John E. McMurry • Robert C. Fay
C H E M I
S T R Y
Fifth Edition
Chapter 15
Applications of Aqueous Equilibria
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
Copyright © 2008 Pearson Prentice Hall, Inc.
Neutralization Reaction
 General Formula
Acid + Base - Water + Salt
Neutralization Reactions
Strong Acid-Strong Base
HCl(aq)
+
NaOH(aq)
H2O(l) +
NaCl(aq)
Assuming complete dissociation:
H1+(aq) + OH1-(aq)
H2O(l)
H3O1+(aq) + OH1-(aq)
2H2O(l)
or
(net ionic equation)
After neutralization: pH = 7
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Chapter
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Strong acid-Strong base
neutralization
 When the number moles of acid and base are mixed
together
[H3O+] = [-OH] = 1.0 x 10-7M
 Reaction proceeds far to the right
Neutralization Reactions
Weak Acid-Strong Base
CH3CO2H(aq)
+
NaOH(aq)
H2O(l)
+
NaCH3CO2(aq)
Assuming complete dissociation:
CH3CO2H(aq) + OH1-(aq)
H2O(l) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH > 7
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Weak acid-strong base
neutralization
 Neutralization of any weak acid by a strong base goes 100%
to completion
 -OH has a great infinity for protons
Neutralization Reactions
Strong Acid-Weak Base
HCl(aq)
+ NH3(aq)
NH4Cl(aq)
Assuming complete dissociation:
H1+(aq) + NH3(aq)
NH41+(aq)
or
H3O1+(aq) + NH3 (aq)
H2O(l) +
NH41+(aq)
(net ionic equation)
After neutralization: pH < 7
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Chapter
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Strong acid-weak base
neutralization
 Neutralization of any weak base by a strong acid goes 100%
to completion
 H3O+ has a great infinity for protons
Neutralization Reactions
Weak Acid-Weak Base
CH3CO2H(aq) + NH3(aq)
NH4CH3CO2(aq)
CH3CO2H(aq) + NH3(aq)
NH41+(aq) + CH3CO21-(aq)
(net ionic equation)
After neutralization: pH = ?
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Weak acid-weak base
neutralization
 Less tendency to proceed to completion than neutralization
involving strong acids and strong bases
The Common-Ion Effect
Common-Ion Effect: The shift in the position of an equilibrium
on addition of a substance that provides an ion in common
with one of the ions already involved in the equilibrium.
• Example of Le Chatelier’s principle
• E.g Adding HCl and NaOH to a solution of acetic acid
would shift the equilibrium to which direction?
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
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The Common-Ion Effect
Le Châtelier’s Principle
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
The addition of acetate ion to a solution of acetic acid
suppresses the dissociation of the acid. The equilibrium shifts
to the left.
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Chapter
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The Common-Ion Effect
The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that
is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium
acetate in enough water to make 1.00 L of solution. Ka = 1.8 x 10-5
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
[H3O1+][CH3CO21-]
Ka =
[CH3CO2H]
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Example
 In 0.15 M NH3, the pH is 11.21 and the percent dissociation
is 1.1%. Calculate the concentrations of NH3, pH and
percent dissociation of ammonia in a solution that is 0.15M
and 0.45 MNH4Cl
Buffer Solutions
Buffer Solution: A solution which contains a weak acid and
its conjugate base and resists drastic changes in pH.
Weak acid
For
Example:
+
CH3CO2H + CH3CO21HF + F1NH41+ + NH3
H2PO42- + HPO42-
Conjugate base
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Buffer Solutions
 Add a small amount of base (-OH) to a buffer solution
 Acid component of solution neutralizes the added base
 Add a small amount of acid (H3O+) to a buffer solution
 Base component of solution neutralizes the added acid
 The addition of –OH or H3O+ to a buffer solution will
change the pH of the solution, but not as drastically as the
addition of –OH or H3O+ to a non-buffered solution
Buffer Solutions
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
Weak acid
Conjugate base
(NaCH3CO2)
Addition of OH1- to a buffer:
CH3CO2H(aq) +
OH1-(aq)
100%
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H2O(l) + CH3CO21-(aq)
Chapter
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Buffer Solutions
Addition of H3O1+ to a buffer:
CH3CO2
1-(aq)
+ H3
O1+(aq)
100%
H2O(l) + CH3CO2H(aq)
Example
 pH of human blood (pH = 7.4) controlled by conjugated
acid-base pairs (H2CO3/HCO3-)
 With addition of H3O+
 With addition of -OH
Buffer Solutions
Buffer capacity
 A measure of the amount of acid or base that a buffer
solution can absorb without a significant change in pH
 Depends on how much weak acid and conjugated base is
present
 For equal volume of solution, the more concentration the
solution, the greater the buffer capacity
 For solution with the same concentration, increasing the
volume increases the buffer capacity
Example
 Calculate the pH of the buffer that results from mixing
60.0mL of 0.250M HCHO2 and 15.0 mL of 0.500M
NaCHO2
Ka = 1.8 x 10-4
Example
 Calculate the pH of 0.100L of a buffer solution that is
0.25M in HF and 0.50 M in NaF,
Ka = 6.3 x 10-4
 What is the change in pH on addition of 0.002 mol HCl
 What is the change in pH on addition of 0.010 moles KOH
 Calculate the pH after addition of 0.080 moles HBr
Example
 calculate the pH of a 50.0 ml buffer solution that is 0.50 M
in NH3 and 0.20 M NH4Cl. For ammonia, pKb = 4.75
 Calculate the pH after addition of 150.0 mg HBr
The Henderson-Hasselbalch
Equation
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
Weak acid
Conjugate base
Acid(aq) + H2O(l)
Ka =
[H3O1+][Base]
H3O1+(aq) + Base(aq)
[H3O1+] = Ka
[Acid]
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[Acid]
[Base]
Chapter
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The Henderson-Hasselbalch
Equation
[Acid]
[H3O1+] = Ka
[Base]
[Acid]
-log([H3O1+]) = -log(Ka) - log
[Base]
[Base]
pH = pKa + log
[Acid]
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Example
 Calculate the pH of a buffer solution that is 0.50 M in
benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate
(NaC7H5O2). Ka = 6.5 x 10-5
 How would you prepare a NaHCO3-Na2CO3 buffer solution
that has pH = 10.40 Ka2 = 5.6 x 10-11