Revised Ana Chem

Chemical and Physical Principles

Analytical Chemistry

ANALYTICAL CHEMISTRY Analytical Chemistry • Branch of Chemistry that deals with the analysis, identification, separation and composition of matter • Involves methods used to identify the substances that are present in a sample (qualitative analysis) and the exact amount of the identified substances (quantitative analysis) Quantitative Methods of Analysis A. Classification of Methods of Analysis 1.

Classical methods a. Gravimetric method – measurement of the mass of a substance that is chemically related to the analyte b. Volumetric Volumetric method – measurement – measurement of the volume of solution necessary to react completely with the analyte

2.

Modern methods Spectr tros osco copic pic metho method d – measurement a. Spec measurement of the electromagnetic electromagnetic radiation radiation produced by the analyte or its interactions with it b. Electroanalytic Meth od – involves – involves measurement of the electrical properties of the analyte such as current, potential or quantity of charge

3.

Other methods – involves – involves the measurement of the properties of the analyte such as heat of reaction ( calorimeter ), ), index of refraction (refractometer ), ), optical polarimeter ) or mass-to-charge ratio (mass spectrometer ) activity ( polarimeter

B. Typical Steps in Analysis 1.

Selection of an appropriate method • In the selection of method of analysis, it is necessary to consider the level of accuracy, complexity and component of the sample, availability of equipment and trained personnel and the time of analysis • Standard procedures are usually available from literature such as Chemical Abstracts, Analytica Chimica Acta, Applied Spectroscopy, Journal of the Association of Analytical Chemists, etc. tc.

2.

Obtaining a representative sample • The American Society for Testing and Materials (ASTM), National Bureau of Standards (NBS) and Association of Official Analytical Chemists (AOAC) are such a few organization organizationss that impose standard standard sampling sampling procedures procedures for analysis of some samples • Three steps are generally followed in obtaining samples: obtaining a gross sample, obtaining a laboratory sample and obtaining an analysis sample • A gross sample is obtained from a bulk sample and obtained in such a manner that it is considered a representative of the bulk sample • A laboratory sample is a fraction of the gross sample weighing several grams wherein further reduction to few milligrams results into an analysis sample

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3.


Measuremen Measurementt of the sample sample • First step in preparing a sample involves measurement of either mass or volume • Solid samples samples are are dried dried in an oven oven usually usually from 110°C-120°C for about 1-2 hours and cooled in a dessicator; analysis is said to be done in a dry basis. Samples that decompose upon heat treatment are analyzed on a wet basis or as-received basis • Replicate samples are taken for analysis to ensure accuracy of the method used and quality of the results. results. Results obtained from these replicate samples are treated using various statistical tests to establish reliability Table 1. 1. Classification of analysis based on sample size Method Sample Mass Sample Volume macro more than 100 mg more than 0.100 mL – 100 mg semi-micro 10 mg – 1 0.050 mL to 0.100 mL – 10 mg m ic r o 1 – 1 less than 0. 0.050 mL – ultra-micro less than 1 mg

4.

Preparation of a solution of the sample • Most methods are designed to process liquid samples specifically solutions, since these samples are homogenous and are easy to handle • The following solvents are commonly employed in preparation of the solutions of the sample: a. Water . Samples of soluble salts readily dissolve in water at room temperature and heating may be done to facilitate dissolution of the sample. b. Non-oxidizing acids. In many instances, some portion of the sample will not dissolve in water and usually the addition of acids render the sample soluble. Hydrochloric acid is a typical non-oxidizing acid along with dilute sulfuric and perchloric acid. c. Oxidizing acids. For more stubborn samples, hot, concentrated sulfuric acid, nitric acid and aqua regia are used. Aqua regia is a mixture of hydrochloric acid and nitric acid in 3:1 volume ratio. Hydrofluoric acid is also used for dissolving silicate ores. d. Fluxing agents. Samples which were not dissolved in aqueous solvents are usually fused with a molten solvent called flux. Fluxing agents may be classified as acidic (K2S2O7, KHF2 and B2O3), basic (Na2CO3, K2CO3, NaOH or KOH) and oxidizing (Na2O2). Fusion is done by mixing a finely ground sample with the solid flux in an inert crucible and heated until the flux melts. • Reagents and chemicals used in the laboratory are classified as follows: a. Commercial or technical reagents . Reagents that undergo superficial purification and not directly used for analysis b. United States Pharmacopoeia (USP Grade) or National Formulary Reagents. Reagents used by pharmacists and unfit for analysis c. Chemical Pure (CP) reagents . Reagents that are more refined compared to technical reagents d. Reagent grade or analytical reagent (AR) or certified reagent . Reagents analyzed by the manufacturer with the analysis found on the label of the container e. Primary standard grade . Chemicals with purity greater than 99.95%

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3.


Measuremen Measurementt of the sample sample • First step in preparing a sample involves measurement of either mass or volume • Solid samples samples are are dried dried in an oven oven usually usually from 110°C-120°C for about 1-2 hours and cooled in a dessicator; analysis is said to be done in a dry basis. Samples that decompose upon heat treatment are analyzed on a wet basis or as-received basis • Replicate samples are taken for analysis to ensure accuracy of the method used and quality of the results. results. Results obtained from these replicate samples are treated using various statistical tests to establish reliability Table 1. 1. Classification of analysis based on sample size Method Sample Mass Sample Volume macro more than 100 mg more than 0.100 mL – 100 mg semi-micro 10 mg – 1 0.050 mL to 0.100 mL – 10 mg m ic r o 1 – 1 less than 0. 0.050 mL – ultra-micro less than 1 mg

4.

Preparation of a solution of the sample • Most methods are designed to process liquid samples specifically solutions, since these samples are homogenous and are easy to handle • The following solvents are commonly employed in preparation of the solutions of the sample: a. Water . Samples of soluble salts readily dissolve in water at room temperature and heating may be done to facilitate dissolution of the sample. b. Non-oxidizing acids. In many instances, some portion of the sample will not dissolve in water and usually the addition of acids render the sample soluble. Hydrochloric acid is a typical non-oxidizing acid along with dilute sulfuric and perchloric acid. c. Oxidizing acids. For more stubborn samples, hot, concentrated sulfuric acid, nitric acid and aqua regia are used. Aqua regia is a mixture of hydrochloric acid and nitric acid in 3:1 volume ratio. Hydrofluoric acid is also used for dissolving silicate ores. d. Fluxing agents. Samples which were not dissolved in aqueous solvents are usually fused with a molten solvent called flux. Fluxing agents may be classified as acidic (K2S2O7, KHF2 and B2O3), basic (Na2CO3, K2CO3, NaOH or KOH) and oxidizing (Na2O2). Fusion is done by mixing a finely ground sample with the solid flux in an inert crucible and heated until the flux melts. • Reagents and chemicals used in the laboratory are classified as follows: a. Commercial or technical reagents . Reagents that undergo superficial purification and not directly used for analysis b. United States Pharmacopoeia (USP Grade) or National Formulary Reagents. Reagents used by pharmacists and unfit for analysis c. Chemical Pure (CP) reagents . Reagents that are more refined compared to technical reagents d. Reagent grade or analytical reagent (AR) or certified reagent . Reagents analyzed by the manufacturer with the analysis found on the label of the container e. Primary standard grade . Chemicals with purity greater than 99.95%

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•


There are several ways of expressing concentration of a particular species in solution. The following are most commonly used to express concentrations of solutions: a.

Weight percent – usually used to express concentration of commercial aqueous reagents

 wt  = weight of solute × 100%   wt  weight of solution

% b.

Volume percent – commonly used to specify the concentration of a pure liquid compound diluted with another liquid

 vol  = volume of solute × 100%   vol  volume of solution

%

For alcoholic beverages, percentage of alcohol is usually expressed in terms of proof as follows:

  vol       vol  

proof = 2 %  c.

Weight/volume percent – used to indicate the concentration of a solid reagent in a dilute aqueous solution

 wt  = weight of solute (gram) × 100%   vol  volume of solution (mL)

%

d. Mole fraction (x) – commonly used in unit operations to express concentrations of solute present in a stream of gas or liquid mole of solute x= mole of solute + mole of solvent e. Molality (m) – temper temperatur ature-i e-inde ndepen penden dentt concen concentrat tration ion term used used conveniently in physicochemical measurements of colligative properties of solutions mole of solute (mol) m = kilogram of solvent (kg) f.

Molarity (M) – most commonly used in titration and denotes the amount of solute, in moles, dissolved in a solvent and diluting to a final volume of 1L in a volumetric flask mole of solute (mol) M = volume of solution (L) Formality (F) – concentration term identical to molarity commonly used for solutions of ionic salts that do not exist as molecule in solid or in solution

g. Normality (N) – once popular unit of concentration still used by some chemists equivalent of solute (eq) N = volume of solution (L)

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An equivalent represents the mass of material providing Avogadro’s number of reacting units. The number of equivalents is given by the number of moles multiplied by the number of reacting units per molecule or atom. m m equivalent of solute = = EW MW f where m = mass of solute [g], EW = equivalent weight [g-equiv – 1], MW = molar mass [g-mol – 1] and f = molar equivalent [equiv-mol – 1] Therefore, normality (N) is related to molarity (M) according to the following equation:

 eq  ×  mol  =  eq   mol   L   L 

N = f × M = 

Table 2. Molar equivalents of solutes Nature of solute Molar equivalent acid number of replaceable H+ base number of equivalent HO – salt net charge of an ion oxidant gain of electron reductant loss of electron h.

p-function - used to express concentrations at a magnitude of 10n where n is any integer less than zero defined as follows: p[ f ( x ) ] = − log ( x ) where x = concentration of the species in molarity, M

5.

Treatment of the sample • Some samples has to be reduced or oxidized prior to analysis or sometimes treated to become colored or converted to a form that it can be readily volatilized • More often, the accuracy of an analysis is affected by the presence of unwanted components called interferences • Interferences can be eliminated by converting it into non-interfering form by a process called masking. Typical masking procedure may be done by converting the interference into a stable complex ion that does not react with the reagents added to the sample • Some separation processes are commonly employed to isolate the analyte from the interferences such as precipitation, electrodeposition, extraction, ion exchange, volatilization and chromatography

6.

Measurement of the analyte • Using classical methods of analysis, results can be accurate up to a few parts per thousand or better, requires relatively large amount of sample and usually applied to measurement of major constituents in a sample • Instrumental methods are generally more sensitive and selective. Analysis is rapid, automated and capable of measuring more than one analyte at a time; however, these techniques are more expensive than classical methods.

117



Table 3. Classification of constituents in a sample Component Relative Amount major greater than 1.00% semi-micro 0.10% – 1.00% micro 0.001% – 0.10% ultra-micro less than 0.001% 7.

Calculation of results and reporting of data • Results of analysis can be expressed depending on the nature of analyte a. Solid Samples. Calculations on solid samples are based on mass. The most common way of expressing results in a macro determination is by % mass or % weight

 wt  =   wt 

%

wt. analyte wt. sample

×100%

Table 4. Concentrations of analyte in solid samples in trace concentrations Unit Definiton Unit mg analyte g analyte  wt  = gram analyte ×103 parts per or pt   thousand g sample kg sample  wt  gram sample parts per million parts per billion

 wt  = gram analyte ×10 6   wt  gram sample  wt  = gram analyte ×109 ppb   wt  gram sample

ppm

µg analyte g sample

or

ng analyte g sample

or

mg analyte kg sample µg analyte kg sample

b. Liquid Samples. Similarly, concentrations of solid or liquid analytes in liquid samples obtained from a macro analysis is usually expressed as % weight by volume or % volume by volume defined as follows:

 wt  = gram analyte ×100%   vol  mL sample

%

 vol  = volume analyte ×100%   vol  volume sample

or % 

Table 5. Concentrations of analyte in liquid samples in trace concentrations (wt/vol and vol/vol) parts per million parts per billion parts per trillion parts per million parts per billion

 wt  = gram analyte × 106   vol  mL sample  wt  = gram analyte ×10 9 ppb   vol  mL sample  wt  = gram analyte ×1012 ppb   vol  mL sample  vol  = vol analyte ×10 6 ppm   vol  vol sample  vol  = vol analyte ×109 ppm   vol  vol sample ppm

118

µg analyte mL sample ng analyte mL sample pg analyte mL sample nL analyte mL sample pL analyte mL sample

or or or or or

mg analyte L sample µg analyte L sample ng analyte L sample µL analyte L sample nL analyte L sample



Gravimetric Methods of Analysis A. Types of Gravimetric Analysis 1.

2. 3.

Extraction method. The analyte in the sample is obtained using as appropriate solvent and the residue from the solution, after evaporation of the solvent, is chemically related to the analyte Precipitation Method. The analyte is converted into a sparingly soluble solid, filtered, washed, dried or ignited and weighed Volatilization Method. The sample is treated to yield a gas that is passed in an absorbing medium; the analysis is based upon the change in mass of the medium

B. Gravimetric Factor and Precipitating Agents 1.

Calculation in gravimetric analysis To calculate the amount of analyte in the sample… mass of final form % analyte = × GF ×100% mass of sample Gravimetric Factor (GF)

GF = 2.

 x mol analyte   ×  molar mass of final form  y mol final form   molar ratio molar mass of analyte

Precipitating agent • An ideal precipitating agent must give an insoluble product that has the following properties: can be easily filtered and washed free from impurities or contaminants has very low solubility to avoid losses during filtration and washing inert towards components of the atmosphere has known composition after subjecting to appropriate heat treatment    

Table 6. Precipitating agents used in precipitation gravimetry Species Precipitated

Final Form

Cl Br I SO4 – 2 As Bi Cd Cu Sn Sb Mg Zn

AgCl AgBr AgI BaSO4 As2O3 Bi2S3 CdSO4 CuO SnO2 Sb2O3 Mg2P2O7 Zn2P2O7

Precipitant

AgNO3 BaCl2

H2S

(NH4)2HPO4

Species Precipitated

Final Form

Al Cr Fe Sn Ba Cd Sr Ca Mg Zn K Hg

Al2O3 Cr2O3 Fe2O3 SnO2 BaSO4 CdSO4 SrSO4 CaCO3 MgCO3 ZnCO3 H2PtCl6 HgS

Precipitant

NH3 H2SO4

(NH4)2C2O4 K2PtCl6 (NH4)2S

C. Theory of Precipitation 1.

Properties of precipitates a. Particle size. Solid particles formed from precipitation may vary accordingly: • Colloidal – tiny particles with size ranging from 0.1 microns to 100 microns in diameter; these particles do not settle readily and cannot be filtered easily 119


• •


Crystalline – particles with size ranging from 100 microns or greater; these particles settle readily and easily filtered Particle size is usually affected by temperature, concentration of reactants, solubility of precipitate and mixing rate

b. Appearance. Precipitates may appear to be colloidal (S) , curdy (AgCl), fine crystal (BaSO4), coarse crystal (PbCl2) or gelatinous (Al(OH)3) c. Relative supersaturation (von Weimarn ratio) relative supersaturation =

Q −S

S where Q = concentration of the solute as precipitation begins and S = solubility of the precipitate • In order to obtain low relative supersaturation and form a crystalline precipitate, Q must be minimized and S must be maximized . The following methods are done to accomplish such conditions: increase the temperature during precipitation (to maximize S) precipitate from dilute solution (to minimize Q) slow addition of precipitating agent with stirring (to minimize Q)   

2.

Mechanism of precipitation Precipitation is assumed to occur in two ways: a. Nucleation • Prevails at high relative supersaturation • Results in the formation of large number of small particles b. Particle growth • Prevails at low relative supersaturation • Results in the formation of small number of large particles

3.

Colloidal precipitates a. Electrical nature of colloidal suspensions • Suspensions, which are stable since these particles are either positively or negatively charged, hence repel each other • By heating, stirring and addition of electrolyte causes this suspension to combine together and form a readily filterable solid • This process of converting a colloidal suspension into a readily filterable solid is called coagulation or agglomeration b. Adsorption of colloids • Ions are attached directly to the solid surface and comprise the Primary Adsorption primary adsorption layer Layer • The charge of this layer is dependent on the charge of the ion present in excess • A charged primary adsorption Counter layer attracts excess oppositely Ion charged ions to form a secondary Layer layer or counter-ion layer • These ions, held by electrostatic forces have higher mobility compared to the ions in the primary layer • The two layers, which constitute an electrical double layer, prevent other particles to come close thus inhibits the formation of larger aggregates 120



c. Factors affecting adsorption • Common Ion Effect. Precipitates have a tendency to adsorb ions identical to it more than any other ions • Paneth-Fajans-Hahn Rule. In cases that there is more than one ion adsorbed, the one having a lower solubility is adsorbed to a greater extent. • Extent of Ionization of the Contaminant. The degree of adsorption increases as the ionization of the contaminant decreases • Effect of Concentration. Greater adsorption of contaminant ion increases as its concentration in the liquid phase increases

Volumetric Methods of Analysis A. Important Terminologies 1. Standard solution – solution of known concentration 2. Standardization – process of determining the concentration of an unknown solution 3. Primary standard – a substance of high purity used for standardization 4. Secondary standard – compound whose purity was established by a chemical analysis and serves as reference material for volumetric analysis 5. Equivalence point – point in titration where the amount of titrant added is chemically equivalent to the analyte in the sample 6. End point – an observable change in a titration process which estimates the equivalence point 7. Titration error – the difference between the actual volume of titrant required to reach the end point and the theoretical volume of titrant required to reach the equivalence point B. Conditions for a Volumetric Analysis 1. The reaction must be rapid and can be represented by a simple balanced equation 2. The reaction must be complete and no side reaction occurs 3. An appropriate indicator must be available in order to detect the end point of the reaction C. Characteristics of a Good Primary Standard 1. High purity and high equivalent weight 2. Stable towards air, high temperature and humidity 3. Soluble in water D. Types of Titration 1. Direct Titration – type of titration where the analyte reacts with the standard solution directly 2. Back Titration – type of titration where an excess standard solution is added and the excess is determined by the addition of another standard solution 3. Replacement Titration – type of titration where the analyte is converted to a product chemically related to it and the product of such reaction is titrated with a standard solution

121



Acid-Base Titration

E.

Theories of acids and bases Theories

1.

Arrhenius theory Svante August Arrhenius (1859-1927)

Acid substances that dissociate in aqueous solution to form aH3O+

Base substances that dissociate in aqueous solution to form bHO –

proton donor

proton acceptor

species that accepts lone pair electrons

species that donates lone pair electrons

Brønsted-Lowry theory Johannes Nicolaus Brønsted (1879-1947) Thomas Martin Lowry (1874-1936)

Lewis theory Gilbert Newton Lewis (1875-1946) a b

hydronium ion, protonated water or solvated proton hydroxide ion

Autoprotolysis or self-ionization reactions • Involves spontaneous reaction of molecules producing a pair of ions • Protic solvents have reactive H + and undergo autoprotolysis

2.

H2O + H2O  H3O+ + HO – NH3 + NH3  NH4+ + NH2 – CH3COOH + CH3COOH

• •

+  CH3COOH2



= 14.0



= 29.8



= 14.5



= 16.7



= 19.1

25 C pK auto 25 C pK auto –

+ CH3COO

25 C pK auto

CH3OH + CH3OH  CH3OH2+ + CH3O –

25 C pK auto

CH3CH2OH + CH3CH2OH  CH3CH2OH2+ + CH3CH2O –

25 C pK auto

In these reactions, a molecule (or an ion) can act as an acid and as a base and is termed as amphoteric Other similar terms are: Amphipatic compounds are those that possess both hydrophilic and lipophilic properties Amphiprotic species are amphoteric molecules that can either accept or donate a proton Ampholytes are amphoteric molecules that contain both acidic and basic groups and commonly exist as zwitterions at a certain pH range







Table 7. Ion product constants for water T,°C KW×1014 T,°C KW×1014 0 0.11 20 0.69 5 0.19 25 1.00 10 0.30 30 1.45 15 0.46 35 2.05

T,°C 40 45 50 100

KW×1014 2.84 3.86 5.18 49.87

Concentrations are expressed in molarity using density of water at each temperature. Source: W.L. Marshall and E.U. Franck, Ion Product of water Substance, 0-1000 °C, 1-10,000 Bars , J. Phys. Chem. Ref. Data 10(2), 1981, pp. 295-304.

Strength of acids and bases

3.

Strong Acids HCl HBr HI 1 2

Bases HNO3 HClO4 1 H2SO4

LiOH NaOH KOH

RbOH CsOH 2 R4NOH

Weak Acids carboxylic acids polyprotic acids metal cations

only the first ionization is complete; dissociation of the second proton has an equilibrium constant of 1.2 quaternary ammonium hydroxide; hydroxide salt of an ammonium cation

122

× 10 – 2

Bases ammonia amines


4.


Calculation of pH • At 25°C, the ion product constant for water, K W is equal to 1.00 × 10 – 14 • At this temperature… pK W = pH + pOH = 14

•

For a dilute solution at 25°C in which the contribution of water to the amount – of H 3O+ and HO in solution and the correction due to activity coefficients are negligible, the following formulas can be used to calculate the pH:

a. Strong acids (SA) and strong bases (SB) • HA + H2O  H3O+(aq) + A – (aq)

•

MOH 

M+(aq)

+

– HO (aq)

b. Weak acids (WA) and weak bases (WB) • HA + H2O  H3O+(aq) + A – (aq)

•

B + H2O  BH+(aq) + HO – (aq)

(SA):

pH = −log[ M SA ]

(SB):

pH = 14 + log[ M SB ]

(WA):

pH = − 12 log[ K a M WA ]

(WB):

pH = 14 + 12 log[ K b M WB ]

c. Hydrolysis of salts • As a general rule, salts coming weak acids or weak bases hydrolyze in water, that is, only the strong conjugate hydrolyzes in water • Acidic salt (AS) is formed from the reaction of a strong acid and weak base HCl(aq) + NH3(aq)  NH4+(aq) + Cl – 1(aq) SA

WB

SCA

SCB

Since only the strong conjugate hydrolyzes in water… NH4+(aq) + H2O NH3(g) + H3O+(aq)  M  K [NH 3 ][H 3O + ] Kh = W = (AS): pH = 7 − 12 log  AS  + Kb [NH 4 ]  Kb 

•

Basic salt (BS) is formed from the reaction of a strong base and weak acid – NaOH(aq) + HCN(aq)  Na+(aq) + CN (aq) + H2O SB

WA

WCB

SCA

Since only the strong conjugate hydrolyzes in water… – – 1 CN 1(aq) + H2O HCN(g) + HO (aq)  M  K [HCN][HO −1 ] Kh = W = (BS): pH = 7 + 12 log  BS  1 − Ka [CN ]  Ka 

• •

Neutral salt (NS) is formed from the reaction of a strong base and strong acid Salts from weak acid and weak base (WAB) will have the following hydrolytic equilibrium expressed by the equation NH4+(aq) + CN – (aq) + H2O NH4OH + HCN  Kh

•

=

KW Ka Kb

=

[HCN][NH 4 OH] + [CN −1 ][NH 4 ][H 2 O] –

 K WKa    Kb 

(WAB): pH = − 12 log 

– 2

Amphoteric salts (HA 1 or HA base that hydrolyzes H3A + H2O  – 1 H2A + H2O  HA – 2 + H2O 

)ionize as a weak acid and also a Brønsted H3O+ H3O+ H3O+

+ + +

H2A – 1 HA – 2 A – 3

Ka1 Ka2 Ka3

 K W K a1 + K a1K a2 [H 2 A −1 ]  1 pH of H 2 A = − log   ≈ − 2 log[ K a1K a2 ]   K a1 + [H 2 A −1 ]  K K + K a2 K a3[HA −2 ]  1 pH of HA − 2 = − 12 log W a2  ≈ − 2 log[ K a2 K a3 ] K a2 + [HA − 2 ]   −1

1 2

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Table 8. Ionization constants of weak acids and bases Weak acids Acetic Benzoic Hydrocyanic Hydrofluoric Hydrogen sulfide Oxalic Phosphoric Phosphorus Sulfuric Sulfurous

Formula CH3COOH C6H5COOH HCN HF H2S HOOCCOOH H3PO4 H3PO3 H2SO4 H2SO3

Ka1 1.75 × 10 – 5 6.30 × 10 – 5 7.20 × 10 – 10 6.70 × 10 – 4 9.10 × 10 – 8 6.50 × 10 – 2 1.10 × 10 – 2 5.00 × 10 – 2

Weak bases Ammonia Aniline Diethyl amine Dimethyl amine Ethyl amine Methyl amine THAM Triethyl amine Trimethyl amine Ethylenediamine Zinc hydroxide

Formula NH3 C6H5NH2 (C2H5)2NH (CH3)2NH C2H5NH2 CH3NH2 (CH2OH)3CN (C2H5)3N (CH3)3N H2NC2H4NH2 Zn(OH)2

Kb1 1.75 × 10 – 5 4.00 × 10 – 10 8.50 × 10 – 4 5.90 × 10 – 4 4.30 × 10 – 4 4.80 × 10 – 4 1.20 × 10 – 6 5.30 × 10 – 4 6.30 × 10 – 5 8.50 × 10 – 5

1.30 × 10 – 2

K a2

1.20 × 10 – 15 6.10 × 10 – 5 7.50 × 10 – 8 2.60 × 10 – 7 1.20 × 10 – 2 5.00 × 10 – 6

K a3

4.80 × 10 – 13

K b2

7.10 × 10 – 8 4.40 × 10 – 5

d. Buffer solutions • Solution that has the ability to resist changes in hydrogen ion concentration upon the addition of small amounts of acid or base (buffer action) • Usually consists of a mixture of weak acid (HA) and its conjugate salt (A – 1) or of a weak base (B) and its conjugate salt (BH+) • Henderson-Hasselbalch equation [H + ][M basic + H + ] [M acidic ] pH = pK A − log KA = + [M basic ] [M acidic − H ] KB

•

=

[HO −1 ][M acidic [M basic

+ HO −1 ]

pH

− HO −1 ]

= pK W − pK B − log

[M acidic ] [M basic ]

Buffer capacity or buffer intensity or buffer index is the number of moles of strong acid or strong base for a liter of solution to cause a unit change in pH d[C HA ] d[C B ] β=− = d[ pH ] d[ pH ] where CHA and CB = number of moles per liter of strong base or strong acid, respectively to cause d[pH]. For a buffer solution containing weak acid and its conjugate salt with concentrations greater than 0.001 M, the buffer capacity is estimated as C HA C A −1 β = 2.303 C HA + C A −1

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To account for the contribution of water to either H 3O+ or HO – 1 in solution, use the following scheme: strong base (SB)

x

2

weak acid (WA) acidic salt (AS)

pH = −log( x )

+ M x − K W = 0

x

pH = 14 + log ( x )

strong acid (SA)

2

+ K x − MK = 0

 KA  K  B K= K W /K B K W /K A

WA

  AS  BS  

WB

weak base (WB) basic salt (BS)

5. Commercial concentrated acids and bases Acids

%wt

Specific Gravity

Molarity

Bases

%wt

Specific Gravity

Molarity

HAc HF HCl HBr HI HNO3 HClO4 H2SO4 H3PO4

99.7 49.0 37.3 48.0 47.0 70.0 70.5 96.5 85.0

1.05 1.17 1.18 1.50 1.50 1.42 1.67 1.84 1.70

17.4 28.9 12.1 8.9 5.5 15.8 11.7 18.1 14.7

NH3 KOH NaOH

29.0 45.0 51.0

0.90 1.46 1.48

15.3 11.7 18.9

6. Primary standards for acid-base titration a. Acidic substances for standardizing basic solutions Name

Formula

Molar Mass

Molar equivalent

Benzoic acid Potassium h ydrogen bis(iodate) Potassium hydrogen o-phthalate Sulfamic Acid

C6H5COOH KH(IO3)2 C6H4(COOH)(COOK) HSO3NH2

122.125 389.915 204.22 97.09

1 1 1 1

b. Basic substances for standardizing acidic solutions Name

Formula

Molar Mass

Molar equivalent

Sodium Carbonate Mercuric Oxide Sodium tetraborate decahydrate Tris(hydroxymethyl)aminomethane

Na2CO3 HgO Na2B4O7⋅10H2O (CH2OH)3CNH2

105.989 216.59 204.22 121.137

2 2 2 1

7. Indicators for acid-base titration Common Name

pKa

pH transition Range

Color change

methyl orange bromocresol green methyl red bromothymol blue m-cresol purple phenolphthalein thymolphthalein

3.46 4.66 5.00 7.10 8.32 9.00 10.0 1.70 8.96

3.1-4.4 3.8-5.4 4.2-6.3 6.2-7.6 7.6-9.2 8.3-10.0 9.4-10.6 1.2-2.8 8.0-9.6

r-y y-b r-o y-b y-p c-r c-b r-y y-b

thymol blue

•

Indicator pH range is pK a±1 and the appropriate indicator for an acid-base titration is one with pKa close to the equivalence point pH 125



8. Applications of acid-base titration a. Determination of nitrogen – Kjeldahl Method • Developed by Johan Gustav Christoffer Kjeldahl (1879-1900) in 1883 Step 1: Digestion involves oxidation of the sample with hot and concentrated sulfuric acid to convert the carbon and hydrogen to CO 2 and H2O, respectively and the nitrogen (amides or amines) to NH4+ for inorganic nitrates and nitrites, sample is reduced to NH 4+ using either Devarda alloy (50% Cu-45% Al-5%Zn) or Arnd’s alloy (60% Cu-40% Mg) HgO and H2SeO3 are added as catalyst while K 2SO4 is added to increase the boiling point of the solution Step 2: Distillation The oxidized solution is treated with NaOH to liberate NH 3 gas – NH4+(aq) + HO 1(aq)  NH3(g) + H2O(l) Glass or Porcelain beads are added to avoid bumping In some modifications, hydrogen peroxide is added to decompose organic matrix formed If mercuric oxide, HgO is used as a catalyst, it is necessary to add Na2S2O3 to precipitate mercuric sulfide Hg2+(aq) + S2O3-2(aq) +2HO-1(aq)  HgS(s) + SO4-2(aq) + H2O(l) Step 3: Titration Ammonia gas is collected in an known excess of standard HCl and the excess is titrated with a standard solution of NaOH using methyl red or bromocresol green as indicator (back titration) NH3(g) + HCl(aq)  NH4Cl(aq) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) The distillate can also be collected using excess of boric acid and the resulting solution is titrated using standard HCl (replacement titration) NH3 (g) + H3BO3 (aq)  NH4+(aq) + H2BO3-2(aq) H2BO3-2(aq) + HCl(aq)  H3BO3(aq) + Cl – 1(aq) + H2O(l) • For conversion of nitrogen content to protein content of selected food products, the following factors apply: 







 







Table 9. Jones factor for protein conversion Animal origin Food

Eggs Meat Milk In general… Cereals Meat Products Dairy Products

Vegetable origin

Factor

6.25 6.25 6.38 5.70 6.25 6.38

Food

Barley Corn Oats Rice Rye Sorghums Peanuts

Factor

5.83 6.25 5.86 5.95 5.83 6.25 5.46

Food

Wheat Whole kernel Bran Beans Castor Soybean Velvet

Factor

5.83 6.31 5.30 5.71 6.25

Source: Food and Agriculture Organization of the United Nations - http://www.fao.o rg/docrep/006/y5022e/ y5022e03.htm

b. Double indicator method for mixture of bases – Warder Titration • The presence of hydroxide, carbonate and bicarbonate in water is also referred to as alkalinity which is a measure of the acid-neutralizing capacity of water • One method requires titration of the mixture to reach the phenolphthalein endpoint with the volume recorded as V0-Ph. On the same solution, methyl red is then added and an additional volume is required to reach the end point recorded as V Ph-MR 126


•

•


Another method requires two duplicate samples. One sample is treated with phenolphthalein and the other with methyl red. Volumes of titrant required to reach the phenolphthalein and methyl red endpoints are recorded as V 0-Ph and V0-MR, respectively. m-Cresol purple can also be used to detect phenolphthalein alkalinity (P) while bromocresol green or methyl orange for the total alkalinity (T) Table 10. Alkalinity relationships Result from titration

NaOH

Na 2CO3

NaHCO3

V0-Ph > VPh-MR V0-Ph < VPh-MR V0-Ph = VPh-MR V0-Ph ≈ 0 ; VPh-MR > 0 V0-Ph > 0 ; VPh-MR ≈ 0

V0-Ph – VPh-MR 0 0 0 V0-Ph

2VPh-MR 2V0-Ph 2V0-Ph = 2VPh-MR 0 0

0 VPh-MR – V0-Ph 0 VPh-MR 0

Note: If analysis involves measurement of volume due to total alkalinity (V conversion: V Ph-MR = V0-MR – V0-Ph

0-MR),

use the

The following relationship can be summarized using the following diagram:

c. Acid number or acid value • Defined as the mass (mg) of KOH that will neutralize the acid produced from water degradative reaction of one gram of fat or oil Acid number =

mL (VKOH )(M KOH )(56.10)

gram of fat or oil

d. Saponification number or Koettstorfer number

• • •

H2C

OOCR 1

HC

OOCR 2

H2C

OOCR3

H2C HO

+ 3 KOH

R1COOK

OH +

CH H2C

OH

R2COOK R3COOK

Defined as the mass (mg) of KOH required to saponify one gram of fat or oil Can be used to determine the approximate molar mass of fat or oil The sample is refluxed with ethanolic KOH and the resulting solution is titrated with standard HCl Sap Value =

mL mL (Vblank − Vwith samplel )(M HCl )(56.10)

gram of fat or oil

Molar Mass of Fat or Oil = 127

168,300 Sap Value



F. Precipitation Titration 1. Important terminologies • Saturated solution – solution that contains the maximum amount of solute dissolved in a given amount solvent at a specific temperature • Solubility – the maximum amount of solute dissolved in a given solvent at a specific temperature • Solutions that contain dissolved solute less than the maximum are called unsaturated while those that contain dissolved solute more than the maximum are called supersaturated solutions 2. Solubility rules for ionic compounds in water at 25 °C Soluble compounds

Insoluble compounds

All nitrates, bicarbonates, chlorates and compounds containing alkali metal ions and ammonium ion. All halides except that of Ag+, Hg22+ and Pb2+ All sulfates except that of Ag+, Ca++, Sr++, Ba++ and Pb++

All carbonates, phosphates, chromates and sulfides except that of alkali metal ions and ammonium ion All hydroxides except that of alkali metal ions and Ba++

3. Solubility product constant (K SP) • Consider an aqueous saturated solution of a sparingly soluble salt represented by the equation: – x (aq)

AxBy (s)  x A+y (aq) + y B

The equilibrium constant for this reaction would be: [A + y (aq) ] x [B − x (aq) ] y K eq = [ A x B y(s) ] However the concentration of the solid A xBy in the solution will be constant (the ratio of the number moles of A xBy and the volume of the solid is constant). Thus, K = K [ A B ] = [A + y ] x [B − x ] y SP

eq

x

y(s)

(aq)

(aq)

Table 11. Solubility product constants at 25 °C COMPOUND

KSP

AgCl AgBr AgI BaCO3 Ba(IO3)2 Al(OH)3 CaCO3 (calcite)

1.82 × 10 5.00 × 10 – 13 8.30 × 10 – 7 5.00 × 10 – 9 1.57 × 10 – 9 3.00 × 10 – 34 4.50 × 10 – 9 – 10

COMPOUND

KSP

Mg(OH)2 Ca(OH)2 PbCl2 PbI2 Hg2Cl2 Hg2Br2 Hg2I2

7.10 × 10 – 12 6.50 × 10 – 6 1.70 × 10 – 5 7.90 × 10 – 9 1.20 × 10 – 18 5.60 × 10 – 23 4.70 × 10 – 29

4. Argentometric titrations • One of the oldest analytical techniques that started in the mid –1800’s • Silver nitrate (AgNO3) is commonly used as titrant and the end point are observed as follows: a. Formation of a secondary colored precipitate – Mohr method • Developed by Karl Friedrich Mohr (1806-1879) in 1865 Titrant: Titration reaction:

AgNO3 – Ag+(aq) + Cl 1(aq)  AgCl(s) white

128


Indicator: Indicator reaction:


K2CrO4 2Ag+(aq) + CrO4 – 2(aq)  Ag2CrO4(s) yellow

• • •

red

In practice, the indicator concentration is kept between 0.002 M to 0.005 M Titration is done at a pH of 8 to avoid precipitation of silver as hydroxide (above pH of 10) and eliminate formation of HCrO 4 – 1 (below pH of 6) which results to consumption of more titrant Usually a low concentration of chromate is desired to detect the end point clearly since a chromate ion imparts an intense yellow color

b. Formation of a colored complex – Volhard method • Developed by Jacob Volhard (1834-1910) in 1874 Titrant: Back titration:

KSCN Ag+(aq) + Cl – 1(aq)  AgCl(s) excess

white

Ag+(aq) + SCN – 1(aq)  AgSCN(s) white


ferric alum, NH4Fe(SO4)2⋅12H2O Fe+3(aq) + SCN – 1(aq)  Fe(SCN)+2(aq) red

• • •

Titration is done in acidic medium using HNO 3 with indicator concentration of about 0.01 M For the titration of chloride, the resulting precipitate is filtered off before the back titration since it reacts with the titrant and is more soluble than AgSCN For the titration of iodide, the indicator is not added until all iodide is precipitated since the dissolved iodide is oxidized by the ferric ion

c. Formation of a colored adsorption complex – Fajans Method • Developed by Kazimierz Fajans (1887-1975) in 1926 Titrant: Titration reaction:

AgNO3 Ag+(aq) + Cl – 1(aq)  AgCl(s):Cl – 1. . . . DCF – 1


fluorescein, dichlorofluorescein or eosin Ag+(aq) + Cl – 1(aq)  AgCl(s):Ag+1:DCF – 1

excess

excess

•

•

• • •

white

white

greenish-yellow

pink

Before the equivalence point, chloride anion adsorbs to the precipitate in the primary adsorption layer and drives the adsorption dye anion away by electrostatic repulsion and the dye imparts a greenish-yellow color in solution As soon as the equivalence point is just exceeded with the presence of excess silver ion, this ion now adsorbs to the precipitate in the primary adsorption layer where the oppositely-charged adsorption dye anion adsorbs to the counter-ion layer and imparts a pink color in solution For titration of chlorides, fluorescein may be used at an optimum pH range between 7-10 while dichlorofluorescein is used in acidic solution of pH greater than 4.4 For bromides, iodides and thiocyanates, eosin is used for titration in acidic medium of pH between 1-2 imparting magenta color at the end point Dextrin is added to prevent excessive coagulation of the AgCl precipitate

129



G. Complexation Titration 1. Important terminologies • Ligand – molecule or ion which possesses at least one unshared pair of electron capable of forming coordinate covalent bond with an ion • Coordination complex or metal complex – formed when a metal ion is bonded to monodentate ligands • Chelate complex – formed when a metal ion is bonded to a polydentate ligand • Chelants – chemicals that form soluble complex molecules with a metal ion which results to inactivation of the ion’s ability to react with other elements to produce precipitates 2. Titration methods involving complexes a. Determination of cyanide – Liebig method • The titration is carried by the dropwise addition of AgNO3 in a solution of a cyanide forming a soluble cyanide complex of silver: – – 2CN 1 + Ag+  Ag(CN)2 1 • The endpoint of the titration is the formation of a permanent faint turbidity: Ag(CN)2 – 1 + Ag+  Ag [Ag(CN)2](s) b. Determination of nickel • An ammoniacal solution of nickel is treated with a measured excess of standard cyanide solution: – – Ni(NH3)6+3 + 4CN 1 + 6H2O  Ni(CN)4 1 + 6NH4OH • The excess cyanide is determined according to the Liebig method c. Titration with ethylenediaminetetraacetic acid (EDTA) • The structure suggests six potential sites (hexadentate) for metal bonding – the four carboxyl groups and two amino groups • Reagents for EDTA titration Free acid, H4Y – can be used as a primary standard when dried for several hours from 130 °C to 145°C and dissolved completely with small amount of base Disodium EDTA dihydrate, Na2H2Y⋅2H2O – analytical reagent grade is commercially available and usually dried at 80°C for 24 hours NH3-NH4Cl buffer solution (pH = 10) – prepared from 17.5 grams of NH4Cl and 142 mL of concentrated NH3 and • Indicators for EDTA titration Eriochrome Black T or Solochrome – used for titrations with pH more than 6.5 since it polymerizes in strongly acidic solutions; color changes between pH range of 7-11 from royal blue to wine red; Calmagite – similar to EBT but color change is sharper and its aqueous solution is stable; used for titration at pH = 10 using NH 3-NH4Cl buffer • Solutions of EDTA combines with any metal ions in a 1:1 ratio 









O

C O

OH

C

CH2

O

O

O

H2C

H2 C

C

C

N HO

N

CH2

+ H2C

H2C C

O

OH

N

H2C

O

H2C

CH2 OH

+n

M

M

H2C

N

O C H2

C H2C O

O C

O

130

C

O


•


Types of EDTA titration Direct titration – solution containing the metal cation is buffered to the desired pH and titrated directly with EDTA using auxiliary complexing agent such as citrate, tartrate or triethanolamine to avoid precipitation of the metal as hydroxide Back titration – in the absence of metal indicator, the solution is treated with excess EDTA, buffered to the desired pH and the excess is determined using standard solution of either sulfates or chlorides of zinc or magnesium Replacement titration – applied for cations that do not react with the metal indicator like Ca+2; in the determination of Ca +2, small amount of magnesium chloride is added to EDTA where Ca+2 initially displaces Mg+2 in the EDTA complex and displaced Mg+2 combines with EBT producing a red complex; when all the calcium is titrated, the liberated Mg+2 is released, combines with EDTA and the endpoint is observed with the formation of blue uncomplexed indicator 





H. Oxidation-Reduction Titration 1. Important terminologies • Oxidation – process which involves increase in oxidation state as a result of loss of electron • Reduction – process which involves decrease in oxidation state as a result of gain of electron • Disproportionation – process in which an element in an intermediate oxidation state yields products in both lower and higher oxidation states • Oxidant or oxidizing agent – substance that accepts electron and undergoes reduction • Reductant or reducing agent – substance that donates electron and undergoes oxidation 2. Oxidation numbers and balancing oxidation-reduction reactions a. Rules in assigning oxidation numbers • An atom in its free or elemental form has oxidation equal to zero • For monoatomic ions, the oxidation number is equal to its charge • Metals have positive oxidation number such as alkali metals (+1), alkaline earth metals (+2), aluminum (+3), zinc (+2) and silver (+1) • Nonmetals usually have negative oxidation numbers: Oxygen is usually – 2, except in peroxides ( – 2) and superoxides ( – 1) Hydrogen is usually +1, except in hydrides ( – 1) Fluorine has – 1 oxidation state; other halogens are usually in the – 1 oxidation state, except when combined with oxygen, they are positive; when different halogens are bound to each other, – 1 is assigned to the more electronegative halogen • The sum of oxidation number of elements in a compound is equal to zero • The sum of oxidation number of elements in a polyatomic ion is equal to the charge of the ion   

b. Balancing oxidation-reduction reactions • A reaction is balanced when the number of atoms of each element and the net charge on both sides are equal

131


•


Step 1: Assign oxidation numbers to each of the species in the reaction NO2 – 1

+

NO3 – 1



+7 – 2

+3 – 2

•

MnO4 – 1

+

+5 – 2

+4

Step 2: Identify oxidation and reduction reactions and indicate the number of electrons lost or gained, respectively NO 2 – 1

Oxidation:

NO3 – 1



+3

+

2e –



MnO2

+5

MnO 4 – 1

Reduction:

3e –

+

+7

•

MnO2

+4

Step 3: Balance the reaction by multiplying a factor on both sides of the reaction so that the numbers of electrons on both reactions are the same Oxidation × 3:

3 NO2 – 1

3 NO3 – 1



+3

Reduction × 2:

+

6e –



2 MnO2

+5

2 MnO4 – 1

+

6e –



NO3 – 1

+7

3

NO2 – 1

+

2

MnO4 – 1

+4

3

+

2 MnO2

•

Step 4: Balance the charges (by adding H or HO ) and number of hydrogen and/or oxygen atoms (by adding H 2O) on both sides of the equation

•

In acidic medium, add H2O to the oxygen-deficient side and supply H + to balance the hydrogen 3 NO2 – 1 + 2 MnO4 – 1 + 2 H+

•

–

+



3 NO3 – 1 + 2 MnO2 + H2O

In basic medium, balance assuming reaction was in acidic medium. Neutralize H+ by adding HO – 1 on both sides of the reaction and simplify 3 NO2 – 1 + 2 MnO4 – 1 + 2 H+ + 2 HO – 1 2 H2O 3 NO2 – 1 + 2 MnO4 – 1 + 3

NO2 – 1

+ 2

MnO4 – 1

+ H2O

  

3 NO3 – 1 + 2 MnO2 + H2O + 2 HO – 1 3 NO3 – 1 + 2 MnO2 + H2O + 2 HO – 1 3 NO3 – 1 + 2 MnO2 + 2 HO – 1

3. Standard electrode potential • The potential of a half-cell reaction with the standard hydrogen electrode (SHE) used as anode when the activities of all reactant and products are taken as unity, that is, 1M concentration and 1 atm partial pressure • Usually listed as standard reduction potential ( εred°) where a positive value implies that the electrode was used as a cathode and the SHE as anode • High value of a reduction potential indicates that the electrode is a good oxidizing agent Nernst equation • Formulated by Walther Hermann Nernst (1864-1941) • Accounts for the effect of concentration on electrode potentials

For a half-cell reduction reaction…Ox  Red + ne – RT a Red ε Red = ε° Red − ln nF a Ox where εRed = actual cell potential [V], ε°Red = standard reduction potential [V], R = 8.314 J-mol – 1-K – 1, T = temperature [K], n = number of electrons that appear in the half-cell reaction [mol], a = activity [ ] and F, Faraday’s constant = 96485.3399 coul-(mol e – ) – 1 At 25°C and for a given cell…

ε cell

= ε°cell −

where Q = reaction quotient [ ] 132

0.05916 n

log Q


•


The equilibrium constant and the standard electrode potential are related as follows: nε ° log K = (T = 298.15 K, ε = 0 and Q = K) 0.05916 V

4. Oxidation-reduction titration methods a. Auxiliary oxidants and reductants • Pre-reductants Jones reductor - consists of zinc metal treated with 2% solution of HgCl2 (amalgamated zinc) and used to reduce Fe +3 (Fe+2), Cu+2 (Cu), TiO+2 (Ti+2), UO2+2 (U+3 or U+4) and Cr+3(Cr+2) Walden reductor - consists of a column filled with silver metal or an insoluble salt of silver and does not reduce Fe +3 and TiO+2 Na2SO3, NaHSO3, or SnCl2 





•

Pre-oxidants NaBiO3, (NH4)2S2O8, K2S2O8, Br2, Cl2, Na2O2 or H2O2 

b. Permanganate titration • Titration is carried out in acidic medium using sulfuric acid • In the presence of HCl, titrant is consumed to oxidize Cl – 1 • Acidic and basic solutions of KMnO4 are less stable than neutral ones and kept in dark-colored bottles to avoid decomposition Titrant: Half-cell reactions: acidic medium basic medium

MnO4 – 1 + 8H+ + 5e –  Mn+2 + 4H2O MnO4 – 1 + 2H2O + 3e –  MnO2 + 4HO –

Primary standards: As2O3 Na2C2O4 Fe metal FeSO4⋅(en)SO4⋅4H2O

H3AsO3 + H2O  H3AsO4 + 2H+ + 2e – C2O4 – 2 2CO2 + 2e – – Fe  Fe+2 + 2e Fe+2  Fe+3 + e –

Indicator: Endpoint:

self-indicating pale pink color that persists for 30 s

 

   

KMnO4

c. Dichromate titration • Titration is carried out in acidic medium only • Titrant is stable towards light and less easily reduced in the presence of organic matter compared to permanganate Titrant: Half-cell reaction: Primary standards: Fe metal FeSO4⋅(en)2SO4⋅4H2O  


K2Cr2O7 – – Cr2O7 2 + 14H+ + 6e  2Cr+3 + 7H2O –

Fe(s)  Fe+2 + 2e – Fe+2  Fe+3 + e

sodium diphenylamine sulfonate N-phenylanthranilic acid first appearance of blue-violet

133



d. Cerium (IV) titration • Titration is carried out in acidic medium using sulfuric acid at concentrations 0.5 M or higher • In the presence of HCl, titrant is consumed to oxidize Cl – 1 Titrant: Half-cell reaction:

Ce(SO4)2 and (NH4)4[Ce(SO4)4]⋅2H2O – Ce+4 + e  Ce+3

Primary standard: Half-cell reaction:

As2O3 H3AsO3 + H2O  H3AsO4 + 2H + + 2e –

Indicator: End point:

ferroin / N-phenylanthranilic acid orange-red to pale blue / yellowish-green to purple

e. Iodimetry: Direct titration with iodine • Titration is carried out in neutral, weak alkaline or weak acidic solutions Titrant: Half-cell reaction:

I2 dissolved in concentrated solution of KI I3 – 1 + 2e –  3I – 1


As2O3 – H3AsO3 + H2O  H3AsO4 + 2H + + 2e


Starch solution Formation of intensely blue-colored complex

f. Iodometry: Indirect titration with iodine • The analyte is an oxidizing agent which reacts with I – 1 added to the solution in excess to liberate I2 equivalent to the amount of analyte present Titrant: Half-cell reaction:

Na2S2O3 2S2O3 – 2  S4O6 – 2 + 2e –


KIO3 or K2Cr2O7 – – 2IO3 1 + 12H+ + 10e  I2 + 6H2O – 2 – + Cr2O7 + 14H + 6e  2Cr+3 + 7H2O


Starch solution Color change from blue to colorless

g. Summary of oxidants and reductants used in titration Oxidants

Half-cell reaction

KMnO4 (acidic) KMnO4 (basic) MnO2 K2Cr2O7 Ce(SO4)2 I2 in KI I2 (satd) I2(aq) KIO3

MnO4 – 1 + 8H+ + 5e –  Mn+2 + 4H2O MnO4 – 1 + 2H2O + 3e –  MnO2 + 4HO – MnO2 + 4H+ + 2e –  Mn+2 + 2H2O Cr2O7 – 2 + 14H+ + 6e –  2Cr+3 + 7H2O Ce+4 + e –  Ce+3 I3 – 1 + 2e –  3I – 1 I2 + 2e –  2I – 1 I2 + 2e –  2I – 1 – 1 2IO3 + 12H+ + 10e –  I2 + 6H2O

Reductants

Half-cell reaction

As2O3 Na2C2O4 Fe metal FeSO4 Na2S2O3

H3AsO3 + H2O  H3AsO4 + 2H+ + 2e – C2O4 – 2 2CO2 + 2e – Fe  Fe+2 + 2e – Fe+2  Fe+3 + e – 2S2O3 – 2  S4O6 – 2 + 2e –

134

ε°

(V)

1.51 1.695 1.23 1.33 1.61 0.5355 0.5345 0.6197 1.20 ε°

(V)

– 0.559 0.49 0.440 – 0.771 – 0.08

Combining ratio

5 3 2 6 1 2 1 1 5 Combining ratio

4 2 2 1 1



5. Application of oxidation-reduction titration a. Iodine number of oils and fats • Measure of the degree of unsaturation of fats or oils • Expressed as the number of centigrams of iodine absorbed by 100 grams of fat or oil • Sample is dispersed in chloroform, treated with solution of iodine monochloride in glacial acetic acid (Wij’s solution) and allowed to react in the dark for 30 min C

•

+ IBr (excess) 

C

+ IBr (unreacted)

C

C

I

Br

KI is added to liberate the unreacted iodine and titrated with standard Na2S2O3 solution IBr (unreacted) + KI  I2 + KBr Iodine Number

=

(V

mL,sample Na 2S2O 3

mL, blank )( M Na S O − VNa S O 2 2

3

2 2

3

)

1 I2 2 Na 2S2 O 3

mg I × 253.80 × 101 cgmg 1 mmol I 2

2

mass sample (g)

b. Peroxide value of oils and fats • Measure of the extent of oxidative rancidity of fats and oils during storage • Expressed as the number of milliequivalent or millimole of peroxide per kilogram of sample • Sample is dissolved in a mixture of chloroform and acetic acid (2:3), bubbled with nitrogen gas to remove remaining oxygen and treated with excess KI to liberate iodine 2 I – 1 + RO2H + H2O  ROH + 2 HO – 1 + I2 Peroxide Value =

(V

mL,sample Na 2S2O 3

mL, blank )( M Na S O − VNa S O 2 2

3

2 2

3

)

1 meq Na 2S2O 3 1 mmol Na 2S2O 3

mass sample (kg)

c. Dissolved oxygen (DO) – Winkler Method • Measure of the amount of oxygen dissolved or carried in a given medium • Measure of the ability to oxidize organic impurities in a body of water • Sample is treated with manganese (II) hydroxide which is converted to a brown precipitate of manganese (III) hydroxide in the presence of oxygen 4 Mn(OH)2 + O2 + 2 H2O  4 Mn(OH)3 • Alkaline iodide-azide solution is added and the precipitate is then dissolved in concentrated H3PO4 to release the iodine – Mn(OH)3 + I 1 + 3H+  Mn+2 + ½ I2 + 3 H2O 1 mmol O ( VNamL S O )( M Na S O ) × 4 mmol Na S O 2

Dissolved oxygen =

2 2

3

2 2

3

2 2

mg O × 132mmol O

2

3

2

volume sample (L)

d. Chemical oxygen demand (COD) • Measure of the amount of oxygen necessary to oxidize all the organic material in a water sample • Expressed as milligrams of oxygen required for oxidation per liter of sample • Sample is refluxed in the presence of HgSO 4, Ag2SO4 /H2SO4 solution and a known excess amount of standard K 2Cr2O7 solution and back titrated with standard (NH4)2Fe(SO4)2 solution 135


COD =

(V

mL,blank Fe +2


K Cr O 6 mmol O mg O − VFemL,+ sample )( M Fe+ ) × 1 mmol × 4 mmol × 132mmol K Cr O O 6 mmol Fe + 2

2

2

2 7 2

2

2

2

2

7

2

volume sample (L)

Molecular Absorption Spectrometry A.

Absorption Process and Beer-Lambert Law If a beam of light passes through a glass container filled with liquid, the emergent radiation is always less powerful than that entering • Consider a block of absorbing matter where a beam of monochromatic radiation of radiant power, P0 strikes the surface perpendicularly and passes through the length of the material, b • The emergent or transmitted radiation will always have less radiant power, P than the entering or incident radiation P0 P • The fraction of incident radiation transmitted by the solution, P/P0 is called transmittance and related to absorbance according to the equation: b P A = −log T = −log P0 Po P

•

•

where A = absorbance, T = transmittance, P 0 = incident radiant power [W], and P = transmitted radiant power [W] dx Beer-Lambert’s law states that the absorbance is directly proportional to the concentration of the absorbing species and to the path length P A = −log T = −log = εbc P0 where ε = molar absorptivity [L-mol 1-cm 1], b = path length [cm] and c = – concentration [mol-L 1] –

B.

–

Quantitative Analysis

1. Standard addition method • Involves addition of several increments of a standard solution to aliquots of samples of the same size and the resulting solution upon adding the color development reagent is then diluted to a fixed volume (VT) and measured for its absorbance • Assume several identical aliquots of the unknown solution of volume Vx were treated with several increments of standard solution of volume Vs of known concentration Cs and diluted to a fixed final volume VT. • If each of these solutions were assumed to obey Beer’s law, the absorbance (AS) of each solution is described by: As

=

εbVSC S VT

+

εbVX C X VT

= kVSCS + kVX C X = mCS + b

where k = εb/VT, m = kVS and b = kCXVX

136



2. Analysis of mixtures • The total absorbance of a solution at a given wavelength is equal to the sum of the absorbances of the components in the mixture At λ1: A1 = ε x1 bC x + ε y1 bC y At λ2:

A2

= εx

2

bC x

+ εy

2

bC y

3. Photometric titration • Plot of absorbance versus volume of titrant where the curves consist of two straight line regions with different slopes • The end point is the intersection of the extrapolated linear regions

REVIEW QUESTIONS AND PROBLEMS 1.

All of the following is used as a hygroscopic material in desiccators except a. CaSO4 b. Mg(ClO4)2 c. P2O5 d. H2SO4

2.

Analytical methods classified as micro analysis uses sample mass ranging from a. < 1 mg b. 1-10 mg c. 10-100 mg d. > 100 mg

3.

Chemical which are tested by the manufacturers showing the actual percentages of impurities and labeled on the containers are called __________. a. reagent grade chemicals c. certified reagent b. analytical reagent d. all of these

4.

Vessels intended to contain definite volumes of liquid at a certain temperature usually 20°C are labeled as follows except a. TC b. C c. In d. TD

5.

What proportion by weight of Na2C2O4 (134) to KHC2O4⋅H2C2O4 (218.2) that must be mixed in a solution so that the normality of the resulting solution as a reducing agent is three times the normality as an acid? a. 0.33 b. 0.65 c. 1.54 d. 3.07

137



6.

Platinum crucibles can be used for the following processes without significant loss except a. Fusion with sodium carbonate, borax or alkali bifluorides b. Evaporation with hydrofluoric acid c. Ignition of oxides of Ca and Sr d. Heating with sulfides

7.

What volume of water must be added to concentrated HCl solution to prepare 100 mL 0.955 M HCl solution? a. 7.9 mL b. 15.8 mL c. 46.0 mL d. 92.1 mL

8.

What grade of water as defined by the British Standard 3978 is suitable for the determination of trace quantities which can be prepared by the distillation of deionized water? a. Grade 1 b. Grade 2 c. Grade 3 d. Grade 4

For numbers 9 to 11 … – A 20% wt/wt aqueous solution of NaCl (58.45) at 25°C has a density of 1.145 g-mL 1. Express the amount of solute in this solution as follows: 9.

% wt/vol a. 17

b. 19

c.21

d. 23

10. molarity (M) a. 0.98

b. 1.96

c. 3.92

d. 5.88

11. molality (m) a. 3.42

b. 5.13

c. 6.84

d. 8.56

12. The following describes colloidal suspensions formed during precipitation except a. These particles are almost invisible to the naked eye b. They settle readily from a given solution c. They are not easily filtered d. none of the these 13. A 1.5176 g sample of a CaCO3 was dissolved in an acidic solution. The calcium was precipitated as CaC2O4⋅H2O (146.11) and the ignited precipitate at 230°C was found to weigh 0.8249 g. What is the percentage of CaO (56.08) in the sample? a. 20.9% b. 23.8% c. 41.8% d. 47.6% For numbers 14 to 16 … To a solution containing a precipitate of PbS, an excess of H 2S was added. 14. What is the ion present in the primary adsorption layer? – a. Pb+ b. H+ c. S

d. HO

15. The charge of this layer is __________. a. negative b. positive

d. any of these

c. neutral

16. The ion present in the counter ion layer is __________. a. Pb+2 b. H+ c. S – 2

–

d. HO –

17. A mixture containing FeCl3 (162.20) and AlCl3 (133.33) only weighs 750.8 mg. The chlorides were precipitated using ammonia and ignited to Fe2O3 (159.69) and Al2O3 (101.96), respectively. The oxide mixture weighs 351.3 mg. Calculate the percentage of Al (26.98) in the sample. a. 15.5% b. 41.2% c. 43.3% d. 58.8% 138



18. Which of the following does not describe the correct way to wash precipitates? a. Minimum volume of washing liquid must be used to wash the precipitate b. Wash with small portions of washing liquid c. For very soluble precipitates, ionic salts containing common ion must be added to the washing liquid d. Gelatinous precipitates requires more washing than crystalline precipitates 19. What weight of Mn ore should be taken so that the percentage of MnO2 (86.94) in the ore would be twice the mass of Mn3O4 (228.82) precipitate obtained in milligram? a. 19.0 mg b. 38.0 mg c. 57.0 mg d. 76.0 mg 20. Process by which an agglomerated colloid return to it dispersed state during washing due to leaching of electrolyte responsible for its coagulation a. nucleation b. coagulation c. agglomeration d. peptization 21. A 5.488-gram sample containing MgCl2 (95.21) and NaCl (58.45) was dissolved in sufficient to give 1L solution. Analysis of chloride content of a 250 mL aliquot resulted in the formation of 2.462 gram AgCl. The magnesium in a second 100 mL aliquot resulted in the formation of 0.2610 gram of Mg 2P2O7 (222.53) after treatment with (NH4)2HPO4. Determine the percentage of NaCl in the sample. a. 12.7% b. 23.2% c. 36.1% d. 40.7% 22. It is the expressed as the volume of a solution chemically equivalent to a mass of a solid reagent a. titer b. aliquot c. molarity d. ppm 23. What is the extent of ionization of the second proton of 0.5 M H2SO4 at 25°C? a. 1% b. 2% c. 4% d. 8% 24. Which of the following aqueous solutions of the same concentration will have the largest pH value at 25°C? a. NH3 b. KCl c. NH4Cl d. NaF 25. Which of the following solutions at 25°C will have the lowest pH value? – a. 0.15 M Na2SO4 (KA2 of H2SO4 = 1.2 × 10 2) b. 5.2 × 10 – 8 M HCl c. 0.05 M NaCN (KA of HCN = 6.2 × 10 – 10) d. 0.01 M NH4Cl (KA of NH4+ = 5.6 × 10 – 10) 26. Which of the following acid-base pairs will result in the formation of a buffer solution when titration is done before the equivalence point? a. NaOH and HCl c. NH3 – HBr b. KOH – HNO3 d. all of these 27. What volume of 0.1025 M HCl must be added to 15.64 mL of 0.0956 M NH3 to produce a solution of pH = 9.00? a. 5.3 mL b. 7.3 mL c. 9.3 mL d. 11.3 mL 28. Which of the following statements is not correct? a. The buffer capacity is always a positive number b. The larger the buffer capacity, the more resistant the solution is to pH change c. In general, alkaline buffering capacity is maximum over a pH range of pKb±1 d. The acid buffering capacity is maximum at pH equal to pKa 139



29. What concentration of acetic acid is needed to prepare a buffer solution with pH = 4.50 and a buffer capacity of 0.10 M-pH – 1? a. 0.067 M b. 0.122 M c. 0.174 M d. 0.225 M 30. Approximately how many grams of NH4Cl (53.45) should be dissolved in a liter of 0.125 F NH3 to reduce the concentration of hydroxide ions to one-thousandth of its original value? a. 79.1 g b. 62.5 g c. 11.7 g d. 7.91 g 31. The conjugate base of H 2PO4 – 1 is __________ a. HPO4 – b. PO4 – c. H3PO4

d. P2O5

For numbers 32 to 35 … Calculate the pH of the resulting solution when 25.0 mL of 0.100 M H 2C2O4 was treated with the following volumes of 0.100 M NaOH: 32. 0 mL a. 1.26

b. 2.84

c. 4.07

d. 5.12

33. 15 mL a. 1.57

b. 1.71

c. 1.86

d. 2.04

34. 25 mL a. 1.44

b. 2.88

c. 5.76

d. 6.64

35. 49.9 mL a. 4.39

b. 7.39

c. 8.37

d. 12.18

For numbers 36 to 38 … The acid dissociation constant of acetic acid in methanol is 3.02 × 10 – 10. Calculate the pH of the following solutions in acetic acid: 36. 0.035 M CH3COOH a. 3.10

b. 5.49

c. 6.90

d. 8.51

37. 0.035 M CH3COOH + 0.070 M CH3COONa a. 4.44 b. 5.05 c. 9.22

d. 9.82

38. 0.035 M CH3COONa a. 2.97

d. 12.38

b. 4.32

c. 11.03

39. In the standardization of an acid solution with primary standard sodium carbonate, why is it necessary to boil the solution before completing the titration? a. to eliminate the reaction product, carbon dioxide and carbonic acid b. to destroy the buffering action of the resulting solution due to the presence of carbonic acid and unreacted hydrogen carbonate c. to achieve a sharper endpoint with methyl red indicator due to the large decrease in pH d. all of the these 40. Calculate the molarity of NaOH solution if 12.25 mL was used to titrate 0.2615 gram of primary standard KHP. a. 0.1045 b. 0.1354 c. 0.2509 d. 0.1697 140



41. What is the best basis for choosing the right indicator for a given acid – base titration from among the following? a. type of acid c. pH at equivalence point b. type of base d. molarity of the acid or base 42. In standardizing a solution of NaOH against 1.431 grams of KHP, the analyst uses 35.50 mL of the alkali and has to run back with 8.25 mL of acid (1mL = 10.75 mg NaOH). What is the molarity of the NaOH solution? a. 0.2118 M b. 0.2044 M c. 0.7831 M d. 0.2598 M 43. In the titration of a weak acid with a strong base, which of the following is the best indicator to be used? a. bromocresol green c. methyl red b. methyl orange d. phenolphthalein 44. A 0.2055-g sample of calcite (impure CaCO3) is treated with 27.18 ml of 0.0712 N HCl and the excess is found to require 5.44 ml of 0.0869 N NaOH for back titration. Calculate the percentage purity of calcite in terms of % wt/wt CaCO3 (100). a. 17.8% b. 35.6% c. 53.4% d. 71.2% 45. Process of determining the nitrogen content of organic materials by mixing the sample with powdered copper (II) oxide and ignited to a combustion tube giving CO2, H2O, N2 and small amounts of nitrogen oxides. a. Kjeldahl Method c. Winkler Method b. Dumas Method d. Wij’s Method 46. A 640 mg sample of P2O5 (141.94) contains some H 3PO4 (97.995) impurity. The sample is reacted with water and the resulting solution is titrated with 0.867 M NaOH requiring 20.7 mL to the thymolphthalein end point. Calculate the percentage of impurity in the sample. a. 0.90% b. 1.80% c. 2.70% d. 3.60% 47. In the titration of phosphoric acid, which of the following statements is true? a. Titration curve contains three inflection points since it is a triprotic acid b. Phenolphthalein indicator is used to detect the second end point. c. It can be treated as a monoprotic or diprotic acid during titration d. all of these For numbers 48 to 50 … A 500-mg sample of each mixture was analyzed for its alkaline content using 0.1025 M HCl via double indicator method. Mixture 1 2 3 4 5 V0-Ph (mL) 4.27 0.01 5.12 6.37 5.63 V0-MR (mL) 10.18 6.19 10.24 6.38 9.04 48. Which of the following mixtures contains NaHCO3? a. Mixtures 2 and 4 c. Mixtures 3 and 4 b. Mixtures 4 and 5 d. Mixtures 1 and 2 49. Calculate the purity of the sample containing NaHCO3 only. a. 1.82% b. 5.22% c. 10.64%

d. 11.13%

50. What is the % wt NaOH for the sample containing a mixture of NaOH-Na2CO3? a. 1.82% b. 5.22% c. 10.64% d. 11.13% 141



51. In the analysis of nitrogen using Kjeldahl Method, which of the following is added to decompose organic matrices present in the sample? a. H2SO4 b. HgO c. K2SO4 d. H2O2 52. A 7.279-gram sample of meat was analyzed for its nitrogen content using Kjeldahl Method. Upon digestion, the ammonia liberated was collected in 250 mL of 0.855 M H3BO3. The resulting solution was titrated with 37.25 mL of 0.3122 M HCl using mixed indicator. Determine the % protein in the sample using 6.25 as factor for meat products. a. 13.98% b. 2.24% c. 19.69% d. 3.14% 53. Which of the following is NOT used as primary standard for the standardization of alkali solutions? a. HgO b. H2C2O4 c. HSO3NH2 d. C6H5COOH For number 54 and 55 … A 1.05 gram sample of butter is refluxed with ethanolic KOH and required 10.4 mL of 0.1875 M HCl to reach the phenolphthalein end point. If blank determination required 27.7 mL of the same standard acid, 54. Calculate the saponification value a. 45 b. 87

c. 125

d. 173

55. Assuming butter comprised mainly of fat, what is its molar mass? a. 475 b. 970 c. 1445 d. 1940 56. Which of the following statements is true about precipitation titration? a. In Volhard method, AgCl is more soluble than AgSCN thus requiring filtration of AgCl prior to back titration b. Titration involving adsorption indicators is slow but has wide range of application c. In Mohr titration, the concentration of the chromate ion in more basic solutions is too low to produce the precipitate near the equivalence point d. all of these 57. In the determination of chloride using Mohr method, what should be the theoretical molar concentration of the chromate ion indicator given that KSP of Ag2CrO4 is 1.1 × 10 – 12? a. 0.002 b. 0.004 c. 0.006 d. 0.008 58. Which of the following is not a correct analytical method – titrant pair? a. Mohr – AgNO3 c. Volhard – AgNO3 b. Fajans – AgNO3 d. Liebig – AgNO3 59. A 1.500-gram sample of impure aluminum chloride was dissolved in water and treated with 45.32 mL of 0.1000 M AgNO 3 using K2CrO4 as indicator. Express the analysis in %AlCl3 (133.33). a. 40.28% b. 13.43% c. 4.48% d. 27.36% 60. In Volhard Method, why is it necessary to carry out titration in acidic solution? a. To prevent precipitation of iron as hydrated as hydrated oxide b. To prevent formation of AgSCN precipitate c. To prevent reduction of halide d. To prevent precipitation of silver as hydrated as hydrated oxide 142



61. A mixture of LiBr (86.845) and BaBr2 (297.22) weighing 800 mg is treated 50.00 mL of 0.1879 M AgNO3 and the excess is found to require 8.76 mL of 0.3179 M KSCN for back titration, using ferric alum as indicator. What is the percentage of BaBr2 in the sample? a. 67.95% b. 32.05% c. 35.62% d. 64.38% 62. In Mohr titration, which of the following statement is CORRECT? a. The indicator is usually kept at a concentration of 0.2-0.5 M so as not to obscure the red precipitate color b. At low pH, part of the indicator is present as HCrO 4 – 1 and less Ag+ are required to reach the endpoint c. At high pH, silver is precipitated as silver hydroxide thus produces error in the amount of titrant added d. all of these 63. A 750.25-gram alloy of nickel was dissolved and treated to remove the impurities. Its ammoniacal solution was treated with 50 mL of 0.1075 M KCN and the excess cyanide required 2.25 mL of 0.00925 M AgNO 3. Determine % Ni (58.69) in the alloy. a. 20.86% b. 37.69% c. 10.53% d. 41.72% 64. Which of the following is true about Liebig method for determination of cyanide? a. The titration process requires an indicator to signal the end of titration b. The addition of excess amount of AgNO 3 produces an i nsoluble compound c. A red complex of silver with cyanide is formed which signals the end of the titration process d. The ratio of silver to cyanide is 2:1 with a permanent faint turbidity as the endpoint 65. A 500-mg sample containing NaCN required 23.50 mL of 0.1255 M AgNO3 to – obtain a permanent faint turbidity. Express the result of this analysis as % CN . a. 15.34% b. 23.01% c. 17.25% d. 30.67% 66. Which of the following ions is best titrated with EDTA at minimum pH less than 7? a. Ca+2 b. Sr+2 c. Mg+2 d. Fe+3 67. Which of the following affects the stability of metal complexes? a. geometrical factors c. chelate effect b. macrocyclic effect d. all of these 68. An EDTA solution was prepared by dissolving the disodium salt in 1L of water. It was standardized using 0.5063 gram of primary standard CaCO 3 and consumed 28.50 mL of the solution. The standard solution was used to determine the hardness of a 2L sample of mineral water, which required 35.57 mL of the EDTA solution. Express the analysis in terms of ppm CaCO 3. a. 89 ppm b. 316 ppm c. 158 ppm d. 269 ppm 69. Which of the following statements is true? a. Multidentate chelating agents form weaker complexes with metal ions b. All metals can be determined with high precision and accuracy using chelometric titration c. Eriochrome black T gives a sharp endpoint for the titration of calcium d. Ca-EDTA complex is more stable than Mg-EDTA complex 143



70. The 300 mg sample of impure Na2SO4 (142.04) was dissolved in sufficient water and the sulfate was precipitated by the addition of 35.00 mL of 0.1022 M BaCl2. The precipitate was removed by filtration and the remaining BaCl 2 consumed 6.79 mL of 0.2467 M EDTA for titration to the Calmagite endpoint. Calculate the purity of the sample. a. 80% b. 85% c. 90% d. 95% 71. A 0.8521 gram sample of an alloy was found to contain Cu (63.55) and Zn (65.41) with small amounts of Pb (207.2) and Hg (200.59). The sample was dissolved in nitric acid and diluted to 500 mL. A 10 mL aliquot was treated with KI to mask the Hg and the resulting solution required 7.06 mL of 0.0348 M EDTA solution. A second 25 mL aliquot was treated with ascorbic acid and the pH was adjusted to 2.00 to reduce Hg+2 and the metallic Hg was removed from the solution. To this solution, thiourea was then added to mask the Cu and the resulting solution required 8.58 mL for titration. The lead ion was titrated in a 250 mL in the presence of NaCN to mask Cu, Zn and Hg and required 3.11 mL for titration. Calculate the percentage of Cu and Hg in the sample of alloy. a. 47% Cu and 3% Hg c. 53% Cu and 7% Hg b. 44% Cu and 5% Hg d. 56% Cu and 5% Hg 72. Commonly, the analyte in a sample is present in two different oxidation states. Prereduction is then necessary before titration. One of the metallic reductors is zinc soaked in a dilute solution of mercuric chloride. This reductor is known as __________. a. Walden reductor c. Lindlars catalyst b. Devarda Alloy d. Jones reductor –

73. At pH = 7 and a pressure of 1 bar, the potential for the half reaction, 2H+(aq) + 2e  H2 (g) is __________. a. 0 V b. – 0.414 V c. – 0.828 V d. – 1.255 V 74. Which of the following is false about iodine as an oxidizing agent in titration? a. Standard iodine solutions have low smaller electrode potential b. Sensitive and reversible indicators are readily available c. Iodine is very soluble in water and losses are minimal d. The solution lacks stability and requires regular standardization 75. What is the molarity of a KMnO4 solution standardized against 1.356 gram Na 2C2O4 (134 g / mol) requiring 25.1 mL of the solution in acidic medium? a. 0.161 M b. 0.403 M c. 1.008 M d. 0.856 M 76. All of the following is used as oxidant in redox titrations except a. KMnO4 b. Cerium (IV) c. K2Cr2O7

d. Iodide

77. A sample of iron ore weighing 385.6 mg was dissolved in acid and passed through a Jones reductor. The resulting solution 52.36 mL of 0.01436 M K 2Cr2O7 for titration to the diphenylamine sulfonic acid endpoint. Calculate % Fe3O4 (231.55 g / mol) in the ore sample. a. 15.05% b. 45.15% c. 90.30% d. 67.98% 78. A sample of pyrolusite weighing 0.2400 gram was treated with excess KI. The iodine liberated required 46.24 mL of 0.1105 M Na 2S2O3 solution. Calculate % MnO2 (86.94) in the sample. a. 46.27% b. 30.85% c. 92.54% d. 76.12% 144

Revised Ana Chem

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