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General Chemistry: Chapter 5
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Chapter 5: Introduction to Reactions in
Aqueous Solutions
Philip Dutton
University of Windsor, Canada

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General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Prentice-Hall © 2002
General Chemistry: Chapter 5
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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

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General Chemistry: Chapter 5
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5.1 The Nature of Aqueous Solutions
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Electrolytes
Some solutes can dissociate into ions.
Electric charge can be carried.



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Types of Electrolytes
Weak electrolyte partially dissociates.
Fair conductor of electricity.
Non-electrolyte does not dissociate.
Poor conductor of electricity.
Strong electrolyte dissociates completely.
Good electrical conduction.
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Representation of Electrolytes using Chemical Equations
MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
A strong electrolyte:
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MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
[Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M
[Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M
Notation for Concentration
In 0.0050 M MgCl2:
Stoichiometry is important.
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Example 5-1
Calculating Ion concentrations in a Solution of a Strong Electolyte.
What are the aluminum and sulfate ion concentrations in 0.0165 M Al2(SO4)3?.
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
Balanced Chemical Equation:
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[Al] = × =
1 L
2 mol Al3+
1 mol Al2(SO4)3
0.0165 mol Al2(SO4)3
0.0330 M Al3+
Example 5-1
0.0495 M SO42-
Aluminum Concentration:
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5-2 Precipitation Reactions
Soluble ions can combine to form an insoluble compound.
Precipitation occurs.
Ag+(aq) + Cl-(aq) → AgCl(s)
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Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →
AgI(s) + Na+(aq) + NO3-(aq)
Net Ionic Equation
AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)
Overall Precipitation Reaction:
Complete ionic equation:
Ag+(aq) + I-(aq) → AgI(s)
Net ionic equation:
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Solubility Rules
Compounds that are soluble:
Li+, Na+, K+, Rb+, Cs+ NH4+


NO3- ClO4- CH3CO2-
Alkali metal ion and ammonium ion salts


Nitrates, perchlorates and acetates
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Solubility Rules
Chlorides, bromides and iodides Cl-, Br-, I-
Except those of Pb2+, Ag+, and Hg22+.
Sulfates SO42-
Except those of Sr2+, Ba2+, Pb2+ and Hg22+.
Ca(SO4) is slightly soluble.
Compounds that are mostly soluble:
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Solubility Rules
Hydroxides and sulfides HO-, S2-
Except alkali metal and ammonium salts
Sulfides of alkaline earths are soluble
Hydroxides of Sr2+ and Ca2+ are slightly soluble.
Carbonates and phosphates CO32-, PO43-
Except alkali metal and ammonium salts
Compounds that are insoluble:
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5-3 Acid-Base Reactions
Latin acidus (sour)
Sour taste
Arabic al-qali (ashes of certain plants)
Bitter taste

Svante Arrhenius 1884 Acid-Base theory.
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Acids
Acids provide H+ in aqueous solution.

Strong acids:


Weak acids:
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Bases
Bases provide OH- in aqueous solution.

Strong bases:


Weak bases:
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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)
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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)
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More Acid-Base Reactions
Limestone and marble.
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)
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Limestone and Marble
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Gas Forming Reactions
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Hematite is converted to iron in a blast furnace.
CO(g) is oxidized to carbon dioxide.
Fe3+ is reduced to metallic iron.
5-4 Oxidation-Reduction: Some
General Principles
Oxidation and reduction always occur together.
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Oxidation State Changes
3+
2-
2+
2-
4+
2-
0
Assign oxidation states:
CO(g) is oxidized to carbon dioxide.
Fe3+ is reduced to metallic iron.
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Oxidation and Reduction
Oxidation
O.S. of some element increases in the reaction.
Electrons are on the right of the equation

Reduction
O.S. of some element decreases in the reaction.
Electrons are on the left of the equation.
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Zinc in Copper Sulfate
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
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Half-Reactions
Represent a reaction by two half-reactions.
Oxidation:
Reduction:
Overall:
Zn(s) → Zn2+(aq) + 2 e-
Cu2+(aq) + 2 e- → Cu(s)
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
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Balancing Oxidation-Reduction Equations
Few can be balanced by inspection.
Systematic approach required.

The Half-Reaction (Ion-Electron) Method
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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)
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Example 5-6
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
Determine the oxidation states:
SO32-(aq) → SO42-(aq) + 2 e-(aq)
Write the half-reactions:
5 e-(aq) +MnO4-(aq) → Mn2+(aq)
Balance atoms other than H and O:
Already balanced for elements.
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Example 5-6
Balance O by adding H2O:
H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq)
5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Check that the charges are balanced: Add e- if necessary.
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Example 5-6
Multiply the half-reactions to balance all e-:
5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq)
16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)
Check the balance!
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Balancing in Acid
Write the equations for the half-reactions.
Balance all atoms except H and O.
Balance oxygen using H2O.
Balance hydrogen using H+.
Balance charge using e-.
Equalize the number of electrons.
Add the half reactions.
Check the balance.
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Balancing in Basic Solution
OH- appears instead of H+.

Treat the equation as if it were in acid.
Then add OH- to each side to neutralize H+.
Remove H2O appearing on both sides of equation.
Check the balance.
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5-6 Oxidizing and Reducing Agents.
An oxidizing agent (oxidant ):
Contains an element whose oxidation state decreases in a redox reaction

A reducing agent (reductant):
Contains an element whose oxidation state increases in a redox reaction.
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Redox
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Example 5-8
Identifying Oxidizing and Reducing Agents.
Hydrogen peroxide, H2O2, 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.
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5 H2O2(aq) + 2 MnO4-(aq) + 6 H+ →
8 H2O(l) + 2 Mn2+(aq) + 5 O2(g)
Example 5-8
H2O2(aq) + 2 Fe2+(aq) + 2 H+ → 2 H2O(l) + 2 Fe3+(aq)
Iron is oxidized and peroxide is reduced.
Manganese is reduced and peroxide is oxidized.
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5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations.
Titration
Carefully controlled addition of one solution to another.
Equivalence Point
Both reactants have reacted completely.
Indicators
Substances which change colour near an equivalence point.
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Indicators
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Example 5-10
Standardizing a Solution for Use in Redox Titrations.
A piece of iron wire weighing 0.1568 g is converted to Fe2+(aq) and requires 26.42 mL of a KMnO4(aq) solution for its titration. What is the molarity of the KMnO4(aq)?
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) →
4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
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Example 5-10
Determine KMnO4 consumed in the reaction:
Determine the concentration:
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
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Chapter 5 Questions
1, 2, 3, 5, 6, 8, 14, 17, 19, 24, 27, 33, 37, 41, 43, 51, 53, 59, 68, 71, 82, 96.