Chem 16 General Chemistry 1 07 Chemical Reactions
Dr. Gil C. Claudio Second Semester 2014-2015 Table of Contents
Contents 1 Writing and Balancing Chemical Equations
1
2 Mass Relationships in Chemical Reactions 2.1 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Limiting Reactants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 3 4
3 Theoretical, Actual, and Percent Yields
4
4 Types of Chemical Reactions 4.1 4.1 Comb Combin inat atio ion n and and Deco Decomp mpos osiition tion Reac Reacti tion onss . . . . . . . . . . . . . . . . . 4.2 Displacement Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Combustion Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 5 6 6
5 Aqueous Reactions 5.1 The Nature of Aqueous Solutions 5.2 Precipitation Reactions . . . . . . 5.3 Acid-Base Reactions . . . . . . . 5.4 Oxidation-Reduction Reactions .
. . . .
6 6 7 8 10
6 Solution Stoichiometry 6.1 Concen centration of Solutions ons . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Dilution of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Stoichiometry try in Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 14 15 15
7 Titrations and Calculations for Acid-Base Titrations
15
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References References of these notes •
General Chemistry, Chemistry, 10th ed, by Ralph H. Petrucci, F. Geoffrey Herring, Jeffy D. Madura, and Carey Bisonnette.
•
Chemistry: The Central Science, Science , 13th 13th ed., ed., by Theo Theodo dore re L. Brow Brown, n, H. Euge Eugene ne LeMay Jr., Bruce E. Bursten, Catherine J. Murphy, Murphy, Patrick M. Woodward, Woodward, and Matthew W. Stoltzfus.
1 Writing riting and Balancin Balancing g Chemical Chemical Equations Equations The Law of Conservation of Mass The law of conservation of mass is the scientific law that states that the total mass of the products of a chemical reaction is the same as the total mass of the reactants, so that the mass remains constant during the reaction.
1
Chemical Equations Chemical reactions are written down in chemical equations. These equations show what substances react–called the reactants–and which substances are formed–called the products. •
The reactants are written on one side, and the products on the other, with an arrow pointing from reactants to products showing the path of the reaction. A + B reactants
→ →
C + D products
Stoichiometric coefficients are the coefficients used to balance an equation. Balancing Chemical Equations by Inspection Chemical equations may be balanced using the following steps:
1. Write the molecular formulae of the reactants and the products, making sure that the chemical formula of each substance is correct before placing any coefficient in the equation. 2. Balance each element in the equation by placing a coefficient before each substance to make the number of each atom in the left (reactants) equal to the number of the same atom on the right (products). 3. After balancing the equation, check the symbol of every element to verify if the coefficients are correct. This is done by multiplying the coefficient by the subscript of each element. The total should be the same on both sides of the equation. Balancing: Some Reminders Take note of the following when balancing the equation: •
The subscripts of a compound are never changed, for changing them would mean changing the identity of the entire molecule.
•
To make things easier (though not always the first step), start with the most complex chemical formula, i.e., the formula containing the greatest number of atoms.
•
Whenever possible, balance polyatomic ions as a unit and not separately.
Balance the Following Equations PHMB 10e, Example 4-1, p 114
1. NH3 + O2 → N2 + H2 O 2. H3 PO4 + CaO → Ca3 (PO4 )2 + H2 O 3. C3 H8 + O2 → CO2 + H2 O 4. NH3 + O2 → NO2 + H2 O 5. NO2 + NH3 → N2 + H2 O States of Matter The state of matter or physical form of reactants and products is shown by symbols in parentheses. •
(g) gas
•
(l) liquid
•
(s) solid
•
(aq) aqueous solution 2
Reaction Conditions The Greek capital letter delta, ∆, means that a high temperature is required. E.g.,
2 Ag 2 O(s)
∆ − − →
4 Ag(s) + O2 (g)
Other more explicit statement of reaction conditions can be written CO(g) + 2 H2 (g)
350◦ C, 340 atm
−−−−−−−−→
ZnO+Cr2 O3
CH3 OH(g)
2 Mass Relationships in Chemical Reactions 2.1 Stoichiometry Stoichiometry Stoichiometry refers to quantitative measurements and relationships involving substances and mixtures of chemical interest.
The balanced chemical equation 2 H 2 (g) + O2 (g)
→
2 H2 O
means that • •
2x molecules H2 reacts with x molecules O2 to produce 2x molecules H2 O molecules ↔ moles
Stoichiometry Examples PHMB 10e, Examples 4-3 to 4-5, pp 117-118
1. How many moles of CO2 are produced in the combustion of 2.72 mol of triethylene glycol, C6 H14 O4 , in an excess of O2 ? 2. What mass of CO2 is formed in the reaction of 4.16 g triethylene glycol, C6 H14 O4 , with an excess of O2 ? 3. What mass of CO2 is consumed in the complete combustion of 6.86 g of triethylene glycol, C6 H14 O4 ? ANSWERS
1. 16.3 mol CO2 2. 7.31 g O2 3. 11.0 g O2 Combining Other Factors PHMB 10e, Examples 4-6 and 4-7, pp 121-122 Given the reaction
2 Al(g) + 6 HCl(aq)
→
2 AlCl3 (aq) + 3 H2 (g)
1. An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3 . A 0.691 cm3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H2 obtained? 2. A hydrochloric acid solution consists of 28.0% HCl by mass and has a density of 1.14 g/mL. What volume of this solution is required to react completely with 1.87 g Al? ANSWERS:
1. 0.207 g H2 2. 23.8 mL HCl solution 3
2.2 Limiting Reactants The Limiting Reactant The Limiting reactant (or limiting reagent) is the reactant present in the smallest stoichiometric quantity in a mixture of reactants. •
The quantity of the product formed is determined by the complete consumption of the limiting reactant.
•
When the masses of the reactants are given, it does not necessarily mean that all reactants are used up in the reaction. Most of the time, one of the reactants is the limiting reagent while the rest are in excess.
•
The limiting reagent is determined by comparing the possible molar combinations of the reactants given the moles of the reactants present. The reagent which is not in excess is the limiting reagent.
Determining the Limiting Reactant PHMB 10e, Examples 4-12 and 4-13, pp 129-131 Phosphorus trichloride, PCl3 , is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. Liquid PCl3 is made by the direct combination of phosphorus and chlorine
P4 (s) + Cl2 (g)
→
PCl3 (l)
1. What is the maximum mass of PCl3 that can be obtained from 125 g P4 and 323 g Cl2 ? 2. What mass of P4 remains? ANSWERS:
1. 417 g PCl3 2. 31 g P4 remaining
3 Theoretical, Actual, and Percent Yields Theoretical Yield Stoichiometry gives us a theoretical yield of the products, that is, the maximum amount of the product when all the limiting reagent is consumed in the reaction. But in reality, we do not always obtain theoretical yields. The quantity of product that is actually produced is called the actual yield. There are many reasons for this: •
Not all products are collected carefully during the experiment
•
the reactants may undergo side reactions
•
or not all of the limiting reagent is consumed.
Percent Yield Thus in real experiment, the amount of products actually obtained in a reaction is called the actual yield, which is oftentimes less than the theoretical yield. The percent yield relates the actual yield to the theoretical yield.
Percent yield =
actual yield × 100% theoretical yield
4
Theoretical, Actual, and Percent Yields PHMB 10e, Example 4-14, p 133 Billions of kilograms of urea, CO(NH2 )2 , are produced annually for use as a fertilizer using the reaction
NH3 (g) + CO2 (g)
→
CO(NH2 )2 (s) + H2 O(l)
The typical starting reaction mixture has a 3:1 mole ratio of NH3 to CO2 . If 47.7 g urea forms per mole of CO2 that reacts, what is the 1. theoretical yield; 2. actual yield; and 3. percent yield? ANSWERS:
1. theoretical yield = 60.1 g CO(NH2 )2 2. actual yield = 47.7 g CO(NH2 )2 3. percent yield = 79.4% Adjusting Reactants PHMB 10e, Example 4-15, p 134 When heated with sulfuric or phosphoric acid, cyclohexanol, C6 H11 OH, is converted to cyclohexene, C6 H10 . The chemical equation is
C6 H11 OH(l)
→
C6 H10 (l) + H2 O(l)
If the percent yield is 83%, what mass of cyclohexanol must we use to obtain 25 g of cyclohexene? ANSWER: 37 g C6 H11 OH
4 Types of Chemical Reactions 4.1 Combination and Decomposition Reactions Combination Reactions A combination reaction is a chemical reaction in which two or more substances combine to form a single product.
A+ B
→
C
For example, magnesium metal burns brilliantly in air to produce magnesium oxide 2 Mg(s) + O2 (g)
→
2 MgO(s)
Decomposition Reaction In a decomposition reaction one substance undergoes a reaction to produce two or more other substances.
C
→
A+ B
For example, many metal carbonates decompose to form metal oxides and carbon dioxide when heated CaCO3
∆ − − →
CaO(s) + CO2 (g)
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4.2 Displacement Reactions Displacement Reactions In a displacement reaction, an element reacts with a compound, displacing an element from it.
A + BX
→
AX + B
Zn(s) + 2 HBr(aq)
→
ZnBr2 (aq) + H2 (g)
Mn(s) + Pb(NO3 )2 (aq)
→
Mn(NO3 )2 (aq) + Pb(s)
Some examples are
4.3 Combustion Reactions Combustion Reactions A combustion reaction is a chemical reaction that proceeds with evolution of heat and usually also a flame. •
most combustion involves reaction with oxygen, as in the burning of a match.
For example, the combustion of propane, C3 H8 , a gas used for cooking, is described by the equation C3 H8 (g) + 5 O2 (g)
→
3 CO2 (g) + 4 H2 O(g)
5 Aqueous Reactions 5.1 The Nature of Aqueous Solutions Solutions A homogeneous mixture (or solution) is a mixture of elements and/or compounds that has a uniform composition and properties within a given sample. •
However, the composition and properties may vary from one sample to another.
The solvent is the solution component in which one or more solutes are dissolved. •
Usually the solvent is present in greater amount than are the solutes and determines the state of matter in which the solution exists.
A solute is a solution component that is dissolved in a solvent. •
A solution may have several solutes, with the solutes generally present in lesser amounts than is the solvent.
Electrolytes An electrolyte is a substance that provides ions when dissolved in water. •
A strong electrolyte is a substance that is completely ionized in solution. E.g., NaCl.
•
A weak electrolyte is a substance that is only partially ionized in solution in a reversible reaction. E.g., CH3 COOH.
A nonelectrolyte is a substance that is essentially non-ionized, both in the pure state and in solution. E.g., CH3 OH.
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5.2 Precipitation Reactions Precipitation Reaction A precipitate is an insoluble solid that deposits from a solution as a result of a chemical reaction. •
A reaction in which one of the products is a precipitate is called a precipitation reaction.
An example is the formation of PbI2 , a water-insoluble yellow solid. 2 KI(aq) + Pb(NO3 )2 (aq)
→
2 KNO3 (aq) + PbI2 (s)
Net Ionic Equations A net ionic equation represents a reaction between ions in solution in such a way that all nonparticipant (spectator) ions are eliminated from the equation. The equation must be balanced both atomically and for net electric charge. Net Ionic Equations: An Example Given the reaction
AgNO3 (aq) + NaI(aq)
→
AgI(s) + NaNO3 (aq)
Since the strong electrolytes dissociate in water Ag+ (aq) + NO3– (aq) + Na+ (aq) + I– (aq) ✘ ✘ ✘ ✘ ✘
✘ ✘ ✘ ✘
→
AgI(s) + Na+ (aq) + NO3– (aq) ✘ ✘ ✘ ✘
✘ ✘ ✘ ✘ ✘
Removing the spectator ions, we get the net ionic equation Ag+ (aq) + I– (aq)
→
AgI(s)
Predicting Precipitation Reactions To predict whether a precipitate forms when we mix aqueous solutions of two strong electrolytes
1. note the ions present in the reactants, 2. consider the possible cation-anion combinations, and 3. use tables or guidelines to determine if any of these combinations is insoluble. A Solubility Guideline Follow the lower-numbered guideline when two guidelines are in conflict. This leads to the correct prediction in most cases.
1. Salts of group 1 cations (with some exceptions for Li+ ) and the cation are soluble. 2. Nitrates, acetates, and perchlorates are soluble. 3. Salts of silver, lead, and mercury(I) are insoluble. 4. Chlorides, bromides, and iodides are soluble. 5. Carbonates, phosphates, sulfides, oxides, and hydroxides are insoluble (sulfides of group 2 cations and hydroxides of Ca2+ , Sr2+ , and Ba2+ are slightly soluble). 6. Sulfates are soluble except for those of calcium, strontium, and barium.
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Acids Sour taste Provide H+ ions React with active metals to give H2 Produce CO2 when added to limestone (CaCO3 ) Neutralize bases
Bases Bitter taste Provide OH– ions Slippery feeling
Neutralize acids
Another Solubility Guideline Soluble ionic compounds
contains NO3– CH3 COOH– Cl– , Br – , I – SO42–
important exceptions None None Compounds of Ag+ , Hg 22+ , and Pb2+ Compounds of Sr2+ , Ba 2+ , Hg22+ , and Pb2+
Insoluble ionic compounds contains S2– PO43– OH–
important exceptions Compounds of NH4+ , the alkali metal cations, Ca2+ , Sr 2+ , and Ba2+ Compounds of NH4+ , the alkali metal cations Compounds of NH4+ , the alkali metal cations, Ca2+ , Sr 2+ , and Ba2+
Predicting Precipitation Reactions PHMB 10e, Example 5-2, p 159 Predict whether a reaction will occur in each of the following cases. If so, write a net ionic equation for the reaction.
1. NaOH(aq) + MgCl2 (aq) 2. BaS(aq) + CuSO4 (aq)
→
→
?
?
3. (NH4 )2 SO4 (aq) + ZnCl2 (aq)
→
?
ANSWER:
1. Mg(OH)2 (s) 2. BaSO4 (s) + CuS(s) 3. all soluble
5.3 Acid-Base Reactions Properties of Acids and Bases Some Common Acids and Bases Acids citrus fruits, vinegar, carbonated beverages, tomatoes, black coffee, gastric fluid, vitamin C, aspirin, ant venom, battery acid, muriatic acid, sulfuric acid Bases household ammonia, soaps and detergents, baking soda, lye, milk of magnesia, oven cleaners, drain cleaners, antacids.
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Acids An acid can be defined as a substance that provides hydrogen ions (H+ ) in aqueous solution. •
Based on the Arrhenius theory of acids. HCl(g) + H2 O(l) HCl(aq)
→
→
H3 O+ (aq) + Cl– (aq)
H+ (aq) + Cl– (aq)
Strong and Weak Acids Strong acids have a strong tendency for producing H+ ions. They are molecular compounds that are almost completely ionized into and accompanying anions when in aqueous solution. •
HCl, HBr, HI, HClO4 , HNO3 , H 2 SO4
Weak acids are molecular compounds that have a weak tendency for producing H+ ions; weak acids are incompletely ionized in aqueous solution.
CH3 COOH(aq)
⇋
H+ (aq) + CH3 COO– (aq)
Bases The Arrhenius definition of a base is a substance that produces hydroxide ions (OH– ) in aqueous solution.
NaOH(aq)
→
Na+ (aq) + OH– (aq)
A strong base is a base that is completely ionized in aqueous solution. •
LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2 , Sr(OH)2 , Ba(OH)2
A weak base is a base that is partially ionized in aqueous solution in a reversible reaction. NH3 (aq) + H2 O(l)
⇋
NH4+ (aq) + OH– (aq)
Ion-Product Constant of Water Dissociation of water
H2 O
⇋
H+ + OH –
The ion-product constant for water, K w K w = [H+ ][OH– ] = 1.0 × 10−14 at 25 0 C (or K w = [H3 O+ ][OH– ]) A solution for which [H3 O+ ] = [OH– ] is said to be neutral. pH pH is the negative logarithm of the hydronium ion concentration. •
“pouvoir hydrogene” which means hydrogen power
Equations of pH •
pH = -log [H3 O+ ] (or pH = -log[H+ ])
•
pOH = -log [OH– ]
•
pH + pOH = 14 [H3 O+ ] > [OH– ] [H3 O+ ] = [OH– ] [H3 O+ ] < [OH– ]
solution is acidic solution is neutral solution is basic 9
at 25◦ C pH < 7.0 pH = 7.0 pH > 7.0
Neutralization Reactions In a neutralization reaction, an acid and a base react in stoichiometric proportions, so that there is no excess of either acid or base in the final solution.
HCl(aq) acid
+ +
NaOH(aq) base
→ →
NaCl(aq) salt
+ +
H2 O(l) water
The net ionic equation is H+ (aq) + OH– (aq)
→
H2 O(l)
Examples of Acid-Base Reactions Mg(OH)2 , a base, is found in milk of magnesia.
Mg(OH)2 (s) + 2 H+ (aq)
→
Mg2+ (aq) + 2 H2 O(l)
Calcium carbonate is present in limestone and marble CaCO3 (s) + 2 H+ (aq)
→
Ca2+ (aq) + H2 O(l) + CO2 (g)
Writing Equations for Acid-Base Reactions PHMB 10e, Example 5-3, p 165 Write a net ionic equation to represent the reaction of
1. aqueous strontium hydroxide with nitric acid; 2. solid aluminum hydroxide with hydrochloric acid. ANSWERS:
1. 2 H+ (aq) + 2 OH– (aq) 2. Al(OH)3 (s) + 3 H+ (aq)
→
→
2 H2 O(l) Al3+ (aq) + 3 H2 O(l)
5.4 Oxidation-Reduction Reactions Oxidation-Reduction Reactions An oxidation-reduction (or redox) reaction is one that involves a simultaneous oxidation and reduction of reactants. •
A reduction process is one in which electrons are gained and the oxidation state of some atom decreases.
•
Oxidation is a process in which electrons are lost and the oxidation state of some atom increases.
An oxidation state relates to the number of electrons an atom loses, gains, or shares in combining with other atoms to form molecules or polyatomic ions. Oxidation State Change Given the reaction below with the oxidation states
Fe2 O3 (s) Fe = +3 O = -2
+
3 CO(g) C = +2 O = -2
∆ − − →
2 Fe(l) Fe = 0
+
3 CO2 (g) C = +4 O = -2
Changes in oxidation state •
Fe gains electrons, from +3 to 0. Thus Fe2 O3 (not just Fe) is reduced.
•
C loses electrons, from +2 to +4. Thus CO (not just C) is oxidized.
•
O does not change its oxidation state.
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Identifying Oxidation Reduction Reactions PHMB 10e, Example 5-4, pp 166-167 Indicate whether each of the following is an oxidation reduction reaction.
1. MnO2 (s) + 4 H+ (aq) + 2 Cl– (aq) 2. H2 PO4– (aq) + OH– (aq)
→
Mn2+ (aq) + 2 H2 O((l) + Cl2 (g)
→
HPO42– (aq) + H2 O(l)
ANSWERS:
1. MnO2 is reduced to Mn2+ . Cl– is oxidized to Cl2 2. This is not an oxidation-reduction reaction. Oxidation and Reduction Half-Reactions Given the redox reaction of zinc metal in copper (II) sulfate solution
Zn(s) + CuSO4 (aq)
→
ZnSO4 (aq) + Cu(s)
the net ionic equation is Zn(s) + Cu2+ (aq)
→
Zn2+ (aq) + Cu(s)
The two half-reactions are oxidation reduction overall
Zn(s) 2+ Cu (aq) + 2 e– Zn(s) + Cu2+ (aq)
→ → →
Zn2+ (aq) + 2 e– Cu(s) Zn2+ (aq) + Cu(s)
Some Assumptions of Redox Reactions
1. The reactions we are dealing with all occur in water solutions. This means that the ionic molecules readily dissociate to their ionic species, and that the ions that are most of the time involved in the reaction. This also means that water can be involved as either a reactant or a product. 2. Redox reactions most of the time occur in either acidic or basic medium, thus the reactions involve either H+ or OH– as either reactants or products. Ion Electron Method: Acidic Medium
1. Separate the net ionic equation into two half-reactions (HR). Each HR contains the species with the same atoms aside from O and H. 2. Balance first all the atoms aside from O and H . 3. Then balance O and H. (a) Balance O by adding H2 O to the side of the HR lacking in O. (b) Balance H by adding H+ to the side of the HR lacking in H. 4. Balance the charge by adding e– to the side with the greater total positive charge. •
Since we are dealing with redox reactions, one of the HR loses electrons (e– at the right side of the HR) and the other HR gains the electrons (e– at the left side of the HR).
5. Balance the electrons by multiplying an integer so that the number of electrons lost is equal to the number of electrons gained. 6. Add the two half-reactions, and cancel all the species that appear on both sides of the reaction. 11
Ion Electron Method: Basic Medium The procedure for balancing redox reactions in basic medium follow the same steps as the one for the acidic medium, with two additional steps in the end.
1. For every H+ in the final reaction, add the same number of OH– to both sides of the reaction. The H+ and OH– in one side of the reaction are then combined to form H2 O. 2. Cancel all excess H2 O present in both sides of the equation. An Example in Acidic Medium In the reaction of sodium oxalate and potassium permanganate in hydrochloric acid medium, the unbalanced chemical reaction can be written as:
HCl(aq) + Na2 C2 O4 (aq) + KMnO4 (aq) → CO2 (g) + KCl(aq) + MnCl2 (aq) + H2 O(l) + NaCl(aq) Removing the spectator ions from the chemical equation and temporarily removing water and the H+ ions, we can simplify the equation to C2 O42– (aq) + MnO4– (aq) → CO2 (g) + Mn2+ (aq) The two half-reactions are: C2 O42– (aq) → CO2 (aq) MnO4– (aq) → Mn2+ (aq) Answer The balanced half reactions are
5 C 2 O42– (aq) → 10 CO2 (aq) + 10 e– 10 e – + 16 H+ (aq) + 2 MnO4– (aq) → 2 Mn2+ (aq) + 8 H2 O(l) Adding the two half-reactions and bringing back the spectator ions, the final answer is 16 HCl(aq) + 5 Na2 C2 O4 (aq) + 2 KMnO4 (aq) → 10 CO2 (aq) + 2 KCl(aq) + 2 MnCl2 (aq) + 8 H2 O (l) + 10 NaCl(aq) An Example in Basic Medium
I2 (aq) + Cl2 (aq) + NaOH(aq) → NaH3 IO6 (aq) + NaCl(aq) + H2 O(l) The net ionic equation is I2 + Cl2 → H3 IO62– + Cl – The two half-reactions are I2 → H3 IO62– Cl2 → Cl– Answer
18 OH– (aq) + 7 Cl2 (aq) + I2 (aq) → 2 H3 IO62– (aq) + 14 Cl– (aq) + 6 H2 O(l) Bringing back the spectator ions I2 (aq) + 7 Cl2 (aq) + 18 NaOH(aq) → 2 NaH3 IO6 (aq) + 14 NaCl(aq) + 6 H2 O(l)
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Examples: Redox Reactions in Acidic Medium
Ni2+ (aq) + IO4– (aq) → Ni3+ (aq) + I– (aq) O2 (g) + Br– (aq) → H2 O(l) + Br2 (aq) Ca(s) + Cr2 O72– (aq) → Ca2+ (aq) + Cr3+ (aq) IO3– (aq) + Mn2+ (aq) → I– (aq) + MnO2 (aq) Cr2 O72– (aq) + Cl– (aq) → Cr3+ (aq) + Cl2 (g) Examples: Redox Reactions in Basic Medium
SO2 (g) + I2 (aq) → SO3 (aq) + I– (aq) Zn(aq) + NO3– (aq) → NH3 (aq) + Zn2+ (aq) ClO– (aq) + CrO2– (aq) → Cl– (aq) + CrO42– (aq) K(s) + H2 O(l) → K+ (aq) + H2 (g) CN– (aq) + MnO4– (aq) → CNO– (aq) + MnO2 (aq) Disproportionation Reactions In a disproportionation reaction, the same substance is both oxidized and reduced.
One example is the decomposition of hydrogen peroxide H2 O2 2 H 2 O2 (aq)
→
2 H2 O(l) + O2 (g)
Another is the disproportionation of S2 O32– in acidic solution S2 O32– (aq) + H+ (aq)
→
S(s) + SO2 (g) + H2 O(l)
Oxidizing and Reducing Agents An oxidizing agent (or oxidant) makes possible an oxidation process by itself being reduced.
A reducing agent (or reductant) makes possible a reduction process by itself becoming oxidized. Identifying Oxidizing and Reducing Agents PHMB 10e, Example 5-8, p 177 Hydrogen peroxide, H2 O2 , is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or a reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent.
1. H2 O2 (aq) + 2 Fe2+ (aq) + 2 H+ (aq) → 2 H 2 O(l) + 2 Fe3+ (aq) 2. 5 H2 O2 (aq) + 2 MnO4– (aq) + 6 H+ (aq) → 8 H 2 O(l) + 2 Mn2 (aq) + 5 O2 (g) ANSWER
1. H2 O2 is an oxidizing agent 2. H2 O2 is a reducing agent
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Activity Series An activity series is a list of metals in order of decreasing ease of oxidation. •
The metals at the top of the table (e.g, the alkali metals and the alkaline earth metals) are called active metals for they are most easily oxidized.
•
The metals at the bottom of the activity series (e.g., transition elements from groups 8B and 1B) are called noble metals for they very stable and form compounds less readily. They are used to make coins and jewelry.
•
The activity series can be used to predict the outcome of reactions between metals and either metal salts or acids.
•
Any metal on the list can be oxidized by the ions of elements below it.
Activity Series of Metals in Aqueous Solution Lithium Potassium Barium Calcium Sodium Magnesium Aluminum Manganese Zinc Chromium Iron Cobalt Nickel Tin Lead Hydrogen Copper Silver Mercury Platinum Gold
Li(s) K(s) Ba(s) Ca(s) Na(s) Mg(s) Al(s) Mn(s) Zn(s) Cr(s) Fe(s) Co(s) Ni(s) Sn(s) Pb(s) H2 (g) Cu(s) Ag(s) Hg(s) Pt(s) Au(s)
→ → → → → → → → → → → → → → → → → → → → →
Li+ (aq) + e – K+ (aq) + e – Ba2+ (aq) + 2 e – Ca2+ (aq) + 2 e – Na+ (aq) + e – Mg2+ (aq) + 2 e – Al3+ (aq) + 3 e – Mn2+ (aq) + 2 e – Zn2+ (aq) + 2 e – Cr3+ (aq) + 3 e – Fe2+ (aq) + 2 e – Co2+ (aq) + 2 e – Ni2+ (aq) + 2 e – Sn2+ (aq) + 2 e – Pb2+ (aq) + 2 e – 2 H+ (aq) + 2 e – Cu2+ (aq) + 2 e – Ag+ (aq) + e – Hg2+ (aq) + 2 e – Pt2+ (aq) + 2 e – Au3+ (aq) + 3 e –
Activity Series: Some Examples Copper metal can be oxidized by silver ions
Cu(s) + 2 Ag+ (aq)
→
Cu2+ (aq) + 2 Ag(s)
Ni(s) reacts with HCl(aq) to form H 2 Ni(s) + 2 HCl(aq)
→
NiCl2 (aq) + H2 (g)
One exception is the oxidation of Cu(s) by nitric acid, where the metal is oxidized not by H+ but by the nitrate ion. Cu(s) + 4 HNO3 (aq)
→
Cu(NO3 )2 (aq) + 2 H2 O(l) + 2 NO2 (g)
This does not happen with HCl(aq).
6 Solution Stoichiometry 6.1 Concentration of Solutions Molarity Molarity is the number of moles of solute per liter of solution.
M = M =
n V
moles solute volume (L) solution →
n = M × V
Not volume of solvent. •
One liter of a solution usually contains either slightly more or slightly less than 1 liter of solvent because the process of dissolution causes the volume of liquid to increase or decrease. 14
Calculating Molar Quantities PHMB 10e, Example 4-8, p 124 A solution is prepared by dissolving 25.0 mL ethanol, CH3 CH2 OH (d = 0.789 g/mL) in enough water to produce 250.0 mL solution. What is the molarity of ethanol in the solution? ANSWER: 1.71 M
6.2 Dilution of Solutions Solution Dilution The principle of dilution is that the same amount of solute is present in concentrated solution (small volume) as in the larger volume of a diluted solution. n = M × V V 1 M1 = V 2 M2 →
Preparing a Solution by Dilution PHMB 10e, Example 4-10, p 126 A particular analytical chemistry procedure requires 0.0100 M K 2 CrO4 . What volume of 0.250 M K2 CrO4 must be diluted with water to prepare 0.2500 L of 0.0100 M K2 CrO4 ? ANSWER: 10.0 mL
6.3 Stoichiometry in Solutions Stoichiometry of Reactions in Solution The appropriate stoichiometric factor is used to solve the stoichiometry of reaction in solution. The only difference is that moles is calculated from molarity. A Reaction in Solution PHMB 10e, Example 4-11, p 128 A 25.00 mL pipetful of 0.250 M K2 CrO4 is added to an excess of AgNO3 (aq). What mass of Ag2 CrO4 will precipitate from the solution?
K2 CrO4 (aq) + 2 AgNO3 (aq)
→
Ag2 CrO4 (s) 2 KNO3 (aq)
ANSWER: 2.07 g
7 Titrations Titrations
and
Calculations
for
Acid-Base
Titrations Titration is a procedure for carrying out a chemical reaction between two solutions by the controlled addition (from a buret) of one solution to the other. •
The equivalence point of a titration is the condition in which the reactants are in stoichiometric proportions. They consume each other, and neither reactant is in excess.
•
An indicator is an added substance that changes color at the equivalence point in a titration.
15
Titration of Vinegar PHMB 10e, Example 5-9, p 179 Vinegar is a dilute aqueous solution of acetic acid produced by the bacterial fermentation of apple cider, wine, or other carbohydrate material. The legal minimum acetic acid content of vinegar is 4% by mass. A 5.00 mL sample of a particular vinegar is titrated with 38.08 mL of 0.1000 M NaOH. Does this sample exceed the minimum limit? (Vinegar has a density of about 1.01 g/mL) ANSWER: 4.53 %, it exceeds the minimum limit Standardizing a Solution PHMB 10e, Example 5-10, p 180 A piece of iron wire weighing 0.1568 g is converted to Fe2+ (aq) and requires 26.24 mL of a KMnO4 (aq) solution for its titration. What is the molarity of the KMnO4 (aq)?
5 Fe 2+ (aq) + MnO4– (aq) + 8 H+ (aq) → 5 Fe 3+ (aq) + Mn2+ (aq) + 4 H2 O(l) ANSWER: 0.02140 M KMnO4
16
