their relative magnitudes.
747
in theory, if the gases were continuously supplied and the pH was maintained in both cells, the reaction would continue indefinitely. The system described above comprises a complete electrical circuit, with the driving force for electron flow through the wire being provided by the chemical (redox) disequilibrium between the two parts of the system. Note that, from a chemical perspective, this scenario is essentially identical to the one considered earlier in which the H2 and O2 were dissolved in a single solution, which was in contact with a single platinum electrode. In both systems, H2 (aq) oxidation and O2 (aq) reduction occur at separate locations on the platinum, and electrons flow from anodic sites through a metallic conductor to the cathodic site, where they are consumed. The major difference between the two systems is that, in the former one, the anodes and cathodes were distributed more or less randomly on the electrode, so that the electron flow occurred simultaneously in all directions; by contrast, in the current scenario, all the anodic sites are in one part of the system, and all the cathodic sites are in another, so the electrons all flow in the same direction. In essence, by separating the solutions and electrodes, we have rectified the electron flow. If, in the system with rectified electron flow, an electrical appliance is placed in the circuit, the Gibbs energy decline when the overall redox reaction proceeds will be expended ‘pushing” the electrons through the appliance. Thus, the chemical reaction could be used to drive an electrical system. This is, of course, the principle of operation of a chemical battery. In such systems, a solution containing a large salt concentration called a salt bridge is sometimes used to link the solutions in the anodic and cathodic cells to ensure that diffusion of salt ions does not become the bottleneck limiting the reaction rate.
12.18.3
Redox Reactions in Systems with Externally Imposed Electrical Potentials
The preceding discussion describes a situation in which electrodes are inserted into solutions to catalyze reactions that are favorable but are kinetically inhibited. We next consider how electrical potentials applied to electrodes and controlled externally can facilitate or impede redox reactions. Continuing with the same example system described above, imagine that an external circuit is used to force the potential of the electrode in the hydrogenated solution to a value that is 1.5 V more positive than that of the electrode in the oxygenated solution. The exact values of the potentials are not critical, but to make the example concrete, we will assume that the electrodes in the hydrogenated and oxygenated solutions have imposed potentials (yext ) of +0.9 and 0.6 V, respectively. Near each electrode, the electrical potential propagates into solution some distance, eventually decaying to zero. Although the profile of potential versus distance from each electrode is important for a variety of practical reasons, all that matters for now is that immediately adjacent to each electrode, the electrical potential is almost that same as in the electrode. Then, if a packet of solution with the bulk composi748
tion is brought near the electrode, its EH value will change from the EH of the bulk solution to EH,bulk + yext , i.e., EH, with imposed = EH, no imposed + yext potential
potential
(12.89)
The EH values in the two solutions near the electrodes in the example system are therefore Hydrogenated solution:
EH,hydrogenated = 0.227V + 0.900V = +0.637V
Oxygenated solution:
EH,oxygenated = +0.802V
0.600V = +0.202V
The key result is that the relative EH values have been reversed, indicating that now, near the electrodes, the electrons in the hydrogenated solution have lower molar Gibbs energy than those in the oxygenated solution. As a result, electrons transfer out of the oxygenated solution (via oxidation of H2 O to O2 ), travel through the electrodes and wire, and enter the hydrogenated solution (via reduction of H2 to H2 O). Thus, the imposition of the electrical potential has reversed the direction of the reaction. The thermal analog of this situation is air conditioning, in which an input of electrical energy drives a process that would otherwise be thermodynamically unfavorable (extracting heat from a cool environment and transferring it to a warmer one). It should be clear that this result (reversal of the direction in which the reaction proceeds) would have been obtained regardless of the absolute values of the imposed potentials; the only requirement is that the imposed potential difference be larger than and in the opposite direction from the natural potential difference that develops because of the different chemical compositions of the two solutions. Where the absolute potentials do come into play is in controlling the kinetics of the electrode reactions, a topic which is beyond the scope of the current discussion. This topic is addressed in many texts on electrochemistry and electrode dynamics.21 In industrial settings, external electrical potentials are applied to drive all sorts of reactions in directions that they would not proceed spontaneously, with the most obvious examples being electroplating and metal refining. The approach is sometimes also used to prevent an undesirable reaction from occurring, such as by imposing a positive potential on pipes to prevent them from corroding. Example 12.14 According to the Nernst equation, the EH of a solution at pH 7 that contains dissolved oxygen in equilibrium with the atmosphere is 0.802 V. What reaction occurs if an electrode is inserted into a solution that has been equilibrated with air if the electrode potential is controlled at 1.0 V? What would the equilibrium pH be if the solution was continuously bubbled with air? 21 An
excellent introduction to this topic is provided by Bockris et al. (Bockris, J.O., Reddy, A.K.N, and Gamboa-Aldeco, M.E. Modern Electrochemistry, Plenum, New York, 3 vol. (1998-2000).
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Solution The electrode surface is at a higher electrical potential than the adjacent solution, so electrons transfer from solution to the electrode. The electrons are withdrawn from water molecules, converting them to dissolved oxygen and H+ via the following reaction: 2 H2 O )* O2 (aq) + 4 H+ + 4 e In the absence of other limitations, this reaction would lower the pH of the solution and increase the dissolved oxygen concentration until the composition of the solution was consistent with the Nernst equation in the form of Equation (12.78b): EH = EH
0.0592 V {Red} log 4 {Ox}
1.0 V = 1.272 V pH = 4.61 +
2 ⇠⇠ {H⇠ 0.0592 V ⇠ 2 O} log 4 {O2 (aq)}{H+ }4
1 log {O2 (aq)} 4
If the solution were bubbled with air, the dissolved oxygen concentration would stabilize at 2.77 ⇥ 10 4 M, the value in equilibrium with atmospheric O2 , so the pH would be 1 pH = 4.61 + log (2.77 ⇥ 10 4 ) = 3.72 4
12.18.4
Redox Electrodes, Reference Electrodes, and the Measurement of EH
If, in the system shown in Figure 12.26, the switch is closed and the gas supply is halted, the pe values of the two solutions will approach one another. As they do so, the driving force for both half-reactions decreases, and once the pe values in the solutions are equal, the driving force disappears. If the electrode surface is a good catalyst for the two half-reactions and the wire provides negligible resistance, the reaction will proceed to equilibrium quickly. On the other hand, if a large electrical resistance is placed in the wire, the driving force for electron transfer will exist, but the reaction will proceed very slowly, because the transfer of electrons will be impeded. This arrangement describes, in simplest terms, the workings of a potentiometer or voltmeter. In such instruments, a known, extremely high resistance R is placed between two points of different electrical potential, and the (very small) current i that flows across the resistance is measured. The magnitude of the current is then used to determine the voltage difference V between the two points by Ohm’s law: V = iR. The potential difference that the voltmeter actually measures is Dy, i.e., the difference in electrical potential between the two electrodes. However, if the electrodes 750
are good catalysts, the pe of each electrode will be almost identical to that in the corresponding solution, so the reading can be equated (approximately) to the difference in electrochemical potential between the two solutions, i.e., to DEH . To the extent that an electrical potential difference exists between either solution and the electrode in it, the measured value of Dy will differ from DEH between the solutions. The difference in potential between the electrode and the solution adjacent to it is called the junction potential. Recall that the H+ /H2 (g) system under standard conditions has pe = EH = 0 by definition. Therefore, if a solution whose EH is to be determined is placed in one cell and the H+ /H2 (g) system under standard conditions is placed in the other, and if the electrodes equilibrate with both solutions rapidly, then the measured Dy (that is, ytest sol’n ystd. H+ /H2 ) will indicate the value of EH in the unknown solution directly. Example 12.15 A system similar to that shown in Figure 12.26 is set up, but with different solutions bathing the electrodes. The solution on the left is at pH 0 and is bubbled with H2 (g) at PH2 = 1.0 atm; i.e., the left portion of the system is a standard H+ /H2 (g) half-cell. (a) If the solution on the right contains 10 4 M T OT OCl and 10 3 M Cl– at pH 7.53 and can be treated as ideal, what are the pe and EH of the solution? (b) A voltmeter is placed in the circuit between the two electrodes. What is the reading on the voltmeter? (Note: In schematics such as Figure 12.26, the convention is to assume that the voltmeter is set up such that its output equals EH of the solution on the right minus EH of the solution on the left.) Solution (a) The redox reaction occurring in the cell on the right is HOCl + H+ + 2e )* Cl + H2 O
pe = 25.1
Because the pH equals pKa for HOCl, we know that {HOCl} = 0.5 (T OT OCl), or 5 ⇥ 10 5 . Substituting this value into the Nernst equation [Equation (12.18)] to find pe, and then using Equation (12.77) to convert the pe to EH , we find pe = pe pe = 25.1
{Cl }{H2 O} 1 log 2 {HOCl}{H+ }
1 (10 3.0 )(1.0) log = 20.68 2 (5 ⇥ 10 5 )(10 7.53 )
EH,HOCl/Cl = (0.0592 V)pe = 1.220 V
751
(b) The voltmeter measures the difference in potentials between the two sides of the system. In the current case, the potential of the right side is 1.220 V, and that of the left side is zero, so the meter will read +1.220 V. Example 12.16 Chloride ion is added to the solution in the right-side cell in the previous example until the meter reads +1.183 V. How much Cl– has been added? Solution The standard potential for the HOCl/Cl– couple can be expressed in volts as EH,HOCl/Cl
= (0.0592 V)pe = (0.0592 V)(25.1) = 1.48 V
Again using the Nernst equation, this time in the form of Equation (12.78b), and keeping {HOCl} = 5 ⇥ 10 5 , we can find {Cl } in the solution as follows: {Cl }{H2 O} 0.0592 V log 2 {HOCl}{H+ } ⇥ 1.183 V = 1.48 V (0.0295 V) log{Cl } EH = EH
{Cl } = 10
1.729
= 0.0186
log{5 ⇥ 10 5 }
log{10
7.53
⇤ }
The Cl– concentration in the solution is 0.0186 M. Since the original solution contained 0.001 M Cl– , 0.0176 M Cl– was added. A system comprising a pH 0.0 solution equilibrated with H2 (g) at 1.0 atm plus the electrode immersed in it is referred to as a standard hydrogen cell or a standard hydrogen electrode (SHE). The SHE is a convenient conceptual construct because, by maintaining the pH and partial pressure of H2 (g) constant, the electrode potential can be held steadily at 0.0 V. However, the need for the specialized gas phase makes the use of such an electrode impractical. It should be apparent, though, that any electrode whose potential can be maintained at a constant, known value would serve the same purpose. For instance, if a solution (call it solution A) had a fixed potential of 100 mV when measured against the standard hydrogen electrode, it could be used in a standard electrode. If the potential of a solution B were determined to be 40 mV greater than that of solution A in a setup like that shown in Figure 12.26, we would know that the potential of solution B relative to an SHE would be 60 mV, i.e., EH,B = 60 mV. An electrode whose potential is fixed (or so strongly buffered that it can be treated as if it were fixed) is called a reference electrode. The buffering can be accomplished by fixing the activities of all the reactants that participate in the redox reaction in the cell (e.g., by adding them in great excess over the amounts that are likely to be generated or consumed) or by equilibrating the solution with a different phase. For 752
instance, the standard hydrogen electrode relies on the presence of H+ in solution at an activity of 1.0 (pH 0.0) and equilibration of the solution with H2 (g) at an activity of 1.0. Historically, the most commonly used reference electrode has been the saturated calomel electrode (SCE), which relies on equilibration of a saturated KCl solution (in which {Cl– } is 2.55) with calomel, Hg2 Cl2 (s), and pure liquid mercury, Hg(l). This combination establishes a fixed pe of 4.14 (EH,SCE = 244 mV) via the following reaction and calculations. Hg2 Cl2 (s) + 2e )* 2 Hg(l) + 2 Cl EH = EH
{Hg(l)}2 {Cl }2 0.0592 V log 2 {Hg2 Cl2 (s)}
0.0592 V (1)2 (2.55)2 log = 0.244 V 2 (1) EH 0.244 V pe = = = 4.14 0.0592 V 0.0592 V EH = 0.268 V
Example 12.17 What potential difference would be recorded by a voltmeter between the solution in Example 12.16 and an SCE? Solution The potential of the solution measured against the standard hydrogen electrode was 1183 mV, and the SCE has a fixed potential of +244 mV. The potential difference between the solution and the SCE will therefore be (1183 244) mV, or 939 mV. Another common reference electrode is the silver/silver chloride electrode, in which both Ag(s) and AgCl(s) are equilibrated with a saturated KCl solution. In this case, the potential is fixed by the following equilibrium: AgCl(s) + e )* Ag(s) + Cl
pe = 3.76
EH = 222 mV
The fixed pe and EH of this type of electrode are 3.36 and 199 mV, respectively. In practice, reference electrodes are enclosed in a glass or plastic tube, which is then inserted into the solution to be analyzed. A small salt bridge is built into the side of the electrode to establish good electrical contact with the solution. A similar arrangement is often used to enclose the other electrode (the working electrode), whose potential is intended to change in response to changing conditions in the sample being analyzed. Sometimes, the reference electrode and working electrode are combined in a single housing, in which case the unit is referred to as a combination electrode. A schematic of a setup for analyzing the EH of a solution using a Pt working electrode and a calomel reference electrode is shown in Figure 12.27. 753
Figure 12.27 A calomel reference electrode (on left) connected to a platinum working electrode for the measurement of EH . If the Pt electrode is selectively responsive to the H+ /H2 (g) couple, and if the solution is equilibrated with H2 (g) at a known partial pressure, the reading on the voltmeter can be related to the pH of the solution.
12.18.5
Practical Considerations and Limitations in the Use of Electrodes
When the EH of a solution is measured using a set of electrodes, some current must flow, so some conversion of oxidized species to reduced species, or vice versa, must occur in the reference electrode. For instance, if the solution being analyzed has an EH that is lower (more reducing) than a silver/silver chloride reference electrode immersed in it, some of the silver ions in solution will be reduced to silver metal. The depletion of silver ions lowers the product {Ag+ }{Cl– } in solution, causing some solid silver chloride to dissolve. However, assuming that the reservoir of AgCl(s) is large enough that some of it remains present, the activities of Ag(s) and AgCl(s) remain at 1.0. Furthermore, the mass of AgCl(s) that dissolves is very small (because of the high resistance in the voltmeter), so the change in Ag+ and Cl– concentrations is negligible, and the EH of the electrode remains effectively fixed at 199 mV. Although the conceptual basis for measuring EH (or, equivalently, pe) is relatively straightforward, in practice the measurements are not as simple as theory predicts. Some of the reasons for this result have already been alluded to, such as the possibility 754
that the junction potential is significant. More commonly, the problem that arises is that the solutions being analyzed contain many chemical species other than the redox couple of interest, including other redox couples. Any of these species might bind (adsorb) to the electrode surface and change its catalytic properties in ways that are specific to the system being analyzed, i.e., in ways that cannot be predicted or corrected for by any simple procedure. Furthermore, the electrode might respond to any of the redox couples in the system. In such a case, the meaning of the measured value of EH is not obvious. This situation might be considered analogous to trying to analyze a system in which {Ac } = {HAc} and {OCl } = {HOCl}. The first equality indicates that the solution is pH 4.76, and the second that it is 7.53. We generally do not concern ourselves with such situations when analyzing acid/base systems, because such systems equilibrate rapidly; i.e., this hypothetical situation is never a realistic possibility. On the other hand, the coexistence in solution of several redox couples, all disequilibrated to different extents, is the norm. As a result, even if the electrode is operating in accordance with theory, the significance of the measured value is often unclear. The ideal situation would be to develop electrodes that respond only to a particular redox couple of interest, and then to use those electrodes to determine the EH or pe characterizing that reaction. Electrodes that approach such behavior have been developed for a number of redox couples. If the electrode is set up in such a way that one of the redox-active species is present at a fixed activity, the measured EH can be used to determine the activity of the other species of that couple. For instance, if the Pt (working) electrode in Figure 12.27 were sensitive to the EH of the H+ /H2 (g) redox couple, and if the sample were bubbled with H2 (g) at PH2 (g) = 1.0 atm, then the EH of the electrode would depend only on the H+ activity in solution. As a result, the setup could be used to determine solution pH. As with the SHE, the actual measurement of pH using such a setup is impractical; however, the same principle is applied in many commercial electrodes. Such electrodes are commonly referred to as ion-selective electrodes, although they are used for analyzing both ions and neutral species. In truth, the electrical potential developed in many modern ion-selective electrodes does not rely directly on a redox reaction. Rather, the working electrode consists of a thin glass membrane across which a potential develops if the solutions on the two sides of the membrane have different compositions. For instance, glass membranes can be prepared that develop a potential difference in response to different H+ activities on the two sides. If the H+ activity on one side is fixed (typically by placing the membrane at the tip of an electrode housing and filling the housing with 1.0 M acid), then the potential across the membrane reflects the H+ activity in the solution in which it is immersed. Thus, the 1 M acid serves the same role as bubbling the solution with hydrogen gas in the idealized (but impractical) pH electrode described above. By modifying the composition of the glass, its potential can be made to respond selectively to ions other than H+ , so that glass electrodes are useful as selective ion electrodes for a number of dissolved species. In other cases, metallic 755
electrodes that respond selectively to the activity of specific ions and that do rely on redox reactions to induce an electrical signal are used.
12.19
SUMMARY
This chapter describes the qualitative importance and quantitative interpretation of redox reactions – reactions in which electrons are transferred from one molecule to another. Because free, hydrated electrons are extremely unstable, all the electrons released by molecules undergoing oxidation are presumed to bind immediately to other molecules, which are thereby reduced. By convention, the electrons are assigned to specific atoms in each molecule. Even though free electrons are unstable, it is convenient to write reactions in which they are reactants or products in order to compare their relative affinity for different molecules. By convention, the equilibrium constant for a half-reaction in which one electron is released from a substance being oxidized is designated e . The value of e characterizes a redox couple in a manner that is analogous to how Ka characterizes an acid/base couple. Specifically, increasing e (corresponding to decreasing pe ) indicates increasing stability of the oxidized species relative to the reduced species, in much the same way that increasing Ka indicates increasing stability of the base relative to its conjugate acid. Because redox conjugate pairs often differ in their acidities, the exact pe value where dominance switches between the oxidized and reduced species can depend on pH as well as pe . The historical development of electrochemical concepts in different fields has led to disparate terminology to describe similar concepts. As a result, electron energy is expressed in terms as varied as the molar Gibbs energy of the electrons (kJ/mol), the pe of the system (dimensionless), and the EH of the system (volts). However, in all cases, the essential idea is that, given the opportunity, electrons transfer from higher- to lower-energy environments, and the energy they release in doing so can be captured and used productively. The equilibrium speciation of a redox couple under a specific set of conditions can be determined with the Nernst equation. Redox speciation can also be represented over a wide range of conditions on log C-pe diagrams, which serve essentially the same role as log C-pH diagrams do in acid/base systems. Because redox speciation often depends on pH as well as pe, another convenient way to display redox speciation is on a pe-pH predominance area diagram. LogC-pe diagrams provide information about the concentration of nondominant species that cannot be shown on a predominance area diagram, but they do not allow one to display speciation as a function of both pH and pe on a single plot, as is possible on a predominance area diagram. Because electrons are often exchanged in multiples of two, the pe range over which the oxidized and reduced species of a redox couple are both significant is often exceedingly small. This fact is reflected in the slopes of the lines representing 756
nondominant species on log C-pe diagrams, which are often steeper than those on log C-pH diagrams. Over the narrow regions of pe where the equilibrium redox speciation is changing dramatically, the solution pe is well buffered, just as solution pH is buffered in regions of the pH scale where the acid/base speciation is changing most rapidly. Although comparisons between acid/base and redox systems are extensive and useful, they must be tempered by the understanding that, whereas most acid/base reactions reach equilibrium rapidly, many redox reactions do not. The extremely high activation energies that impede the progress of redox reactions mean that although equilibrium calculations can correctly describe the thermodynamically stable state, the system might reside in a grossly different state for long periods. The most important example of a system that is thermodynamically unstable but kinetically stable is pure water in contact with the atmosphere. If H2 (aq) and O2 (aq) are equilibrated with the gas phase via Henry’s law, an enormous driving force favors their conversion to water molecules, but that reaction proceeds at a negligible rate under normal environmental conditions. Similarly, oxidation of organic matter by O2 is strongly favored thermodynamically, but that reaction does not proceed at a significant rate in the absence of catalysts such as enzymes. Because the Gibbs energy released when a reaction proceeds is directly related to the extent of disequilibrium, redox reactions tend to be more energetic than other environmentally significant reactions. The energy release accompanying these reactions can be thought of as reflecting the different energies of the electrons in the various species. By mediating redox reactions, catalyzing their progress, and capturing the energy released as they proceed, organisms obtain the energy needed to power virtually all metabolic activity. The ease with which electrons flow through electrically conductive materials like metallic wires provides a mechanism for converting the chemical energy associated with redox reactions to electrical energy, and vice versa. This property allows us to develop batteries that rely on redox reactions to drive electrical appliances, and to use electrical current to drive redox reactions in solution, potentially causing those reactions to proceed in the opposite direction from the one that they would follow without the electrical energy input. The linkage between redox reactions and electricity is also the basis for the operation of electrodes to measure redox potentials in solution.
12.20
PROBLEMS
Please re-read the note preceding the Chapter 8 Problems on p.457. The same guidelines apply to the following problems. 1. Determine the average oxidation state of C in each of the following compounds. (a) Methane (CH4 ) 757
(b) (c) (d) (e)
Formaldehyde (CH2 O) Carbon monoxide (CO) Lactic acid (CH3 CHOHCOOH) Glutamine (H2 NCOCH2 CH2 CHNH2 COOH)
2. Write the oxidation and reduction half-reactions and a balanced overall reaction for each the following processes. (a) Conversion of HOCl + HS to Cl + SO24 (b) Conversion of CH3 OH + O2 (aq) to (CO2 (g) + H2 O(l)) (c) Conversion of MnO4 + Mn2+ to MnO2 (s) 3. Determine the following values for a solution with pH = 7.0 and pe = 8.0. (a) the (Fe3+ )/(Fe2+ ) ratio (b) the (NO3 )/(NH+ 4 ) ratio (c) the concentration of Fe2+ , if the solution is in equilibrium with solid ferric hydroxide 4. What concentration of HOCl is required to oxidize 20 mg/L NH3 N to N2 ? 5. Chromium hydroxide [Cr(OH)3 (s)] is present in a contaminated groundwater with pH = 7.0 and pe = 4.0. Can the solution contain a significant concentration of Cr(VI) species? 6. Identify the electron donor (reductant) and the electron acceptor (oxidant), and calculate Dpe , DGr , and DEH for the following reactions. peo for conversion of I2 (aq) to I– is 10.471. 1 (a) Mn2+ + H2 O + O (aq) )* MnO2 (s) + 2 H+ 2 2 (b) Cl2 (aq) + 2 I– )* 2 Cl– + I2 (aq) 1 1 (c) Fe2+ + O (aq) + H+ )* Fe3+ + H2 O 4 2 2 7. Determine DEH for a reaction in which Ag+ is reduced to Ag(s) by oxidation of Br– to Br2 (aq) in an ideal solution at 25 C containing 10 4 M Ag+ , 10 3.7 M Br– , and 10 5 M Br2 (aq). Is the reaction thermodynamically favorable under these conditions? 8. In flue-gas desulfurization, SO2 (g) is captured in water, where it hydrolyzes to H2 SO3 , a relatively strong acid. The sulfite ion, SO2– 3 , can be oxidized to sulfate by reacting with dissolved oxygen. What is the equilibrium pe in an ideal solution at pH 9.0 and with a (SO24 )/(SO23 ) ratio of 9? What minimum O2 (aq) concentration is required to make the sulfite oxidation thermodynamically favorable under these conditions? 758
9. Prepare a log C-pe diagram showing the concentrations of SO2– 4 , S(s), H2 S(aq), 2+ + 4 Cu , and Cu , for a system at pH 8.0 that contains 10 M T OT S and 10 3 M T OT Cu. Show data for pe values from 15 to +15 and log C values from 20 to 3. 10. Determine the standard potential of the following nitrification reaction. Would the standard potential change if the reaction were written with NH+4 as the starting material, rather than NH3 (aq)? If so, how much? If not, why not? NH3 + 2 O2 (aq) )* NO–3 + H+ + H2 O 11. Determine the ratio of total dissolved arsenite [As(III)] to total dissolved arsenate [As(V)] at a pH of 6.0 and a pe of 5.0, considering all the acid/base reactions that those species undergo. 12. What is the theoretical oxygen demand of cresol, C7 H8 O, in mg O2 per mg C? 13. A water sample in which ethanol (C2 H5 OH) is thought to be the major oxidizable species is analyzed and determined to have a COD of 710 mg/L. (a) Write a balanced redox reaction for the oxidation of ethanol by H2 CrO4 in the test apparatus. (b) What is the ethanol concentration in the original solution, in mg/L? 14. A solution at pH 7.0 contains 10 5 M T OT Zn and 10 contact with both of the corresponding pure metals.
4M
T OT Fe(II) and is in
(a) Determine the equilibrium constant and the value of Dpeo for the following reaction: Zn2+ + Fe(s) )* Zn(s) + Fe2+ (b) Calculate the equilibrium pe in the solution. (c) What reaction, if any, would you expect to proceed in this system? 15. Permanganate (MnO–4 ) and chlorine (Cl2 ) are both strong oxidizing agents. (a) Write a balanced reaction for oxidation of Mn2+ by Cl2 , if the products are MnO–4 and Cl– . What is the equilibrium constant for this reaction? (b) Is permanganate a stronger or weaker oxidizing agent than chlorine under standard conditions? 16.
(a) Which of the six species in the following redox couples is the strongest oxidant, and which is the strongest reductant: Al3+ /Al(s), Cu2+ /Cu(s), and Pb4+ /Pb2+ ? 759
(b) If an ideal solution at redox equilibrium contained 10 4 M each of Al3+ , Cu2+ , Pb4+ , and Pb2+ , and it also contained 1 mg/L each of Al(s) and Cu(s), in which direction would the following reactions proceed? 2 Al3+ + 3 Cu(s) )* 2 Al(s) + 3 Cu2+ 2 Al3+ + 3 Pb2+ )* 2 Al(s) + 3 Pb4+ Pb4+ + Cu(s) )* Pb2+ + Cu2+ (c) What would be the equilibrium composition and pe of the solution in part (b)? (Hint: try estimating the solution composition, including the pe, by imagining that the system is made by adding all the inputs in their most oxidized form, followed by the available electrons. Assume that, when the electrons are added, they bind first to the oxidized species with the greatest affinity for them, and then to other species with progressively less affinity. Finally, test whether the system speciation estimated in this way is consistent with the Nernst equation.) 17. Lead (Pb) is a powerful neurotoxin. In the past, elemental lead was used extensively in water distribution systems as the primary component of pipes, in solder, and in alloys in metal valves. (a) The water adjacent to some pure Pb(s) in a distribution system is at pH 8.4 and contains 75 µg/L Pb2+ . Estimate the pe of the solution. (b) How much dissolved oxygen would be present in the solution described in part (a) at equilibrium? (c) If the solution in part (a) contained 2 mg/L O2 (aq), would oxidation of more Pb(s) or reduction of Pb2+ be favored? What would DGr be for the reaction that is favored (written with a stoichiometric coefficient of 1 for Pb)? 18. The pore water in some lake sediments has pH= 7.90 and EH = 250 mV, and the O2 (aq)/H2 O and S(VI)/S( II) redox couples have reached equilibrium in the water. (a) If the water contains 10 5.10 M SO2– 4 , what are the concentrations of the dissolved sulfide species? (b) What is the dissolved oxygen concentration (mol/L) in the solution? (c) What T OT Cu(II) would be present if the solution were in equilibrium with CuS(s)? (d) What would T OT Cu(I) be in the solution in part (c)? 19. Bacteria oxidize organic matter to gain energy to drive their metabolic reactions. What is the maximum amount of energy that the organisms can gain by consuming dissolved oxygen to oxidize 10 4 moles of T OT Ac in a pH 760
8.0 solution containing 10 2 M T OT Ac, if the solution is in equilibrium with atmospheric oxygen and carbon dioxide? What about at pH 6.0? 20. During turnover of a lake, reduced bottom water is mixed with oxidized surface water. Assume that bottom and surface waters with the following compositions mix in a 1:1 ratio. Bottom water: T OT Fe(II) = 1.5 ⇥ 10 (SO24 ) = 1.0 ⇥ 10
Surface water: (O2 (aq)) = 3 ⇥ 10
(NO3 ) = 1.2 ⇥ 10
3
4 4
3
M; T OT S( II) = 3 ⇥ 10
M; (SO24 ) = 1.3 ⇥ 10
3
4
M;
M;
M
(a) Prepare a log C-pe diagram for the mixed solution. Use Visual Minteq to simulate the speciation in the solution, and assume the pH and ionic strength are fixed at 7.50 and 0.005 M, respectively. Consider the following possible oxidation states: Fe(II) and Fe(III); S( II) and S(VI); and N( III), N(0), and N(V). Also, consider Fe(II) OH and Fe(III) OH complexes, but ignore all other complexes and any solids that might form. Your plot should cover the ranges 10 < pe < 15 and 2 > log C > 14. (Note: If Visual Minteq returns an error message when you scan across the whole pe range of interest, try running it twice — once from pe 0 to pe 15, and then from pe 0 to pe 10.) (b) Add lines to the diagram for O(0) and O( II) species. Note that, if O2 (aq) is chosen as a component, Visual Minteq automatically includes the O2 (aq)/H2 O redox couple in the calculations, so you need not (and should not) specify inclusion of that couple using the “Redox” pulldown menu. (c) Write the T OT e equation for the mixed solution, and find the solution composition at equilibrium. 21. A wastewater at pH 7.5 contains 45 mg/L SO4 S. (a) What fraction of the sulfate can be reduced to sulfide without causing the partial pressure of H2 S in equilibrium with the solution to exceed 10 6 atm? Assume the solution is well-buffered at pH 7.5. (b) If the H2 S partial pressure reached 10 6 atm and the solution contained 0.1 mg/L T OT Zn, would you expect wurtzite [ZnS(s)] to precipitate? 5 22. A lake water at pH 6.5 contains 1 ⇥ 10 3 M SO2– 4 and 10 M T OT [S( II)]. Assuming the solution is at equilibrium with respect to redox reactions, determine the pe of the water. What is the equilibrium value of PO2 ?
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23. Re-run the simulation in Example 12.12, but without the assumption that the pH remains constant. What are the equilibrium pe and pH of the water in this case? Are any solids supersaturated in the solution? 24. A well water at 25 C and pH 8.40 contains 4 mg/L T OT Fe(II) and is in equilibrium with siderite [FeCO3 (s)], with which it was in contact underground (no solids are in the water collected in the well). When the water is brought to the surface, it is aerated, thereby oxidizing essentially all of the Fe(II) to Fe(III) and causing precipitation of ferrihydrite [Fe(OH)3 (s)]. The aeration also equilibrates the solution with atmospheric CO2 . Assume that the solution behaves ideally. (a) What is the alkalinity of the untreated water? If you solve the problem manually, assume that the only significant dissolved Fe(II) species at pH 8.40 are Fe2+ and FeOH+ , and that the concentrations of HCO–3 and CO2– 3 are negligible at the alkalinity endpoint. (b) Write a balanced reaction for oxidation of Fe2+ by dissolved oxygen to form ferrihydrite, and determine the alkalinity of the treated water (i.e., after oxidation of the Fe(II) and equilibration of the solution with atmospheric CO2 ). Consider only the dissolved species, not the solids, as possible contributors to the alkalinity. Keep in mind that both the Fe2+ and the FeOH+ in the original solution are oxidized to Fe(OH)3 (s). (c) Assuming that the pH after the oxidation step is <5.0, approximately how much CO2 is exchanged between the solution and the gas bubbles during the aeration step, and in what direction is the exchange? Express your answer in moles of CO2 exchanged per liter of water. 25. A groundwater that is well buffered at pH 7.7 contains 5 ⇥ 10 3 M T OT CO3 , 3 ⇥ 10 6 M T OT As(III), 3 ⇥ 10 7 M T OT As(V), and no dissolved oxygen. In general, it is much easier to remove As(V) than As(III) from solution, so you wish to oxidize the As(III) in preparation for a subsequent treatment step to remove the arsenic from the water. You have chosen to explore whether that oxidation might be accomplished by addition of 1.6 ⇥ 10 6 M potassium permanganate (KMnO4 ), which you hope will undergo reactions like the following: + 2+ + 5 H3 AsO3 + 2 KMnO4 ! 5 HAsO2– 4 + 2 K + 2 Mn + 3 H2 O + 4 H
(a) Develop a log C-pe diagram showing the major species of all the redoxactive groups in the system at pH 7.7. Consider pe values from 10 to 15 and log C values from 10 to 4. (Note: To answer the subsequent parts of the question, it will be useful to re-draw the log C-pe diagram for smaller pe ranges, as needed.) 762
(b) What is the equilibrium pe of the system after the KMnO4 has been added? Assume that the O2 (aq)/H2 O redox couple is not active. If the system reaches equilibrium, will the treatment objective be achieved? (c) Repeat part (b) assuming that the O2 (aq)/H2 O couple participates in the redox reactions. Will the treatment objective be achieved in this case? How will the composition of the equilibrium solution differ from that in part (b)? 26. Mercuric salts were a common waste product from the manufacture of plutonium in nuclear weapons production facilities, with concentrations approaching 10 2 M in some solutions. Wastewater containing these mercuric salts was disposed of in unlined lagoons, and the water from these lagoons sometimes leaked into the local aquifer. Under reducing conditions in the aquifer, the pe of the water could decline, resulting in the reduction of the Hg(II) to dissolved or liquid elemental mercury [Hg(aq) or Hg(l), respectively]. Determine the equilibrium speciation of Hg in an aquifer at a pe of 8.0 and a pH of 7.0, if T OT Hgdiss is 10 4 M. Consider Hg(0) and Hg(II) species. Would precipitation of Hg(l) be thermodynamically favorable in this solution? GHg(aq) = 37.3 kJ/mol. 27. Reference electrodes are sometimes constructed based on the Cu(s)/CuSO4 redox couple by immersing a copper wire in a solution that contains crystals of CuSO4 · 5 H2 O (chalcanthite, with log Ks0 = 2.64). A porous frit or membrane allows dissolved species to exchange between the solutions inside and outside the electrode. As long as chalcanthite is present inside the electrode, the activities of the redox species (and therefore the redox potential) in the reference solution remain constant, which is a key requirement for an effective reference electrode. In the following calculations, assume T = 25 C and that the solutions are ideal. (a) What are the pH and the composition of a solution made by equilibrating chalcanthite with pure water, ignoring redox reactions? Use Visual Minteq, and allow the program to determine the ionic strength and activity coefficients. (You might need to indicate that the initial solution contains small, equal concentrations of T OT Cu(II) and T OT SO4 .) (b) What is the redox potential of the electrode solution, assuming that the Cu(s) equilibrates with Cu(II) species in solution, but that the redox reactions do not significantly change the total dissolved Cu(II) concentration? (c) Test the assumption in part (b) by computing the equilibrium concentration of T OT Cu(I) at the computed pe. Similarly, determine if a significant fraction of the T OT SO4 would be converted to sulfite (i.e., S(IV)) species.
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28. You wish to explore whether iron filings could be used to remove dissolved mercury from a pH 1.5 solution that contains 180 µg/L T OT Hg(II), 10 mg/L T OT Fe(II), and 10 3 M T OT Cl( I). In theory, the added iron could react with either the Hg(II), reducing it to Hg(l), or with H+ , reducing it to H2 (aq), which could then migrate out of solution. (a) What are the dominant Fe(II) and Hg(II) species in the initial solution? (b) Compare the Gibbs energies of reaction for reactions of the filings (i.e., metallic iron) with each of the possible oxidants (H+ and Hg2+ ) at 25 C, assuming that H2 (aq) is in equilibrium with the atmosphere. Repeat the calculation for reaction of the filings with the form of Hg(II) that you determined to be dominant in part (a). Give all answers in kJ per mole of Fe. Which reaction is most favorable from a thermodynamic perspective? 29. A soil solution contains 10 5 M T OT Mn(II), 10 2 M T OT CO3 , and 4 mg/L dissolved O2 , and is at pH 7.5. The dissolved manganese is present primarily as either free Mn2+ ions or as MnCO3 (aq) complexes (K1 = 104.70 ). (a) Determine whether the solid rhodocrocite [MnCO3 (s)] is undersaturated, saturated, or supersaturated in the solution. (b) The Visual Minteq database includes the reaction shown below for formation of the mineral manganite. What are the chemical formula for manganite and the oxidation state of Mn in that solid? Mn2+ + 2 H2 O )* Manganite + 3 H+ + e– (c) Will either manganite or rhodocrocite be present if the soil solution reaches redox equilibrium? If so, how much and which solid(s) will be present? If not, how much Mn2+ would have to be present to induce precipitation of one of the solids? In either case, assume that the pH is well buffered. 30. A metal undergoes the following two redox half-reactions: Me3+ + e– )* Me2+ EHo = 0.53 V Me3+ + 3 e– )* Me(s)
EHo = 0.87 V
What potential would be measured if a cell with an SHE were connected to a cell containing an electrode made of Me(s) in a solution of 0.1 M Me2+ at 25 C? Assume ideal solution behavior and that the ions can migrate between the cells through an ion-permeable membrane. Which electrode is the cathode and which is the anode? 31. Arsenic(V) can be removed from drinking water readily by adsorption to hydrous ferric oxide [FeOOH(s)], but As(III) is much more difficult to remove. One treatment approach that has been explored is to pass the water through a bed containing pyrolusite [MnO2 (s)] pellets prior to contacting the water with 764
FeOOH(s), in the hope that the MnO2 (s) will oxidize the As(III) to As(V). The following reaction is included in the Visual Minteq database for formation of pyrolusite: Mn2+ + 2 H2 O )* Pyrolusite + 4 H+ + 2 e–
log K = 41.38
(Remember that Visual Minteq shows the stoichiometry for formation of solids, but the log K for their dissolution. The sign of log K given here has been reversed from that in the database, so that it applies to the reaction as shown.) Consider a solution that is well buffered at pH 8 and that initially contains 10 µg/L T OT Mn(II), 25 µg/L T OT As(III) and 10 µg/L T OT As(V). The solution is then dosed with 0.1 mg/L pyrolusite. Assume that all the dissolved Mn(II) is present as Mn2+ . (a) Write a balanced reaction for oxidation of As(III) to As(V) by reduction of pyrolusite to Mn(II). Use pyrolusite as one reactant, and the dominant dissolved species in each oxidation state for the other reactants and products. What pe would be required for each half-reaction to be at equilibrium? Do you expect the desired reaction to proceed? (b) Compute the composition of the system once it reaches equilibrium, assuming that water is not redox-active. 32. Prove using the Gibbs phase rule that a solution cannot be in redox equilibrium and also be in gas/liquid equilibrium with the atmospheric partial pressures of O2 and H2 . Consider a system in which the total pressure and temperature are fixed at 1 atm and 25 C, respectively. 33. In systems where methane is being generated by fermentation of organic matter, the overall process is typically modeled as a two-stage sequence that generates acetate and molecular hydrogen in the first stage and then consumes them in the second (via two separate reactions). CO2 (g) is also generated as part of the process, and the various gases bubble out of the solution. For example, the reaction sequence for fermentation of butyrate (C3 H7 COO– , pKa = 4.82) is shown below. 1st stage: C3 H7 COO– + 2 H2 O )* 2 CH3 COO– + 2 H2 (g) + H+ 2nd stage:
CH3 COO– + H+ )* CH4 (g) + CO2 (g) CO2 (g) + 4 H2 (g) )* CH4 (g) + 2 H2 O
(a) If the hydrogen partial pressure in the system becomes too large, the first reaction is impeded, and the process can fail. For a system at pH 6.8 that contains 2.0 ⇥ 10 3 M T OT But and 1.0 ⇥ 10 2 M T OT Ac, what is the maximum partial pressure of H2 (g) that will allow the first reaction 765
to proceed? The standard molar Gibbs energy of formation of butyrate is 350.4 kJ/mol; all other values needed to solve the problem can be obtained from the text. (b) If the process generates gas bubbles that contain 45% CO2 (g) and 55% CH4 (g), what is the minimum partial pressure of H2 (g) that is needed for the second reaction in the second stage to proceed? (c) What is the alkalinity of the untreated solution in mg/L as CaCO3 ? How does the overall reaction affect the alkalinity of the solution (increase, decrease, or no change), assuming that the process is working properly so that virtually all of the acetate and H2 (g) being generated by the first reaction are consumed at the same rate by the latter reactions? 34. Plot a log C-pe diagram for bromine species in a solution containing 0.25 mg/L T OT Br at pH 7.5. Consider Br– , HOBr, OBr– , and BrO–3 . What would the equilibrium speciation of Br be in a solution initially containing 0.25 mg/L Br– and 25 mg/L Cl– , if HOCl were added at a dose of 2 mg/L as Cl2 . 35. The following graph indicates the speciation of the As(V)/As(III) and the Fe(III)/Fe(II) redox couples at pH 7, for a system containing 2 ⇥ 10 4 M T OT Fe and 1 ⇥ 10 4 M T OT As, and in which no solids precipitate. To minimize clutter, only the two most dominant species of each element in each oxidation state are shown. All the lines continue without changing slope as they extend beyond the pe region shown.
(a) Identify the strongest oxidizing agent and strongest reducing agent among the species shown.
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(b) A solution containing 2 ⇥ 10 4 M FeCl2 , 5 ⇥ 10 5 M Na3 AsO3 , and 5 ⇥ 10 5 M Na3 AsO4 is well-buffered at pH 7. What will the equilibrium pe and solution composition be? Assume that redox reactions of water are kinetically inhibited. 36. An aluminum pot is filled with tap water at pH 7.0 that contains 25 mg/L Cl– and 1 mg/L T OT OCl as Cl2 . What is T OT Aldiss in the solution after it equilibrates? 37. A solution that is buffered at pH 7.0 contains 10 mg/L T OT Cr(VI). Ferrous sulfate (FeSO4 ) is added to the solution to provide 10 mg/L of T OT Fe(II). Write a redox reaction between Fe2+ and HCrO–4 that generates Fe(OH)3 (s) and Cr(OH)3 (s) as products. (a) After 1 mg/L of Fe(II) has reacted, both solids are present in the solution. What is the molar Gibbs energy of reaction for the redox reaction under these conditions? (b) Compute EH of each half-cell reaction, and DEH of the overall redox reaction under the conditions specified in part (a). (c) What is the concentration of T OT Fe(II) at equilibrium? (d) Repeat part (b) for the equilibrated solution. 38. Use Visual Minteq to find the equilibrium composition of a solution made by adding 1.0 g nantokite [CuCl(s)] to 1.0 L of pure water. Consider the possibility that the nantokite might partially or completely dissolve, and also that it might disproportionate to form Cu(II) and Cu(0). Allow the program to determine the equilibrium pe, pH, and ionic strength. (Hint: if the program returns an error message, try including a very small, but nonzero, input concentration of Cu2+ [e.g., 10 12 M].) 39. Hexavalent chromium (Cr(VI)) can exist as either a monomer (Hx CrOx–2 4 ) or 2– a dimer (Cr2 O2– ). H CrO is chromic acid, and Cr O is dichromate ion. 7 4 2 7 2 Cr(VI) is used extensively in several industries, including metal processing and leather tanning; it also used to be used as an oxidant to destroy organic matter in a number of applications (e.g., COD tests, cleaning laboratory glassware), although those uses are diminishing. Because Cr(VI) is a carcinogen, its discharge into receiving waters and sewage systems is tightly regulated, and efforts are being made to reduce or eliminate its use in many of these applications. In situations where Cr(VI) is still used, the most common approach for removing it from wastewater is to reduce it to Cr(III), a much less toxic and less soluble form of Cr. The reduction is sometimes accomplished by addition of Fe(II) to the water. 767
(a) Prepare a log C-pe diagram for a pH 5 system containing 1.1 ⇥ 10 T OT Cr and 3 ⇥ 10 3 M T OT Fe.
3M
(b) Write the T OT e equation for a system made by adding 3 ⇥ 10 3 M FeCl2 to a pH 5.0 solution that contains 1.1 ⇥ 10 3 M Cr(VI) and no Cr(III). Assuming the system is well-buffered at this pH, what are the approximate values of pe and T OT Cr(VI) at equilibrium? 40. Recall from Chapter 11 that chelating agents are sometimes added to solutions to prevent the precipitation of metals. Although such chelating agents might be advantageous in a production process, their presence in the waste stream produced by such processes is often problematic. You have been asked to recommend a treatment process to remove copper from a solution at pH 1.0 that contains 5 ⇥ 10 3 M each of Cu2+ and T OT EDTA. Determine the minimum concentration of T OT Cudiss that could be attained by increasing the solution pH to any desired value between 1 and 13 by precipitation of Cu(OH)2 (s). Compare that concentration with the concentration that could be achieved by adding metallic aluminum to the solution to reduce the Cu2+ and precipitate it as Cu(s), if the solution remains at pH 1.0. Consider the acid/base chemistry of Cu2+ , Al3+ , and EDTA in the first part of the problem, and the Cu(II)/Cu(0) and Al(III)/Al(0) redox couples in the second part. (Note: this question is based on a waste treatment process developed and used in the 1990s by the Boeing Co., in which scrap aluminum was added to a waste solution to precipitate strongly chelated Cu(II). The process worked well, but had to be abandoned when it was determined that, under some circumstances, HNO3 in the waste was being reduced to the potentially toxic gases NO(g) and NO2 (g).)
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13
ADSORPTION REACTIONS
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Contents 13.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 TWO VIEWS OF THE INTERFACE AND ADSORPTION EQUILIBRIUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 QUANTITATIVE REPRESENTATIONS OF ADSORPTION EQUILIBRIUM: THE ADSORPTION ISOTHERM . . . . . . . . . . . . 13.4 MODELING ADSORPTION WITH VISUAL MINTEQ . . . . . . 13.5 ADSORPTION OF IONS IN THE PRESENCE OF A SURFACE ELECTRICAL POTENTIAL . . . . . . . . . . . . . . . . . . . . . 13.6 MODELING SURFACE-ADSORBATE ELECTRICAL INTERACTIONS IN VISUAL MINTEQ . . . . . . . . . . . . . . 13.7 SURFACE PRECIPITATION . . . . . . . . . . . . . . . . . . . . . 13.8 SUMMARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9 PROBLEMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.1
INTRODUCTION
13.1.1
Chapter Overview
770 774 778 797 812 827 835 840 841
Adsorption – the accumulation of molecules at the interface between two phases – plays a critical role in the transport, bioavailability, and fate of contaminants as well as many naturally occurring trace compounds in aquatic systems. Adsorption can, for example, cause contaminants to attach to the surfaces of minerals in a soil column, to microorganisms in wastewater treatment plants, or to granules of activated carbon in a home water purifier, thereby preventing the contaminants from passing through the system with the water. In addition to altering the fates of solutes, adsorption reactions can affect the properties and behavior of the solids to which the solutes bind, including their surface charge, hydrophobicity, and tendency to collide with other particles and grow into larger aggregates. Adsorption also plays an important role in many phase transfer reactions, such as transfer of oxygen and carbon dioxide across the 770
air/water interface, and solid dissolution and precipitation reactions. Finally, many oxidation/reduction reactions take place on surfaces, in which case adsorption of the reactants is a necessary preliminary step.
Many contaminants in aquatic systems are adsorbed on the surfaces of particles suspended in the water. This photograph shows the formation of the Ohio River by the confluence of the relatively clear Allegheny River and the much more turbid Monongahela River in Pittsburgh, PA. The fate of the particles after the waters mix will dictate the bioavailability and ecological impacts of any contaminants associated with them. (Mark E. Gibson, Visuals Unlimited.)
In engineered water treatment systems, adsorption often provides the most costeffective means of removing contaminants from solution, especially when the target concentration is extremely low. For example, many trace organic contaminants can be removed by adsorption onto activated carbon, a manufactured solid with an extremely large specific surface area (surface area per unit mass of solid). Ion exchange, in which certain types of ions are removed from solution by adsorption and are replaced by others that the solid releases, is used in applications ranging from home water treatment systems (e.g., for water softening) to large industrial processes for producing ultrapure process water or recovering metals from waste streams. In other water treatment applications, iron or aluminum ions are added to the water and precipitated as hydroxides, thereby generating surfaces to which other metals or natural organic matter bind; the solids and associated contaminants are then settled or filtered out of the water before it enters the distribution system. 771
Figure 13.1 provides a dramatic example of the effect that adsorption can have on the behavior of contaminants. This figure shows the fraction of the total metal adsorbed onto the surface of ferrihydrite, an iron oxide solid that is present in soils and the sediments of many water bodies, and that is also generated when ferric chloride (FeCl3 ) is added to water in treatment plants.
Figure 13.1 Commonly observed trends for the adsorption onto ferrihydrite (10 3 M T OT Fe) as a function of solution pH. (a) adsorption of cations; (b) adsorption of anions.
For both cations and anions, the solute shifts from being almost completely bound to the solid to almost completely in solution over a pH range of one to two units, but the shift is in opposite directions for the two groups: Whereas adsorption of cations increases with increasing pH, adsorption of anions decreases. This dramatic shift is commonly referred to as a pH-adsorption edge. The location of the edge depends on the identities and concentrations of both the solid and solute. The typical sequence of increasing binding strength for cationic metals (corresponding to a steady shift to lower pH for the adsorption edge) is:
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Ca2+ < (Cd2+ , Ni2+ ) < (Zn2+ , Co2+ , Cu+ ) < Cu2+ < Pb2+ < (Cr3+ , Fe3+ , Hg2+ ) In this chapter, after a brief introduction to adsorption terminology, we explore the equilibrium relationship between dissolved and adsorbed solutes and learn how chemical equilibrium software like Visual Minteq can be used to model adsorption reactions.1
13.1.2
Terminology
As has been noted, adsorption is the accumulation of a chemical species at an interface. Although adsorption can occur at any interface, we restrict our consideration here to the boundary between an aqueous solution and a particle. In such cases, adsorption might be thought of as resulting from a reaction in which a solute binds to the surface of the solid, but does not become incorporated into the solid structure. As a result, adsorbed molecules can be considered to be “half in solution and half out.” The reverse of the adsorption reaction (i.e., the release of an adsorbed substance to the bulk solution) is referred to as desorption. The substance that adsorbs is called the adsorbate, and the solid to which it binds is called the adsorbent. Adsorption can be distinguished conceptually from absorption, which is the accumulation of a substance in the interior of a different phase. However, in practice, distinguishing between adsorption and absorption of a solute is difficult, and the two processes are frequently treated jointly; in such cases, the preceding terms are sometimes written as sorption, sorbate, and sorbent, respectively. The amount of material sorbed per unit amount of adsorbent is called the adsorption density and is usually represented in equations by the letter q or the Greek letter G. The adsorption density can be quantified as the adsorbed mass per unit surface area (with units such as mg/m2 ) or per unit mass of adsorbent (e.g., mg adsorbate/g adsorbent). The amount adsorbed can be normalized to the volume of solution (e.g., mg adsorbed per liter) by multiplying the adsorption density by the concentration of surface area or adsorbent in the system (m2 /L or g/L, respectively). Adsorption reactions, like other reactions, tend to approach equilibrium. Once equilibrium is attained, the adsorption and desorption reactions proceed at equal rates, and the system composition remains stable unless it is perturbed by some other process. The equilibrium relationship between the adsorption density and the dissolved adsorbate concentration at a given temperature is called an adsorption isotherm. The surfaces of suspended solids can sometimes acquire an electrical charge, much as dissolved species do, and this surface charge can enhance or impede adsorp1 Much of the material in this chapter is based on Chapters 7 and 8 of Water Quality Engineering: Physical/Chemical Processes, by Benjamin, M.M. and Lawler, D.F., J. Wiley & Sons, Hoboken, NJ (2013).
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tion of ions from solution. If the tendency for a species to adsorb depends strongly on its identity and not simply on the surface charge, the primary driving force for the reaction is attributed to specific chemical interactions between the adsorbate and the surface, and the adsorbate is referred to as being specifically or chemically adsorbed (or chemisorbed). Chemically adsorbing molecules can bind to a surface even under conditions where electrostatic interactions oppose adsorption (much as Cl– might bind to the negatively charged species CuCl–3 to form CuCl2– 4 ). By contrast, physically adsorbed or nonspecifically adsorbed species are attracted to the surface primarily or exclusively via electrostatic interactions.
13.2
TWO VIEWS OF THE INTERFACE AND ADSORPTION EQUILIBRIUM
The fact that adsorbed species are in contact with both the bulk solution and the adsorbent solid, but are not fully in either phase, has led to the development of two broad groups of models to describe adsorption reactions and the corresponding equilibrium relationships. One group of models relies on an analogy between adsorbed molecules and those that are truly dissolved, while the other group focuses more on the similarities between adsorbed molecules and molecules that have been completely removed from solution and have formed a separate phase. These two groups of models differ in how they define the concentration, activity, and activity coefficient of adsorbed species.
13.2.1
Adsorption as a Surface Complexation Reaction
In models that emphasize the similarity between adsorbed and dissolved species, binding of the adsorbate to the adsorbent is viewed as similar to the binding of a proton to a base, a metal to a ligand, or an electron to an oxidant. The metal/ligand analogy is the one that is most often made, with the adsorbent sometimes described as a collection of surface ligands and the adsorption reaction referred to as a surface complexation reaction. This conceptual view is shown schematically in Figure 13.2. The surface ligands are presumed to correspond to specific sites on the adsorbent surface, so the model is sometimes referred to as a site-binding model. In some of the more complex versions of such models, different adsorbed molecules are envisioned to reside at different locations in the interfacial zone (i.e., different distances from the surface). An extreme example of an adsorbent that fits this model is an ion exchange resin. Such resins typically consist of a crosslinked, polymeric skeleton containing a high concentration of identifiable, charged sites. When the resin is exposed to an aqueous solution, adsorbable ions can bind to these sites. If the solution composition later changes, these adsorbed ions can be replaced by other ions from solution, hence the name ion exchange. 774
Figure 13.2 Schematic representation of an adsorptive solid/solution interface according to the site-binding model. The adsorbed molecules are located in the interfacial plane. The asterisks represent binding sites on the surface.
To model adsorbents that have binding sites in fixed locations on the surface, the number of sites per unit mass of solid is estimated and multiplied by the concentration of solid in the system, yielding the concentration of potential adsorption sites in units comparable to those of dissolved species (e.g., moles of sites per liter of solution). We represent this quantity as T OT ⌘S, where S designates a surface site and the symbol ‘⌘’ designates the bond to the underlying solid. In an aqueous system, some of these sites are occupied by water (⌘SH2 O), and others by various adsorbates (⌘SA, ⌘SB, etc.). Because the amount of water adsorbed is never of interest in aquatic systems, sites occupied by water are often referred to as unoccupied sites and represented as ⌘S. A surface complexation (i.e., adsorption) reaction and the corresponding equilibrium constant can be written for occupation of surface sites by each adsorbate in the system, e.g., Soluble A + Hydrated surface site )* Adsorbed A + water A(aq) + ⌘SH2 O )* ⌘SA + H2 O Kads =
{⌘SA}{H2 O} {⌘SH2 O}{A} 775
(13.1) (13.2)
Although Kads has the form of an equilibrium constant and is widely referred to as such, the values used to compute it are frequently concentrations, rather than activities. This approach is especially common in empirical studies, where the key results are the effectiveness of an adsorbent at removing a problematic adsorbate from solution. In such cases, Kads is reported as a dimensional quantity, with units that depend on those that are used to quantify the concentrations of adsorbate and adsorbent. On the other hand, in studies where the goal includes the testing of fundamental models for adsorption processes, Kads is treated as a true equilibrium constant, in which case we need to establish conventions for quantifying the activity of surface species; i.e., we need to choose a standard state for these species. Because the surface complexation model is based on an analogy between adsorbed species and those in true solution, a logical choice is to define the standard states in the same way for the two groups of species. We therefore choose the standard concentration for adsorbed species to be 1.0 mol/L and the reference state to be infinite dilution. As explained in Chapter 2, if the infinite dilution reference state is adopted, the activity coefficient of a species A equals 1.0 if each molecule of A has the same reactivity as it would if it interacted only with water molecules. Because adsorbed
Ion exchange resins are sometimes used to adsorb contaminants in water treatment systems. The reaction in which the adsorbate binds to a site on the resin is akin to that by which metal ions bind to ligands in solution. This figure shows a column packed with ion exchange resin on the left, and magnified images of resin beads on the right (Courtesy of Arup Sengupta and Jinze Li.)
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species reside partially in solution and partially on the solid, this criterion is expanded to specify that infinite dilution applies to the solid surface as well as the solution. That is, in the infinite dilution reference state for adsorbed species, each adsorbed molecule is as reactive as it would be if all the other molecules in the solution and all the other adsorbed molecules were H2 O. The major factor controlling activity coefficients of dissolved species is electrostatic interactions with nearby molecules, and the same is true of adsorbed species. However, the electrical environment that adsorbed species experience depends on the properties and composition of both the solution and the solid surface. An approach for estimating the effects of both of these regions on adsorbate reactivity is presented later in the chapter. Except in that discussion, we will assume that adsorbed species behave ideally (g⌘SA = 1.0).
13.2.2
Adsorption as a Phase Transfer Reaction
The alternative to the surface complexation model for adsorption treats the interfacial region as a different phase from the solution, and considers adsorbed species to have left the aqueous phase. In this model, the adsorbed molecules are thought of not as bonding to specific surface sites, but rather as moving around the surface freely in much the same way that molecules move about in a gas phase. Correspondingly, even if the adsorption density is small, adsorbate molecules are considered to occupy the entire surface phase. This model is depicted in Figure 13.3.
Figure 13.3 Schematic representation of adsorbed species according to the phase transfer model. Adsorbed molecules are envisioned to move freely in the twodimensional interphase region near the surface.
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Quantitative modeling using this paradigm is based on the phase transfer reaction A(aq) )* ⌘ A(interphase) Kp =
{⌘ A(interphase)} {A(aq)}
(13.3) (13.4)
for which the equilibrium constant Kp is sometimes referred to as a partition coefficient. As is apparent from Equation (13.4), Kp bears greater similarity to a Henry’s law constant than to a metal-ligand complexation constant. As a result, the activity of adsorbed A in this model is sometimes quantified by reference to a so-called surface pressure. Briefly, the idea is that if one could encircle some of the adsorbed molecules, their motion would exert a force along the perimeter of the circle, and that force could be quantified as the surface pressure, in units of force per length. Logically enough, the surface pressure increases with increasing adsorption density, just as the pressure in a three-dimensional gas phase increases when the concentration of molecules in a fixed volume of the gas increases.
13.2.3
Which Model and Which Equilibrium Expression are Best?
Both the surface complexation and gas transfer models represent idealized cases; the real situation is not likely to conform to either of these models completely or to be identical in all adsorptive systems. Furthermore, each model has both a great deal of flexibility and some unavoidable limitations, so that neither can be declared universally better than the other. However, strong preferences have developed for modeling certain adsorptive systems by one model or another. When hydrophilic adsorbates such as inorganic ions bind to mineral adsorbents, strong bonds are envisioned to form at specific surface sites, so the site-binding model is preferred. Conversely, sorption of hydrophobic adsorbates to nonmineral adsorbents (e.g., synthetic organic compounds like solutes or pesticides binding to the surfaces of organic solids or activated carbon) is thought to conform more closely to the gas transfer model for adsorption. Because the focus of this text is primarily on hydrophilic solutes, and because the surface complexation model is more easily integrated into chemical equilibrium programs like Visual Minteq, that model is emphasized here. Keep in mind, though, that the gas transfer model is also widely used in environmental engineering; the mathematical formulation of that model is elaborated in a number of publications.2 2 See,
for example, Chapter 7 in Water Quality Engineering: Physical/Chemical Treatment Processes, by Benjamin, M.M., and Lawler, D.F. (Wiley, Hoboken, NJ, 2013).
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13.3
QUANTITATIVE REPRESENTATIONS OF ADSORPTION EQUILIBRIUM: THE ADSORPTION ISOTHERM
13.3.1
Approaches for Developing Isotherm Equations
As noted previously, any equation that relates the adsorption density to the concentration of dissolved adsorbate in a system that has reached equilibrium is called an adsorption isotherm. Adsorption isotherms are typically derived empirically by gathering data for the adsorption density q as a function of the dissolved concentration of the adsorbate and then attempting to fit the data to simple equations. Alternatively, isotherm equations can be derived from theory, based on models for the interactions among the surface, dissolved adsorbate, and adsorbed molecules. Once the isotherm has been developed, it can be used to predict the distribution of adsorbate between the surface and solution for other conditions that have not been studied experimentally. For instance, assuming the system reaches equilibrium, the adsorbent dose csolid needed to reduce the dissolved adsorbate concentration from some initial value cinit to a lower target value cfin can be computed as follows: q=
cinit cfin csolid
csolid =
cinit
(13.5a) cfin
q
(13.5b)
Equation (13.5a) is simply a restatement of the definition of the adsorption density, where the numerator represents the concentration of adsorbed adsorbate (computed as the difference between the initial and final concentrations in solution), and the denominator is the concentration of adsorbent. This equation is then manipulated in Equation (13.5b) to find the adsorbent dose required to achieve the desired change in dissolved adsorbate concentration. Example 13.1 Graphical representations of adsorption isotherms for binding of CrO2– 4 onto three different minerals are shown in Figure 13.4. You wish to reduce the concentration of CrO2– 4 in a wastewater from 0.2 to 0.02 mmol/L (roughly 10 to 1 mg Cr/L) by sorption, using the smallest possible dose of adsorbent. (a) Which of the three adsorbents should be used? (b) What adsorbent dose (g/L) is required? 779
Figure 13.4 Adsorption isotherms for binding of CrO2– 4 onto three different adsorbents
Solution (a) When the solution and solid have equilibrated, ceq will be 0.02 mmol/L, so solid C, which has the largest adsorption density in equilibrium with this concentration, is preferred. (b) The adsorption density on adsorbent C when it is in equilibrium with a dissolved concentration of 0.02 mmol/L is approximately 0.65 mmol CrO24 /g solid. Therefore, applying Equation (13.5b), we can find the required dose of adsorbent as csolid =
13.3.2
(0.20 0.02) mmol CrO24 /L = 0.277 g solid/L 0.65 mmol CrO24 /g solid
Surfaces with Uniform Sites: The Langmuir Isotherm
As has been noted, this chapter focuses on use of the surface complexation model to characterize adsorption. This model can be used to derive several commonly used isotherm equations, each of which is based on a different assumed distribution of surface site properties. We next consider surfaces with three such distributions: (1) uniform sites at all locations; (2) a mixture of two or three types of sites, each with its own binding constant for adsorbates; and (3) a mixture of sites with a wide range of binding constants, but with a specific relationship between the binding constant for a particular type of site and the surface concentration of those sites. A more general discussion that includes consideration of any possible mixture of surface sites is provided by Benjamin and Lawler (see footnote 1). 780
Systems containing only one adsorbate. If all the sites on an adsorptive surface are identical, the same value of Kads applies regardless of which site an adsorbate binds to. In that case, adsorption of A is fully characterized by Equation (13.2). Noting that the activity of water is 1.0, and assuming that the other reactants behave ideally, we can rearrange that equation to obtain Kads (⌘S)(A) = (⌘SA)(1.0)
(13.6)
where ⌘S represents sites that are unoccupied (i.e., occupied by water molecules), and (⌘S) and (⌘SA) represent the concentrations of the designated species in moles per liter. If A is the only adsorbate in the system, we can also write a mole balance on sites as follows, based on the idea that all the surface sites are either unoccupied or occupied by A: [T OT ⌘S] = (⌘S) + (⌘SA) (13.7) Using this expression to substitute for (⌘S) in Equation (13.6), we obtain Kads {[T OT ⌘S]
(⌘SA)}(A) = (⌘SA)
Kads [T OT ⌘S](A) = (⌘SA)[1 + Kads (A)] (⌘SA) =
Kads (A) [T OT ⌘S] 1 + Kads (A)
(13.8)
Dividing both sides of Equation (13.8) by the adsorbent concentration in g/L, the left side becomes the adsorption density qA (moles adsorbed per gram of adsorbent). The ratio of T OT ⌘S to the adsorbent concentration is the maximum possible adsorption density of A, commonly represented as qmax , so Equation (13.8) can be written in the form of an adsorption isotherm as qA =
Kads (A) qmax 1 + Kads (A)
(13.9)
Equation (13.9) is called the Langmuir isotherm. The isotherm incorporates two constants — one (qmax ) establishing the maximum adsorption density, and the other (Kads ) establishing the affinity of the adsorbent for the adsorbate. The isotherm is plotted in Figure 13.5. Under conditions where Kads (A) ⌧ 1, the denominator in Equation (13.9) is approximately equal to 1, so the adsorption density is directly proportional to the dissolved adsorbate concentration: qA ⇡ Kads qmax (A). On the other hand, if Kads (A) 1, the denominator is approximately equal to Kads (A), and qA ⇡ qmax . Other factors being equal, an increase in the adsorption equilibrium constant increases the slope of the isotherm curve at low concentrations of dissolved adsorbate, and an increase in 781
Figure 13.5 Graphical representation of the Langmuir isotherm. For the isotherm shown, Kads = 200 L/mol and qmax = 10 4 mol/g.
qmax increases both the slope of the isotherm at low (A) and the adsorption density at high (A). To test whether experimental adsorption data fit the Langmuir isotherm, the data are often manipulated so that they can be plotted according to either of the following linearized versions of Equation (13.9): 1 1 1 1 = + qA qmax Kads (A) qmax
(13.10)
(A) 1 1 = (A) + qA qmax qmax Kads
(13.11)
Based on the preceding equations, if a plot of 1/qA versus 1/(A) or of (A)/qA versus (A) is linear, the data are consistent with the Langmuir isotherm. However, a plot based on Equation (13.10) tends to compress the data at relatively high values of (A) into a small region near the y axis, while one based on Equation (13.11) does the same to data at relatively low values of (A). Under the circumstances, it is preferable to use nonlinear regression, rather than either of the linearized equations, to test data for their fit to the unmodified isotherm equation [Equation (13.9)] and to evaluate the isotherm parameters. The fractional occupation of surface sites is often of interest in adsorptive systems. This parameter can be evaluated for systems that obey the Langmuir isotherm by dividing both sides of Equation (13.9) by qmax , yielding Fraction of surface sites occupied = 782
qA Kads (A) = qmax 1 + Kads (A)
(13.12)
This result is compared with the fraction combined for various other types of reactions in Table 13.1. The only difference among the equations characterizing acid/base, metal/ligand, redox, and adsorptive systems relates to conventions for writing the reactions: Ka and e are defined based on dissociation reactions, whereas K1 and Kads are defined based on association reactions. Comparisons based on other indicators of speciation would also exhibit strong similarities. Table 13.1 Fractional occupation of binding sites in acid/base, metal/ligand, redox, and adsorption reactions
Fraction of weak base sites protonated as a function of {H+ }
K
a HA )* H+ + A–
K
Fraction of ligand sites occupied as a function of {Men+ }
1 Me + L )* MeL
Fraction of oxidant sites occupied as a function of {e– }
* Ox + e– Red )
Fraction of surface sites occupied as a function of {A}
e
K
ads ⌘S + A )* ⌘SA
{HA} Ka 1 {H+ } = T OT A 1 + Ka 1 {H+ } {MeL} K1 {L} = T OT L 1 + K1 {L}
1 {e } {Red} = e 1 T OT X 1 + e {e }
{⌘SA} Kads {A} = [T OT ⌘S] 1 + Kads {A}
Thus, at least for adsorptive systems that follow the Langmuir isotherm, treating surface sites as ordinary, dissolved ligands allows adsorption to fit comfortably into our existing paradigm for analyzing speciation in any solution. The equations and some of the terminology used to describe adsorption are a bit different from the corresponding equations and terminology that describe soluble equilibria, but this is more a reflection of historical differences in the way the systems were studied than of any fundamental differences among them. Competitive Langmuir adsorption. The scenario described in the preceding section — a system containing a single adsorbate, and in which all the sites on the adsorbent are identical — represents the simplest of all possible cases of interest. The first modification to that scenario that we will consider is one in which the solution contains one adsorbent and two potential adsorbates, each of which follows a Langmuir isotherm when it is the only adsorbate in solution, but with the adsorbates having different affinities for the surface sites. For this scenario, an expression like Equation (13.6) applies to each adsorbate. Writing these expressions and then taking their ratio, we obtain (⌘SA) =
Kads, A (⌘S)(A) (H2 O)
(13.13)
(⌘SB) =
Kads, B (⌘S)(B) (H2 O)
(13.14)
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(⌘SB) Kads, B (B) = (⌘SA) Kads, A (A) (⌘SB) =
(13.15)
Kads, B (B) (⌘SA) Kads, A (A)
(13.16)
The molar site balance for this scenario is [T OT ⌘S] = (⌘S) + (⌘SA) + (⌘SB)
(13.17)
Rearranging this equation and substituting for (⌘SB) from Equation (13.16), we obtain (⌘S) = [T OT ⌘S]
(⌘SA)
= [T OT ⌘S]
(⌘SA)
(⌘SB) Kads, B (B) (⌘SA) Kads, A (A)
(13.18)
Then, substituting this result into Equation (13.13) and setting the activity of H2 O to 1.0 yields
(⌘SA) =
n Kads, A [T OT ⌘S]
(⌘SA) = Kads, A (A)[T OT ⌘S]
(⌘SA)
o
Kads, B (B) Kads, A (A) (⌘SA)
1.0 Kads, A (A)(⌘SA)
(A)
Kads, B (B)(⌘SA)
(⌘SA) {1 + Kads, A (A) + Kads, B (B)} = Kads, A (A)[T OT ⌘S] (⌘SA) =
Kads, A (A) [T OT ⌘S] 1 + Kads, A (A) + Kads, B (B)
(13.19)
Finally, dividing through by the concentration of adsorbent yields the following isotherm equation for adsorbate A in a system where it competes with species B for surface sites: Kads, A (A) qA = qmax (13.20) 1 + Kads, A (A) + Kads, B (B) The analogous expressions for adsorbate B are (⌘SB) =
qB =
Kads, B (B) [T OT ⌘S] 1 + Kads, A (A) + Kads, B (B)
Kads, B (B) qmax 1 + Kads, A (A) + Kads, B (B) 784
(13.21)
(13.22)
Generalizing the preceding result to the case where j adsorbates compete for the sites leads to the following adsorption isotherm for any species i: qi =
Kads, i (i) qmax 1 + Â ( j)
(13.23)
all j
The numerator of Equation (13.23) is the same as in the equation for noncompetitive adsorption [Equation (13.9)], and the denominator differs only in that it contains a term of the form Kads, j ( j) for each adsorbate in the system instead of just for the target adsorbate. Thus, each additional adsorbate in the system increases the denominator of Equation (13.23) and decreases the amount of i adsorbed. As would be expected, the larger the activity of the competing adsorbates ( j) and the greater their tendency to sorb (Kads, j ), the larger is their effect on sorption of i. While we have not encountered equations that look much like Equation (13.23) in previous chapters, this is strictly because of the ways in which we have chosen to represent the relationships. In fact, if we considered a solution containing Cd and Zn along with a single ligand (say, EDTA) and carried out calculations similar to those above, we would obtain for the concentrations of Cd EDTA and Zn EDTA: (Cd EDTA) =
Kl,Cd (Cd2+ ) T OT EDTA 1 + K1,Cd (Cd2+ ) + K1,Zn (Zn2+ )
(13.24a)
(Zn EDTA) =
Kl,Zn (Zn2+ ) T OT EDTA 1 + K1,Cd (Cd2+ ) + K1,Zn (Zn2+ )
(13.24b)
The similarity of Equations (13.24a) and (13.24b) to Equations (13.19) and (13.21) is apparent, indicating that, as in single-adsorbate systems, competitive Langmuir adsorption bears a strong similarity to conventional reactions involving only soluble species. Although Equations (13.24a) and (13.24b) could legitimately be described as the equations that describe “competitive complexation”, those equations are virtually never derived or shown explicitly. Ironically, this is so because competition among metals for whatever ligands are available is assumed to always be relevant and therefore not particularly noteworthy. By contrast, adsorption processes are often viewed with a specific target adsorbate in mind, and experiments are often conducted to compare adsorption of that species in systems where it is the sole adsorbate with its adsorption in more complex systems containing multiple adsorbates. Because of the focus on a particular adsorbate in such cases, equations describing competitive adsorption are of primary interest and hence have received substantial attention.
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Example 13.2 The Langmuir isotherm for sorption of chromate onto adsorbent C in Example 2– 13.1 has qmax = 0.96 mmol CrO2– 4 /g solid and Kads = 104 L/mmol CrO4 . The value of Kads for sorption of sulfate onto the solid is 20 L/mmol SO2– 4 . qmax is the same for the two adsorbates when expressed as moles of adsorbate per gram of solid. If the solution to be treated contains 2.0 mmol/L sulfate in addition to the chromate, how large an adsorbent dose is needed to meet the CrO2– 4 removal target? Solution The isotherms for the two adsorbates can be written based on Equation (13.23) as qCrO2 = 4
qSO2 = 4
(104 L/mmol)(CrO24 ) (0.96 mmol/g) 1 + (104 L/mmol)(CrO24 ) + (20 L/mmol)(SO24 ) (20 L/mmol)(SO24 ) (0.96 mmol/g) 1 + (104 L/mmol)(CrO24 ) + (20 L/mmol)(SO24 )
In addition to the adsorption isotherms, we know that the following mass balance 2– equations on CrO2– 4 and SO4 must be satisfied: T OT CrO4 = (CrO24 ) + qCrO2 csolid 4
T OT SO4 =
(SO24
) + qSO2 csolid 4
Since the target equilibrium concentration of CrO2– 4 is specified, we have four equations in four unknowns (the concentration of dissolved SO2– 4 , the adsorption densities 2– of CrO2– and SO , and the adsorbent dose). When these equations are solved simul4 4 taneously, we obtain csolid = 1.59 g/L
qCrO2 = 0.11 mmol/g
(SO24 ) = 0.74 mmol/L
qSO2 = 0.79 mmol/g
4
4
SO2– 4
Approximately two-thirds of the adsorbs, occupying a significant fraction of the surface sites. As a result, almost six times as much adsorbent is required to achieve the treatment goal as in the noncompetitive system analyzed previously.
13.3.3
Adsorption in Systems with Nonuniform Surface Sites
Although the Langmuir isotherm is adequate for describing many systems, it fails in many others. To understand the reasons for these deviations from idealized behavior, 786
we need to shift our focus from the similarities between adsorption and complexation by dissolved ligands to the ways in which these reactions might differ. In developing the Langmuir isotherm, we assigned a single value to Kads for the reaction between a given adsorbate and all surface sites. The analogous situation in solution is to assign a single value of K1 to the complexation reaction of Cd2+ with Cl– or with EDTA, which seems eminently reasonable. However, adsorptive surfaces are not uniform phases in the same way that a solution is: different locations on a surface might have different properties based on their local environment, such as their proximity to the edge of a crystal face or a defect in the solid structure. Furthermore, even on a large, perfectly flat crystal face, different surface atoms might be attached to the underlying bulk solid via different numbers or types of bonds. Any of these factors might cause different sites to have different affinities for adsorbate molecules; i.e., a single surface might contain both Cl– -strength and EDTA-strength binding sites for Cd2+ , and perhaps many other sites of intermediate strength. Thus, it might be appropriate to treat the surface of even a pure solid phase as a collection of many different ligands rather than as a group of identical ligands. Second, in the Langmuir analysis, we implicitly assumed that binding of adsorbates to the surface had no effect on the value of Kads at other sites. Again, the analogous assumption applied to dissolved ligands — that formation of a Cd-EDTA complex in one part of the solution has no effect on K1 for this reaction elsewhere in the system — seems intuitive. However, adsorbent sites are not independent, and a reaction at one point on the surface can affect the chemistry of other parts of the surface, in at least two ways. First, if an ion adsorbs, it can redistribute electrical charge on the surface, causing other ions to be attracted to the surface either more or less strongly than they would have been if the ion had not adsorbed. And, second, because adsorbates all share the same surface, the likelihood that they will interact directly with one another is greater than if each complex were an independent molecule in solution. Thus, the Langmuir model represents the surface sites as a collection of identical ligands that can bind to adsorbates to form noninteracting complexes, and we anticipate that both the assumptions of site uniformity and noninteraction might be violated at times. In the following sections, we explore the possible consequences of a nonuniform population of sites. Other sources of deviations from Langmuir adsorption are explored later in the chapter. The multisite Langmuir isotherm. Evidence for the presence of more than one type of surface site on an adsorbent often derives from competitive adsorption experiments. The site balance in the preceding analysis of competitive adsorption [Equation (13.17)] assumed that all adsorbates compete for a single population of uniform sites.
787
If two adsorbates have different values of qmax (in mol/m2 or mol/g) when they are the only adsorbate in the system, then at least some surface sites must be available to one of the adsorbates but not the other. In such cases, the system is sometimes represented as having a few distinct populations of sites that are not all accessible to all adsorbates, e.g., one group available only to adsorbate A, one available only to adsorbate B, and one available to both A and B. If a surface contains groups of distinct sites, it might be modeled simply by extending the Langmuir analysis to represent the surface as several ligands, rather than a single ligand, similar to a solution containing a mix of metal-binding ligands. A concentration could be established for each type of site, along with an equilibrium binding constant for each adsorbate that can bind to it, and the system could be modeled following the same steps we used to develop the isotherms for single-site surfaces. Note that, although a hypothetical surface might have, say, three different types of sites, in this model each of those sites is assumed to behave independently; that is, each site behaves as if the other sites did not exist, so that the Langmuir isotherm is satisfied for each site. Nevertheless, the adsorption reactions onto all the sites are linked, because all the sites are equilibrated with the same aqueous solution [and, in particular, with the same value of (A)], and because the mass balance on T OT A includes terms for sorption onto all the sites. Complex natural adsorbent mixtures such as soils are often represented in this way, and in some cases attempts are made to associate the different sites with different mineral phases in the mixtures. Although the number of different types of sites which could be present on a given solid surface is, in theory, unlimited, adsorption can often be modeled by considering just two or three types of sites, implying that just a few site-types dominate adsorption in these systems. The most common multisite Langmuir models represent the surface as having two groups of sites: a small number of sites that bind the adsorbate strongly, and a much larger number where the binding is considerably weaker. The overall isotherm for a single adsorbate in such a system is a linear addition of the isotherms for the two sites, as follows: qA,tot = qA,1 + qA,2 =
qmax,1 Kads,A,1 {A} qmax,2 Kads,A,2 {A} + 1 + Kads,A,1 {A} 1 + Kads,A,2 {A}
(13.25)
Equation (13.25) is easily generalized to any arbitrary number N of surface sites as follows: N N qmax,i Kads,A,i {A} qA,tot = Â qA,i = Â (13.26) i=1 i=1 1 + Kads,A,i {A} The adsorption density on each type of site and the overall adsorption density on a hypothetical, two-site surface are shown in Figure 13.6. 788
Figure 13.6 Adsorption of one adsorbate on a surface with two types of sites, each of which is occupied in accord with the Langmuir isotherm. Part (b) of the figure shows the data at very low q and (A).
Example 13.3 Davis et al. (1998)3 modeled adsorption of Zn2+ onto quartz using a two-site Langmuir isotherm based on the following reactions and equilibrium constants. ⌘Sst H + Zn2+ )* ⌘Sst Zn+ + H+
⌘Swk H + Zn2+ )* ⌘Swk Zn+ + H+ 3 Davis,
log Kst = 0.85 log Kwk = 2.40
J.A., Coston, J.A., Kent, D.B., and Fuller, C.C. (1998) “Application of the surface complexation concept to complex mineral assemblages.” Environ. Sci. Technol. 32, 2820-2828.
789
The surface concentrations of strong (st) and weak (wk) sites were reported to be 0.033 and 3.84 µmol/m2 , respectively. Derive Langmuir isotherms for sorption to the two sites at pH 5. According to the model, what fraction of the total adsorbed zinc is strongly adsorbed when the equilibrium, dissolved Zn2+ concentration is 10 7 , 10 6 , 10 5 , or 10 4 , if all the reactants behave ideally? Solution Substituting the value (H+ ) = 10 5 into the equilibrium constant expressions yields the following conditional adsorption equilibrium constants: pH5
Kads, st =
(⌘Sst Zn+ ) Kst 100.85 = = = 105.85 (⌘Sst H)(Zn2+ ) (H+ ) 10 5.0
pH5
Kads, wk =
(⌘ Swk Zn+ ) Kwk 10 = + = 2+ 10 (⌘Swk H)(Zn ) (H )
2.40 5.0
= 102.60
The surface concentrations of strong and weak sites given in the problem statement correspond to the qmax values for the sites, so the conditional Langmuir isotherms can be written as ✓ ◆ 105.85 (Zn2+ ) µmol qst = 0.033 2 m 1 + 105.85 (Zn2+ ) ✓ ◆ 102.60 (Zn2+ ) µmol qwk = 3.84 2 m 1 + 102.60 (Zn2+ ) By substituting (Zn2+ ) values of 10 7 , 10 6 , 10 5 , and 10 4 , we can find the adsorption densities on the two types of sites and the fraction of the adsorbed zinc that is bound to strong sites, as summarized below. (Zn2+ ) 10 10 10 10
7 6 5 4
qst , µmol/m2
qwk , µmol/m2
qst /qtot
0.0022 0.0137 0.0289 0.0325
0.0002 0.0015 0.0152 0.1470
0.93 0.90 0.66 0.18
When the activity of dissolved Zn2+ is low, most of the adsorbed Zn is bound to strong sites. However, as (Zn2+ ) increases, the strong binding sites approach full occupation and most of the incremental adsorption occurs on the weaker sites, so that at the highest value of (Zn2+ ) considered, more than 80% of the adsorbed Zn is on the weaker sites.
790
The approximation is sometimes made that, if a surface contains several different types of sites, the sites are occupied sequentially from strongest to weakest. However, as in the preceding example, we are often interested in sites that not only have different binding constants, but also are present at vastly different concentrations. In such cases, both groups of sites can make significant contributions to the total adsorption density over a wide range of the dissolved adsorbate concentrations. Under the circumstances, the assumption of sequential occupation could be misleading, so it is generally better to recognize that the different types of sites are actually occupied simultaneously than to rely on the simplifying assumption that they are occupied sequentially. Surface with a semicontinuous distribution of site types. Although it is often possible to represent experimental data satisfactorily by modeling the surface as a collection of a few site types, there are also many systems where such an approach does not succeed. Often, such surfaces behave as though they have sites with a virtually continuous distribution of affinities for the adsorbate. A heterogeneous surface might have sites with any distribution of adsorptive equilibrium constants. However, empirically, many systems behave as though they comprise a very small number of sites with very strong affinity for the adsorbate (large equilibrium constant), and steadily increasing numbers of sites with steadily decreasing affinities. One mathematical function that fits this pattern and that has found widespread application represents the concentration of sites with a given Kads as a function that decreases geometrically as Kads increases. If the surface is assumed to have sites with Kads values from 0 to some maximum value Kmax , the predicted isotherm for a system containing a single adsorbate A conforms to the following simple equation: qA = kf {A}n (13.27) The relationship in Equation (13.27), known as the Freundlich isotherm, has been shown to fit many experimental data sets quite well.4 Like the single-site Langmuir isotherm, a Freundlich isotherm is fully defined by two fitting parameters: kf , which describes the adsorption density under standard conditions (qA = kf when (A) = 1) and is therefore an indicator of the affinity of the adsorbent for the adsorbate, and n, which indicates how dramatically the binding strength changes as the adsorption density changes. If n < 1, the average adsorptive binding strength decreases with increasing surface coverage; the lower the value of n, the more dramatic the decrease. A value of n = 1 indicates that the binding strength is the same on all sites, so the equation reduces to a linear isotherm [qA = k(A)]. Although the derivation of Equation (13.27) from the assumed, geometric rela4 In much of the literature, the exponent in Equation (13.27) is shown as 1/n, rather than as n. Therefore, it is important to understand which convention is being used when interpreting reported values of n.
791
tionship between log Kads and surface site concentration is valid only for 0 < n 1, data are sometimes reported that conform to Equation (13.27) with a value of n > 1, at least over a limited range of (A) values. Such a result implies that the affinity of the surface for the adsorbate increases as the adsorption density increases, which might occur, for example, if newly arriving adsorbate molecules are more strongly attracted to previously adsorbed molecules than to the bare surface. Graphical representations of the Freundlich isotherm for a few values of kf and n are shown in Figure 13.7.
Figure 13.7 Graphical representation of the Freundlich isotherm on (a) linear and (b) logarithmic coordinates. The values of kf shown have units of µmol/g/(mol/L).
Figure 13.7 shows the cumulative adsorption onto all surface sites in a system that follows the Freundlich isotherm. Based on the development of the isotherm equation, the total adsorption density is thought to represent the summation of the adsorption 792
densities onto a wide variety of sites, with the Langmuir isotherm applying to each group of sites individually. Consistent with this idea, for any given value of (A), some sites of all types are occupied, with the high-affinity sites being closer to saturation (i.e., complete coverage) than the low-affinity sites. If only one adsorbate is present in a system and its sorption is characterized by a Freundlich isotherm, the adsorbate’s partitioning between the solid and solution can be determined by combining the mass balance on adsorbate with the isotherm equation, as follows: T OT A = (A) + qA csolid
(13.28a)
T OT A = (A) + kf (A)n csolid
(13.28b)
If the parameters of the isotherm equation (i.e., kf and n) and any two of the three other parameters in Equation (13.28b) [T OT A, (A), and csolid ] are known, the equation can be solved for the remaining parameter. Example 13.4 Freundlich isotherms are commonly reported for sorption of metals onto mineral adsorbents, as shown in Figure 13.8 for sorption of Cd2+ onto ferrihydrite [Fe(OH)3 (am)]. A system containing a fixed concentration of 10 4 M Fe(OH)3 (am) at pH 7.2 is titrated with Cd(NO3 )2 over a range of T OT Cd from 10 7 to 10 5 M. Plot the concentrations of dissolved and adsorbed Cd2+ and the fraction adsorbed as a function of T OT Cd.
Figure 13.8 Isotherms for sorption of Cd2+ onto ferrihydrite at three pH’s. Adapted from Benjamin and Leckie (1981).5 5 Benjamin,
M.M., and Leckie, J.O. (1981) “Multiple-site adsorption of Cd, Cu, Zn, and Pb on amorphous iron oxyhydroxide.” J. Colloid Interface Sci., 79, 209-221.
793
Solution The Freundlich isotherm characterizing the system at pH 7.2 can be determined from the graph to be log qCd = 1.44 + 0.66 log(Cd2+ ) qCd = 27.5(Cd2+ )0.66 where q is in moles of Cd per mole Fe(OH)3 (am). The distribution of Cd at any value of T OT Cd can be determined by inserting the isotherm equation and csolid into Equation (13.28b), as follows: T OT Cd = (Cd2+ ) + 27.5(Cd2+ )0.66 (10 4 ) Solving this equation for (Cd2+ ) over a range of T OT Cd values yields the desired information, which is plotted in Figure 13.9. Both adsorbed and dissolved Cd2+ increase as Cd is added to the system, but the dissolved concentration increases more rapidly than the adsorbed portion, so the fraction sorbed decreases.
Figure 13.9 Computed distribution of Cd2+ in the example system.
One problematic feature of the Freundlich isotherm is that it places no upper limit on the adsorption density. As a result, the isotherm is bound to fail to reproduce experimental data at very large activities of the dissolved adsorbate. An upper bound on the adsorption density is sometimes incorporated into the isotherm empirically by rewriting the isotherm equation as a hybrid of the conventional Langmuir and Freundlich isotherms, as follows: qA =
qmax,A (kL-F {A})n 1 + (kL-F {A})n
(13.29)
Equation (13.29) is commonly referred to as the Langmuir-Freundlich isotherm. 794
In systems where adsorption is characterized by the Freundlich isotherm, competitive effects can be significant if even a small fraction of the surface is occupied, because the sites that are preferentially occupied by the competing adsorbate might be the ones that bind the target adsorbate most strongly. That is, if a solid has only a small number of strong binding sites, a relatively small concentration of a strongly competing adsorbate might occupy almost all those sites. In that case, even though a large number of available sites remain on the surface, the competitive effect would be significant, since the target adsorbate would have to bind to relatively weak sites. To model such a situation, a logical conceptual approach would be to apply the competitive Langmuir model to each group of sites. Use of such an approach is reasonably simple if the adsorbates are characterized by Freundlich isotherms with the same n value. In such a case, the kf values provide a direct measure of the relative affinity of the surface for the competing adsorbates. That is, if nA = nB and kf,A /kf,B = 2, we can infer that every site on the surface has twice the attraction for species A as for species B. If, however, the adsorbates are characterized by different values of n, then A might outcompete B for sorption on some sites, and B might outcompete A for sorption on others. In such a case, additional assumptions or measurements relating the two isotherms are required to model competitive interactions. The additional information required can be represented as an “interaction parameter”, aAB . In a system with m competing adsorbates, the values of m 1 interaction parameters are needed. The resulting isotherm equation for the competitive system is qA = kf,A cA
Â
j=A,B,C,...
aA j cA
!nA
1
(13.30)
where A, B, C,... are the various adsorbates competing for the surface. The value of the interaction parameter for an adsorbate with itself (e.g., aAA ) is 1.0, so Equation (13.30) reduces to the conventional Freundlich isotherm for a system containing only one adsorbate.6 A simpler, but less fundamental approach for incorporating the influence of competing adsorbates is to modify the isotherm expression as follows: qA = kf {A}m {B}b {C}c ...
(13.31)
where B and C are competing adsorbates present in the solution. If negative values are assigned to the exponents b, c, etc., the net effect is that an increase in the activities of those species leads to a decrease in the adsorption density of adsorbate A. This 6 For additional details on this approach, see Sheindorf, C., Rebhun, M., and Sheintuch, M. (1981) “A Freundlich-type multicomponent isotherm.” J. Colloid Interface Sci., 79, 136-142.
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approach has no basis in theory and might be considered incomplete, in that it postulates nothing about the actual adsorption of the competing species. Nevertheless, it is a mathematically simple way to model the decline in the adsorption of one species when the activities of some other species are altered. Equation (13.31) is sometimes referred to as the Extended Freundlich isotherm. Comparison of multisite and Freudlich isotherms. The preceding discussion describes the two most common approaches for modeling adsorption onto surfaces that have sites with different affinities for an adsorbate. The multisite Langmuir model allows one to model any distribution of surface site concentrations and adsorption binding constants, but it suffers from the fact that each postulated type of site increases the number of adjustable parameters in the model by two. It is rarely possible to resolve experimental data well enough to justify proposing more than two types of sites on a given surface, and even then the evaluation of the four constants must be viewed somewhat skeptically. On the other hand, the Freundlich model presupposes a great deal of variability in adsorption binding constants, but postulates a specific form for the relationship between the binding constants and the site concentrations. Because the Freundlich isotherm is a two-parameter model, it does not possess quite as much flexibility as multisite Langmuir models. Nevertheless, it often provides a remarkably good fit to experimental data over a range of several orders of magnitude in adsorbate concentration, and it is therefore frequently used to describe equilibrium adsorption relationships. It should be emphasized that these two isotherms have found popular acceptance because they are mathematically simple and because they seem to fit experimental data, not because there is some fundamental physical or chemical reason to expect them to apply. Unfortunately, a false distinction between these isotherms has become widely accepted, whereby the Langmuir isotherm is viewed as theoretically based and the Freundlich isotherm as strictly empirical. Such a distinction is incorrect on both counts: both the Langmuir and Freundlich isotherms can be derived by starting from a conceptual (theoretical) view of the surface and the adsorption process, but in both cases arbitrary assumptions must be made about the distribution of Kads values. Thus, in both cases, empirical data are needed to justify that the assumptions are applicable to a given situation. Other functional relationships could be (and have been) postulated between surface site densities and binding constants, and any such relationship would be perfectly acceptable as the basis of an isotherm if the experimental data supported it. Correspondingly, the fact that a data set fits a given equation should not be taken as proof that the conceptual model for that isotherm is correct. 796
13.4
MODELING ADSORPTION WITH VISUAL MINTEQ
13.4.1
Single- and Multi-Site Langmuir Adsorption
Conceptually, incorporating Langmuir-type adsorption into Visual Minteq is straightforward — we just input the surface sites as though they were conventional ligands (or groups of ligands), and the reactions as though they were conventional complexation reactions. No modifications are required to account for competitive adsorption, because any competition is implicitly accounted for when the adsorption reactions are input, in the same way that it is accounted for when two or more dissolved metals “compete” for Cl– , EDTA, or other ligands in solution. In practice, though, the inputs used to characterize adsorption reactions in Visual Minteq are slightly different from those for reactions between dissolved species, because adsorption data are usually specified using different parameters than are used for conventional complexation reactions. In Visual Minteq, the core adsorptive site in all surface complexation modeling is represented as an oxide ion at the surface of the solid. When it is unoccupied, the site is presumed to have a 1 charge, so we represent it here as ⌘SO .7 The site is presumed to be capable of binding one or two protons to form ⌘SOH and ⌘SOH+2 , respectively, and the singly-protonated, neutral surface species (⌘SOH) has been selected as the default component for this group. Therefore, that species is used as a reactant in all adsorption reactions. For example, the acid/base reactions of the surface sites are represented in the software as ⌘SOH + H+ )* ⌘SOH+ 2
(13.32)
⌘SOH )* ⌘SO + H+
(13.33)
The equilibrium constants for these reactions depend on the particular surface being modeled. It is possible to specify that a given adsorbent has multiple, different site-types, in which case the default component names for the second, third, and subsequent site-types are ⌘SOHs2, ⌘SOHs3, etc. Such components are presumed to undergo the same types of acid/base reactions as ⌘SOH does. For species formed from these components, Visual Minteq places the species charge before the ‘s’ (e.g., ⌘SO– s2). To simulate Langmuir adsorption using the program, we first click on the “Adsorption” link on the home page and select “Surface complexation reactions”. In 7 In Visual Minteq, the bond between the surface site and the solid is represented by an equal sign (=). Because the software does not display equations, this representation is unambiguous. However, in this text, concentrations of adsorbed species appear in expressions that do have equal signs (e.g., mass and site balances). Therefore, to avoid ambiguity, generic adsorptive sites are designated here by a triple-bar (⌘).
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response, the program opens a page where we can input the number of surfaces (i.e., adsorbents) that are present in the system of interest. For this discussion, we will assume that the system contains a single adsorbent, so we can type the numeral ‘1’ in the box. Once we enter a value for the number of surfaces, a corresponding number of fields appears, each with a pulldown menu from which we can select one of several surface complexation models. These models differ only in terms of how they account for electrostatic interactions between the adsorbate and the surface. We will consider such interactions shortly, but for now we focus on systems in which they are unimportant and therefore choose the nonelectrostatic model (NEM) from the pulldown list. The remainder of this screen contains fields that we will fill in with information about the specific system of interest. However, it is convenient to first define some properties that characterize the adsorbents, independent of the solution in which they are suspended. Information about these properties is stored in a separate “Complexation database” for each adsorbent. For example, the properties that Visual Minteq assigns by default to surface sites when the nonelectrostatic model is used can be seen by clicking on the button for the NEM database (nem std.vdb), which opens the “Surface/complexation database editor”. This database includes one default surface component (⌘SOH, shown at the top, center of the page) and two default surface species (⌘SOH+2 and ⌘SO– , shown in the “Present database species” box at the left). The stoichiometry and equilibrium constants for forming these species via Reactions (13.32) and (13.33) can be seen by clicking on the species’ name; the default log K values are 7.3 and 8.9, corresponding to pKa1 = 7.3 and pKa2 = 8.9 for the acid ⌘SOH+2 . The default surface species can be modified or deleted, and new components or species can be added as needed. Before making any such changes, however, it is useful to save the database with a new name, so that the default database is not modified. To illustrate the simulation of adsorption reactions in Visual Minteq, we will model the results from a study by Papini et al.8 of adsorption of Pb2+ and H+ onto a volcanic soil. In this study, the soil was represented as a single adsorbent containing three site-types, with estimated surface concentrations of 2.33 ⇥ 10 2 , 3.24 ⇥ 10 2 , and 6.74 ⇥ 10 2 mol/g, respectively. In the original study, Pb2+ and H+ were assumed to compete for ⌘SO– and ⌘SO– s2 sites, while ⌘SO– s3 sites were assumed to be capable of binding only H+ . The adsorption reactions that were considered are shown in Table 13.2a, and these reactions are rewritten in tableau format in Table 13.2b.9 For this exercise, we will explore the adsorption of lead as a function of pH, in a system containing 10 4 M T OT Pb and 100 mg/L of the adsorbent. 8 Papini, M.P., Kahie, Y.D., Troia, B., and Majone, M. (1999) “Adsorption of lead at variable pH onto a natural porous medium: Modeling of batch and column experiments.” Environ. Sci. Technol., 33, 4457-4464. 9 The user is free to assign names arbitrarily for surface species. Following the pattern for the components, the convention here is to add s2, s3, etc., after the specification of the chemical formula and charge of the species.
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Table 13.2 Adsorption reactions relevant to the example problem, shown (a) in conventional chemical format and (b) as a Visual Minteq tableau.
Log K
Reaction 1 2 3 4 5 6 7
⌘SO– + H+ )* ⌘SOH
4.95
⌘SO– + Pb2+ )* ⌘SOPb+
4.70
⌘SOHs2 + H+ )* ⌘SOH+2 s2
7.58
⌘SO– s2 + H+ )* ⌘SOHs2
⌘SO– s2 + Pb2+ )* ⌘SOPb+ s2 ⌘SO– s3 + H+ )* ⌘SOHs3
⌘SOHs3 + H+ )* ⌘SOH+2 s3
9.65
9.91 11.71 4.70
Species
H+
Pb2+
⌘SOHs2
⌘SOHs3
log K
⌘SO–
⌘SOH
1
0
1
0
0
4.95
⌘SOPb+
1
1
1
0
0
0.25
⌘SO– s2
1
0
0
1
0
9.65
⌘SOH+2 s2
1
0
0
1
0
7.58
⌘SOPb+ s2
1
1
0
1
0
0.26
1
0
0
0
1
11.71
⌘SOH+2 s3
1
0
0
0
1
4.70
⌘SO– s3
To begin, we follow the steps just described to specify that the system has one adsorbent, choose NEM as the adsorption model, click on the button for the default database (nem std.vdb), and rename the database (e.g., as Papini.vdb). We then modify the default parameters of the database to conform to the system of interest. For example, the default model assumes that the adsorbent surface has only one site-type. To model the current surface, we need to add two new surface components, which we do by clicking the “Add component” button at the top of the page. By default, these new components are given the names ⌘SOHs2 and ⌘SOHs3.10 10 The
database must be saved immediately after a new component or species has been added, and also whenever a change has been made to one of the other fields on the database page; if more than one change is made and the database is then saved, only the last change input will be saved. Note also that the program returns an error message if the “Save” command is given without highlighting one of the species names; nevertheless, the Save command appears to be implemented correctly, so the error message can be ignored.
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The species ⌘SOH+2 and ⌘SO– are included by default. However, Papini et al. assumed that the ⌘SOH sites were monoprotic, capable of undergoing deprotonation but not protonation. We therefore delete the species ⌘SOH+2 (by highlighting it in the list of species and clicking the “Delete this species” button), and we change log K for the deprotonation reaction to the value shown in Table 13.2b ( 4.95), saving the changes after each step. Next, we add the surface species ⌘SOPb+ to the database. To do so, we click the “Add new species” button and assign the new species a name. We then enter the stoichiometry for forming the species from surface and dissolved components. The first component that is specified must be the surface site. The program forces this choice by showing “Surface comp.” (i.e., a surface site component) as the only option in the “Which component type” pulldown menu. All three surface components are listed in the “Component” pulldown menu; because we wish to define a new species formed with component ⌘SOH, we choose that component and enter a stoichiometric coefficient of 1. If we then click “Next component”, the “Which component type?” options change so that we can choose from among surface, inorganic, and organic components. The new species we are forming contains only the surface component and one Pb2+ ion, so we click “Inorganic comp.”, select Pb2+ , and insert the stoichiometric coefficient of 1. If the new species included yet more components, we would repeat this process until they all had been specified. In this example, after specifying that Pb2+ is part of the species, we click the “Add to list” button to assign the species an identifying number; we can then enter log K and the standard enthalpy of the formation reaction in the appropriate fields and re-save the file. A similar procedure is used to input the corresponding information for the other surface species. A screenshot of the database page after all the surface species have been defined is shown in Figure 13.10. After defining the properties of the adsorbent(s) in the system in appropriate databases, we “Quit” the database page to return to the “Surface complexation” page and fill in the information related to the specific solid/liquid suspension of interest. (Note: migrating to the “Surface complexation” page without using the “Quit”’ command to leave the database page sometimes causes the program to freeze.) In particular, we complete the fields for the concentration (g/L) and specific surface area (m2 per g of solid) of each adsorbent, and the concentration of each type of surface site (mmoles of sites per liter of solution). Of these parameters, only the total concentration of sites is important for the current problem; the other two parameters are relevant only in systems where electrostatic interactions are considered. In the current case, the adsorbent has three types of sites, and we need to specify the concentration of each of those site-types independently. To do so, we click the “Add site” button twice; when this is done, the second and third site-types identified on this page are automatically associated with the properties of ⌘SOHs2 and ⌘SOHs3 in the designated database. The concentrations of the three site-types can then be computed. For example, for the first site-type,
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Figure 13.10 Screenshot of the Papini.vdb database page after all the surface species have been defined.
✓
T OT ⌘SOH = 2.33 ⇥ 10
◆ ✓ ◆ sites ⇣ g ⌘ mmol 3 mmol 0.1 10 = 2.33 g L mol L (13.34)
2 mol
The corresponding values of T OT ⌘SOHs2 and T OT ⌘SOHs3 are 3.24 mmol/L and 6.74 mmol/L, respectively. We know the solid concentration (0.1 g/L) and can fill in this value, although, as noted above, it has no impact on the subsequent calculations. We do not know the specific surface area and so we just assign an arbitrary value of 500 m2 /g for that parameter. The box for “Counter-ion accumulation”, is relevant only in selected cases when electrostatics are considered, so it should not be checked for the current simulation. In some cases, especially when the adsorbent is a well-defined pure solid, we have an estimate for its surface site density, defined as the number of sites per unit surface area on the solid. In such cases, the box near the bottom of the “Surface complexation” page can be clicked, and the surface site density can be input instead of the surface site concentration. When either of these parameters is input, the other one can be computed from 801
✓
◆ ✓ ◆✓ ◆✓ ◆ Surface site Surface site Solid Specific = concentration density concentration surface area
(13.35)
The program carries out this calculation and allows the user to toggle between displaying the surface site concentration and the surface site density, by checking or unchecking the box near the bottom of the page. An image of the upper portion of the “Surface complexation” page for the example system is shown in Figure 13.11.
Figure 13.11 Screenshot of the upper part of the “Surface complexation” page for the example system. Values for the solid concentration and the specific surface area must be specified, but these values are used only when surface electrostatic interactions are taken into account and are irrelevant when the adsorption model is the NEM.
Once all the fields on the “Surface complexation”’ page have been populated, we can return to the program home page and save the file. Doing so links the “Surface complexation” page to the newly defined Papini.vdb database.11 We can then fill in the rest of the information needed for the simulation (T OT Pb, range of fixed pH 11 Note that, as of this writing, the newly defined database can be accessed only by opening a saved input file (⇤.VDA) that is linked to it, and that linkage is not made unless the fields on the “Surface complexation” page are fully populated. Therefore, it is important to save the input file as soon as the “Surface complexation” page has been completed. Alternatively, one might populate the “Surface complexation” page (even if only with arbitrary values) before defining the adsorbent properties on the database page, and then save both the database and the input file. That way, any subsequent changes to the database will be saved whenever the page is saved.
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values to be investigated, results to be saved in “Selected sweep”, etc.) and run the program. The output is summarized in Figure 13.12. In addition to the data related to the effects of pH on lead distribution, detailed summaries of the speciation of each component and the partitioning of lead between the dissolved and adsorbed phases for any individual run are available via the “View species distribution” and “Equilibrated mass distribution” links.
Figure 13.12 Speciation of (a) surface and (b) Pb in the example system.
Figure 13.12a shows the acid/base speciation of the surface sites. Because T OT Pb is less than 5% of the concentration of any of the individual site-types and less than 1% of the total concentration of surface sites, the presence of the lead has very little effect on the concentrations of protonated and deprotonated sites as a function of pH. The distribution of lead as a function of pH is considerably more complex. At very low pH, almost all the lead is dissolved, because H+ outcompetes Pb2+ for the 803
available surface sites. However, as the pH increases, competition from H+ diminishes (because the H+ activity declines). As a result, about one-half of T OT Pb is adsorbed at pH 3.0, and from pH ⇠4 to ⇠11, virtually all of the lead is adsorbed, with the total dissolved lead passing through an extremely low minimum of ⇠ 10 9 M at pH near 8.5. Despite the fact that the bulk distribution of the lead between the solution and the surface changes little over this range, the detailed surface speciation does change dramatically, shifting from almost entirely on the ⌘SO sites to almost entirely on the ⌘SOs2 sites as the pH increases from ⇠5.5 to 6.5. Only when the pH approaches 12 does a substantial amount of lead return to solution, primarily as the trihydroxo complex Pb(OH)–3 . Example 13.5 An adsorbent that has a surface site density of 3 sites/nm2 and a specific surface area of 240 m2 /g is present at a concentration of 60 mg /L in a suspension at pH 6.0. All the surface sites are assumed to be identical. The Langmuir adsorption equilibrium constants for binding of Ca2+ , Zn2+ , and Cu2+ are 102.5 , 106 , and 108 , respectively. What fraction of each adsorbate is adsorbed at equilibrium if the solution contains 0.01 M T OT Ca, 10 5 M T OT Zn, and 10 6 M T OT Cu? What fraction of the surface sites is occupied by each adsorbate? Compare the results from Visual Minteq with those that obtained using Equation (13.23). Solution Although the names that Visual Minteq assigns to surface groups imply that adsorption sites are oxide groups that can undergo acid/base reactions, the software is equally applicable to sites that have no interactions with protons. For instance, in this problem, we can indicate that ⌘SOH groups do not undergo protonation or deprotonation by deleting ⌘SO– and ⌘SOH+2 as surface species. The three surface complexation reactions are then input by specifying, for each adsorbate, the stoichiometry and equilibrium constant for a reaction of the type ⌘SOH + Me2+ )* ⌘SOHMe2+ The characteristics of the adsorbent can be input on the “Surface complexation” page, placing a check in the box at the bottom so that we can input the adsorbent’s surface site density in sites/nm2 instead of the site concentration in mmol/L. After specifying the surface site density, if we uncheck that box, the software indicates that the site concentration is 7.17 ⇥ 10 2 mmol/L. When we run the program, the output summary shows the concentrations of all dissolved and adsorbed species at equilibrium. A more concise summary of some of the key results can be seen by clicking the “View species distribution” button; this summary is shown in Figure 13.13. 804
Figure 13.13 Computed equilibrium speciation of surface sites, calcium, copper, and zinc in the example system
About 80% of all the surface sites are occupied, mostly by calcium (65%) and zinc (13%). Even though calcium is the dominant adsorbed species, the vast majority (99.5%) of the total calcium remains in solution. Copper is almost entirely (99.9%) removed from solution, but occupies only 1.4% of the total surface sites. The Visual Minteq calculations can be checked against those obtained using Equation (13.23), which is the generalized result for competitive Langmuir adsorption. For example, substituting (⌘SOHZn)/csolid and (T OT ⌘SOH)/csolid into Equation (13.23) for qZn and qmax , respectively, the concentration of adsorbed zinc can be computed as (⌘SOHZn) [T OT ⌘SOH/csolid ] KZn (Zn2+ ) = csolid 1 + KCa (Ca2+ ) + KZn (Zn2+ ) + KCu (Cu2+ ) (⌘SOHZn) =
[T OT ⌘SOH]KZn (Zn2+ ) 1 + KCa (Ca2+ ) + KZn (Zn2+ ) + KCu (Cu2+ )
All the terms on the right side of this equation are provided in the Visual Minteq output. Inserting these values into the equation, the concentration of adsorbed zinc is computed to be 9.37 ⇥ 10 6 mol/L, consistent with the reported output for that value; the same as true for the concentrations of the other adsorbed species and of ‘unoccupied’ sites, which are identified as ⌘SOH in Visual Minteq. Thus, Equation (13.23) could have been used to obtain the same result as is provided by the software. However, to get the result, we would have had to solve the equation simultaneously with the mass balances on all three adsorbates, taking into account not only the adsorbed metals and the free Me2+ ions, but also dissolved, hydrolyzed species such 805
as CuOH+ . Clearly, using the software is a much more efficient way to solve the problem.
13.4.2
Adsorption that Follows the Freundlich Isotherm
Visual Minteq can also be used to compute the distribution of adsorbates in systems characterized by Freundlich isotherms and to account empirically for competitive adsorption by the Extended Freundlich isotherm.12 Here, we explore how the modeling of such systems differs from that of systems that obey other (and, in particular, Langmuir) isotherms. When Visual Minteq is used to model conventional surface complexation reactions, it utilizes input about the stoichiometry and equilibrium constant for each reaction occurring in the system, and the isotherm is an outcome of the analysis. By contrast, the Freundlich isotherm combines the stoichiometric and equilibrium constant data for many different reactions (specifically, the reactions between an adsorbate and many different site-types) into a single expression with a different functional form from the equations for adsorption onto individual site-types. In particular, for adsorbates that obey Freundlich isotherms, the exponent on the term for a given component in the equilibrium expression (the isotherm) can differ from the stoichiometric coefficient for that component in the adsorption reaction. As a result, the isotherm parameters must be input explicitly when modeling Freundlich adsorption, rather than be derived from the stoichiometry and equilibrium constants. To illustrate this process, we will first use the software to model noncompetitive, Freundlich adsorption of chromate species onto a hypothetical solid, and then demonstrate how – competition from SO2– 4 and OH can be incorporated into the analysis. Consider a suspension at pH 7.0 that contains 2.0 g/L of an adsorbent, and assume adsorption of CrO2– 4 onto this solid at pH 7.0 has been reported to obey the following Freundlich isotherm: ✓ ◆ (Ads-CrO4 ) mol 0.7 = 5 CrO24 (13.36) qCrO4 = csolid kg where (Ads-CrO4 ) and csolid are the concentrations of adsorbed CrO4 and the adsorbent in mol/L and kg/L, respectively. Because the activity of CrO4 is dimensionless, the units of kf are the same as those of qCrO4 . 12 Modeling of competitive adsorption among species that obey Freundlich isotherms is much more common and more fully developed in the literature that uses the gas transfer model of adsorption, rather than surface complexation model. In that literature, competition is most often simulated according to ideal adsorbed solution theory (IAST). See, for example, Crittenden, J.C., Luff, P., Hand, D.W., Oravitz, J.L., Loper, S.W., and Arl, M. (1985) “Prediction of multicomponent adsorption equilibria in background mixtures of unknown composition.” Water Research, 19, 1537-1548, or Benjamin, M.M. (2009) “New conceptualization and solution approach for the ideal adsorbed solution theory (IAST).” Environ. Sci. Technol. 43, 2530-2536.
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Imagine that we wish to use Visual Minteq to simulate the partitioning of chromium in this system for T OT CrO4 concentrations ranging from 10 6 to 10 4 M in a solution with an estimated ionic strength of 0.01 M. The procedure for doing so involves three steps: first, we input the isotherm parameters into a database where information about adsorption of any adsorbate onto the solid can be stored; then, we input information about the adsorbent and adsorbate concentrations in the system of interest on a separate page that is specific to this problem; and, finally, we input conventional information that is needed for any Visual Minteq simulation (e.g., pH, ionic strength, T OTi, etc.) in the usual way. To establish the database for the adsorbent, we utilize the “Adsorption” pulldown menu on the home page to migrate to “Adsorption isotherms”, where we select “Freundlich” as the adsorption model. The Visual Minteq representation of the Freundlich isotherm then appears. Clicking on the “Change/edit database” button on this page opens the default database (Freundlich templ.iso). To avoid accidentally making permanent changes to this default database, we can click “Save changes to new database” and assign the blank database a new name (e.g., CrO4 example1.iso). (An error message might be generated, but the database will be created nevertheless.) We then input the isotherm information. The boxes at the top of the page allow us to choose the units of the adsorption density (in this case, mol/kg) and the parameter that is used to quantify the activity of the dissolved adsorbate (identified as the 1st component). The choices for this latter parameter are the activity and the total dissolved concentration. We are simulating a chemical reaction, so activity would seem to be the appropriate choice in all cases. However, Freundlich isotherms are commonly reported in the literature using the total dissolved concentration of the adsorbate (T OTi) as “ci ”. To facilitate modeling of those scenarios, Visual Minteq provides the option to use ci in this way. In the current example, the isotherm equation is given in terms of {CrO2– 4 }, so “Activity” is the correct choice. Note that, regardless of the units chosen for the adsorption density, ci is calculated using mol/L units (either for T OTi or in the calculation of ai ). We next input the isotherm parameters, using components to specify the participating chemicals. To do so, we click “Add new species”, which causes several new input boxes to open on the bottom portion of the page. We then assign a name to the adsorbed species (e.g., Ads-Cr) and define how that species is made from the components. In this example, the adsorbed species consists of one CrO2– 4 ion and nothing else, so we use the pulldown menu and make that selection.13 We then indicate that the exponent on CrO2– 4 in the isotherm equation is 0.7 by inserting that value into the 13 When
the surface complexation model is used, information about the concentration of surface sites is required, so the surface sites are included as a component needed to generate the adsorbed species. However, when a predefined isotherm is used to model adsorption, only the concentration and surface area of the bulk solid are required, so the surface sites are not included explicitly as a component.
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“Stoic.” box.14 If, for example, the isotherm had been written in terms of the activity + of HCrO–4 instead of CrO2– 4 , we would click “Next component” and add H as another component of the surface species. However, in the current example, the surface species contains only CrO2– 4 , so we are done inputting the information for Ads Cr, and we click “Add to list”. At this point, the program transfers the new information to the center portion of the page and provides a “Species ID number”. We can then input “log KF” and the stoichiometric coefficient for the mass balance (M.B.) calculations. KF is given in the isotherm equation as 2, so log KF is 0.7. The M.B. coefficient is 1.0 because, 2– even though the exponent on CrO2– 4 in the isotherm equation is 0.7, 1.0 CrO4 ion is removed from solution each time the adsorption reaction proceeds. A screenshot of the database page after all the relevant information has been input is shown in Figure 13.14. Once all the fields on this page have been populated, we click “Quit” to return to the “Adsorption isotherms” page. The “Adsorption isotherm” page includes confirmations of our choices for the meaning of ci and the units of adsorption density and provides opportunities to input the adsorbent and total adsorbate concentrations. The adsorbent concentration is actually input as 1/csolid , which is labeled as the liquid-to-solid (L/S) ratio and must be expressed in units of liters of solution per kg of solid. Thus, for the example scenario (2 g/L, or 0.002 kg/L), we input a value of 500.
Figure 13.14 The Visual Minteq Adsorption database page with the fields populated to simulate Freundlich adsorption according to the isotherm in Equation (13.36). 14 The idea here is that the isotherm equation is a pseudoequilibrium constant expression for a reaction * Ads CrO4 . The equilibrium constant is called a mass action (M.A.) of the type: 0.7CrO2– 4 + Ads ) expression, so the exponent is defined as the mass action (M.A.) stoichiometric coefficient.
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The total adsorbate concentration can be input in two ways. One approach is to use the pulldown menu near the bottom of the page to select the adsorbable components that are present in the solution and, after each selection, click the “Add” button. The adsorbable components that are shown include all those that were used to define adsorption isotherms on the database page. In the current example, CrO4 is the only adsorbable component in the database, so it is the only entry in the pulldown menu, and we select it.15 The total concentration of the adsorbable component in the system can then be input by computing the T OTi/csolid ratio and inputting that value in the adjacent box; for the lowest concentration of interest in the current example, and for the units currently in use for T OTi and csolid , this value is T OT CrO4 /csolid =
10 6 mol/L = 5 ⇥ 10 0.002 kg/L
4 mol
kg
Once this ratio is input on the “Adsorption isotherms” page, the corresponding value of T OT CrO4 is computed and appears on the “List of components” page. Alternatively, after using the “Add” button to identify the adsorbable species that are present in the system, we can click the “Save and back to main menu” and input T OT CrO4 directly on the home page without computing the T OTi/csolid ratio; in this case, the program computes the T OTi/csolid ratio and displays that value on the “Adsorption isotherms” page. If a series of simulations is to be run, the initial value of T OTi can be input on the “Multi-problem” page rather than the home page. The fixed pH (7.0) and ionic strength (0.01 M) are also input on the home page in the usual way. For the current problem, we wish to explore a range of T OT CrO4 values, so we migrate to the “Multi-problem” page and indicate that we wish to run 100 simulations in which T OT CrO4 is varied from an initial value of 10 6 M, with an increment of 10 6 M between simulations. The isotherm plot derived from the whole series of runs is shown in Figure 13.15. Having used the isotherm equation to predict the behavior of chromate in sys– tems where it is the only adsorbate, we next consider the effect of SO2– 4 and OH on chromate adsorption, modeling those ions as competing adsorbates and utilizing the Extended Freundlich equation [Equation (13.31)]. Recall that this empirical equation incorporates the concentrations or activities of the competing adsorbates into the isotherm expression for the target adsorbate without explicitly modeling the adsorption behavior of the competing species. For this simulation, we will consider a system containing 10 6 M T OT CrO4 and 9 ⇥ 10 4 M T OT SO4 at pH’s ranging from 5 to 10, assuming that chromate adsorption can be modeled by the following isotherm expression: 15 The
adsorption isotherm database is simply a repository of isotherm equations involving a given solid, so it can include equations for several different adsorbates. If the database is called upon, only the isotherms applicable to the specific problem under consideration are used.
809
Figure 13.15 Isotherm plot for the simulated Freundlich adsorption of chromate onto a hypothetical solid in a noncompetitive system. System conditions are described in the main text.
qCrO4 = 10
10.1
CrO24
0.7
= 10
10.1
CrO24
0.7
= 108.1 CrO24
0.7
OH ✓
Kw {H+ }
H+
1.3
1.3
◆
1.3
SO24
SO24 SO24 0.2
0.2
0.2
(13.37a) (13.37b) (13.37c)
Most of the inputs required to simulate this system in Visual Minteq are identical to those we used previously to simulate the noncompetitive system. The main differences are that we select the “Extended Freundlich” isotherm on the “Adsorp+ tion isotherms” page, we include the activities of SO2– 4 and H when inputting the isotherm equation, and we instruct the program to simulate a range of pH values instead of T OT CrO4 values on the “Multi-problem” page. A screenshot of the “Database” page is shown in Figure 13.16, and the output from the simulation is presented in Figure 13.17. The result for sorption in a noncompetitive system at pH 7.0 from Figure 13.15 is shown for reference. The results indicate that the competition reduces CrO4 sorption by more than half. The preceding discussion provides an extensive but far from comprehensive introduction to the adsorption modeling capabilities within Visual Minteq. Readers are encouraged to explore the other available options, referring to the Help menu or other online assistance as needed.
810
Figure 13.16 Screenshot of the Visual Minteq “Database” page for simulating sorption of CrO4 using the Extended Freundlich isotherm.
Figure 13.17 Concentrations of dissolved and adsorbed CrO4 in simulated system modeling competitive sorption of CrO4 with the Extended Freundlich isotherm.
811
13.5
ADSORPTION OF IONS IN THE PRESENCE OF A SURFACE ELECTRICAL POTENTIAL
13.5.1
Surface Charge, Activity Coefficients for Adsorbed Species, and Apparent Equilibrium Constants
Activity coefficients were introduced in Chapter 2 as a tool to account for the effect of the chemical environment on species’ reactivity, and that concept was further developed in Chapter 4. We also saw in Chapter 2 that activity coefficients of ions depend strongly on the solution ionic strength, and that they can be estimated in low ionic strength solutions with the Debye-H¨uckel equation, and in higher ionic strength solutions by various modifications of that equation. The Debye-H¨uckel equation was originally derived by using the basic principles of physics to determine the strength of the electrical field that develops around a dissolved ion and the extent to which other ions are drawn to or repelled from that region in response. The electrical interactions between the central ion and the surrounding ones were then quantified, and the extent to which those interactions decrease the system’s Gibbs energy and increase its stability were expressed in terms of an activity coefficient, g. Note that, even though we assign values of g to the individual species, in truth it is the whole system that is stabilized by the ionic interactions. That is, what we actually know is that Gtot,sys has declined; our assignment of that change to a particular part of the system is (somewhat) arbitrary. A similar situation applies to adsorption, in that electrical interactions between adsorbed species and their surroundings can stabilize or destabilize the system. However, for adsorbed species, those interactions could be either with ions in solution or with the solid surface. Specifically, if the surface carries an electrical charge, it will generate an electrical potential that extends some distance away from the surface; if an ionic adsorbate resides in that zone, it will be exposed to the potential, and its Gibbs energy will be altered accordingly (increasing if the ion and the potential have the same sign, and decreasing if they have opposite signs). As is the case for solutes in bulk solution, to quantify these interactions, we need to choose baseline (reference) conditions. For adsorbed species, the baseline for interactions with the solution is the same as for fully dissolved species (typically, infinite dilution), and the universal choice for the baseline interactions with the solid are those that occur when the electrical potential (y) to which the adsorbed species is exposed is zero. Thus, an adsorbate bound to an uncharged surface in an infinitely dilute solution is assigned an activity coefficient of 1.0. As noted, in real systems, adsorptive surfaces often do carry a finite charge and therefore have a nonzero electrical potential. We have already seen that such charges can arise via adsorption reactions that form, for example, ⌘SOH+2 , ⌘SO– , ⌘SOCu+ , and similar species. In addition, in some solids, structural ions can be replaced by 812
other ions that are of similar size, but different charge. This phenomenon, referred to as isomorphous substitution, is particularly common in clays, where Al3+ and Si4+ ions in the clay lattice are replaced by ions of lower positive charge, leaving the particle with a net negative charge (and the nearby solution with a positive charge). The surface charge of various pure particles is shown as a function of pH in Figure 13.18. The pH at which the surface is neutral (which occurs for oxides when the concentration of ⌘SOH+2 equals that of ⌘SO– , if H+ is the only adsorbate) is a characteristic of the solid and is referred to as the point of zero charge (PZC or pHPZC ). The PZCs of several metal oxides, clays, and other minerals are presented in Table 13.3.
Figure 13.18 Surface charge of various particles, computed based on the sorption of H+ and OH– ions (by comparison of acid/base titration curves in the presence and absence of the solid). From Stumm, W., and Morgan, J.J. Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd Ed., Copyright c 1996. Reprinted by permission of John Wiley & Sons, Inc.
The profile of y between the surface (where it is designated y ) and the bulk solution (where y = 0 by definition) is critical for characterizing adsorption reactions, because different adsorbed ions might reside at different locations in this region and therefore be exposed to different values of y. Unfortunately, because of the difficulty of probing this extremely thin region, the profile of y vs. distance (x) is not known with certainty; three idealized profiles that have been proposed are shown in Figure 13.19. We will postpone a discussion of the details of these profiles, and simply assume for now that some y vs. x profile approximately like one of those shown in the figure exists. The contribution of electrostatic interactions to the molar Gibbs energy of an ion that is exposed to an electrical potential y, and the corresponding ion activity coefficient, are given by
813
DGi, elec = zi Fy ✓ ◆ zi Fy gi = exp RT
(13.38) (13.39)
where DGi,elec is the change in Gi caused by the electrical interactions, zi is the charge on i (including sign), and F is the Faraday constant.16 Consistent with the preceding discussion, if zi and y are of opposite sign, DGi, elec is negative and gi is < 1, corresponding to an increase in the stability of i. Table 13.3 PZCs of some oxides, clays, and other solids*
Material
PZC
Material
PZC
a-Al2 O3
9.1
d -MnO2
2.8
a-Al(OH)3
5.0
b -MnO2
7.2
g-AlOOH
8.2
SiO2
2.0
CuO
9.5
TiO2
6.2
Fe3 O4
6.5
Feldspars
2.0 - 2.4
a-FeOOH
7.8
Kaolinite
4.6
a-Fe2 O3
8.5
Montmorillonite
2.5
Fe(OH)3 (ferrihydrite)
8.5
Albite
2.0
MgO
12.4
Chrysotile
> 10
*Values are approximate; different methods for preparing the solids can yield PZC values that differ by a few tenths of a unit. From Stumm, W. Chemistry of the Solid-Water Interface. Wiley, New York, 1992.
Based on the analogy between surface sites and dissolved ligands, the standard state concentration for adsorbed species is typically chosen to be 1.0 mol/L in surface complexation modeling. Thus, the activity of adsorbed species (and, by extension, of unoccupied sites) is defined by {⌘SA} = g⌘SA
[⌘SA] [⌘SA] = g⌘SA = g⌘SA [⌘SA] [⌘SA]std.state 1.0 mol/L
16 Equation
(13.40)
(13.38) is essentially identical to Equation (12.74), which expresses the relationship between the electrical potential energy of a species and the electrical potential to which it is exposed. Similarly, Equation (13.39) is an application of Equation (4.86), which expresses the activity coefficient of dissolved ions in terms of the change in Gibbs energy caused by the electrical interactions that they experience.
814
Figure 13.19 Profiles of y versus distance from the surface, as envisioned in three surface complexation adsorption models (not to scale; the d layer is much thicker than the or b layer). The profiles show the magnitude of y; the sign could be either + or . [After Westall and Hohl, Adv. Colloid Interface Sci., 12, 265294 (1980)]
where the final equality is based on the implicit assumption that [⌘SA] is expressed in mol/L units. Substituting the expression for gA from Equation (13.39) into Equation (13.40), and assuming (as we will for the remainder of the chapter) that concentrations of adsorbed species are expressed in mol/L units, we find ✓ ◆ zi Fy {⌘SA} = [⌘SA] exp (13.41) RT Finally, we can use this expression to write the equilibrium constant for a generic adsorption reaction in which a species binds to an unoccupied site (which we represent here as a site that is occupied by a water molecule) as ⌘SH2 O + A )* ⌘SA + H2 O
⇠ ⇠ {⌘SA}⇠ {H⇠ g⌘SA [⌘SA] 2 O} = ⇠[⌘SH O]{A} {⌘SH2 O}{A} ⇠ g⌘SH ⇠⇠ 2 2O ✓ ◆ [⌘SA] zA Fy = exp [⌘SH2 O]{A} RT
(13.42)
Kads, A =
(13.43)
where g⌘SH2 O is 1.0 because zH2 O = 0. If, as is commonly assumed, the electrical potential at the site of adsorption is the only factor that affects the activity coefficient of the adsorbed species, Kads,A is 815
a true equilibrium constant, as valid as any equilibrium constant for a reaction that occurs entirely in solution, or between dissolved species and a solid or gas phase. For this reason, such equilibrium constants are commonly referred to as intrinsic adsorption constants, Kint . Unfortunately, though, Kint values cannot be determined unambiguously, because the value of y at the location where adsorbed A resides can only be estimated based on model calculations. Equation (13.43) is therefore sometimes rearranged as follows to define an experimentally measurable apparent adsorption constant, Kapp : ✓ ◆ [⌘SA] zA FyA Kapp, A ⌘ = Kint, A exp (13.44) [⌘SH2 O]{A} RT From Equation (13.44), it is clear that Kint, A can be defined as the apparent adsorption constant under conditions where yA = 0, and Kapp,A can be defined as the conditional adsorption constant applicable to a particular value of yA .17 We also note, based on Equation (13.44), that 8 if zA yA < 0 > <> Kint [⌘SA] Kapp = = = Kint (13.45) if zA yA = 0 [⌘S]{A} > : < Kint if zA yA > 0 If Kapp is approximately equal to Kint , the implication is that electrical interactions have only a small effect on the adsorption reaction, and that adsorption is dominated by chemical interactions, i.e., that A is chemically adsorbed. On the ⇣ (specifically) ⌘ zA Fy other hand, if Kapp is approximately equal to exp , the extent of adsorption RT is controlled primarily by electrical interactions, and adsorption is physical (nonspecific); in this case, the extent of adsorption depends primarily on the charge of the adsorbate, not its chemical properties.
13.5.2
Distribution and Binding of Ions in the Interfacial Layers
Modern modeling of ionic adsorption represents the interfacial region as comprising at least two, and in some cases three, distinct layers where ions can accumulate. The layer closest to the solid is referred to as the surface or naught ( ) layer; the adjacent layer on the solution side is called the beta (b ) layer; and the layer farthest from the surface is called the diffuse (d) layer. The bulk solution is assumed to start at the outer edge of the diffuse layer. The surface and beta layers have finite thickness, but adsorption models typically treat all the ions in each layer as residing in a single plane running through the center 17 The intrinsic and apparent adsorption constants are often defined based on the concentration, rather
than the activity, of dissolved A. For the remainder of the chapter, we assume that dissolved A behaves ideally, so that the values of Kint,A and Kapp,A do not depend on whether {A} or [A] is used in the calculation.
816
of the layer; these planes are referred to as the naught and beta planes, respectively. Ions in these layers are assumed to be specifically adsorbed, i.e., to form chemical bonds to the solid. Molecular-scale models of these layers typically represent ions in the naught plane as having lost at least some of their waters of hydration so that they can bind directly to the surface, and ions in the beta plane as retaining their hydration sphere and therefore being separated from the surface by a water molecule (Figure 13.20).
Figure 13.20 A conceptual representation of interfacial structure comprising a solid (open circles) and three adsorbed molecules. One adsorbed molecule has lost the water molecules from its primary hydration sphere and binds directly to the surface; a second one has retained its waters of hydration, but still binds to the surface through those waters to form an outer-sphere complex (gray circles); and one is attracted to the surface via electrostatic interactions, but resides in the diffuse layer, not binding directly to the surface at all. (From G. Sposito The Chemistry of Soils. Copyright c Oxford University Press, Inc., New York, NY. Reproduced by permission.)
The diffuse layer begins at the outer edge of the beta layer at a plane referred to as the d plane. The diffuse layer is assumed to be much thicker than the inner layers and to contain ions distributed throughout its thickness. These ions are concentrated or depleted relative to bulk solution strictly due to electrostatic interactions with the surface; i.e., they do not bind directly to the solid, and hence are nonspecifically adsorbed. To emphasize the contrast between the two inner layers and the diffuse layer, the former are sometimes referred to as compact layers. The model that considers ions to be distributed among three layers is called the triple-layer model. The constant capacitance model and the diffuse layer model both posit that adsorbates reside in only two layers – the surface layer (at the naught plane) and the diffuse layer – but they make different assumptions about the profile of y across these two layers. Generic profiles for y as a function of distance from 817
the surface for these three models were shown in Figure 13.19. The two-layer models were developed earlier than the triple-layer model, and based on those earlier models, the whole interfacial region near a charged surface is commonly referred to as the electrical double layer (EDL). All the models assume that, taken as a whole, the electrical double layer must be electrically neutral, so s + sb + sd = 0 (13.46)
where s is the charge density in a given layer, equal to the total charge per unit surface area, with typical units of Coulombs per square meter (C/m2 ). A conceptual schematic based on the triple-layer model of the surface of an iron oxide solid suspended in a solution containing several potential adsorbates is provided in Figure 13.21. The schematic is consistent with the common assumptions that H+ and strong adsorbates bind in the naught layer, weak adsorbates bind in the beta layer, and very weak adsorbates bind only in the diffuse layer (i.e., nonspecifically). The profile of potential in the surface and beta layers is usually assumed to be linear, with different slopes in the two layers. The basis for this assumption is an analogy between the adsorptive system and two parallel-plate capacitors in series in a conventional electrical system. Based on this analogy, the ratio of Dy to s in each layer of adsorbed charge is referred to as the capacitance, C, of that layer. The capacitances of the surface and beta layer are commonly designated as C1 and C2 , respectively, so: C1 = C2 =
s y yb
yb sd sd = yd yd yb
(13.47) (13.48)
C1 and C2 are usually expressed in units of coulombs per volt per unit area (farads [F] per unit area).18 The capacitances C1 and C2 are difficult to measure, and in any case the idea that all the adsorbates are aligned in a plane and that the charge is distributed uniformly on that plane is clearly an idealization. Therefore, C1 and C2 are usually used as fitting parameters to improve the match of model calculations to experimental data. Values on the order of 1 to 3 F/m2 for C1 and 0.2 to 1 F/m2 for C2 are often reported, although values outside those ranges have also been found. The relationship between the charge density in the diffuse layer (sd ) and the potential at the d plane was derived in 1910 for an idealized system that treated the 18 A
single parallel-plate capacitor has equal and opposite charges on the two plates. When two capacitors are placed in series, the middle plate has a charge that is the opposite of the sum of the charges on the outer plates. Thus, in the current case, the model consists of a capacitor near the surface with a charge density of s and one farther from the surface with a charge density of sd .
818
Figure 13.21 Schematic of the interfacial structure as envisioned in the triple-layer model. The charges shown on species in the naught plane represent the sum of the charge from that species plus the Fe atom on the surface; i.e., they represent the total charge expressed at the site. The space between the beta plane and the beginning of the diffuse layer is occupied by hydration waters of the ions in the beta and d planes. Some models do not include a beta layer, in which case the d plane is separated from the surface layer only by the waters of hydration between the ions in the surface plane and the d plane.
ions as point charges, and this relationship is still widely used. Except for geometric considerations, the analysis is just like that for the distribution of charge (i.e., ions) surrounding an ion in solution (that is, the analysis that leads to the Debye-H¨uckel equation).19 For a solution of symmetric electrolytes (salts in which the magnitude of the charge on the cations, |z+ |, is the same as that on the anions, |z |) at 25 C, the result is |z|Fyd sd = 0.1174c0.5 (13.49) b sinh 2RT where sd is in C/m2 , and cb is the electrolyte concentration in bulk solution in mol/L. Equation (13.49) is known as the Gouy-Chapman equation. 19 In
fact, the relationship describing the adsorptive system was derived 13 years before the DebyeH¨uckel equation.
819
Nonspecifically adsorbed ions have no intrinsic chemical affinity for the surface sites, so Kint = 1, and Kapp = exp( zi Fy/RT ). This result can be combined with Equation (13.49) to yield an expression for y as a function of distance from the d plane. For most situations of interest in natural aquatic systems, yd < 25 mV, and y decays approximately exponentially with distance from the d plane toward the bulk solution, i.e.: yx ⇡ yd exp( kx) for yd 25 mV (13.50)
The characteristic distance for this decay (the distance needed for y to decay by a factor of e) is commonly designated by the Greek letter k and is often referred to as the Debye length. k varies with the inverse square root of the ionic strength (i.e., with I 0.5 ) and ranges in natural waters from ⇠10 nm for 10 3 M NaCl to ⇠0.4 nm for seawater. The ionic distribution for a system containing 0.01 M NaCl (k ⇡ 3 nm) is shown in Figure 13.22.
Figure 13.22 Distribution of Na+ and Cl– ions in the diffuse layer near an interface with yd = 25 mV, in a solution of 0.1 M NaCl.
Figure 13.19 shows y vs. x profiles based on the triple-layer model, the constant capacitance model, and the diffuse layer model. In the constant capacitance model, the potential is assumed to decline linearly from y to zero across the surface layer, so that neither a beta layer nor a diffuse layer is present (or, one might say, they are present, but yb = yd = 0). The diffuse layer model assumes that both a compact surface layer and a diffuse layer are present, but that the potential does not decay between the naught plane and the d plane. These latter two models can thus be viewed as limiting cases of the triple-layer model. Some key similarities and differences among these models are summarized in Table 13.4. 820
Table 13.4 Comparison of the charge distribution and electrical potential profile near the surface in three surface complexation adsorption models
Model
Charge Distribution
Constant Capacitance
• Specifically adsorbed species in naught plane • No b layer • Charge equal and opposite to that in the plane provided by nonspecifically sorbed ions, all in d plane
Diffuse layer
• Specifically adsorbed species in naught plane • No b layer • Charge equal and opposite to that in the naught plane provided by nonspecifically sorbed ions, starting in the d plane and distributed throughout the d layer
Triple-layer
• High-affinity specifically adsorbed species bind directly to solid (no intervening H2 O) in naught plane • Lower-affinity specifically adsorbed species bind in b plane, separated from surface by an H2 O molecule • Charge equal and opposite to that in the combined naught and b planes provided by nonspecifically sorbed ions, starting in d plane and distributed throughout d layer
Both the triple-layer model and the diffuse-layer model have been used extensively to model adsorption of trace metals onto oxide adsorbents.20 These and other models often perform approximately equally well at reproducing experimental data, 20 The diffuse-layer model was first proposed by Stumm and coworkers [Stumm, W., Huang, C.P., and Jenkins, S.R. (1970) “Specific chemical interaction affecting the stability of dispersed systems.” Croat. Chem. Acta, 42, 223-244 (1970) and 43: 409-420 (1973)]. The triple-layer model was first fully elaborated by Davis and coworkers [Davis, J.A., James, R.O, and Leckie, J.O. (1978) “Surface ionization and complexation at the oxide/water interface. I. Computation of electrical double layer properties in simple electrolytes.” J. Colloid Interface Sci., 63, 480-499, and Davis, J.A., and Leckie, J.O. (1978) “Surface ionization and complexation at the oxide/water interface. II. Surface properties of amorphous iron oxyhydroxide and adsorption of metal ions.” J. Colloid Interface Sci., 67, 90-107]. Both have since been used in numerous adsorption modeling efforts.
821
but the allocation of the overall adsorptive affinity between chemical and electrostatic contributions, and therefore the Kint values, differs in the different models. As a result, Kint values reported in the literature should only be used in conjunction with the same model that was used to obtain them.
13.5.3
Putting It All Together: The Equations Describing Ionic Adsorption
Based on the preceding discussion, a general approach for computing the distribution of ions between a charged surface and solution can be developed as follows, for a system in which the total concentrations of the ions in the system, the total site concentration, and the Kint values are all known. First, assumptions are made about the locations in the interfacial region where different ions reside. If an ion can bind in both the naught and beta layers, separate Kads values are required for the two reactions. Then, the adsorption density of each ion (Gi , mol/m2 ) is computed based on the mass balances on the ions, the surface site balance, and the Kint values. These calculations are identical to those presented earlier in the chapter, and indicate what the equilibrium state of the system would be if y were zero throughout the interfacial region. A schematic representation of the naught layer based on this initial estimate for the system state is shown on the left side of Figure 13.23. Next, the amount of charge in each layer of the interfacial region is computed, based on the number of anions and cations in that layer; this calculation is represented by arrow 1 in Figure 13.23. The charge in each layer is then converted to a charge density, s (arrow 2 in the figure). The charge densities in the different layers can be related to the change in electrical potential (Dy) across the layers via Equations (13.47) through (13.49), so the y vs. x profile can be computed (arrow 3). Once this profile is established, Equation (13.44) can be used to compute the apparent adsorption equilibrium constants for the ions in the various layers. A revised estimate of the adsorption density for each adsorbate in each layer can then be made using the revised Kapp, i values, and the whole sequence can be repeated iteratively until a set of system conditions (ci , Gi , s in each layer, and y vs. x) is found that satisfies all the equations simultaneously. Consider, for instance, a system in which an adsorbate A2+ can bind in both the naught and beta layers of an oxide adsorbent, via the following reactions: ⌘SOH + A2+ )* ⌘SOA+ + H+ 2+
⌘SOH + A
(13.51) +
+
+ H2 O )* ⌘SO H2 O A + H
(13.52)
Note that, in Reaction (13.53), adsorbed A2+ resides in the beta layer, but the H+ that is released as part of the reaction comes from the naught layer. Therefore, the net 822
Figure 13.23 Schematic of the relationships among adsorbed charge, surface charge, density, and the profile of electrical potential near the interface, and the links between discrete and mean field parameters.
changes in charge are
1 in the naught layer, and +2 in the beta layer.
We assume that oxide groups in the naught layer can undergo protonation or deprotonation, as described previously, that the (intrinsic) equilibrium constants for all the reactions have been evaluated previously, and that the total concentrations of the adsorbate and surface sites (T OT A and T OT ⌘S, respectively) are known. To characterize the surface environment, the values of C1 and C2 also must be known or estimated (somewhat analogous to the need to know the ionic strength if activity coefficients in solution are to be considered). Finally, for this exercise, we will assume that the pH is fixed, although in principle one could leave the pH unspecified and compute it from the T OT H equation. What, then, would be the speciation in the system at equilibrium? For the scenario described, the system has 12 unknowns. These unknowns and the 12 equations available to solve for them are listed in Table 13.5. The only equation that might look unfamiliar is the equilibrium mass action expression for Reaction (13.53). This expression can be derived by noting that the overall reaction is the sum of two half-reactions, one for desorption of H+ from the naught layer, and a second one for adsorption of A2+ into the beta layer, as follows: 823
⌘SOH )* ⌘SO– + H+
(13.53)
⌘SO– + A2+ + H2 O )* ⌘SO H2 O A+
(13.54)
⌘SOH + A2+ + H2 O )* ⌘SO H2 O A+ + H+
(13.53)
Using Equation (13.44) to write the Kapp expressions for these reactions, we obtain the following result for Kapp(13.53) , which is shown as KA,app,b in Table 13.5:
Kapp(13.53) =Kapp(13.54) Kapp(13.55) ⇢
[⌘SO ]{H+ } = exp [⌘SOH] ⇢ =
✓
( 1)Fy RT
[⌘SO H2 O A+ ] ⇠ exp ⇠ [⌘SO ]{A2+ }⇠ {H⇠ 2 O}
✓
◆
⇥
(+2)Fyb RT
◆
✓ ◆ F[y 2yb ] [⌘SO H2 O A+ ]{H+ } exp [⌘SOH]{A2+ } RT
(13.55)
Since the number of equations matches the number of unknowns, a unique solution to the problem is accessible. The algebra is most easily carried out with the aid of chemical equilibrium software, as described shortly. If one were trying to determine Kint values for the various reactions from experimental data, the same set of equations would be used, but the known information would include the equilibrium speciation under a range of conditions, instead of the Kint values. Kint would then be identified as the value that gave the best fit between Kapp computed using the equations in the table and the experimentally determined values. A limited data bank of Kint values is available for adsorption of simple species (mostly uncomplexed metals) onto a few common adsorbents.21 The main application of surface complexation models has been to explore systems in which an adsorbable species is present at a relatively trace concentration in a solution that also contains weakly adsorbing ions (e.g., Na+ , Cl– , HCO–3 , etc.) at concentrations that are orders of magnitude larger than the ion of primary interest. In all such cases, the vast majority of the adsorption in the compact layers is attributable to the strongly binding adsorbate, while the diffuse layer (where all binding is nonspecific) is populated primarily by the major ions from bulk solution. Under the circumstances, it is common to ignore any trace species in the diffuse layer, and to consider only the species in the compact layer(s) as being adsorbed. 21 The most extensive such compilations are for sorption of metal ions and oxyanions onto ferrihydrite (Dzombak, D.A., and Morel, F.M.M. Surface Complex Modeling. Wiley, New York, NY, 1990) and onto the aluminum oxide mineral gibbsite (Karamalidis and Dzombak, D.A. Surface Complex Modeling: Gibbsite. Wiley, Hoboken, NJ, 2010). The Kint values reported in these compilations are based on the diffuse layer model applied to a surface with two site-types.
824
825
Mass action expression for binding A in the b layer (1):
Mass action expressions for species that bind in the naught layer (3):
MB on ⌘S:
H2 O A2 ]
KA,app,b =
✓
Fy RT
◆ F(y 2yb ) [⌘SO H2 O A2+ ]{H+ } = KA,int,b exp 2+ [⌘SOH]{A }{H2 O} RT
KA,app,
[⌘SOA+ ]{H+ } = = KA,int, exp [⌘SOH]{A2+ }
H2 O A2+ ]
H2 O A2 }
+ T OT ⌘S = [⌘SO ] + [⌘SOH] + [⌘SOH+ 2 ] + [⌘SOA ] + [⌘SO ✓ ◆ [⌘SOH]{H+ } Fy Ka1,app = = Ka1,int exp RT [⌘SOH+ 2] ✓ ◆ [⌘SO ]{H+ } Fy Ka2,app = = Ka2,int exp [⌘SOH] RT
T OT A = [A2+ ] + [⌘SOA+ ] + [⌘SO
s , sb , sd
Charge densities:
Equations (12) Mass balances (2): MB on A:
y , yb , yd
+ {⌘SO }, {⌘SOH}, {⌘SOH+ 2 }, {⌘SOA }, {⌘SO
{A}
Potentials:
Activity of surface species:
Activity of soluble species:
Unknowns (12)
Table 13.5 Unknowns and equations available to solve for speciation in a system where species A2+ can sorb in either the naught or the b layer, in a system with known pH
826
Charge-potential relationships (3):
Charge balance in each layer (3): [⌘SO
H2 O A2+ ]
[⌘SO ]
sd = (s + sb ) s C1 = y yb sd sd C2 = = yb yd yd yb zFyd sd = 0.1174I 0.5 sinh RT
sb = 2[⌘SO
+ s = [⌘SOH+ 2 ] + [⌘SOA ]
Table 13.5 – Continued from previous page H2 O A2+ ]
13.5.4
Summary of Adsorption Under Conditions Where Electrostatic Interactions are Significant
To summarize, electrical interactions between charged surfaces and ionic adsorbates can cause the activity coefficients of adsorbed species to deviate from 1.0 and therefore alter the apparent adsorption equilibrium constant for the ions. The magnitude of these effects depends not only on the identity of the adsorbate and the surface, but also on the composition of the system as a whole. The apparent adsorption equilibrium constant can be viewed as the combination of a constant intrinsic value and a variable electrostatic contribution. Ions in the diffuse layer can be treated using the same model construct, but without any chemical binding component (i.e., with Kint = 1.0). At equilibrium, the overall charge around an adsorbent particle, including the fixed charge in the particle’s interior, the charge of specifically adsorbed ions, and the charge in the diffuse layer, must be zero. Therefore, if an ion adsorbs, other processes must combine to expel some like-charged species from the interfacial region and/or bring oppositely charged species into the interfacial region. These processes need not occur at the point where the adsorbing ion resides, but they must take place in the interfacial region. Therefore, adsorption of an ion is always associated with either adsorption or desorption of another ion; it never involves exchange exclusively with water molecules. Sorption reactions that lead to a net change of surface charge distribution (i.e., that have nonzero values of Dz in the interfacial layers) affect and are affected by the electrical potential near the surface; sorption reactions that leave the charge in each interfacial layer unaltered are not affected by that potential, even if the individual species moving to and from the surface as part of the reaction are charged. A number of models exist for the structure and properties of charged interfaces and the binding of ions to surfaces. Modern analytical tools are beginning to provide a great deal more insight into these phenomena. Over time, this new information as well as a continually expanding data bank of intrinsic adsorption constants will enhance the capability to predict adsorption of ions in such systems a priori.
13.6
MODELING SURFACE-ADSORBATE ELECTRICAL INTERACTIONS IN VISUAL MINTEQ
The basic framework for using Visual Minteq to model surface complexation reactions that include electrostatic interactions is identical to that presented earlier for surface complexation modeling of such reactions without electrostatics. The additional information that we need to provide includes the identity of the electrostatic model we wish to use (which establishes the general pattern of the y vs. x profile) and the assumed location of each specifically adsorbed ion (i.e., whether it binds in 827
the naught or beta layer). Once this information is provided, the program carries out iterative calculations to solve all the relevant equations simultaneously. When a solution is obtained, it satisfies not only all the mass balances and equilibrium constant expressions, but also the equations that characterize the electrical potential profile from the surface to the bulk solution. Visual Minteq includes options to use most of the commonly applied models for the distribution of ions in the interfacial region, and also includes parameter values (e.g., specific surface area, Kint values) for adsorption onto a few widely studied adsorbents.22 Here, we explore the use of Visual Minteq to model the pH-dependent adsorption of Cu(II) and As(V) onto hydrous ferric oxide (HFO, also sometimes identified as ferrihydrite or amorphous iron oxide) over the pH range from 2 to 12. We will consider an example system containing 2 ⇥ 10 6 M T OT Cu(II), 10 6 M T OT As(V), 10 2 M NaCl, and an HFO concentration of 10 3 M as Fe. In the Visual Minteq database, HFO is assumed to have a molecular formula of Fe2 O3 · H2 O, which corresponds to an iron content of one mol Fe3+ per 89 g HFO. Therefore, the HFO concentration in the example system is 0.089 g/L. Although the program has the capability to compute the extent to which the solid dissolves and to adjust the concentration of total available surface sites accordingly, HFO is exceedingly insoluble, so we will make the simplifying assumption that all the Fe(III) is present as HFO throughout the pH range considered. HFO is commonly modeled as having two site-types, with surface densities of 1.41 ⇥ 10 3 mol/g HFO and 3.53 ⇥ 10 5 mol/g HFO, respectively. These two sitetypes are usually assumed to be identical in terms of their acid/base properties and their affinities for anionic adsorbates, but the low-density sites are assumed to bind cationic adsorbates much more strongly than the high-density sites do. Values for the site densities, Ka ’s, and adsorption equilibrium constants for several metal ions and oxyanions are included in the Visual Minteq database. These values are based on the compilation of Dzombak and Morel and are applicable in conjunction with the diffuse layer model.23 The relevant reactions and Kint values for the current problem are shown in Table 13.6. Consistent with the preceding discussion, the log Kint values shown in the table for the acid/base reactions and for adsorption of As(V) are the same for binding to both site-types, but those for adsorption of Cu(II) are larger for binding to the 22 Keep
in mind that these values apply only for the specific adsorbent and specific model for which they are reported. Thus, for example, a value of Kint for adsorption of Zn2+ onto Al(OH)3 (s) reported for the triple-layer model cannot be used in conjunction with the diffuse layer model, even though the adsorbent, adsorbate, and adsorption stoichiometry in the two models are identical. 23 Dzombak, D.A., and Morel, F.M.M. Surface Complexation Modeling: Hydrous Ferric Oxide, Wiley, New York (1990). In this text, the authors present a model that they suggest applies not only to adsorption, but also to the transition from adsorption to surface precipitation at large adsorption densities. They refer to this model as the “generalized two-layer model”. In systems where the adsorption density is low and surface precipitation is not expected, this model is identical to the diffuse layer model.
828
low-density sites. Note also that only one surface complex with Cu(II) is envisioned for each site-type, but four different surface complexes containing As(V) are envisioned. Most often, the number and stoichiometry of surface complexes included in a simulation are chosen based on an effort to fit experimental data, and not based on independent evidence that such complexes actually exist. That is, for the adsorbates considered in this example, only one surface complex of Cu(II) is postulated, because the available experimental data could be well fit using only one such complex; on the other hand, multiple complexes of As(V) are postulated because it was difficult to match the experimental observations in any other way. Table 13.6 Adsorption reactions relevant to the example problem24
Reaction 1 2 3 4 5 6 7 8 9 10 11 12 13 14
⌘SOH + H+ )* ⌘FeOH+2
⌘SOHs2 + H+ )* ⌘FehOH+2
Log K int
Dz
7.29
+1
7.29
+1
⌘SOH )* ⌘FeO– + H+
8.93
1
8.93
1
⌘SOH + Cu2+ )* ⌘FeOCu+ + H+
0.6
+1
2.89
+1
+ * ⌘SOH + AsO3– 4 + 3 H ) ⌘FeH2 AsO4 + H2 O
30.98
0
30.98
0
– + * ⌘SOH + AsO3– 4 + 2 H ) ⌘FeHAsO4 + H2 O
25.84
1
25.84
1
19.5
2
19.5
2
11.92
3
11.92
3
⌘SOHs2 )* ⌘FehO– + H+
⌘SOHs2 + Cu2+ )* ⌘FehOCu+ + H+ + ⌘SOHs2 + AsO3– 4 +3H
)* ⌘FehH2 AsO4 + H2 O
– + * ⌘SOHs2 + AsO3– 4 + 2 H ) ⌘FehHAsO4 + H2 O 2– + * ⌘SOH + AsO3– 4 + H ) ⌘FeAsO4 + H2 O
2– + * ⌘SOHs2 + AsO3– 4 + H ) ⌘FehAsO4 + H2 O 3– * ⌘SOH + AsO3– 4 ) ⌘FeOHAsO4 + H2 O
3– * ⌘SOHs2 + AsO3– 4 ) ⌘FehOHAsO4 + H2 O
24 The
representation of HFO surface species in Visual Minteq is different from that for species adsorbed on other solids. The default surface components for the high- and low-site-density site-types are represented as ⌘SOH and ⌘SOHs2, respectively, as is conventional. However, when the sites undergo reactions, Visual Minteq changes the ‘S’ to ‘Fe’ for the high-density sites and to ‘Feh’ (without the ‘s2’) for the low-density sites. Thus, for example the surface acid/base groups for the high- and low-density sites are represented as ⌘FeOH+2 / ⌘SOH/ ⌘FeO– and ⌘FehOH+2 / ⌘SOHs2/ ⌘FehO– , respectively. Complexes with other adsorbates are also shown using Fe and Feh instead of S.
829
To model this system in Visual Minteq, we first input T OT Cu, T OT AsO4 , T OT Na, and T OT Cl in the usual way, and save the file. We then migrate to the “Adsorption/Surface complexation” page, specify that only one surface (i.e., one adsorbent) is present, and select HFO (Dzombak & Morel) as the adsorption model to be used. At that point, several parameter boxes appear, in one of which the program inserts the default specific surface area of 600 m2 /g. If we then type the solid concentration (0.089 g HFO/L) into the appropriate box, the program uses the default values of the site densities for the two site-types to compute the concentration of each site-type in mol/L and fills in the corresponding fields. Because the stoichiometry and log Kint values for all the reactions of interest are in the program’s database, at this point we could proceed directly to the steps for specifying the pH range to be investigated and the “Selected output” to be generated, and run the program. However, for this exercise, we will first look at how the information about adsorption electrostatics can be input if it is not already present. To do that, we migrate to the “Complexaton database” (in this case, feo-dlm 2008.vdb). The page that opens includes information on all the adsorption reactions that are stored in the Visual Minteq database for HFO. The page that appears initially indicates that this adsorbent has two site-types, represented by the uncharged surface components ⌘SOH and ⌘SOHs2. The page also shows a list of all the species that are in the database and that can form by reactions of soluble components with these site-types. If the user highlights any of these species, the initially empty boxes on the screen are populated with the parameters that describe the formation of that species from Visual Minteq components. A screenshot of this page when the species ⌘FeAsO2– 4 is highlighted is shown in Figure 13.24. The information on the page is essentially identical to that provided in Reaction 11 of Table 13.6, except that the component list includes an entry identified as PSIo(1), with a stoichiometric coefficient of 2. In other words, although the actual reaction that is envisioned is Reaction 11, Visual Minteq treats the reaction as though it were ⌘ SOH + AsO34 + H+
H2 O
2 PSIo(1) )* ⌘ FeAsO24
⌘ SOH + AsO34 + H+ )* ⌘ FeAsO24 + H2 O + 2 PSIo(1)
(13.56a) (13.56b)
where Equation (13.56a) is written to show only components as reactants and the surface species as the only product, and Equation (13.56b) shows the same reaction in a more conventional format. PSIo(1) can be considered a ‘pseudocomponent’ that informs the program how the reaction affects the surface charge, and it is the key input that the program uses to incorporate electrostatics into the analysis. The meaning of the parameter is that, when the given surface species forms, the charge in the naught layer decreases by two (because the reaction describes simultaneous adsorption of one AsO3– 4 and one 830
Figure 13.24 Screenshot of the Sorption/complexation database page for the example, showing the stoichiometry for formation of the surface complex ⌘FeAsO2– 4 .
H+ ion into that layer). The corresponding values for all the reactions of interest in the example system are shown in the final column of Table 13.6. In the diffuse layer model, all the adsorbed ions reside in the naught plane, so all the charge carried by the adsorbates resides there as well. If the triple-layer model is used, another pseudo-component [PSIb(1)] is defined and included in the reactions forming surface species, with a coefficient equal to the change in charge in the beta layer when the reaction proceeds.25 Using these coefficients, the program can compute the total charge densities in the naught and beta layers and can then determine the charge density in the diffuse layer from Equation (13.46). It then uses this information to assure that the equations describing system electrostatics are solved simultaneously with the mass action equations when it computes the equilibrium speciation in the system. Values for PSIo(1) [and for PSIb(1) when the triple-layer model is invoked] are included for all reactions that are in the Visual Minteq database. To add a new surface 25 Note
that, despite their names, PSI (1) and PSIb(1) are not the same quantities as y and yb ; they are merely the names for the pseudocomponents used to indicate how a reaction affects the electrical charge in a given interfacial layer.
831
species to the database, we follow the same procedures as were described previously for adding a surface species in the absence of electrostatic effects, with one exception: if electrostatic effects are considered, the “Which component type” pulldown menu includes an option to specify an “Electrostatic comp.” By choosing this option, the user can add the pseudo-component PSIo(1) to the reaction to indicate how the reaction affects the surface charge. If no such entry is included, the program calculations are identical to those carried out using the nonelectrostatic model (i.e., with the NEM database). Again, because all the surface reactions of interest for this particular example are included in the default Visual Minteq database, we need not make any modifications to the database. We can therefore “Quit” this page, specify the pH range to explore and the desired output in the usual way, and run the simulation. Figure 13.25 shows the acid/base speciation of the surface, along with the computed surface potential. As we would expect, for both site-types, the more protonated surface species dominate at low pH, and protons are released from the surface as the pH increases. The unprotonated, singly protonated, and doubly protonated sites account for almost all the sites in the high-surface-density group, because the total concentrations of the adsorbates are not large enough for those species to occupy a significant fraction of the total sites, even at 100% adsorption. Perhaps surprisingly, the intersection of the ⌘FeOH+2 and ⌘SOH curves (i.e., pKa1,app ) is at pH ⇠4.4, well below the corresponding pKa1,int of 7.29. In other words, ⌘FeOH+2 is a good deal more acidic than one would expect based on its intrinsic acidity constant.
Figure 13.25 Surface acid/base speciation and y in the example system.
The explanation for this behavior lies in the value of y , which is also plotted in the figure. This value is positive at low pH and declines steadily with increasing pH, passing through zero at pH 8.10. (The slight decline in y when the pH de832
creases below 2.5 is explained shortly.) Thus, when the pH equals pKa1,int , the surface potential is positive, lowering the surface’s affinity for additional H+ ions. As a result, H+ ions are less likely to attach to an ⌘SOH group than they otherwise would be, and [⌘SOH] at pH=pKa1,int is greater than [⌘FeOH+2 ]. As the pH is lowered below pKa1,int , the increasing H+ activity in solution increases the driving force for protonation of the surface sites, but with each additional site that becomes protonated, the value of y becomes more positive, making it even more difficult for additional H+ ions to bind. In the end, the calculations indicate that the ⌘FeOH+2 concentration does not equal that of ⌘SOH until the pH declines to 4.4. A similar phenomenon applies to the ⌘SOH / ⌘FeO– couple, except that in this case the negative surface potential causes the surface groups to remain protonated under conditions where H+ would otherwise be released. As a result, the concentrations of the acid and its conjugate base are not equal until the pH is raised to ⇠11.9, whereas pKa2,int is 8.93. The same general observations apply to the acid/base speciation of the lowsurface-density sites, except that the surface complexes of those species with Cu and As can account for a significant fraction of the total sites, so the sum of the concentrations [⌘FehO– ] + [⌘SOHs2] + [⌘FehOH+2 ] does not approximate T OT ⌘SOHs2. The net effect is that, for both groups of sites, the range of dominance of the neutral surface species (⌘SOH and ⌘SOHs2) is greatly expanded compared to the case where solid surface potential is less sensitive to pH. The dominant forms of copper and arsenic in the system are shown in Figure 13.26. As is typically the case, over the range of interest in natural waters, adsorption of copper increases and that of arsenic decreases with increasing pH. At pH greater than about 6, virtually all the copper is adsorbed. The adsorbed copper is primarily on the high-affinity, low-surface-density sites, and it occupies approximately 30% of those sites. Copper adsorbed on the low-affinity sites occupies a negligible fraction of those sites, but nevertheless accounts for about 25% of the total copper on the surface. At low pH, most of the copper remains in solution as Cu2+ ; if higher pHs were considered, most of the adsorbed copper would return to the solution 2+ as Cu(OH)–3 and Cu(OH)2– 4 complexes. The sorption of Cu always brings additional positive charge into the naught layer, so sorption is impeded by the surface potential when y is positive and enhanced by it when y is negative. The partitioning of arsenic in the system is, in many ways, the mirror image of that for copper. Almost all of the arsenic is adsorbed from the lowest pH considered up to pH ⇠10, but it then desorbs as the pH is raised to ⇠12. Although the overall distribution of the arsenic between the solid and solution is unchanging over most of the pH range, the speciation of the surface complexes does change, with the dominant surface species becoming progressively less protonated as the pH increases. The surface potential has the opposite effect on arsenic adsorption as it does on copper adsorption. 833
Figure 13.26 Distribution of major species of (a) copper and (b) arsenate in the example system.
Note that, when either copper or arsenic adsorbs, the electrostatic effect depends not just on the charge of the adsorbate, but on the net change in the surface charge (i.e., Dz ) accompanying the reaction. For example, at pH 7, the dominant reaction occurring when copper adsorbs is Cu2+ + ⌘SOHs2 )* ⌘FehOCu+ s2 + H+
(13.57)
Although the copper brings two positive charges to the surface in this reaction, one positive charge is simultaneously expelled from the surface (as H+ ), so the net electrostatic effect reflects the interaction between y and a single positive charge. This means, for example, that if copper adsorbed at a pH where the dominant surface site was doubly protonated, the value of y would have no bearing on the 834
tendency for the reaction to proceed, because the primary adsorption reaction would not affect the amount of adsorbed charge: + + * Cu2+ + ⌘SOH+ 2 s2 ) ⌘FehOCu s2 + 2 H
(13.58)
In Reaction (13.59), the electrostatic energy associated with the adsorption of Cu2+ is exactly balanced by that associated with release of two H+ ions. It is for this reason that the key input to Visual Minteq relating to the electrostatic effects is the change in surface charge associated with the reaction, and not just the charge on the adsorbing ion.
13.7
SURFACE PRECIPITATION
Thus far, we have implicitly assumed that adsorbed species reside far enough from one another that they do not interact, other than their indirect effect on one another via the charge that they bring to the surface. In this final section, we consider the transition from individual adsorbed species to formation of a surface precipitate, a transition that is bound to occur when the adsorption density of certain adsorbates becomes large. This scenario is depicted schematically in Figure 13.27. Empirically, this process causes the amount of adsorbate bound to the surface to increase steadily in response to increasing dissolved adsorbate concentration, but then, rather than approaching some maximum value, to continue to increase without bound. In truth, the latter trend does not reflect additional adsorption, but rather the formation of the new precipitate. If the precipitate is a metal hydroxide and solution pH is fixed, the solubility product imposes a limit on the amount of the adsorbate that can be dissolved. As a result, addition of an increment of metal to a system where a surface precipitate has formed leads to precipitation of virtually all the added metal. If adsorption is quantified simply by analyzing the amount of metal that is removed from solution, the precipitation reaction can easily be mistaken for adsorption, causing the apparent adsorption density to increase indefinitely. Few experimental studies have investigated conditions that span the regions where both adsorption and surface precipitation occur. In those studies, although the endpoints of the process (adsorption at low dissolved adsorbate concentrations, and precipitation at high concentrations) are reasonably well defined, the transition between the two states has not been clearly identified. In some cases, changes in the properties of the adsorbent provide evidence for the formation of surface precipitates. For instance, Figure 13.28a shows the electrophoretic mobility (the movement of colloidal particles when they are exposed to an electrical field) of pure SiO2 (s) and pure Co(OH)2 (s) as a function of pH. In a system containing SiO2 (s) and dissolved Co2+ , the mobility of the particles is similar to that of pure SiO2 (s) at low pH, but it gradually deviates from that pattern between pH 4 and 8. This deviation is presumed to be caused by adsorption of 835
Figure 13.27 Schematic showing on adsorbate A that is first bound to discrete sites on the adsorbent, but that then forms a surface precipitate covering much of the solid, so that the surface of the final product behaves almost like that of a particle of pure A(s).
Co2+ to the SiO2 (s) surface, reducing the negative surface charge. At pH 8.0, however, a dramatic change occurs, and the electrophoretic mobility approaches that of pure Co(OH)2 (s). The logical explanation for this change is that Co(OH)2 (s) has precipitated on the SiO2 (s) surface, causing the particle to behave almost like pure Co(OH)2 (s). Figure 13.28b demonstrates another effect of surface precipitation. This figure indicates that, in the absence of Co2+ , chromate (CrO4 ) sorption onto Fe(OH)3 (s) decreases as pH increases from ⇠6 to ⇠9 (curve Cr/C). However, when Co2+ is present along with the CrO4 and Fe(OH)3 (s), the Co2+ binds to the Fe(OH)3 (s) as the pH increases from 7 to 8 (curve Co/A), and the CrO4 adsorbs along with it (curve Cr/A and triangles). The sorption of the CrO4 at higher pH in the presence of Fe(OH)3 (s) and Co2+ is similar to the pattern for sorption of CrO4 onto Co(OH)2 (s) in a system with no Fe(OH)3 (s) (curve Cr/B), suggesting that the adsorption of CrO4 at pH > 7.5 in the Fe(OH)3 (s) + Co2+ system is caused by precipitation of a surface layer of Co(OH)2 (s). Note that, in this case, the data suggest that a Co(OH)2 (s) surface precipitate forms under conditions where Co(OH)2 (s) would not precipitate from bulk solution. The most widely cited attempt to model the transition between adsorption and surface precipitation was proposed by Farley et a1.27 and is based on the formation of an ideal solid solution or mixed solid at the surface. The solid solution is postulated 26 James,
R.O., and Healy, T.W. (1972) “Adsorption of hydrolyzable metal ions at the oxidewater interface. II. Charge reversal of SiO2 and TiO2 colloids by adsorbed Co(II), La(III), and Th(IV) as model systems.” J. Colloid Interface Sci. 40, 53-64, and Benjamin, M.M. (1983) “Adsorption and surface precipitation of metals on amorphous iron oxyhydroxide.” Environ. Sci. Technol. 17, 686-692. 27 Farley, K.J., Dzombak, D.A., and Morel, F.M.M. (1985) “A surface precipitation model for the sorption of cations on metal oxides.” J. Colloid Interface Sci. 106, 226-242.
836
Figure 13.28 (a) The electrophoretic mobility of particles in a system containing 0.002 g/L SiO2 (s) alone, 10 4 M T OT Co alone, or both SiO2 (s) and Co2+ . In the presence of the SiO2 (s) particles, the Co2+ can adsorb or precipitate onto the SiO2 (s) surface. (b) Removal of Co(II) and Cr(VI) from solution in systems with and without ferrihydrite. Explanations for the labels in part (b) are provided in the main text. Part (a): after James and Healy (1972); part (b): after Benjamin (1983).26
to form at very low adsorption densities and to become gradually more enriched with adsorbate as the apparent adsorption density increases. In the model, a fixed portion of the adsorbent is assumed to participate in formation of the mixed solid. The activity of the adsorbate in the idealized mixed solid phase is presumed to be proportional to its mole fraction in that phase. Defining qsp as the apparent adsorption density attributable to surface precipitation (i.e., the mass of precipitated adsorbate per unit mass of adsorbent in the system), the following equation describing qsp as a function of the dissolved adsorbate concentration can be derived by combining the solubility product expression with a mass balance: qsp =
f (ceq /c0 eq ) f K 0 ceq = 1 ceq /c0 eq 1 K 0 ceq
(13.59)
where f is the mass of adsorbent that contributes to the mixed solid phase, normalized to the adsorbent surface area (e.g., mg /m2 ); c0eq is the adsorbate concentration that would be in equilibrium with the pure precipitated solid (e.g., if the precipitating 2 solid is a metal hydroxide, c0eq = Ks0,Me(OH)2 (s) /{OH } ); and K 0 = 1/c0eq . The form of Equation (13.59) is similar to the Langmuir isotherm, except that the sign in the denominator is opposite in the two expressions (plus in the isotherm, minus in the expression for surface precipitation). The equation includes two constants: one that describes the amount of the adsorbent that participates in the reaction 837
forming the solid solution (f ) and is analogous to qmax in the Langmuir isotherm, and one related to the thermodynamic driving force for binding, analogous to Kads . Equation (13.59) indicates that at low ceq , qsp is a linear function of ceq , but that as ceq approaches 1/K 0 , qsp gets very large (Figure 13.29). The equilibrium dissolved adsorbate concentration, ceq , can never exceed c0eq , since c0eq is the adsorbate concentration in equilibrium with the pure solid; as ceq approaches c0eq , the composition of the mixed surface solid approaches that of a pure precipitate of the adsorbate. Since ceq cannot exceed c0eq , almost all the adsorbate that is added to solution under these conditions precipitates, and the apparent adsorption density grows without bound.
Figure 13.29 The apparent adsorption density attributable to surface precipitation as a function of the equilibrium concentration of dissolved adsorbate, according to the Farley et al. model for surface precipitation. For this plot, f = 1, K 0 = 0.001 (both in arbitrary units). As the product K 0 ceq approaches 1.0, the surface precipitate approximates a pure precipitate of the adsorbate, and qapp increases without bound. The dashed line corresponds to qapp = f K 0 ceq .
The total apparent adsorption density (qapp ) can be computed as qsp + qL , where qL , is the true adsorption density on the original solid, which is presumed to follow a Langmuir isotherm. Figure 13.33a shows the contributions of qsp and qL to qapp for two hypothetical systems, one in which adsorption is relatively strong and one in which it is weak. In both cases, adsorption accounts for most of the adsorbate binding at low ceq (ceq less than ⇠200 and ⇠500 in the systems with weak and strong adsorption, respectively), and surface precipitation dominates at high ceq . Assuming that the adsorption and surface precipitation processes occur in parallel (e.g., on different parts of the surface), the apparent isotherm that one would obtain from experiments is the sum of the Langmuir and surface precipitation isotherms, which is shown for each system in Figure 13.33b. In the system with strong Langmuir binding, the empirical isotherm looks much like a simple Langmuir isotherm at low 838
Figure 13.30 Apparent adsorption isotherms for systems in which both Langmuir adsorption and surface precipitation can occur simultaneously. (a) Isotherms for adsorption and surface precipitation separately. (b) Apparent isotherms for simultaneous adsorption and surface precipitation.
to moderate ceq , and like the isotherm for surface precipitation at higher ceq . Thus, in this case, the isotherm provides a fairly clear indication of the transition from an adsorption-dominated system to one dominated by surface precipitation. However, in the system with weaker adsorption, the curve is qualitatively similar to one for surface precipitation alone. Farley et al. suggested that much of the metal ion adsorption data in the literature might be characterized by relatively strong adsorption and might fall in the region of Figure 13.33b labeled as the pseudo-Freundlich region. (The absolute values of ceq and qapp in this region depend on the particular metal ion and adsorbent under study, but the shape of the curve would be the same for any system in which adsorption 839
is relatively strong.) They also showed that the relationship between qapp and ceq in this region can be fit quite well by a Freundlich isotherm. The implication is that a system that obeys a Freundlich isotherm might actually be undergoing a combination of Langmuir-type adsorption and surface precipitation, rather than adsorption onto a surface with a wide range of site affinities (the classical explanation for such behavior, presented earlier in the chapter). Visual Minteq incorporates the possibility of surface precipitation and uses the model of Farley et a1. as the basis for the calculations. To include this feature, the user must define a new pseudocomponent that accounts for the solid solution. Instructions for using this feature are available in the program’s Help menu.
13.8
SUMMARY
Adsorption, or the binding of molecules at interfaces, is a critical process in natural waters and engineered water treatment systems. As a practical matter, adsorption is important because it can determine whether the fate of a chemical species is more closely linked to the movement of the solids or the solution in a system. Adsorption also often has dramatic effects on the rates of redox and mineral precipitation and dissolution reactions, the availability of the adsorbing species to organisms, and the interactions among colloidal particles suspended in a solution. For all these reasons, adsorption reactions have important geological and ecological implications. Adsorption of hydrophilic species (most inorganic chemicals, natural organic matter, and other ionizable organic species) is commonly modeled as a surface complexation reaction, with surface sites behaving essentially as complexing ligands. Often, both the adsorbates and the surface ligands can participate in acid/base reactions, causing the net amount of sorption to depend strongly on solution pH. Competition for surface sites among adsorbates, competition for adsorbate between the surface and dissolved ligands, and electrostatic interactions at the surface all play important roles in determining the distribution of adsorbate molecules between the solution and the adsorbent particles. The net result of all the factors affecting adsorption can often be synthesized into a relatively simple equation relating the adsorption density to the concentrations of adsorbate and surface sites in the system. Such equations are called adsorption isotherms, and they are often useful for characterizing the sorption process over a limited range of conditions. However, there are fundamental reasons (e.g., the heterogeneity of all surfaces, the effects of adsorbates on the surface electrical potential, interactions among adsorbed species) why simple isotherms are expected to fail to describe such systems completely. By incorporating more of the complexities of adsorption reactions into models, the range of applicability of those models can be extended. Acid/base chemistry, both in solution and at the surface, is a central component of the sorption of ioniz840
able adsorbates onto inorganic surfaces, and its incorporation into adsorption models is now fairly well established. Attempts to account for the relationships among electrical potential near the surface and the adsorption of charged species in adsorption modeling are also reasonably well developed, but are still being modified to take into account new information about surface speciation and structure. Models accounting for surface heterogeneity and interactions among adsorbed species (including formation of surface precipitates) are less developed. Visual Minteq provides the option to use any of several adsorption models to analyze the equilibrium speciation in a system. Some of these, like the HFO model, include data for the surface properties of particular adsorbents, and the available Kint values for sorption to that solid; others contain only default values for the surface acid/base reactions. The use of the program to model the distribution of adsorbates in some simple systems has been demonstrated here, and these examples can be used to simulate many systems of interest. The program can be used to simulate many other scenarios (e.g., ones in which the adsorbent forms by precipitation of a solid in the system, so that the amount of adsorbent changes as a function of pH and other system parameters), and readers are encouraged to explore these capabilities on their own. Because of the central role of adsorption in so many environmental processes, it remains an area of intense research interest. The many topics being addressed currently include adsorption onto mixed assemblages of solids, which are much more common in nature than are the pure solids that have been studied most extensively in the past; the adsorption of complex polyelectrolytes like NOM and its effect on sorption of other ions; the factors that control whether and to what extent adsorbed species catalyze redox reactions; and the microscopic structure of the interfacial region. These efforts are likely to dramatically enhance our understanding of and ability to model adsorption reactions in the future.
13.9
PROBLEMS
Please re-read the note preceding the Chapter 8 Problems on p.457. The same guidelines apply to the following problems. 1. Stillings et al. (1998)28 report the following data for adsorption of oxalate species onto a natural aluminosilicate mineral. The data were collected in solutions at pH 3 to 5, but the authors found that the total adsorption of oxalate was independent of pH in this range. (a) Estimate best-fit parameters to model the data according to the Langmuir and Freundlich isotherms. 28 Stillings,
L.L., Drever, J.I., and Poulson, S.R. (1998) “Oxalate adsorption at a Plagioclase (An47 ) surface and models for ligand-promoted dissolution.” Environ. Sci. Technol. 32, 2856-2864.
841
(b) Use the isotherms derived in part (a) to estimate the adsorption density at an equilibrium dissolved oxalate concentration of 10 mmol/L. Which estimate would you use if you needed to predict qOx for those conditions? {Ox}diss, eq , mmol/L
qOx , µmol/m2
{Ox}diss, eq , mmol/L
qOx , µmol/m2
0.00 0.65 0.94 1.00 1.54 1.65 1.93
0.00 8.03 6.50 2.60 16.74 11.52 10.93
3.45 3.49 3.54 6.50 6.53 6.81
16.84 18.17 18.07 23.69 23.38 18.20
2. In many water treatment plants, contaminants are removed from the feed solution as it flows through a column packed with adsorbent particles; a similar process can occur as groundwater flows through soil particles. Such systems are sometimes modeled as a series of batch reactors containing adsorbent, with flow of water represented by movement from one batch reactor to the next. The compound methyl-isoborneol (MIB) can give water an earthy-musty odor, even at concentrations of just a few ng/L. Gillogly et al.29 reported that MIB sorption from a pure water (no competing adsorbates) onto activated carbon (AC) can be characterized by q = 9.6c0.5 , where q is in ng MIB/mg AC and c is in ng/L. Compare the equilibrium dissolved MIB concentration in a system in which 5 mg AC/L is added to a single batch reactor with that in a system in which the water flows through four batch reactors in series, each containing 1.25 mg AC/L, if the initial MIB concentration in the water is 100 µg/L. 3. In a study of sorption of arsenate and sulfate onto an ion exchange resin, both adsorbates obey Langmuir isotherms. When the resin is initially loaded with Cl– and only one of the adsorbates is present, Kads is 0.12 L/µmol for T OT AsO4 and 1.20 L/µmol for T OT SO4 , and qmax is 150 µmol/g for both adsorbates. Compare the dose of resin required to reduce the arsenate concentration from 1.0 to 0.07 µmol/L in solutions initially containing negligible and 150 mg/L T OT SO4 . 4. Equation (13.23) suggests that if multiple adsorbents are present in a system, each one behaves independently, and they are linked only through the fact that they must all equilibrate with the same solution. However, there are cases 29 Gillogly,
T.E.T., Snoeyink, V.L., Elarde, J.R., Wilson, C.M., and Royal, E.P. (1998). “14 C-MIB adsorption on PAC in natural water.” Journal AWWA 90(1), 98-108.
842
where the assumption of independence among adsorbents fails. For instance, Figure 13.31 shows a case where adsorption decreased when a second adsorbent was added. Suggest some possible explanations for this observation.
Figure 13.31 Fractional removal of 10 6 M phosphate from two solutions containing 10 4 M Fe(OH)3 (s). Phosphate does not sorb onto SiO2 (s) under these conditions, and the addition of SiO2 (s) to the system containing Fe(OH)3 (s) causes sorption of phosphate to decrease [from Anderson and Benjamin30 ].
5. Data for sorption of Cd onto ferrihydrite (Fe(OH)3 (s)) are shown in Figure 10.5 for solutions with ionic strength 0.7 M and containing substantial con2+ centrations of NO–3 , Cl– , and/or SO2– forms complexes with Cl– and 4 . Cd – 2+ SO2– 4 , but not with NO3 . One hypothesis to explain the data is that free Cd can adsorb to the solid, but complexed Cd species cannot. In carrying out the following calculations, assume that the ionic strength affects the activity of solutes, but not that of adsorbed species. Based on the data point where 50% of T OT Cd is adsorbed in the absence of the ligands, estimate the apparent equilibrium constant for the following reaction: ⌘FeOH + Cd2+ )* ⌘FeOCd+ + H+ . Then, use this value of Kads to model the pH-sorption edge for the solution containing 0.5 M Cl , assuming that Cd Cl complexes cannot adsorb and that, in the pH range of interest, essentially all the surface sites are singly protonated (i.e., ignore the species ⌘FeO– and ⌘FeOH+2 ). Compare the predictions with the experimental results. 6. In ‘ligand-promoted dissolution’ processes, dissolution of a solid is thought to proceed via adsorption of the ligand, complexation of a metal ion in the 30 Anderson, P.R., and Benjamin, M. (1990) “Surface and bulk characteristics of binary oxide suspen-
sions.” Environ. Sci. Technol. 24(5), 692-698.
843
solid’s surface layer, and release of the complex into solution. Often, the rate of dissolution is proportional to the adsorption density of the ligand, as shown in Figure 13.32a. However, in other cases, almost no dissolution occurs until enough complexing agent has been added to saturate the available surface sites, as shown in Figure 13.32b.
Figure 13.32 (a) Enhancement in the rate of dissolution of Al2 O2 (s) attributable to the presence of four carboxylic acids [from Furrer and Stumm31 ]. (b) Increase in the dissolution rate of ferrihydrite by addition of P3 O10 . qmax is 0.10 mol P3 O10 /mol Fe. The data points at the far right progress upward as more P3 O10 is added to the system, all of which remains in solution because q = qmax [from Lin and Benjamin32 ]. 31 Furrer, G., and Stumm, W. (1986) “The coordination chemistry of weathering: I. Dissolution kinetics of d -Al2 O3 and BeO.” Geochimica Cosmochimica Acta 50(9), 1847-1860. 32 Lin, C.F., and Benjamin, M.M. (1990) “Dissolution kinetics of minerals in the presence of sorbing and complexing ligands.” Environ. Sci. Technol. 24(1), 126-134.
844
By writing out all the reactions that you expect to take place in such systems and considering the different ways that these reactions might compete with one another, suggest an explanation for the different types of behavior. (Hint: tripolyphosphate species (Hx P3 O5–x 10 ) are much stronger complexing agents and also much stronger adsorbates that the other ligands shown.) 7. Ferrihydrite is to be used to treat a photo processing waste solution that con– + tains Ag+ , S2 O2– 3 , Na , and NO3 . Any of these ions can adsorb to the ferrihydrite surface, with the former two binding much more strongly than the latter two. Write the reactions that would have to be considered to compute the equilibrium composition of the solution, and prepare the input table to model the system using the Triple Layer Model within Visual Minteq. Consider acid/ base reactions of the surface, sorption of the four free ions, complexation of Ag+ with S2 O2– 3 , and sorption of AgS2 O3 (aq) complexes. Assume that free – + Ag+ and S2 O2– 3 ions adsorb in the ‘o’ layer; that Na and NO3 adsorb in the b layer, and that when AgS2 O3 (aq) sorbs, the S2 O3 portion of the molecule resides in the ‘o’ layer, but the Ag portion resides in the beta layer. 8. Figure 13.33 shows linearized fits to the Langmuir isotherm according to Equation 13.11 for sorption of Cd onto goethite (a-FeOOH) at pH 7.0 and various temperatures. Estimate the molar enthalpy of the adsorption reaction and predict the adsorption density of Cd in equilibrium with a dissolved concentration of 5 ⇥ 10 5 M at 2 C.
Figure 13.33 Linearized Langmuir isotherms for adsorption of Cd2+ onto goethite [after Johnson33 ].
33 Johnson,
B.B. (1990) “Effect of pH, temperature, and concentration on the adsorption of cadmium on goethite.” Environ. Sci. Technol. 24(1), 112-118.
845
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The following collection of equations is provided as a reference and to help readers identify locations in the text where the equations are presented and discussed. Note that many of these equations are applicable only under limited conditions that are specified in the text. The equations should not be used without reviewing those limitations and making sure they are met in the pariticular case of interest. Pi =
nG,i RT = cG,i RT Vtot
(1.5)
nG,i RT Ptot
(1.6)
1 ci z2i 2 all ions
(1.17)
Vi = I=
k = Â ki = Â li ci zi i
(1.21)
i
gA =
aA cA /cA
(2.1d)
log10 (g D-H ) = Az2 I 1/2 log10 (g Ext.D-H ) = Az2 log10 (g Davies ) = Az2
(
(2.4)
I 1/2 1 + BaI 1/2
I 1/2 1 + I 1/2
0.3 I
(2.5) )
log10 (g Pitzer ) = log10 (g Ext.D-H ) + Â Bi j I m j j
867
(2.6) (2.7)
Table 2.3. Relationships among Q, K, and the direction of chemical reactions
Q
log(Q/K) < 0
Not at equilibrium
Reactants ! Products
Q>K
log(Q/K) > 0
Not at equilibrium
Reactants
Q=K
log(Q/K) = 0
At equilibrium
No net reaction
K|T1
{H+ } OH H2 O ✓ ◆ DH r 1 1 = R T1 T2
then:
Koverall = ’ Ki
Kw = ln If Roverall = Â Ri , i
K|T2
(2.11) (2.17) (2.21)
i
k = kAr exp
✓
ln k = ln kAr kf0 {C} {D} = 0 {A} {B} kr
◆
(3.3)
EAr 1 R T
(3.4)
EAr RT
eq
⌘K
(3.16b)
{C} {D} Q kr0 {C} {D} rr {A} {B} = = = K kf0 /kr0 kf0 {A} {B} rf K|T2 K|T1 ln
K|T2 K|T1
DEAr = exp R =
✓
DEAr R
✓
◆✓
1 T1
1 T1
dH = dU + d (PV ) 868
Products
1 T2 1 T2
◆
◆
(3.17)
(3.19a)
(3.19b)
(4.32a)
DH = DU + D (PV )
(4.32b)
H = U + PV + H ⇤
(4.33)
H = U + PV
(4.34)
DSsurr = DG = DH
DHsys T
(4.36)
T DS
(4.39)
DG = T DStot G⌘H
(4.40)
(T S) + G⇤
(4.41)
∂G ∂ ni
(4.42)
µchem,i = Gi ⌘
dG = Â Gi dni
(4.45)
dG = Â dGi = Â Gi dni
(4.48)
all i
all i
Xi ⌘
all i
∂X ∂ ni
(4.51) n j6=i ,P,T ⇤
Gi = H i
T Si + Gi
(4.53)
Gi = Gi + RT ln ai
(4.66)
DGi = RT ln ✓
ai,2 ai,1
◆ kJ Gi = Gi + 5.71 log ai mol ✓ ◆ ai,2 kJ DGi = 5.71 log mol ai,1 869
(4.67)
(4.68) (4.69)
0
g i = exp @
D Ginter,i
comp
0
gi = exp @
0
RT D Hi
comp
RT
DGi = V i DP
DGi = Gi,T2
Gi,T1 =
(4.87)
1 A
(4.88) (4.89)
P2 P1
◆ 1 Gi,T1
T2 T1
(4.86)
A
ci comp = exp @ ci RT
✓
A
1
D GS,i
DGi = RT ln
1
(4.90)
T2
ZT2
T1
H i + G⇤i dT T2
 ni dGi = 0
(4.93) (4.98)
all i
dG =
dG ⌘ DGr = aGA dh DGr = a GA
= Â ni Gi
bGB + cGC + dGD
b GB + c GC + d GD
(4.112)
(4.114a) (4.114b)
i
DGr =DGr + RT ln
{C}c {D}d
{A}a {B}b
DGr = DGr + RT ln Q ✓
(4.115) (4.116)
◆ kJ DGr = DGr + 5.71 log Q mol
(4.117)
DH r = Â ni H i
(4.118)
870
DSr = Â ni Si DGr = DH r DGr
(4.119)
T DSr =0
eq
(4.123)
DGr = RT ln K K = exp
DGr RT
✓
◆
DGr =
(4.120)
kJ 5.71 mol
(4.126a)
!
(4.126b)
log K
Q K ✓ ◆ kJ Q DGr = 5.71 log mol K DGr = RT ln
ZKT2
KT1
ln
1 d ln K = R
DH r,25 KT2 = KT1 R
ZT2
T1
C
(4.130) (4.131)
DH r dT T2
✓
1 T1
1 T2
(4.127)
(4.142)
◆
DGr,overall = Â DGr,i
(4.143) (4.148)
i
Ka ⌘
H3 O+ {A } {H+ } {A } = {HA} {HA} H2 O Kb = Ka 1 Kw
(5.1) (5.14)
Ka Kb = Kw
(5.15a)
pKa + pKb = pKw
(5.15b)
871
log
{A } = log H+ + log Ka = pH {HA}
pKa
{H+ } 1 = + Ka + {H } (Ka /{H+ } ) + 1 Ka 1 a1 = = + Ka + {H } 1 + ({H+ }/Ka ) a0 =
(5.22) (5.23)
d log [A ] d log [HA] = +1 d pH d pH pKa,eff = pKa + log {H+ }
ai =
n
Â
i
(5.17)
(5.27)
gA gHA
(5.31)
i
’ Ka j j=0
+
H
k
k
’ Ka j j=0
k=0
(5.41)
!
nA, j Comp A + nB, j Comp B + ... + nN, j Comp N )* 1Spec j
(6.4)
Meq
T OTieq = ni,1 c1 + ni,2 c2 + ... + ni,M j cMeq = T OTiin = ni,1 c1 + ni,2 c2 + ... + ni,Min cMin =
f⌘
ALK =
equiv H+ added during titration
L
 ni, j c j
Min
 ni,k ck
=
L
= T OT Hafter
ALKCO3 ⇡ [HCO3 ]init + 2[CO23 ]init 872
(6.18)
k=1
Base added (equiv/L) T OT A (equiv/L) mol H+ added during titration
(6.14)
j=1
(8.6)
T OT Hbefore
(8.9)
(8.15a)
= (a1 + 2a2 )T OT CO3
(8.15b)
pH 4.5
ALKadded i = T OT Hi |as added dT OT Hstrong dpH
b ⌘
(8.26)
⇥ bw = 2.303 H+ + OH
⇤
(8.40b)
i)2 ai a j
(8.45)
bHA = 2.303 (T OT A) a0 a1 n n 1
bHn A = 2.3(T OT A) Â Â ( j j>i i=0
cG,i = Pi = Ptot
c0G,i
=
MWi
RT cG,i RT
(8.17)
ni Pi = VG RT cG,i
=
N
 cG,i
(9.1) = yi
N
(8.40a)
(9.4)
 cG, j
j=1
j=1
Henry’s Law:
cG,i = HcL,i
(9.9)
Raoult’s Law:
PG,i = Pvap,i xi
(9.10)
H0 ⌘ 1 H0 = gL,i
aG,i aL,i
gG,i
cL,i cG,i
(9.11) !
H
cG mG /VG = = Hcc cL mL /VL ✓ ◆ HT2 DH r 1 1 ln = HT1 R T1 T2 log gL,i = ki I 873
(9.13c) (9.17) (9.18) (9.19)
R g L,i =
1
(9.53)
xi,sat0 n
n
bn = K1 K2 K3 ...Kn = ’ Ki
(10.12)
bi = bi Kwi
(10.14)
Ki = Ki Kw
(10.15)
i=1
⇤
⇤
m
n
T OT Me = (Mex+ ) 1 + Â Â bL j ,i (L j )i j=1 i=1
Ks0 =
(Men+ ) (OH )n (Me(OH)n (s))
= (Men+ )(OH )n eq
F =C
P+2
pe = log
{Red}1/ne 1 {Ox}1/ne {e } 874
(12.13)
!
1 1 Ox + e )* Red ne ne log K(12.15) = log
(11.3)
(12.12)
{Ox}1/ne {e } {Red}1/ne
{Red}1/ne 1 {Ox}1/ne {e }
eq
(10.37)
(11.50)
1 1 Red )* Ox + e ne ne e ⌘ K(12.12) =
!
(12.14b)
(12.15) !
= pe
(12.16)
1 {Red} log ne {Ox}
pe = pe
1 log Q ne
DpeOx’t/Red’t = DpeOx’t/Red’t pepH ⌘ pe
nH pH ne
(12.21) (12.28)
nH (7.0) ne
(12.30)
1 {Red} log ne {Ox}
(12.31)
pe (W) ⌘ pe7.0 = pe pe (W) = pe (W)
(12.18)
log{O2 (aq)} = 86.00 + 4pH + 4pe
(12.62)
log{O2 (g)} = 83.12 + 4pH + 4pe
(12.63)
log{H2 (aq)} = 3.10
(12.64)
log{H2 (g)} = 2pH
2pH
2pe
2pe
✓ ◆ kJ pe 5.71 mol
Ge =
EH = (0.059 V) pe 0.059 mV {Red} log ne {Ox} ✓ ◆ kJ 5.71 ne DpeOx’t/Red’t mol
EH = EH DGr =
(12.73) (12.77) (12.78b) (12.83)
{⌘SA}{H2 O} {⌘SH2 O}{A}
(13.2)
{⌘ A(interphase)} {A(aq)}
(13.4)
Kads = Kp =
(12.65)
q=
cinit cfin csolid
875
(13.5a)
qA =
Kads (A) qmax 1 + Kads (A)
Fraction of surface sites occupied = qi =
qA Kads (A) = qmax 1 + Kads (A)
Kads, i (i) qmax 1 + Â ( j)
(13.9) (13.12) (13.23)
all j
qA = kf {A}n
(13.27)
✓
(13.39)
◆ zi Fy gi = exp RT ✓ ◆ zi Fy {⌘SA} = [⌘SA] exp RT ✓ ◆ [⌘SA] zA FyA Kapp, A ⌘ = Kint, A exp [⌘SH2 O]{A} RT s + sb + sd = 0
C1 = C2 =
s y yb
sd sd = yd yd yb
sd = 0.1174c0.5 b sinh
|z|Fyd 2RT
for yd 25 mV
876
(13.44) (13.46)
(13.47)
yb
yx ⇡ yd exp( kx)
(13.41)
(13.48)
(13.49) (13.50)
Index A Absolute temperature, 16 Acid(s). See also multiprotic acid; phosphoric acid; titration about, 221, 223 acidity constants at 25°C, 225t5.1 carboxylic, 227, 229–230, 232, 269 conjugate, 533, 567 haloacetic, 19 Lewis, 221, 221n2, 223 organic, 227, 229–233 polyprotic, 231 strength of, 219, 222–223, 234, 245, 283 strong, 234, 246, 266–269 strong and weak, 222–229 titration with strong, 386–392, 388t8.1, 389f8.4, 391f8.5 titration with weak, 402–405, 403f8.12 titrations of strong, 379–381, 383, 385– 392 weak, 222, 229, 231, 246, 261–262, 269, 271, 274, 278, 280 weak, monovalent, 409, 409f8.14 Acid dissociation constant (Ka), 223, 283 Acid dissociation reactions about, 70 of acetic acid, 311–316 acid and bases, 221, 224, 226, 231, 252, 261, 279 Acid(s) and base(s) graphical solutions for sets of simultaneous equations, 292–293 schematic representation of donation of H+ to water, 222f5.4 strong, 266–268 structure of, 221–222, 222f5.4 titration of solutions containing unknown, 406–408, 407f8.13 weak, 269–274, 279–280
Acid/base. See also conjugate acid/base pairs behavior of NOM, 439 buffer intensity of water and weak, 443–450, 444t8.7, 446f8.26, 450f8.27 buffering, 556 conjugate pairs, 222–229, 229f5.7, 329– 330, 342, 679 equilibria in nonideal solutions, 280–283 problems, comparison of approaches for solving, 341–342 problems, numerical approaches for solving, 266 reactions, 84–85, 217–219 reactions vs. redox reactions, 689–690t12.4, 692, 692f12.4 speciation of, 219, 383, 389f8.4, 390 species, 330, 336, 343 titrations of natural organic matter, 438–442, 438f8.23, 441f8.24, 442f8.25 Acid/base reactions, 789 Acid/base speciation as a function of pH, 238–242, 241t5.2, 242f5.13 simple spreadsheet analysis for determining, 274–279, 276t5.4 visualizing, 292, 297 Acidic and basic volatile species, 515 Acidic gases, 508, 516 Acidic or basic gas(es) partial pressure of, 494 as pH buffers, 508 pH of solution, alter, 516 pH of solutions in equilibrium with, 494–502, 495t9.3, 496t9.4, 497f9.8, 498f9.9, 499f9.10, 502f9.11 solution equilibrated with, 515 solution pH, sensitivity to, 516
877
Acidic waters from mines, 549 Acidity about, 22–23 defined, 415 mineral, 415 reactions, combining, 233–238 reactions, reversed, 233 strong, 439 weak, 439 Acidity constant (Ka) acids and bases, 223–224, 234, 237, 245, 247–249, 259, 266 at 25°C for some important acids, 225t5.1 shift in apparent acidity constant caused by nonideality, 248f5.17 values of a0 and a1 of monoprotic acids, 241t5.2 Acid-neutralizing capacity (ANC), 414 Activated carbon, 760–761, 766, 768 Activated complex about, 81, 81n2, 82, 82n3 Gibbs energy of the, 83 model (EAr), 81n2, 82, 82f3.1, 83, 92, 202–204, 203f4.10 Activated sludge systems for treating domestic wastewater, 489 Activation energy (E*), 80–83, 85, 87–88, 121, 666, 744 Activation energy barrier, 80, 81n2, 88, 744 Activity as an intensive property, 56–57 of solids dispersed in water, 56–57 Activity coefficient(s) about, 43–46, 72 for adsorbed species, 769, 804, 815, 817 of Ca2+ in a solution prepared by dissolution of CaCl2, 54f2.2 dimensionless quantities of, 45 equations for predicting in aqueous solutions, 52t2.2a of major ions in seawater, 64–67 predicting from knowledge of solution composition, 49–56, 52t2.2a of symmetric electrolytes, 50–51
Activity quotient, 67 Adsorbable organic halogen (AOX), 19 Adsorbate(s) about, 772–773, 840 hydrophilic, 777 reactivity, 776 Adsorbed species activity coefficients for, 776, 811, 822, 824 infinite dilution reference state for, 776 nonspecific, 773 Adsorbent(s) about, 772–775, 777–783 mineral, 777, 792 Adsorption activity coefficients for adsorbed species, 776, 811, 822, 824 apparent adsorption constant (Kapp), 815, 816n17, 821, 827 apparent adsorption density, 835, 837– 838, 838f13.29 apparent adsorption isotherms, 838, 839f13.30 charge distribution and electrical potential profile near surfaces, 820t13.4 constant (Kapp), apparent, 816, 816n17, 821, 827 constant capacitance model, 816, 819, 821t13.4 defined, 772 density, 772, 776, 777–781, 787, 789f13.6, 790–791, 792f13.7, 793–794 density, apparent, 835, 837–838, 838f13.29 diffuse layer model, 816, 819, 813n21, 821t13.4, 823, 824n21, 824, 8270–828, 828n22–23, 831 environmental processes and, 841 equations describing ionic, 821–824, 823f13.23 interfacial structure as envisioned in the triple-layer model, schematic of, 819f13.21
878
intrinsic adsorption constants (Kint), 815, 827 ionic distribution, 820, 820f13.22 of ions in presence of a surface electrical potential, 811–827 ions in the interfacial layers, distribution and binding of, 815–821, 817f13.20, 820f13.22, 821t13.4 multisite Langmuir isotherm, 789f13.6 phase transfer model, schematic of adsorbed species according to the, 776–777, 777f13.3 phase transfer model vs. site-binding model, 777 site-binding model, schematic of an adsorptive solid/solution interface according to the, 775f13.2 surface charge and, 769, 772–773, 812, 813f13.18, 823f13.23, 827, 831, 833–834, 836 as surface complexation reaction, 773–776, 775f13.2 in systems with nonuniform surface sites, 785–795 Visual Minteq, modeling surfaceadsorbate electrical interactions in, 827–834, 829t13.6, 831f13.24, 832f13.25, 834f13.26 Visual Minteq, modeling with, 796–810 where electrostatic interactions are significant, 824–825t13.5, 827 Adsorption equilibrium quantitative representations of, 778–795 two views of the interface and, 773–778 Adsorption isotherm apparent, 838, 839f13.30 defined, 772 Freundlich isotherm, 790–791, 792f13.7, 792–793, 793f13.8, 794f13.9, 794–795 Freundlich isotherm, Extended, 795 Freundlich isotherm vs. multisite, 796 in systems with nonuniform surface sites, 785–795
isotherm equations, approaches for developing, 778–780, 780f13.4 Langmuir adsorption, competitive, 782–785 Langmuir isotherm, multisite, 789f13.6 Langmuir isotherm, multisite, 787–791, 789f13.6 Langmuir isotherm: surfaces with uniform sites, 779–785, 782f13.5 Langmuir-Freundlich isotherm, 793 surface with a semicontinuous distribution of site types, 790–795 Adsorption models, surface complexation, 815f13.19 Adsorption reaction(s). See also desorption reactions about, 140, 769–772 defined, 772 equilibrium and, 772 equilibrium constant and, 814–815 phase transfer reaction, adsorption as a, 776–777, 777f13.3 site-binding model, 773, 775f13.2, 777 surface complexation reaction, adsorption as a, 773–776, 775f13.2 terminology of, 772–773 Adsorptive solid/solution interface, 775f13.2 ALK. See alkalinity (ALK) Alkalinity (ALK). See also titration about, 4, 22–23, 25 acidity and, conceptual basis and definition, 413–416, 415f8.17 assessing contribution of different species to, 416–423, 417t8.5, 421f8.18 carbonate, 415 in carbonate-dominated systems, 427–434, 431f8.19, 432f8.20, 434f8.21 caustic, 415 computed from titration data or based on solution composition, 416 concept, summary of, 437 as conservative quantity, implications of, 426–427
879
contribution of added species to, 423–426, 425t8.6 defined, 414, 415n6, 454 endpoint, 416–419, 422–423, 429, 436, 454, 512, 513n14 European standard for analyzing alkalinity (ISO 9963-2), 414 phenolphthalein, 415 titration, 513 total, 415, 418, 457, 455 Visual Minteq, calculations embedded in, 435–437, 435f8.22 Alum. See hydrated aluminum sulfate Aluminum hydroxide [Al(OH)3(am)], 591–593, 592f11.5, 597 Aluminum sulfate [Al2(SO4)3], 379 Amine [NH2] groups, 438 Amines organic, 232 schematics of (a) primary amine, (b) secondary amine, and (c) protonated secondary amine, 233f5.11 Amino acids, 232, 233f5.12 Ammonia [NH3] acids and bases, 229, 230f5.8, 245, 245f5.16 molecule, inorganic, 232 species, 304 titrations, 379, 391, 398–399, 409–410, 417t8.5, 418, 442, 452–453, 458 Ammonium ion, 17, 225, 229, 232 Ammonium [NH4–N], 17 Anaerobic degradation of organic matter, 270 environments, 3 sediments, wastewaters, and sludges, 470 waste solution, 333–334 ANC. See acid-neutralizing capacity (ANC) Anion(s). See also cation(s) adsorption reactions and, 771, 772f13.1, 818, 821 in basics of aquatic chemistry, 3n1, 9, 14–15, 27, 29t1.2 defined, 3n1 ion size parameter, 53t2.2b Anodes, 744–745, 747–746
Anodic cell, 748f12.26 AOX. See adsorbable organic halogen (AOX) Apparent adsorption constant (Kapp), 808, 808n17, 814, 820 density, 828, 830–831, 831f13.29 isotherms, 831, 832f13.30 Aragonite, 581t11.1 Arrhenius model for reaction kinetics, 91–92 Arrhenius’s equation, 83, 121 Arsenic acid [H3AsO4], 308, 677t12.3, 684 Arsenous acid [H3AsO3], 677t12.3, 684 Association reaction, 545 Avogadro’s number, 11n7 B Bacteria oxygen and metabolism of, 672 respiration, 432f8.20 Base(s). See also conjugate base; titration Brönsted, 221 deprotonated, 333–334 equivalent, 384, 384n1 inorganic, 229, 231 Lewis, 221, 221n2 strength of, 234, 283 strong, 268 strong monoprotic, 405 titration curves in solutions with NaOH, 388, 389f8.4, 390 titration curves in solutions with mixtures of bases, 391f8.5 titrations of strong, 379, 381–383, 385–392 weak, 269, 283 Basic gases, 516 Basicity, 234 constant (Kb), 283 reaction, 260 Benzene, 46–48 Henry’s constant, 479t9.2, 481, 481f9.2 Raoult’s law, 474 standard state conditions, 46–48 Beta layer, 821, 825f13.5, 827, 831. See also naught layer
880
Beta plane, 816. See also naught plane Binding sites adsorbents and, 774, 775f13.2 at solid/solution interface, 775f13.2 in acid/base, metal/ligand, redox, and adsorption reactions, 783t13.1 Bioavailability, 529 Biochemical oxygen demand (BOD), 18n11 Bioenvironmental chemistry of metals, 567 Bisphenol A (BPA) about, 101–104, 101n9 decay of TOTBPA by reaction with chlorine at pH 7.5, 104f3.7 molecular structure of, 102f305 BOD. See biochemical oxygen demand (BOD) Boltzmann constant (kB), 93 Bond energy, 80 Bond rearrangement, 88 Bonding, intermolecular chemical, 79–81, 81n2, 85–87, 114, 121 Boundary layers, 116n12, 117, 118f3.12, 119 BPA. See bisphenol A (BPA) Bromoform (CHBr3), 19–20, 198, 198f4.9, 487f9.4 Brönsted acids, 221, 223 Brönsted bases, 221 Buffer, defined, 442, 454 Buffer intensity definition, 442–443 of multiprotic acids, 451–453, 451t8.8, 452f8.28, 456–467 of water and weak acid/base groups, 443–450, 444t8.7, 446f8.26, 450f8.27 Buffering acid/base, 556 in solutions equilibrated with weak acid and weak base gases, 508–512 system, acid/base, 553 Buffers. See also metal ion buffers acid/base, 556
C Cadmium, hydroxo complexes of, 534 Cadmium hydroxide [Cd(OH)2(s)] about, 578–579, 582t11.1, 615 log C-pH diagram for, 586f11.1, 587, 588f11.2 system in equilibrium with, 584–585, 586f11.1 Calcite [CaCO3(s)], 506. See also calcium carbonate Calcium carbonate [CaCO3], 85 alkalinity, 23, 25 precipitation, 149–150 species concentration and, 22–23, 30 standard molar Gibbs energy of formation, 156 titration and alkalinity, 385, 419, 422 dissolution, 589 Visual Minteq and, 589f11.3, 590f11.4, 626–627 Calomel electrode, saturated, 753, 754f12.27 Capacitance (C), 819 Carbon dioxide [CO2] about, 8–9, 25, 85, 237–238 gas/liquid equilibrium, 512–515, 514f9.16, 515f9.17 Carbonate acid/base group, 414, 418, 427, 430, 437, 558 Carbonate alkalinity, 415 Carbonic acid [H2CO3] about, 7–8, 18, 25 acids and bases, 219, 225, 228–229, 229f5.7, 237–238, 249–251, 251f5.18 redox chemistry, 671, 671f12.1 Carboxyl (COOH) groups, 438–439 Carboxylic acid, 227, 229–230, 232, 269 Catalyst(s) chemical reactions and, 80, 85 electrode, 750–751 enzymes, 664, 666, 744, 757 Cathodes, 744–745, 747–748 Cathodic cell, 746f12.26 Cation(s). See also anion(s) adsorption reactions and, 772, 772f13.1, 818, 821, 836n27
881
in basics of aquatic chemistry, 3n1, 9, 14–15, 27, 29 defined, 3n1 values of ion size, 53t2.2b Caustic alkalinity, 415 Characteristic time, 105–106 Charge balance, 13–14 Charge conservation equation, 327, 341–342 Charge conservation rule, 667t12.1, 669 Charge equivalents, 13, 21–22 Chelated metals, 541, 567 Chelates, 539–540 Chelating agents, 539–540, 540f10.4, 567 Chelation, 540. See also complexation Chelators, 539, 567 Chemical activity about, 39, 39n1, 72 quantifying, 40–41 (See also standard state) of a substance, 49 Chemical equilibrium about, 2, 38, 58–59 dynamic molecular kinetic model, 59 models of, 57–68 relationships among Q, K, and direction of chemical reactions, 67t2.3 thermodynamic model, 59 Chemical equilibrium software, 310. See also Visual Minteq Chemical oxygen demand (COD) test, 18n11, 672–674 Chemical potential dissolution vs. reaction with the solvent, 202 as driving force for chemical reactions, 179–204 energy, 121, 143, 147–149, 164, 189 energy changes during a hypothetical elementary reaction, 203f4.10 enthalpy and entropy changes in chemical reactions, 184 Gibbs energy of composite reactions, 199–202
Gibbs energy of reaction and K, dependence on temperature and pressure, 195–199, 198f4.9 Gibbs energy of reaction and system condition, summary of relationships between, 188t4.5 Gibbs energy of reaction and the driving force for chemical change, 179–183 Gibbs energy of reaction to the equilibrium constant, relationship of, 185–187 Gibbs energy of reaction under nonequilibrium conditions, 187–195, 188t4.5, 191f4.7, 194f4.8 thermodynamics, kinetics, and the activated complex model, 202–204, 203f4.10 van’t Hoff relationship for transfer of bromoform from solution to the gas phase, 198f4.9 Chemical potential of i, 148–149, 204. See also molar Gibbs energy (Gi) Chemical reactions combination of parallel and series reactions, 69f2.3 combining, 69–71, 92–97 relationships among Q, K, and the direction and reversibility of, 90t3.1 Chemical reactivity. See also Gibbs energy about, 179, 204 characterizing, 38–49 Chemically adsorbed, 773 Chemisorbed, 773 Chemistry of metals in aqueous systems, 576 Chlorine, 3, 13, 19, 19n12 Chlorine dioxide, 13 Chlorine-based disinfectants, 19, 22 Chloroform, 477, 479t9.2, 487f9.4 Chromium [Cr], 3–4, 188–191 Closed systems, 143, 471, 513. See also open systems COD test. See chemical oxygen demand (COD) test Combination electrode, 753 Compact layers, 816, 823
882
Complexation, 535t10.2, 540–541, 546, 567 Components defined, 310 expressing input and equilibrium concentrations in terms of, 316–327, 318t6.4, 322f6.10 schematic showing formation of four species from combinations of three components, 311f6.9 Computer software for chemical equilibrium modeling. See also Visual Minteq about, 354–355, 375 acid/base problem, computer algorithm for solving simple, 355–363, 356t7.1, 357t7.2, 360f7.1, 361f7.2, 362f7.3 chemical equilibrium software to develop log C-pH diagrams, 368–372, 370f7.7, 371f7.8–360f7.9, 372f7.10 Davies equation, 359, 361 flowchart of calculations used to compute equilibrium speciation, 360f7.1 Solver® routine embedded in Excel®, 276 for systems in which equilibrium activities of one or more species are known, 363–368, 364t7.3, 365f7.4, 366f7.5, 367f7.6 TOTH equation and, 310, 342–343 Concentration(s) characteristic time and, 105–106 chemical change and, 40 chemical reaction rates and, 86–87 of compounds: hardness, alkalinity, and oxidant concentrations, 20–24 defined, 57 of dissolved constituents in natural water bodies, 4t1.1 equivalent, 45 equivalents per liter, 12–15 of gas/liquid interface, 39 gas-phase, 15–16, 16n10 gradient, 39 in mass or moles per unit volume, 39 mass/volume and mass/mass units for solute, 10–12 molal, 12
molar, 12, 94, 478, 483 mole fraction, 41–42, 45–47 order of the reaction and, 87 rate constant and, 88 rate expressions for elementary reactions, 94 reaction rate constant and, 87, 97–99, 100t3.2, 101 reference state, 44–45 sequential reactions, concentration vs. time profiles, 107, 107f3.9 solutes in terrestrial waters, ranges of several, 8, 9f1.4 of a species controls activity, 40–41 standard state, 41–43, 45, 64–65, 72, 154 standard state conditions for solutes, 43t2.1, 46 of a substance dissolved in another, 57 temperature, pressure, and overall composition of the phase affect, 41 total solute concentration: TDS and ionic strength, 24–29 in units of moles per liter [A], 11–12, 17, 26 Conductivity, 24, 27–30 Conjugate acids about, 232, 234, 240, 246, 533, 567 Conjugate acid/base pairs. See also acid/base about, 226, 227f5.6, 234, 240, 246, 251, 308 relationships among phosphate species, 227f5.6 Conjugate base about, 234, 246, 248, 260, 274, 283 Constant capacitance model, 815f13.19, 816, 819, 821t13.4 Contaminant(s) in aquatic systems, 770 concentration, 3 in water, 770–771, 775 Coordination number, 530 Corrosion of brass, 716 prevention, 749
883
of sewer pipe, 228 of statues, buildings, and structures, 503 of water supply pipes, 664 Cross-boundary transfers, 39n3
Disinfectants, chlorine-based, 19, 22 Disinfection by-products (DBPs), 19, 19n12 Disproportionation reaction, 714 Dissociation acid dissociation constant (Ka), 223, 283 acid dissociation reactions, 221, 224, 252, 279 of any acid, 226 constant for carbonic acid, 237 constant for water (Kw), 283 multiple acid dissociation reactions, 224 multiple dissociation reactions of acid or base, 260–261 reactions, 545, 775 reactions of carbonic acid, 249 reverse of acid dissociation reaction, 233 of water (Kw reaction), 219, 221, 226, 233, 235, 259–260, 308, 311–315, 314t6.2, 315t6.3 Dissolution of salts, 6–9, 7f1.3 Dissolution process, 6n4 Dissolved organic carbon (DOC), 4, 18, 30, 194–195 Dissolved organic matter (DOM), 439 Distribution coefficient, 60, 475, 478. See also partition coefficient DOC. See dissolved organic carbon (DOC) DOM. See dissolved organic matter (DOM) Dominant species, identifying, 327–341, 329t6.5, 336f6.12, 337f6.13 Donor atoms, 538
D Davies equation acids and bases, 248, 248f5.17, 249, 281–282 chemical equilibrium, 52t2.2a, 55–56, 62–63 computer software and, 359, 361 DBPs. See disinfection by-products (DBPs) Debye-Hückel equation, 811n19 Debye–Hückel limiting law, 51, 52t2.2a Dechlorinating organics, 675 Deffeyes diagram, 431f8.19, 434 Deprotonated acid/base couples, 382 acid/base groups, 331 bases, 333–334 Deprotonation reactions, 451 Desorption, 765, 815, 834 Desorption reactions, 765. See also adsorption reactions DIC. See dissolved organic carbon (DIC) Diffuse layer, 815–816, 817f13.20, 818, 819f13.21, 818, 820f13.22 Diffuse layer model, 815f13.19, 816, 819, 821n20, 821t13.4, 823, 824n21, 824, 828n22–23, 831 Diffusion-controlled, 88 Dimers, 577 Dipoles, 26, 30 Disequilibrium change in electron energy in a reaction in terms of extent of, 736t12.9 chemical (redox), 748 chemical reaction and, 67, 73 Gibbs energy of reaction and, 736t12.9 Gibbs energy released and, 757 of the H2/H2O/O2 system, 724 in phase-transfer reactions, 115 ratio Q/K as indicator of, 90, 94, 122 between two phases, 118–119
E E° definition of, 679–680 ranges in natural aquatic systems, typical, 680f12.2 EDTA. See ethylene diamine tetra-acetic acid (EDTA) EH scale (electron energy and activity), 729–731, 731f12.19 Electrical conductivity, 24, 27 Electrical double layer, 817, 821n20 Electrical fields, 26–27 Electrical interactions, 7, 7f1.3
884
Electrical potential about, 140 energy, 139–142, 141f4.2, 142–143 energy of an object (PEelec), 729–730 energy per mole of a species, 729–730 ion-selective electrodes and, 755 potentiometer or voltmeter and, 750–751 “virtual” electrical potential, 730 Electrical resistance, 27–28 Electrochemical cell, schematic of, 746f12.26 Electrochemical energy, 747 Electrochemical potential, 747, 751 Electrochemical reactions anodes and cathodes, 744–745, 745f12.25 driving force and energy change in, quantifying the, 745–748, 746f12.26 electrochemical cell, schematic of, 746f12.26 electrodes, practical considerations and limitations of, 754–756 Nernst equation and, 746, 702–752 redox electrodes, reference electrodes, and measurement of EH, 750–754, 754f12.27 redox reactions in systems with externally imposed electrical potentials, 748–750 Electrochemistry about, 665, 749, 749n21 redox reactions and, 744–756 Electrode(s) about, 750–751 combination, 753 dynamics, 749, 749n21 ion-selective, 755 reference, 752–754, 754f12.27 saturated calomel, 753, 754f12.27 silver/silver chloride, 753–754 standard hydrogen, 752–753 working, 754f12.27, 755 Electron(s) chemical bond formation and, 667
electron energy change in a reaction in terms of extent of disequilibrium, 736t12.9 energy of exchangeable electrons in a system, 736t12.9 free electrons (e–), 670, 674–675, 709, 729n13, 746, 756 hydrogen atoms and, 5 hydrogen ions and, 5, 5f1.1, 6 hydrogen-oxygen bonds and, 5 in oxidation half-reaction, 695 oxidation-reduction reactions, 13, 23–24 oxidized species of redox couples and, 681 pe° and, 689 redox half-reactions and, 671–676, 760, 686 redox reactions and, 663, 669n2, 670, 670t12.2 redox reactions with both electrons and protons, 692–703 redox reactions with exchange of multiple electrons, 703–710 redox reactions with no H+ exchange and, 688 redox species and, 689 water molecule and, 5f1.1, 6, 6f1.2 Electron acceptors, 663, 739 Electron donors, 663 Electron energy and activity (EH scale), 729–731, 731f12.19 Electron (e–) equivalents, 13 Electron pairs, unshared, 645 Electronegative atom, 227 Electroneutrality, 632, 634 Electrophoretic mobility, 837, 837f13.28 Electrostatic bonds, 5 Electrostatic interactions about, 7, 9, 50 adsorption reactions, 776, 812, 814, 817f13.20, 824, 841 Visual Minteq and, 797, 799, 802f13.11, 820 Electrostatic unit (esu), 5
885
Elementary reactions. See also nonelementary reactions about, 39, 39n2, 121–122 Arrhenius’s equation, 83 bimolecular, 88 chemical reactions at the molecular level, 92–93 kinetics of, 86–88 Michaelis-Menten kinetics and, 113 “power law” equations and, 95 rate constants for, 93 rate expressions for, 95 rates of, 86 reacting species concentrations and, 83 reaction rates and, 83 sequence of, 79 temperature, effect of, 91–92 Endothermic, 82 Energy changes of redox reactions about, 728–737 EH scale: alternative way of expressing electron energy and activity, 729–731, 731f12.19 molar Gibbs energy change of redox reactions, 732–736, 734f12.20, 736t12.9 molar Gibbs energy of electrons, 728–729 schematic of relationship between electron energy levels and speciation of various redox couples, 731f12.19 Energy content, 145, 147 Enthalpy (H). See also molar enthalpy (Hi) about, 80–83 in chemical reactions, 184 of composite reactions, 199–202 defined, 144–145, 145n4, 146 molar, 80, 82f3.1, 92, 121 total enthalpy of the system, 145, 197 of volatilization, 486, 506 Entropy (S) about, 81, 82n3, 133, 133n1 of composite reactions, 199–202 defined, 133–134, 133n1
Gibbs energy and, 144–148, 147n6 molar, 82f3.1, 150–151, 153, 161, 184, 203, 205 of reaction, 184, 205 standard molar, 154 in the standard state, 41–43 (See also reference state) universal, 132–133 Environmental Protection Agency (EPA), 683 Environmental reaction rates pH–dependence of important, 86f3.3 Enzyme-based reactions, classic kinetics of, 111f3.11 Enzymes, 666, 668, 746, 759. See also catalyst EPA. See Environmental Protection Agency (EPA); U.S. Environmental Protection Agency (EPA) Equilibrium algebraic equations to compute species activity at, 314, 314t6.2, 315t6.3 chemicals present in the system at, 310–311 composition of a system, 2 composition of any mixture of acids and bases, procedures for determining, 333 concentrations in terms of components and component mass balances, expressing input and, 316–341, 317n3 concentrations of all species in solution, 297–300, 299f6.3 conditions at, 292, 297–298, 301, 342, 736t12.9 criterion for, 300 dominant acid/base species at, 329–331 problems, determining species and relevant equations for solving, 256–265 software, chemical, 310 speciation, 293, 308, 319, 327, 341–342 TOTH equation and dominant species at, 325, 325n4
886
Equilibrium constant (K) about, 59–64, 89, 109, 116, 788–789 acid/base reactions and, 796 for acids and bases, 293, 293n1, 304, 307, 313–316, 319, 341 adsorption adsorption reactions and, 780, 790, 814–815 apparent adsorption constant (Kapp), 815, 816n17, 821, 827 for composite reactions, 71 defined, 88–91 for deprotonation of an acid, 533–534 for dissociation of water, 92 for dissolution of four forms of iron oxyhydroxide, 630t11.3 for dissolution of a metal hydroxide solid, 579–580 dissolution reactions in Visual Minteq, 605, 606f11.13 effect of temperature on, 68, 91–92 for formation of a complex of the type MeLj, 534 intrinsic adsorption constants (Kint), 815, 827 isotherm equation is a pseudoequilibrium constant expression, 808n14 Kads, 775, 780 Langmuir adsorption, 803 for metal-ligand complexes, 533 of nonelementary reactions, 92–97 for an overall reaction, 73 partition coefficient (Kp), 776 phase rule and, 633n10 for reaction between an uncomplexed ion and i ligands to form a complex of the type MeLi, 533 for reaction between metal hydroxide solid and a soluble hydroxo complex, 584 for reaction between two dissolved species, adding one ligand to a complex with i – 1 ligands to form a complex with i ligands, 533 reaction quotient vs., 95
surface complexation reaction and, 767 temperature and, 91–92 temperature and total pressure and, 632 in terms of energy change at standard state, 738t12.9 van’t Hoff equation, 68, 73 Visual Minteq and, 805, 827–828 Equilibrium pH. See also pH about, 299, 307 algebraic and spreadsheet analyses to determine, 293 charge balance and, 294, 305, 308 from the graph and charge balance, 308 log C-pH diagram and, 293, 336, 336f6.12 log C-pH diagram and PCs to solve, 319–323 mass balance and, 303, 303f6.6 TOTH equation and PC to identify, 342 TOTH equation to determine, 323–327 Equilibrium species, 417t8.5. See also species charge conservation equation and, 327 charge on the known input chemicals and, 327 inputs and, 318t6.4 total concentration of each component in all, 318 Equivalence points about, 409–413 in carbonate system, 411–413, 411f8.15, 412f8.16, 412t8.3, 413t8.4 defined, 409–411, 409f8.14 Equivalent acid, 384, 384n1, 416 base, 384, 384n1 ionic conductance, 28–29 Equivalent weight (EW) about, 384–385 of Ca2+, 21–22 of CaCO3, 23 of Cl2, 24 of Mg2+, 22 of ozone, 24 Equivalents (equiv), 384
887
Equivalents per liter (equiv/L), 12–13 Error in the charge balance (ErrorCB), 14–15 ErrorCB. See error in the charge balance (ErrorCB) esu (electrostatic unit), 5 Ethylene diamine tetra-acetic acid (EDTA), 536, 539, 540f10.4 European standard for analyzing alkalinity (ISO 9963-2), 414 EW. See equivalent weight (EW) Exchange of equal-charged ions, 841 Exothermic, 82 Extended Debye–Hückel equation about, 51, 52t2.2a, 52t2.2b, 53 activity coefficients of various ions according to, 54t2.2(a) for dissolved ions in fresh water, 55 values of ion size parameter for use in, 53t2.2b Extent of reaction, 191
multisite vs., 795 Visual Minteq and, 805–810, 808f13.14, 810f13.15, 811f13.16–f13.17 G Gas(es) absorption, 471 as neutral molecules, 9 dissolution, 471 stripping, 471 (See also volatilization) Gas/liquid equilibrium, 578 about, 470–471, 481 basic concepts and terminology for, 471–481 buffering in solutions equilibrated with weak acid and weak base gases, 508–512 CO2 dissolution, alkalinity, and acidity, 512–515 effect of ionization of dissolved gases, 491–494, 492f9.6 gas-phase composition, effect on, 482–483 H, various ways of interpreting, 481f9.2 Henry’s Law and, 472–481, 474f9.1, 477f, 478t9.1, 479t9.2, 481f9.2, 482–67, 485, 490, 494, 519–524 ideal gas law, 471–472 pH of solutions in equilibrium with acidic or basic gases, 494–502, 495t9.3, 496t9.4, 497f9.8, 498f9.9, 499f9.10, 502f9.11 Raoult’s Law and, 472–474, 474f9.1, 475–477, 477f, 485, 520–524 in Visual Minteq, 502–508 of volatile organic compounds, 476 volatility and Henry’s constant, factors affecting, 483–490, 484f9.3 Gas-phase concentration(s) about, 15–16 titrations and, 471–472, 475, 478, 480 Gas/solution equilibrium, 576–577 Gas-transfer processes modeling, 119 Geochemical cycles, 379
F Ferrihydrite [Fe(OH)3(am)] adsorption reactions, 771, 772f13.1, 792–793, 793f13.8, 814t13.3, 824n21, 827, 837f13.28 Ks0 value of, 581t11.1 solid polymorphs, relative solubilities of, 581t11.1, 628, 630t11.3 Fick’s law, 116, 116n13 Filterable residue, 25 First law of thermodynamics, 145, 147 Force acting on an object, 138–139 Formation constant, 533. See also stability constant Fossil fuel combustion, 470 Free electrons (e–), 670, 674–675, 709, 729n13, 746, 756 Free metal ions, 530, 541, 553, 556 Freundlich isotherm. See also Langmuir isotherm about, 791–796, 792f13.7, 792–793, 793f13.8, 794f13.9, 794–795 Extended, 795 Langmuir-Freundlich isotherm, 793
888
Geochemical redox sequence, 737–740, 738f12.21 Gibbs energy (G). See also molar Gibbs energy; molar Gibbs energy (Gi); potential energy about, 664 activity coefficient of dissolved ions and, 806n16 change in, 146 change in the total system, 149–150 of composite reactions, 199–202 electrical energy per mole and, 732 energies and potential energy, 133–134, 149, 204–205 entropy and, 144–148, 147n6 Gibbs free energy, 423 molar, 148, 151 Nernst equation and, 733 partial molar, 148 per mole, 730 of reaction and K, dependence on temperature and pressure, 195–199, 198f4.9 of reaction and system condition, summary of relationships, 188t4.5 of reaction and the driving force for chemical change, 179–183 of reaction in terms of energy of electrons associated with the two half reactions, 736t12.9 and reaction kinetics, 81–82, 82f3.1, 82n3, 83 of reaction under nonequilibrium conditions, 187–195, 188t4.5, 191f4.7, 194f4.8 of reaction’s relationship to the equilibrium constant, 185–187 Gibbs energy of reaction and equilibrium constants for some redox half-reactions, 677–678t12.3 molar, equation for, 666 standard, equation for, 665 in terms of extent of disequilibrium, 736t12.9
Gibbs phase rule about, 629–641 chemical equilibrium software, 629, 633 to determine the number of degrees of freedom in various solutions, 633–634 violation of, 633 Gibbs-Duhem equation, 170 Gibbs-Helmholtz equation, 168, 176 Gouy-Chapman equation, 818 Gran titration technique, 103n4 Gravitational field, 137 potential, 136–138 potential energy, 133, 133n2, 134–139, 135f4.1, 141f4.2, 142, 147, 153 Gypsum, 581t11.1 H H+ equivalents, 13, 23 H+ ion, 218 Half-reactions. See redox half-reaction(s) Half-saturation constant, 110 Half-time (half-life) of a reaction, 99, 106 Haloacetic acids [XnH3nC–COOH], 19 Hardness of water, 21–22 Heat content, 145 of molecules, 80 Heavy metals, 3 Henry’s constant about, 515 defined, 473, 478 factors affecting volatility and, 483–490 halogenated VOCs, temperature dependence of Henry’s constant for several, 487t9.4 hydrogen-bonded water molecules surrounding a hydrophobic solute, schematic of, 484f9.3 hydrophobic or hydrophilic nature of solute, 484–485, 484f9.3 molecular structure and, 486 oxygen solubility in water as a function of temperature and salt content, 488f9.5
889
Hydronium ions [H3O+], 221, 222f5.4, 223, 225t5.1, 229 Hydrophilic (water-loving) about, 6–7, 30 adsorbates, 777 solutes, 777 Hydrophobic (water fearing), 6–7, 30, 47 Hydroxide ion(s) about, 567 complexes with, 530–536 electron density between a metal ion and a water molecule in the inner hydration sphere, schematic of the shifts in, 532f10.2 stability constants for complexation of metals by OH–, 535–536t10.2 Hypochlorous acid [HOCl] about, 13, 19, 60, 85, 391 acids and bases, 217, 225, 243, 245, 248, 274 Hypochlorous acid/hypochlorite pair, 243, 243f5.14
partial pressure and, 478t9.1, 480, 488, 493 solution composition and, 487–490, 488f9.5 of some environmentally important gases, 479 temperature and, 486–487, 487f9.4 units commonly used for, 478t9.1 vapor pressure of pure compound, 485–486 volatility and, 481, 481f9.2, 483–490 volatilization reaction and, 475, 486 Henry’s law about, 472–483, 485, 490, 494, 519–524 equation, 473 equilibrium values of H2(aq) and O2(aq), 721 gas/liquid equilibrium, effect of ionization of dissolved gases on, 491–494, 492f9.6, 494f9.7 gas-liquid equilibrium and, 474f9.1, 477f liquid-phase activity coefficient in, using the Raoult’s Law standard state convention, 520–524 log C-pH diagram for a solution in equilibrium with atmospheric CO2(g), 492f9.6 partial pressure and, 490–491 reactions for O2 and H2, 722–723 reference state, 476, 520 (See also infinite dilution; pure phase of the species) Heterogeneous (multi-phase) reaction, 40, 113–120, 118f3.12 Heterogeneous reactions, 85 Homogeneous (one-phase) reaction, 40 Hydration sphere, outer, 530 Hydrogen bonds, 5–6, 6f1.2, 30 Hydrogen cell, standard, 752 Hydrogen electrode, standard, 752–753 Hydrogen-oxygen bonds, 5 Hydrolysis about, 219, 221, 234, 283 of water, 220–221 Zn-induced, 234–235
I IAP. See ion activity product (IAP) Ideal behavior of solutes, 263–264, 342–343 Ideal gas about, 15–16 partial pressure of, 471–472 Ideal gas law about, 15, 41–42 gas/liquid equilibrium and, 471–472 Ideal solution, 46 Industrial wastewater treatment, 631 Infinite dilution. See also Henry’s law, reference state about, 476, 478 reference state, 42, 46, 54f2.2, 55, 64 Infinite dilution reference state, 42, 776 Inner coordination sphere, 530 Inner hydration sphere. See also hydration sphere, outer about, 530 ligand replaces water in, 538–539 ligands in, 532
890
OH– ligands in, 533 water molecules in, 532f10.2, 533 Inner sphere complex, 538 Intensive properties, 56 Interfacial structure, 809f13.20, 810f13.21 Intermolecular potential energy, 164 Internal energy (U), 144–145, 159 Intramolecular potential energy (Gintra), 159, 164 Intrinsic adsorption constants (Kint), 815, 827 Ion activity product (IAP), 615–617, 636, 638n11 Ion exchange resins, 775 Ion pairs, 538 Ionic adsorption equations, 821–824, 823f13.23 composition, 13–14, 27 conductance at 25°C and infinite dilution, equivalent, 29t1.2 strength, 24, 26–27, 27n13, 248–249, 248f5.17, 252, 256f5.19, 263, 281–282 Ions in interfacial layers, distribution and binding, 815–821, 817f13.20, 820f13.22, 821t13.4 Ion-selective electrodes, 755 Irreversible reactions, 90, 98, 101, 104–106, 113, 224n4 ISO 9963-2 (European standard for analyzing alkalinity), 414 Isotherm(s). See also adsorption isotherm; Freundlich isotherm; Langmuir isotherm apparent adsorption, 838, 839f13.30 equation is a pseudoequilibrium constant expression, 808n14 Langmuir-Freundlich, 793 surface precipitation, 838
K kilopascals (kPa), 16 Kinetic energy, 40, 50 Kinetic inhibition of redox reactions, 664, 744 kPa. See kilopascals (kPa) L Langmuir adsorption competitive, 782–785 equilibrium constant, 803 Visual Minteq and, 796–804, 799t13.2, 801f13.10, 802f13.11, 803f13.12, 805f13.13 Langmuir isotherm. See also Freundlich isotherm multisite, 786–790, 789f13.6 surfaces with uniform sites, 778–784, 782f13.5 Langmuir-Freundlich isotherm, 793 Law of thermodynamics, second, 132–133, 145 Layer(s) diffuse, 815–816, 817f13.20, 817, 819f13.21, 818, 820f13.22 electrical double, 817, 821n20 interfacial, 827 naught, 815, 817, 821–822, 825t13.5, 827, 831, 833 surface, 815–816, 819f13.21, 819, 837 Leachate, 398– 400, 399t8.2, 400f8.10, 402f8.11, 429 Lewis acid, 221, 221n2, 223 Lewis base, 221, 221n2 Ligand(s) (L) as buffers, 553 exchange, 538, 567 in inner hydration sphere, 532 inorganic ligands and some metals to which they bind, 532t10.1 metal-ligand complex, 530 protonated, 583 surface, 773, 840 systems containing ligands other than hydroxide, 587
J Junction potential, 751, 755
891
Ligand complexes, mixed about, 539 bond structure of the strong chelating agent EDTA, in its fully deprotonated form (EDTA4 ), 540f10.4(a) chelating agents and, 539–541, 540f10.4, 541f10.5 Ligands, complexing metal speciation in systems containing, 541–544t10.3, 545t10.4 stability constants for complexation of metals by OH–, 535–536t10.2 Log C-log L (ligand) diagrams, 541, 567 Log C-pe diagrams redox speciation and, 687–689, 688f12.3 Log C-pH diagram(s) acid/base problems, to solve complex, 293 acid/base problems, to solve weak, 293–308, 294f6.1, 295t6.1, 296f6.2 acid/base speciation, for visualizing, 292–293 benefit of, 342, 345 for dissolved metals, 536–538 effects of nonideal solute behavior on, 247–249, 248f5.17 equilibrium concentrations of all species in solution, 297–300, 299f6.3 equilibrium pH and, 293, 319–323, 336, 336f6.12 graphical solutions for sets of simultaneous equations, 292–293 for monoprotic acid, 243–246, 243f5.14, 245f5.16, 249–256 for multiprotic acids, 249–256, 251f5.18, 253t5.3a, 254–255t5.3, 256f5.19, 545 for open system in equilibrium with atmospheric CO2(g); where solution alkalinity is a function of pH, 514f9.16, 515f9.17 proton condition vs. TOTH equation for solving problems with, 310 for solution in equilibrium with atmospheric CO2(g), 492f9.6, 497f9.8
TOTH or PC equation to solve problems, 327, 333 to understand how speciation varies with pH, 333 Visual Minteq and, 394, 413, 421 Logarithmic plot of values of two individual terms (CL and CU) and the value of their sum, 346f6.14 Log C-log (CO2) diagram, 602f11.11 Log-log diagrams, 345–350, 346f6.14 London-van der Waals interactions, 485n5 M Mass action expression, 800n14 Mass balance about, 293, 293n1 on acid/base groups, 343 adsorption reactions, 787, 792, 804, 807, 812, 825f13.5, 827, 837 charge balance and, 300, 303 equations, 785 equilibrium pH and, 303, 303f6.6 for generic system with m different ligands, 549 on H (TOTH equation and proton condition), 308–310, 342–343 on H+, 293, 323, 342 on H2O and equilibrium speciation, 319 log C-pH diagram and, 342 stoichiometric coefficients for, 325n4 Mass concentration, 478 Mass fraction, 11, 30 Master variable, 244 Maximum contaminant level (MCL) about, 4, 19, 683 of arsenic, 308n2 MCL. See maximum contaminant level (MCL) Mechanical potential, 142 Mechanical potential energy, 142–145, 147, 159, 164, 204 Me–H2O bond, 566 Me–L complexes, 648 meq/l. See milliequivalents per liter (meq/l) Metabolic reactions, 553
892
Metal(s) about, 528–529 bioenvironmental chemistry of, 567 complexation, 664, 710, 717, 775 complexes, formation and structure of, 529–530 free (uncomplexed), 578 hydroxide solids, 579, 584, 593 ions, formation of solids containing, 577–578 oxidation states, 529 oxyanions, 533 precipitates, insoluble, 663 precipitation, 664 recovery, 761 in solution, 530 Metal ion buffers. See also buffers about, 553–556 titration curve showing response of log (Cu2+) to changes in TOTCu, for system characterized in Figure 10.11, 555f10.12 Metal ions (Me), 85 formation of solids containing, 577–578 free, 530, 541, 553, 556, 600 multitude of reactions of, 647 species containing two or more, 642n12 supersaturated, 648 unshared electron pairs and, 647 Metal speciation acid/base vs. metal/complex systems, comparison of, 545t10.4 as a function of pH and the concentration of another ligand in a system, 567 metal–ligand complexes, stability constants for some, 542–544t10.3 in solutions equilibrated with a solid, 583–587, 586f11.1 in systems containing complexing ligands, 541–544t10.3, 542–544t10.3, 545t10.4, 546f10.6, 548f10.7 Visual Minteq output summary for Zn speciation in seawater, 551f10.9 Visual Minteq summary of (a) components and (b) species for
analysis of Zn speciation in seawater, 550f10.8 Metal/complex chemistry, 553 Metal-ligand complex(es) about, 530, 533 stability constants for, 533–534, 542–544t10.3 Metastable solid, 628 mg/L. See micrograms per liter (mg/L) Michaelis-Menten equation, 110–113 micrograms per liter (mcg/L or µg/L), 11 milliequivalents per liter (meq/L), 13 milligrams per liter (mg/L), 11 MINEQL family of programs, 355 Mineral acidity, 415 Mineral adsorbents, 777, 792 Mixed solid, 837 Mixed-ligand complexes. See ligand complexes, mixed Molal concentration (m), 12 Molar concentration (M), 11, 12, 94, 478, 483 Molar enthalpy (Hi) about, 80, 82f3.1, 92, 121, 150–151, 184, 203–204 of formation, 153–154 of reaction, 121 Molar entropy about, 82f3.1, 150–151, 153, 161, 184, 203, 205 standard, 154 Molar Gibbs energy. See also chemical potential of i; Gibbs energy about, 148, 154, 160, 196, 204–205 activated complex model and, 82f3.1 change accompanying redox reactions, 732–736, 734f12.20, 736t12.9 comparing the magnitude of factors that affect, 178–179, 178t4.3 dependence on pressure (P) and temperature (T), 168–175, 169t4.2 dependence on system composition, temperature, and pressure, 158–166 dependence on system composition at constant T and P, 162–166, 165f4.4
893
dependence on system composition, contributing factors, 159–162 derivation of dependence on pressure (P), 169–175 derivation of dependence on temperature (T), 175–177 of different species, comparison of, 166–168, 167f4.5 electron energy levels for a half reaction and energy changes accompanying an n-electron reduction reaction, equations describing, 738t12.9 of electrons, 728 electrostatic interactions and, 812 equation, 666, 729 of formation, 153–154, 155f4.3, 204 of formation, standard, 154 of formation of all elements at 25° C, when present in the same phase that they are in under standard conditions, 167f4.5 of formation of some species of interest, standard, 156t4.1 of formation of various compounds at 25°C, 168f4.6 Gibbs-Duhem equation, 170 partial, 148 of a reaction, 205 of reaction for conversion of bichromate dichromate of water, 171 Molar Gibbs energy, molar enthalpy, and molar entropy alternate datum convention for, 153–158, 155f4.3, 156t4.1 datum choices for, 151 molar Gibbs energy of formation, schematic showing the, 155f4.3 molar Gibbs energy of formation of some species of interest, standard, 156t4.1 under nondatum conditions, 151–153 quantifying, 151–158 Molar potential energy, 163 Molar thermodynamic quantities, other, 150
Molar volume, 150 Mole, 11n7 Mole fraction about, 11–12, 16, 30, 41–42, 45–47, 472–473, 476–478, 520–524 Mole of stoichiometric reaction, 96–97 Molecular kinetic energy, 80 Molecular kinetic model, dynamic, 59 Moles per liter (mol/L), 11–12, 17, 26 mol/L. See moles per liter (mol/L) Monomer, 577, 642n12 Monoprotic acid about, 241t5.2, 242, 242f5.13 log C-pH diagram for, 243–246, 243f5.14, 245f5.16 Monoprotic acid/base group, 274, 280 Multidentate ligands, 539 Multiple acid dissociation reactions, 224 Multiple dissociation reactions of acid or base, 260–261 Multiprotic acid, 252 buffer intensity of, 451–453, 451t8.8, 452f8.28, 456–467 titration curves for, 392–393, 393f8.6 Multiprotic acid/base groups, 274, 279 Multiprotic acids log C-pH diagram for, 249–256, 251f5.18, 253t5.3a, 254–255t5.3, 256f5.19 log C-pH diagrams for, 249–256, 251f5.18, 253t5.3a, 254–255t5.3, 256f5.19 Muriatic acid, 219 N NAPL. See nonaqueous phase liquid (NAPL) Natural organic matter (NOM) acid/base behavior of, 439 acid/base titrations of, 438–442, 438f8.23, 441f8.24, 442f8.25 acidic functional groups in, 551–553 acids and bases, 217, 218f5.1, 231, 231f5.10, 232 adsorption and, 763 effect of pH on binding of, 218f5.1
894
Natural water bodies, dissolved constituents in, 4t1.1 Natural waters, 3, 10 Naught layer, 815, 817, 821–822, 825t13.5, 827, 831, 833. See also beta layer Naught plane, 816, 818f13.21, 819–820, 831. See also beta plane NEM. See nonelectrostatic model (NEM) Nernst equation chemical potential energy of electrons and, 731, 731f12.19 defined, 736t12.9 electrochemical reactions and, 746, 750–752 equation, 736t12.9 equilibrium speciation of a redox couple and, 756 Gibbs energy and, 731 partial pressure and, 746 pe° and, 683–684, 686, 695, 733, 736t12.9 pe°(W), defined, 694–695 schematic of relationship between electron energy levels and speciation of various redox couples, based on the, 731f12.19 Neutral complexes, 536n1 molecules, 85 solutes, 7, 26 species, 493, 515–516 Neutralization of acidic runoff, 383 Nitrate [NO3–N], 17–18 Nitrite [NO2–N], 17–18 Nitrogen (N2) in organic compounds, 17 speciation, 20–24 species, 17 NOM. See natural organic matter (NOM) Nonaqueous phase liquid (NAPL), 475, 524 Nonelectrostatic model (NEM), 797–798 Nonelementary reactions. See also elementary reactions about, 92–93, 121–122 deriving equilibrium constants of, 93–95
kinetics and equilibrium constants of, 92–97 kinetics of, 86, 95–97, 104 reaction model for enzyme-based reactions, 109–113 Nonequilibrium states, 58 Nonideal behavior, 252, 280 solute behavior on log C-pH diagrams, 247–249, 248f5.17 solutions, acid/base equilibria in, 280–283 Nonpolar molecules, 7 Nonpolar solutes, 30 Nonuniform surface sites, 785–795 Normality (N), 13 O Ohm’s law, 750 Open systems, 471, 509, 512. See also closed systems Order of a reaction, 87, 95, 98 Organic acids, 227, 229–233 Organic acids and bases, 229–233 Organic amines, 232 Organic carbon, dissolved, 4 Oxidant. See also oxidizing agent about, 735 defined, 663 disinfection of water with, 694 as electron acceptors, 663 kinetic inhibition and, 744 oxygen, 672–674 strong, 682, 714, 738–739 subscript Ox/Red, 685n5, 686 weak, 681, 737, 739 Oxidant concentrations, 23 Oxidation defined, 663 redox reactions and, 663 Oxidation and reduction of water, 719–728 Oxidation number(s), 529 of an atom in a molecule, average, 670 of C, 669, 669f12.1
895
conventions for assigning charge to various atoms, 667, 669f12.1 defined, 663, 663n1 determining, 669–670, 667t12.1, 669f12.1 Oxidation state(s) of atoms in a molecule, average, 670 defined, 663, 663n1 metal oxidation states, 529, 531f10.1 pe and, 698, 738f12.21 Visual Minteq and, 712n8–9 Oxidation-reduction reaction (“redox” reaction), 13, 23, 85, 140, 193, 761 Oxide particles, 217 Oxidizing agent, 663, 672, 737, 739. See also oxidant Ox/Red subscript, 685n5, 686 Oxyanions, 824n21, 828 Oxygen dissolution reaction, 668–669 Ozone, 473
pe° computing from species activities, 683–685 definitions of, 680–682, 693 Nernst equation and, 683–684, 686, 695, 733, 736t12.9 pe°pH, 693 pe°(W), defined, 694–695. See also Nernst equation ranges in natural aquatic systems, typical, 680f12.2 for reaction between the dominant redoxactive species in a system, 698 Pentachlorophenol, 382–383 pe-pH predominance area diagrams and, 740–743, 741t12.10, 742f12.22, 743f12.23–f12.24 space, 740 pH. See also equilibrium pH acid/base speciation as a function of, 238–242 acids or bases, responds to addition of, 737 adsorption edge, 771 buffers, 555 in carbonate-dominated systems, 427–434, 431f8.19, 432f8.20, 434f8.21 chemical equilibrium and, 29, 60 dependence of important environmental reaction rates, 86f3.3 effect on binding of natural organic matter, 218f5.1(b) environmental chemical reactions, 85 of ideal solutions containing weak acids and bases, 269–274 ionic conductance and, 29 of pure water, 163 effect of dissolution and acid/base reactions of various gases on, 502f9.11 reaction kinetics, 98, 102–103, 103f3.6 of solutions containing only water and strong acids or bases, 266–268
P p convention, 11 Partial pressure(s) about, 15–16, 16n9, 30 of an acidic or basic gas, 494 of an ideal gas, 471–472 atmospheric, 747 Henry’s constant and, 478t9.1, 480, 488, 493 Henry’s Law and, 490–491 oxygen dissolution reaction and, 668 Raoult’s law and, 474, 474f9.1 standard hydrogen cell and, 754 Visual Minteq and, 504, 504f9.12, 505, 505f9.13, 506–507, 507f9.14 Partial volume, 15–16 Partition coefficient, 60, 475, 515, 776. See also distribution coefficient Pascal (Pa), 16n9, 471n1 PCBs. See polychlorinated biphenyls (PCBs) pe. See log C-pe diagrams PE. See potential energy (PE)
896
of solutions in equilibrium with acidic or basic gases, 494–502, 495t9.3, 496t9.4, 497f9.8, 498f9.9, 499f9.10, 502f9.11 speciation and net calculated charge as a function of, 276t5.4 speciation of systems containing multiprotic weak acids and bases, 279–280 of a system with known inputs, determine, 307–308 Phase boundary, 39, 39n3, 85, 116n12, 120 Phase-transfer model schematic of adsorbed species according, 777f13.3 site-binding model vs., 778 Phase-transfer reactions adsorption reactions, 776–777, 777f13.3 reaction kinetics, 113–120, 118f3.12 Phenolphthalein alkalinity, 415 Phosphate speciation as a function of pH, 253t5.3a, 254–255t5.3b, 256f5.19 conjugate acid/base relationships among, 227 Phosphoric acid [H3PO4] about, 40 acids and bases, 219, 224–225, 234, 239, 256f5.19 Physically adsorbed species, 766 Pitzer Equation (Specific Interaction Model), 52t2.2a Planck constant (h), 93 Point of zero charge (PZC), 814t13.3 Polarity, 7 of water molecules, 30 Polymerization process, 577 Polymers, 577, 648 Polymorphs, 628–629 Polynuclear complexes, 577 Polyprotic acids, 231 Potential energy (PE). See also Gibbs energy about, 133–143, 145n5, 146–148, 159, 181, 190
chemical, 143, 147–149, 164, 189 in chemical systems, overview of, 143–423 electrical, 139–142, 141f4.2, 142–143 Gibbs, 204 gravitational, 133, 133n2, 134–139, 135f4.1, 142, 147, 153 intermolecular, 164 intramolecular, 159 mechanical, 142–145, 147, 159, 164, 204 molar, 163 Potentiometer, 752. See also voltmeter ppm, 16n10 ppmm, 16n10 ppmV, 16n10 Precipitation acid, 503 and dissolution processes in an example system, 409f11.17 effect of gas/liquid equilibrium on pH, 482n3 of the most stable solid, 648 reaction, equilibrium constant for the, 578 of solids other than hydroxides, 593–598 of solids, predominance area diagrams of possible, 641–647, 645t1.4, 646f11.28–625f11.29, 647f11.30 Predominance area diagrams about, 557–558 equilibrium relationships used to develop a predominance area diagram for the Pb2+-Cl–-OH– system, 561–562t10.6 pe-pH, 742–745, 743t12.10, 744f12.22, 745f12.23–716f12.24 ‘predominance line diagram’ showing pH ranges for dominance of various carbonate species, 558f10.15 predominance relationships among carbonate species, 559t10.5 soluble species and, 642, 647 two-dimensional slices through a threedimensional log C-pH–log L diagram, 557f10.14 Predominance line diagram, 558, 558f10.15
897
Pressure (P), 133 activity of any chemical phase and, 56 activity of any solid or gas phase and, 56 gas-phase concentrations and, 15–16, 30 Gibbs phase rule and, 629, 632 infinite dilution reference state and, 42 molar Gibbs energy and, 158–166, 168–175, 169t4.2 partial pressure (Pi), 15, 72 reactivity of chemicals and, 40–41 reference state and, 41, 43t2.1 standard, 41, 43t2.1 in the standard state, 41–43 surface, 769 Proton accepting species, 443 acceptors, 221–222 donating species, 443 donors, 221 “unprotected,” 5 Proton condition (PC) about, 385–386, 421 log C-pH diagrams and, 310 mass balance and, 308–310, 342–343 TOTH equation, 310, 319, 321–322, 322f6.10, 343, 345 Protonation acid-base groups, 390, 439–440 reactions, 408 Protonation/deprotonation reactions during a titration, 438, 438f8.23 reactions over a wider pH range, 442, 442f8.25 psi. See pounds per square inch (psi) PZC. See point of zero charge (PZC)
liquid-phase activity coefficient in Henry’s Law using the Raoult’s Law standard state convention, 520–524 partial pressure and, 474, 474f9.1 reference state, 476–477, 520 volatility and, 485 Rate expression coefficients in a, 97 for elementary reactions, 95 for nonelementary reactions, 94 reaction kinetics and, 88, 94–98, 98n7, 99, 100t3.2, 101–102, 106, 110, 113, 115, 121–22 Rate-limiting reaction step, 105–107, 111, 113–114, 119, 122 Reaction kinetics characteristic profiles for decay of a reactant undergoing an irreversible, first order reaction in a batch reactor, 99f3.4 experimental evaluation of, 97–104 rate expression and, 97–98, 98n7, 99, 100t3.2, 101–102, 106, 110, 113 Reaction mechanism, 79. See also reaction pathway Reaction pathway(s) about, 79, 104–113, 122 classic kinetics of enzyme-based reactions, 111f3.11 classical, model, 104–113 sequential reactions: concentration vs. time profiles, 107f3.9 water flowing through a series of tanks, 105f3.8 Reaction quotient (Q) chemical reactions, 72–73 oxidation numbers, 666, 686 reaction kinetics, 89, 94–95, 121 Reaction rate constant (k). See also Arrhenius’s equation about, 83, 87 effect of temperature on, 83, 84f3.2 first-order, for oxidation of ferrous iron, 86f3.3
R Radon, gaseous, 489 Raoult’s Law about, 472–477, 485, 520–524 equation, 474 gas-liquid equilibrium and, 474f9.1, 477f
898
Reaction reversibility. See also irreversible reaction about, 88–91, 90t3.1, 104, 109, 121 Reactivity of chemicals, factors affecting, 38–40 defined, 71–72 Recarbonation, 506–508 Redox chemistry, 663–669, 667t12.1, 669f12.1 conjugate pair, 679, 692, 756 couples, 676, 681–682, 687, 692f12.4, 695, 701, 755 electrodes, 750 equilibria, 740 Redox half-reaction(s) about, 671–674 catalysts and, 672 defined, 672 electrical potential (EH) and, 747 electrochemical reactions and, 745, 747, 750 electrons and, 671–676, 678, 686 equilibrium constants for, 674–678t12.3, 678–679 H2 oxidation and O2 reduction, 744 Nernst equation and, 686 pe values for, 686, 736 pe° values for, 685, 732 pe°pH for redox couples, 738 pe(W) values for, 677t12.3, 695 redox reaction electrical potential (EH), 748–750 for reduction of oxygen, 672 Visual Minteq and, 711, 714 Redox potential (EH). See also electrochemical potential about, 138, 729–730, 730n14, 747–749 electrochemical reactions and, 746–747 electrodes, between, 27 electrodes, near, 749 Gibbs energy and, 730 Gibbs energy of electrons and, 729 ion-selective electrodes and, 755 redox half-reactions, 747–750 redox reactions, 748–750 “virtual” electrical potential, 730
Redox reaction electrical potential (EH), 748–750 Redox reaction(s). See also oxidation-reduction reaction acid/base reactions vs., 689–690t12.4, 692, 692f12.4 balancing, 670–671, 670t12.2 combining, 685–687 electrochemical reactions, anodes and cathodes, 744–745, 745f12.25 electrochemistry and, 744–756 energy changes accompanying, 728–737 involving exchange of both electrons and proton, 692–703 involving exchange of multiple electrons, 699–703, 700f12.6, 702f12.7 kinetic inhibition of, 664, 744 modeling with Visual Minteq, 710–719 with no H+ exchange, 687–689, 688f12.3 in systems with externally imposed electrical potentials, 748–750 Redox speciation and log C-pe diagrams, 687–688, 688f12.3 Redox systems choosing components and preparing the tableau, 705–710, 705f12.8, 706f12.9, 708t12.5, 709f12.10 equilibrium speciation in, computing, 703–710 governing equations and preparation of log C-pe diagrams, 703–705 Redox titrations geochemical redox sequence and, 737– 740, 738f12.21 to infer identities and concentrations of unknown species in a solution, 421 Redox-active species, 671, 698, 719, 755 Reducing agents, 663 Reductants, 663, 681 Reduction reaction, 681, 711, 720t12.6, 736t12.9, 744n19. See also oxidation reaction Reference electrodes, 752–754, 754f12.27
899
Reference state. See also standard state about, 41–42, 43t2.1, 49 activity coefficient and, 50 infinite dilution, 42, 46, 52t2.2a, 54f2.2, 55, 64–67 of pure liquid benzene, 46–48 of pure organic liquid, 47 in seawater, 64 of a solid, 578–579 of solutes, 47 standard-state concentration and, 72 Reverse(d) of acid dissociation reaction, 233 acidity reaction, 233 Reversible reaction. See also irreversible reaction about, 88–91, 90t3.1, 104, 109, 121
dispersed in solution, 56 dissolution reaction, 770 equilibrium with multiple solids, general criteria for, 620–626 flow chart for determining equilibrium speciation in a system with known TOTMe, in which Me(OH)2(s) might precipitate and only hydroxo complexes form, 610f11.14 formation of nonhydroxo complexes in systems with, 598–604 metal speciation in solutions equilibrated with a, 583 metastable, 628 polymorphs, relative solubilities, 628–629 precipitation of, other than hydroxides, 593–598 solid polymorphs, relative solubilities of, 628–629 systems in which more than one solid must be considered, 616–626 systems in which only one solid need be considered, 608–616, 610f11.14 Solid/solution equilibrium, 576 modeling with Visual Minteq, 587–593, 589f11.3, 590f11.4, 593f11.6 Solid/solution interface, 774 Solubility product, 579–593, 600, 616–617, 628, 638–639, 643 Ks0 values of some solids of interest, 581–582t11.1 Soluble complexes, 577–578, 602, 648 Soluble Me-L chemistry, summary of, 566– 567 Soluble species dominant, 642–644 predominance area diagrams and, 642, 647 Solutes hydrophilic, 770 in terrestrial waters, concentration of, 8–9, 9f1.4 in water, behavior of, 6–9
S Salting out, 487 Salt(s) about, 8, 8n5, 304, 334 bridge, 750, 755 dissolution of, 6–9 ions, 13, 25, 326n5, 329–330, 341 pillars in the Dead Sea, 10 Saturation index, 615–616 Seawater, 532, 546, 546n3, 549, 550f10.8, 551f10.9 Second law of thermodynamics, 132–133, 145 Site-binding model defined, 773 phase transfer model vs., 777 site-binding model, schematic of adsorptive solid/solution interface according, 775f13.2 Soda ash, 580. See also sodium carbonate Software. See chemical equilibrium software; Visual Minteq Solid(s) chemical activity of pure, 578–579 determining whether a solid will precipitate under given conditions, 608–629
900
Solution pH acids and bases, 217, 239–240, 242f5.13, 256, 267–268, 274–275 titrations and buffers, 379, 381, 385, 388t8.1, 389–390, 394, 400, 402f8.11, 404–405, 410, 424–426, 433, 435, 437 Solver® routine embedded in Excel®, 276 Sorbate, 773 Sorbent, 773 Sorption, 773 Speciation of an ideal solution containing TOTA, 444t8.7 of any monoprotic acid, 245 of carbonic acid, 249 defined, 2 dependence on pH, graphical display of, 246 equilibrium, 217, 293, 308, 319, 327, 341–342 equilibrium solution, on Visual Minteq, 395f8.7 Log C–pH graphs and the effect of a pH change on, 430, 434f8.21 and net calculated charge as a function of pH, 276t5.4 simple spreadsheet analysis for determining acid/base, 274–279, 276t5.4 of a solution of water and a strong acid or base at equilibrium, 266–268 of systems containing multiprotic weak acids and bases, pH and, 279–280 in systems with nonideal behavior, 252, 256f5.19, 280–283 TOTH equation to solve for, 386 of weak acid/base groups, 389f8.4, 390 Species. See also equilibrium species acid/base, 330, 336, 343 defined, 310 identifying dominant, 327–341, 329t6.5, 336f6.12, 337f6.13 neutral, 493, 515–516 nondominant, 558
protonation/deprotonation of dissolved, 508 redox-active, 671, 691n, 698, 719, 755 Specific conductance, 27 Specific Interaction Model (Pitzer Equation), 51, 52t2.2a, 54f2.2(b) Specific rate of substrate utilization, 110 Stability constant(s) for complexation of metals by OH–, 535t10.2 for ligands, rules for naming, 538–539 of metal-ligand complexes, 533 for metals and ligands 4t10.3 Stagnant-layer model, 117, 118f3.12 Standard concentration, 41, 72 Standard hydrogen cell, 752 Standard hydrogen electrode, 752–753 Standard Methods for the Evaluation of Water and Wastewater, 414, 414n5, 415 Standard state, 578–579. See also reference state about, 40–41, 49 concentration of any species in, 41, 45, 64 conditions, alternative choices for, 46–49 environmental conditions, 41–43 for solutes, 43t2.1, 46 for species in various phases, 42, 43t2.1 Standard state concentration, 475–476, 504, 516 Standard state conditions, 578 State function, 136, 138 State variable, 136, 144–146, 148 Stoichiometric coefficient(s) acids and bases, 311f6.9, 312–316, 318–320, 324–325, 325n4 chemical reactivity and, 59, 69, 87, 95–96, 106 “crossover” point and, 693t12.4 gas/liquid equilibrium, 495t9.3, 496t9.4, 498, 498t, 500, 500t, 501, 510 reaction kinetics and, 87, 95–96, 106 redox reactions, algorithm for balancing, 670t12.2 Stoichiometric reaction, mole of, 96–97
901
Stoichiometric softening, 580, 590, 590f11.4 Stoichiometry of solids, 589n2 Supersaturated solids, 617–618, 619f11.17, 620, 622, 622f11.18, 623f11.19 Supersaturated solution, 475, 491, 511, 580, 589, 591n3, 609, 616, 620. See also undersaturated solution Surface layer, 815–816, 819f13.21, 819, 823 ligands, 773, 840 precipitation, 835–839, 836f13.27, 836, 836n26, 837–838, 838f13.29, 839, 839f13.30 pressure, 776 site density, 800–801, 803 Surface charge about, 769, 772–773, 812, 813f13.18, 823f13.23, 827, 831, 833–834, 836 of various particles, computed based on the sorption of H+ and OH– ions, 812 Surface complexation adsorption models, 774–777, 779, 815f13.19 Visual Minteq and, 796–797, 799–803, 802f13.11, 802n11, 803, 805, 806n12, 813, 815f13.19 Surface complexation reaction, 773–776, 775f13.2 Surface electrical potential, 841 adsorption of ions in the presence of, 811–827 Suspended solids, 773
ideal solution and, 46 infinite dilution reference state and, 42 molar Gibbs energy and, 158–166, 168–175, 169t4.2 plot of the rate constant vs. inverse absolute temperature, 84f3.2 reaction rate constants and, 83–84 reaction rates and, 121 reactivity of chemicals and, 40 reference state, 41, 43t2.1 standard, 41, 43t2.1 in the standard state, 41–43 system equilibrium and, 59 van’t Hoff equation and, 92, 121 Thermal motion, 7 Thermomechanical systems, 133 Three-dimensional diagrams, 557, 557f10.14 TIC. See total inorganic carbon (TIC) Titration. See also alkalinity; log C-pH diagram(s) about, 379, 453–454 acid/base titrations of natural organic matter, 438–442, 438f8.23, 441f8.24, 442f8.25 acids and bases, titration of solutions containing unknown, 406–408, 407f8.13 alkalinity, 513 alkalinity, assessing contribution of different species to, 416–423, 417t8.5, 421f8.18 alkalinity, contribution of added species to, 423–426, 425t8.6 alkalinity and acidity, conceptual basis and definition, 413–416, 415f8.17 alkalinity as a conservative quantity, implications of, 426–427 alkalinity calculations embedded in Visual Minteq, 435–437, 435f8.22 alkalinity computed from titration data or based on solution composition, 416 alkalinity concept, summary of, 437 data, quantitative interpretation of, 384–393
T TDS. See total dissolved solids (TDS) Temperature (T), 133 activation energy and, 87 activity of any chemical phase and, 56 activity of any solid or gas phase and, 56 conductivity of a solution and, 29 enthalpy and, 81 entropy and, 81 equilibrium constant and, 68, 73, 91–92 Gibbs energy and, 81, 83 Gibbs phase rule, 629, 632
902
data, units for quantifying, 384–385 equivalence points, defined, 409–411, 409f8.14 equivalence points in the carbonate system, 411–413, 411f8.15, 412f8.16, 412t8.3, 413t8.4 Gran titration technique, 103n4 pH and alkalinity in carbonatedominated systems, 427–434, 431f8.19, 432f8.20, 434f8.21 reactions occurring during a, 380–384, 380f8.1, 382f8.2 of strong base with a strong acid, 385–386, 385n2, 386f8.3 Visual Minteq, alkalinity calculations embedded in, 435–437, 435f8.22 Visual Minteq simulating evaporation or extraction of pure water from a solution, 401–402, 402f8.11 Visual Minteq simulations varying only TOTH, 394–396, 395f8.7, 396f8.8– 385f8.9 Visual Minteq simulations varying TOTi of more than one component, 397–400, 399t8.2, 400–401f8.10 Visual Minteq to simulate titrations, 393–402 of weak base with a strong acid, 386–392, 388t8.1, 389f8.4, 391f8.5 Titration curves about, 453–454 for addition of strong base to solutions containing HAc, HOCl, or NH4Cl, 409 effect of acidity of water on, 405–406 of natural waters, 439 of NOM isolated from a “pristine” groundwater source, 438f8.23, 439–440 of NOM with two types of acid/base groups, 442, 442f8.25 of sample of unknown composition and of a blank (pure water) with strong acid or base, 406, 407f8.13
for systems with multiprotic acids, 392–393, 393f8.6 TOC. See total organic carbon (TOC) Total alkalinity, 415, 418, 457, 455 Total dissolved solids (TDS), 4t1.1, 24–26 Total inorganic carbon (TIC), 18, 323–324, 326 Total organic carbon (TOC), 18, 323–324 Total organic halogen (TOX), 19–20 TOTH equation about, 293, 310 advantages of using, 342 computer software and, 310, 342–343 equilibrium pH, to determine, 323–327, 342 gas/liquid equilibrium and, 494–497, 500–501, 512–513 for identifying dominant species, 327–342 proton condition (PC) and, 310, 319, 321–323, 322f6.10, 343, 345 titrations, 385–387, 388t8.1, 411, 421, 423, 433, 444 TOX. See total organic halogen (TOX) Transition state model, 81. See also activated complex Triple-layer model, 815f13.19, 816–817, 819f13.21, 819–820, 821n20, 821t13.4, 828n22, 831 Trisodium phosphate [Na3PO4], 224 U Uncomplexed ions, 530. See also free metal ions Undersaturated solid, 623f11.19 Undersaturated solution, 475, 491, 580, 610, 615–617, 619, 619f11.17, 620, 622, 639. See also supersaturated solution Universal energy, 145, 147 Universal entropy, 132–133 Universal gas constant, 16 U.S. Environmental Protection Agency (EPA), 217
903
log C-pe diagrams prepared with, 711–714, 714f12.11 modeling of systems with potential solids, 626–628 modifying the component and species databases in, 373–374 Multi-problem/Sweep page, 370f7.7, 394, 398–399, 401f8.10, 419, 429, 441, 441f8.24, 458, 548, 714, 715f12.11 Parameters/Specify alkalinity menu, 550 Parameters/Specify pe and Eh link, 713, 713n11 partial pressure and, 504, 504f9.12, 505, 505f9.13, 506–507, 507f9.14 pseudospecies used to allow both species in the redox couple to be components, 710–711, 714f12.11 redox pulldown menu, 712 redox reactions modeling with, 710–719 simulating evaporation or extraction of pure water from a solution, 401–402, 402f8.11 simulations varying only TOTH, 394–396, 395f8.7, 396f8.8–385f8.9 simulations varying TOTi of more than one component, 397–400, 399t8.2, 400–401f8.10 Solid phases and excluded species, 588–589 solid/solution equilibrium modeling with, 587–593, 588f11.2, 589f11.3, 590f11.4, 592f11.5, 593f11.6 Sorption/complexation database page, screenshot of, 823f13.24 Specify infinite solid phases, 588 Specify possible solid phases, 611, 613 surface acid/base speciation, 832f13.25 surface complexation adsorption models and, 769–797, 799–801, 802f13.11, 802n11, 803, 805, 806n12, 803, 815f13.19 surface complexation page, 800
V Van’t Hoff equation, 68, 73, 92, 121, 197, 203, 205, 486n7 Visual Minteq. See also computer software adsorption, modeling, 796–810 adsorption reactions relevant to the example problem, 829t13.6 alkalinity calculations embedded in, 435–437, 435f8.22 alkalinity factors determination and inserting them into database of, 454–455 chemical equilibrium problems, solving, 355, 355n1–2, 359, 363–365, 368–369, 370f7.7, 371f7.8–363f7.9, 372, 372f7.10, 375 “Counter-ion accumulation, “ 793 database, adding new solids to, 604–608, 606f11.13 Database management/Gases page, 505, 505f9.13, 506 Database management/Solid phases 584, 606f11.13 electrostatic interactions and, 797, 799, 802f13.11, 827 Equilibrated mass distribution page, 507 equilibrium constant and, 805, 827–828 equilibrium constant dissolution reaction in, 605, 606f11.13 equilibrium constant for dissolution of a solid, 589, 589f11.3 equilibrium pe and solution composition, solving for, 714–719, 715f12.12, 716f12.13, 718f12.14 Freundlich isotherm, 805–810, 808f13.14, 810f13.15, 811f13.16–f13.17 gas/liquid equilibrium in, 502–508 Gibbs phase rule, 629, 633 input screen for gases, 504f9.12 Langmuir adsorption, single- and multisite, 796–805, 799t13.2, 801f13.10, 802f13.11, 803f13.12, 805f13.13 log C-pH diagram using, 548
904
surface-adsorbate electrical interactions, modeling, 827–834, 829t13.6, 831f13.24, 832f13.25, 834f13.26 titrations, to simulate, 393–402 Volatile organic compounds, 476 Volatile solutes, 475 Volatile species about, 472–476, 480–485, 493–494, 497 acidic and basic, 515 Volatility about, 478 defined, 471 Henry’s constant and, 481, 481f9.2, 483–490 hydrogen-bonded water molecules surrounding a hydrophobic solute, schematic of, 484f9.3 hydrophobic or hydrophilic nature of solute, 484–485, 484f9.3 molecular structure and, 486 oxygen solubility in water as a function of temperature and salt content, 488f9.5 Raoult’s law and, 485 solution composition and, 487–490, 488f9.5 temperature and, 486–487, 487f9.4 temperature dependence of Henry’s constant for several halogenated VOCs, 487t9.4 vapor pressure of pure compound, 485–486 Volatilization about, 471–472, 475, 483, 490 enthalpy of, 486, 506 reaction, 472, 503–504, 506, 516 reaction and Henry’s constant, 475, 486 Voltmeter, 752–756, 756f12.27 Volume fraction, 15–16
Water activities of dissolved and gaseous H2 and O2 in equilibrium with water for a system at pH 7.0 and 25° C, 724f12.16 dissociation of water (Kw), 219–221, 220f5.2–214f5.3 equilibrium among H2O, O2(aq), and H2(aq), Figure 12.9 modified to include, 725f12.17 hydrated H+ ion [H3O+], hydrogenbonded to two other water molecules, 220f5.3 log C-pe diagram for the H2/H2O/O2 redox system, 791–723, 720t12.6, 721f12.15, 722f12.16 molar Gibbs energy of, 171 molecules, polarity of, 30 oxidation and reduction of, 791–728 redox activity of water when analyzing pe of a solution, 724–728, 725f12.17, 726t12.7, 727f12.18, 728t12.8 redox equilibrium between H2O and the atmosphere, impossibility of, 723–724 redox reactions of water, representations of, 720t12.6 softening, 763 treatment processes, 2, 23, 232 value of Kw from 0 to 100°C, 220f5.2 water oxidation reaction, 720, 722t12.6 water reduction reaction, 720, 722t12.6 Water molecules arrangement of water molecules and ions around a dissolved ion, 50f2.1, 53 hydrogen-bonded, 6f1.2 schematic representation of the structure, and charge distribution, 3, 5, 5f1.1 structure and interactions among molecules, 3–6 tendency to orient themselves around ions facilitates dissolution of salt particles, 7f1.3
W Wastewater, 626–628 domestic and industrial, 3 treatment operations, 593 treatment plants, 631, 760
905
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