1. Chapter 5: Introduction to Reactions in Aqueous Solutions
Chemistry 140 Fall 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 5: Introduction to Reactions in Aqueous Solutions
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
(modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)
General Chemistry: Chapter 5
Prentice-Hall © 2002

2. General Chemistry: Chapter 5
Contents
5-1 The Nature of Aqueous Solutions
5-2 Precipitation Reactions
5-3 Acid-Base Reactions
5-4 Oxidation-Reduction: Some General Principles
5-5 Balancing Oxidation-Reduction Equations
5-6 Oxidizing and Reducing Agents
5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations
Focus on Water Treatment
General Chemistry: Chapter 5
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3. General Chemistry: Chapter 5
Chemistry 140 Fall 2002
Acids
Acids provide H+ in aqueous solution.
Strong acids:
Weak acids:
HCl(aq)
H+(aq) + Cl-(aq) → →
HCl + H2O
H3O+(aq) + Cl-(aq) → ←
CH3CO2H(aq)
H+(aq) + CH3CO2-(aq)
Strong acids completely ionize in solution
Weak acids partially ionize in solution
Compare to the electrolyte strengths. → ←
CH3CO2H + H2O
H3O+(aq) + CH3CO2-(aq)
General Chemistry: Chapter 5
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4. General Chemistry: Chapter 5
Chemistry 140 Fall 2002
Bases
Bases provide OH- in aqueous solution.
Strong bases:
Weak bases:
NaOH(aq)
Na+(aq) + OH-(aq) →
H2O
Strong bases completely dissociate (or nearly so)
Primarily hydroxides of group 1 and some group 2 metals
Certain substances produce ions by reacting with water, not just dissolving in it.
NH4 is a weak base because the eaciton does not go to completion.
MOST bases are weak bases → ←
NH3(aq) + H2O(l)
OH-(aq) + NH4+(aq)
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5. Recognizing Acids and Bases
Chemistry 140 Fall 2002
Recognizing Acids and Bases
Acids have ionizable hydrogen ions.
CH3CO2H or HC2H3O2
Bases have OH- combined with a metal ion.
KOH
or are identified by chemical equations
Na2CO3(s) + H2O(l) → HCO3-(aq) + 2 Na+(aq) + OH-(aq)
Ionizable protons are usually inidicated separately in a formula.
General Chemistry: Chapter 5
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6. General Chemistry: Chapter 5
Neutralization
Perhaps the most significant reaction between acids and bases is the ability to cancel the properties of the other.
Neutralization reactions result in the production of water and an aqueous solution of the salt
ie. HCl + NaOH 
Overall equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Cl-(aq) + Na+(aq)
Net Ionic equation
H+(aq) + OH-(aq)  H2O(l)
General Chemistry: Chapter 5
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7. More Acid-Base Reactions
Chemistry 140 Fall 2002
More Acid-Base Reactions
Milk of magnesia Mg(OH)2
Mg(OH)2(s) + 2 H+(aq) → Mg2+(aq) + 2 H2O(l)
Mg(OH)2(s) + 2 CH3CO2H(aq) →
Mg2+(aq) + 2 CH3CO2-(aq) + 2 H2O(l)
Equation 1 is reaction with strong acid
Equation 2 is reaction with weak acid
Note the characteristic formation of water.
General Chemistry: Chapter 5
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8. More Acid-Base Reactions
Chemistry 140 Fall 2002
More Acid-Base Reactions
Limestone and marble.
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)
But: H2CO3(aq) → H2O(l) + CO2(g)
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
This equation shows CO32- acting as a base instead of OH-.
Our definition must be expanded from simple Arrhenius theory (Bronsted-Lowry and Lewis)
General Chemistry: Chapter 5
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9. General Chemistry: Chapter 5
Limestone and Marble
General Chemistry: Chapter 5
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10. General Chemistry: Chapter 5
Gas Forming Reactions
Gas Forming Reactions
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11. 5-4 Oxidation-Reduction: Some General Principles
Hematite is converted to iron in a blast furnace.
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D
Oxidation and reduction reactions always occur together.
Fe3+ is reduced to metallic iron.
CO(g) is oxidized to carbon dioxide.
General Chemistry: Chapter 5
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12. General Chemistry: Chapter 5
Redox Reactions
Oxidation and reduction must occur together
Oxidation-reduction reactions are often referred to as redox reactions
A reaction where a substance gains an O (oxygen) is oxidation
A reaction where a substance loses an O is reduction
LEO says GER or OIL RIG (involves change in O.S. of some element)
Loss of electrons = oxidation (O.S. goes up)
Gain of electrons = reduction (O.S. goes down)
General Chemistry: Chapter 5
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13. Oxidation State Changes
Assign oxidation states: 3+ 2- 2+ 2- 4+ 2-
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D
O.S. of Fe3+ decreases to Fe0
(gain of e- = reduction)
O.S. of C2+ increases to C4+
(loss of e- = oxidation)
Fe3+ is reduced to metallic iron.
(Loss of O = reduction)
CO(g) is oxidized to carbon dioxide.
(gain of O = oxidation)
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14. Oxidation and Reduction
O.S. of some element increases in the reaction.
Electrons are on the right of the equation.
Loss of electrons.
Reduction
O.S. of some element decreases in the reaction.
Electrons are on the left of the equation.
Gain of electrons.
General Chemistry: Chapter 5
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15. Zinc in Copper Sulfate Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
General Chemistry: Chapter 5
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16. Electrons Lost = Electrons Gained
Half-Reactions
Represent a reaction by two half-reactions.
Oxidation:
Zn(s) → Zn2+(aq) + 2 e-
2 electrons on each side of the equation cancel out
Reduction:
Cu2+(aq) + 2 e- → Cu(s)
Overall:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Electrons Lost = Electrons Gained
General Chemistry: Chapter 5
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17. Balancing Oxidation-Reduction Equations
Few can be balanced by inspection.
Systematic approach required.
The Half-Reaction (Ion-Electron) Method
Initial split into oxidation and reduction half-reactions which happen simultaneously.
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18. Example 5-6
Balancing the Equation for a Redox Reaction in Acidic Solution.
The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution..
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
Determine the oxidation states: 4+ 7+ 6+ 2+
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
General Chemistry: Chapter 5
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19. Example 5-6 Write the half-reactions: SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq)
Balance atoms other than H and O:
Already balanced for elements.
SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq)
General Chemistry: Chapter 5
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20. Example 5-6 Balance O by adding H2O: H2O(l) + SO32-(aq) → SO42-(aq)
MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Balance hydrogen by adding H+:
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq)
8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Check that the charges are balanced: Add e- if necessary.
H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2e-
5e- + 8 H+(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
General Chemistry: Chapter 5
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21. Example 5-6 Multiply the half-reactions to balance all e-:
[ H2O(l) + SO32-(aq) → SO42-(aq) + 2 H+(aq) + 2e- ] x 5
[ 5e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l) ] x 2
5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e H+(aq) 6 3
16 H+(aq) + 10 e- + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)
Add both equations and simplify (eliminate things common on both sides):
5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)
Check the balance (atoms and charges)!
General Chemistry: Chapter 5
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