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5.1 The Nature of Aqueous Solutions
Water




Inexpensive
Can dissolve a vast number of substances
Many substances dissociate into ions
Aqueous solutions are found everywhere
◦ Seawater
◦ Living systems
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General Chemistry: Chapter 5
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Electrolytes
 Some solutes can
dissociate into ions.
 Electric charge can be
carried.
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General Chemistry: Chapter 5
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Types of Electrolytes
Strong electrolyte dissociates completely.
 Good electrical conduction.
Weak electrolyte partially dissociates.
 Fair conductor of electricity.
Non-electrolyte does not dissociate.
 Poor conductor of electricity.
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General Chemistry: Chapter 5
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Representation of Electrolytes using
Chemical Equations
A strong electrolyte:
MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
A weak electrolyte:
→ CH3CO2-(aq) + H+(aq)
CH3CO2H(aq) ←
A non-electrolyte:
CH3OH(aq)
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Three Types of Electrolytes
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General Chemistry: Chapter 5
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Solvation: The Hydrated Proton
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General Chemistry: Chapter 5
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Relative Concentrations in Solution
MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)
In 0.0050 M MgCl2:
Stoichiometry is important.
[Mg2+] = 0.0050 M
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[Cl-] = 0.0100 M
General Chemistry: Chapter 5
[MgCl2] = 0 M
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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?.
Write a Balanced Chemical Equation:
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
Identify the Stoichiometric Factors :
2 mol Al3+
1 mol Al2(SO4)3
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3 mol SO421 mol Al2(SO4)3
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EXAMPLE 7-3
Aluminum Concentration:
[Al] =
0.0165 mol Al2(SO4)3
1L
2 mol Al3+

1 mol Al2(SO4)3
mol Al3+
= 0.0330
1L
Sulfate Concentration:
0.0165 mol Al2(SO4)3
3 mol SO42[SO42-] =

=
1L
1 mol Al2(SO4)3
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General Chemistry: Chapter 5
0.0495 M SO42-
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5-2 Precipitation Reactions
 Soluble ions can combine
to form an insoluble
compound.
 Precipitation occurs.
 A test for the presence of
chloride ion in water.
Ag+(aq) + Cl-(aq) → AgCl(s)
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General Chemistry: Chapter 5
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Silver Nitrate and Sodium Iodide
AgNO3(aq)
NaI(aq)
AgI(s)
Na+(aq) NO3-(aq)
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General Chemistry: Chapter 5
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Net Ionic Equation
Overall Precipitation Reaction:
AgNO3(aq) +NaI(aq) → AgI(s) + NaNO3(aq)
Complete ionic equation:
Spectator ions
Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →
AgI(s) + Na+(aq) + NO3-(aq)
Net ionic equation:
Ag+(aq) + I-(aq) → AgI(s)
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Solubility Rules
 Compounds that are soluble:
 Alkali metal ion and ammonium ion salts
Li+, Na+, K+, Rb+, Cs+
NH4+
 Nitrates, perchlorates and acetates
NO3-
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ClO4-
CH3CO2-
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Solubility Rules
 Compounds that are mostly soluble:
 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.
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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 and Bases
Some Definitions
Arrhenius
 An acid is a substance that, when dissolved in water,
ionizes and increases the concentration of hydrogen
ions, H+.
HCl → H+ + Cl A base is a substance that, when dissolved in water,
increases the concentration
of hydroxide ions, OH-.
But, Not all bases contain OHNaOH → Na+ + OH-
Brønsted-Lowry
 An acid is a proton donor.
 A base is a proton acceptor.
e.g. ammonia, NH3, is a base.
A Brønsted-Lowry acid…
…must have an ionizable or removable (acidic)
proton.
A Brønsted-Lowry base…
…must have a pair of nonbonding electrons.
In an acid – base reaction, the acid donates
a proton (H+) to the base.
If a substance can be either…
…it is amphiprotic.
H2O
HCO3
What Happens When an Acid Dissolves in
Water?
 Water acts as a BrønstedLowry base and abstracts
a proton (H+) from the
acid.
 As a result, the conjugate
base of the acid and a
hydronium ion are
formed.
Self-ionization of water
H2O + H2O ⇄ H3O+ (aq) + OH- (aq)
Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases
and a few act as acids.
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
This is referred to as autoionization.
For ALL aqueous solutions:
Kc = [H3O+] [OH-]
At 25C, Kw = [H3O+]·[OH-] = 1.0·10-14 =
Kw (at 25 ºC)
[H3O+] = [OH-] the solution is….
[H3O+] > [OH-] the solution is….
[H3O+] < [OH-] the solution is….
Acids
Acids provide H+ in aqueous solution.
Strong acids completely ionize:
HCl(aq)
→
H+(aq) + Cl-(aq)
Weak acid ionization is not complete:
CH3CO2H(aq)
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←
→
H+(aq) + CH3CO2-(aq)
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Acids
 Arrhenius defined acids as substances that
increase the concentration of H+ when dissolved in
water.
 Brønsted and Lowry defined them as proton
donors.
Bases
Bases provide OH- in aqueous solution.
Strong bases:
NaOH(aq)
→ Na+(aq) + OH-(aq)
H2O
Weak bases:
NH3(aq) + H2O(l)
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←
→
OH-(aq) + NH4+(aq)
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Acids
There are only seven
strong acids:
•
•
•
•
•
•
•
Hydrochloric (HCl)
Hydrobromic (HBr)
Hydroiodic (HI)
Nitric (HNO3)
Sulfuric (H2SO4)
Chloric (HClO3)
Perchloric (HClO4)
Acid-Base Reactions
In an acid-base
reaction, the acid
donates a proton
(H+) to the base.
Recognizing Acids and Bases.
 Acids have ionizable hydrogen ions.
 CH3CO2H or HC2H3O2
 Bases have OH- combined with a metal ion.
KOH
or can be 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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Electrochemical Reactions
In electrochemical reactions, electrons are
transferred from one species to another.
Mg (s) + HCl (aq) →
Oxidation Numbers
In order to keep track
of what loses electrons
and what gains them,
we assign oxidation
numbers.
Oxidation and Reduction
 A species is oxidized when it loses electrons.
 Here, zinc loses two electrons to go from neutral zinc
metal to the Zn2+ ion.
Oxidation and Reduction
 A species is reduced when it gains electrons.
 Here, each of the H+ gains an electron, and they
combine to form H2.
Oxidation and Reduction
 What is reduced is the oxidizing agent.
 H+ oxidizes Zn by taking electrons from it.
 What is oxidized is the reducing agent.
 Zn reduces H+ by giving it electrons.
Assigning Oxidation Numbers, ON
1. Elements in their elemental form have an
oxidation number of 0.
Mg Cl2 Pb
1. The oxidation number of a monatomic ion is the
same as its charge.
Zn2+ F- Na+
In combination with other elements:
3. Nonmetals tend to have negative oxidation
numbers
 Oxygen has an oxidation number of −2,
except in the peroxide ion, which has an
oxidation number of −1.
MgO
Na2O

HF
Hydrogen is −1 when bonded to a metal and
+1 when bonded to a nonmetal.
NaH
Assigning Oxidation Numbers

Fluorine always has an oxidation number of
−1.
HF NaF

The other halogens have an oxidation number
of −1 when they are negative; they can have
positive oxidation numbers, however, most
notably in oxyanions.
ZnBr2 CuCl2
Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
HCl H2O K2O
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the ion.
CO32- ClO4-
Examples
Determine the oxidation number of sulfur in
the following compounds:
H2S
SO42S8
S2O32-
Electrochemical Reactions
In a spontaneous Oxidation-Reduction (redox) Reaction
electrons are transferred and energy is released
as the reaction proceeds:
 Zn strip dissolves…
 The blue color due to
Cu2+ fades.
 Copper metal is
deposited.
Oxidation and Reduction Half-Reactions
A reaction represented by two half-reactions.
Oxidation:
Zn(s) → Zn2+(aq) + 2 e-
Reduction:
Cu2+(aq) + 2 e- → Cu(s)
Overall:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
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Corrosion and…
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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Oxidation State Changes
 Assign oxidation states:
D
Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)
Fe3+ is
CO(g) is
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to metallic iron.
to carbon dioxide.
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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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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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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
Determine the oxidation states:
4+
6+
7+
2+
SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq)
Write the half-reactions:
SO32-(aq) → SO42-(aq) + 2 e-(aq)
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)
Balance hydrogen by adding H+:
H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) + 2 H+(aq)
8 H+(aq) + 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)
Check that the charges are balanced:
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General Chemistry: Chapter 5
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)
6
3
Add both equations and simplify:
5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) →
5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l)
Check the balance!
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EXAMPLE 7-3
3+
2+
H2O2(aq) + 2 Fe2+(aq) + 2 H+ → 2 H2O(l) + 2 Fe3+(aq)
Iron is oxidized and
peroxide is reduced.
7+
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+ →
2+
8 H2O(l) + 2 Mn2+(aq) + 5 O2(g)
Manganese is reduced and
peroxide is oxidized.
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