CHE 116 Prof. T.L. Heise
Chapter Seventeen: Additional
Aspects of Aqueous Equilibria
 Water
is the most important solvent
on this planet.
 Aqueous solutions encountered in
nature contain many solutes
 Many equilibria take place in these
solutions
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Common Ion Effect is still important in
Acid/Base equilibria.
Using a salt that contains a
common ion will cause an acid base
equilibria to shift just as we saw in
Chapter 15, using Le Chateliers
Principle
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Consider:
HC2H3O2(aq)  H+(aq) + C2H3O2(aq)
- if adding NaC2H3O2, complete
dissociation into Na+ and C2H3O2ions will occur.
- additional C2H3O2- ions causes
the reaction to shift left, using up
H+ ions, and reducing acidity
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2Ka = 4.5 x 10-4 = [H+][NO2-]
[HNO2]
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2I
0.085
0
0.10
D
-x
+x
+x
E
0.085-x x
0.10+x
Copyright T. L. Heise 2001 - 2002
6
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2Ka = 4.5 x 10-4 = [x][0.10+x]
[0.085-x]
Copyright T. L. Heise 2001 - 2002
7
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2Ka = 4.5 x 10-4 = [x][0.10+x]
[0.085-x]
*make assumption
Copyright T. L. Heise 2001 - 2002
8
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2Ka = 4.5 x 10-4 = [x][0.10]
[0.085]
Copyright T. L. Heise 2001 - 2002
9
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
HNO2  H+ + NO2Ka = 4.5 x 10-4 = [x][0.10]
[0.085]
x = 3.83 x 10-4
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the pH of a
solution containing 0.085 M nitrous
acid, HNO2 (Ka = 4.5 x 10-4), and
0.10 M potassium nitrite, KNO2.
[H+] = 3.83 x 10-4
pH = -log [H+]
pH = 3.42
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
Copyright T. L. Heise 2001 - 2002
12
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
Ka = 1.8 x 10-4 = [CHO2-][H+]
[HCHO2]
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
I
0.050
0
0.10
D
-x
+x
+x
E
0.050-x
x
0.10+x
Copyright T. L. Heise 2001 - 2002
14
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
Ka = 1.8 x 10-4 = [x][0.10+x]
[0.050-x]
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
Ka = 1.8 x 10-4 = [x][0.10+x]
[0.050-x]
* make assumption
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
Ka = 1.8 x 10-4 = [x][0.10]
[0.050]
Copyright T. L. Heise 2001 - 2002
17
CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
HCHO2  CHO2- + H+
Ka = 1.8 x 10-4 = [x][0.10]
[0.050]
x = 9.0 x 10-5
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
pH will depend on nitric acid,
which is a strong acid with 100%
dissociation.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
Sample exercise: Calculate the formate
ion concentration and pH of a solution
that is 0.050 M in formic acid, HCHO2
(Ka = 1.8 x 10-4), and 0.10 M in HNO3.
pH = -log[H+]
pH = -log[0.10]
pH = 1.0
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CHE 116 Prof. T.L. Heise
The Common Ion Effect
The common ion effect is equally
important in the consideration of a
basic solution!
NH3(aq) + H2O  NH4+(aq) + OH-(aq)
- adding NH4Cl will cause a shift to the
left and a decrease in OH- ion,
increasing acidity.
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Solutions like those just discussed,
containing weak conjugate acid-base
pairs, resist drastic changes in their
pH levels.
These solutions are called buffers
human blood is an extremely
important buffer system
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Buffers resist changes in pH because
they contain both an acidic species to
neutralize OH- and a basic species to
neutralize H+.
The species must not actually react with
each other in a neutralization
reaction, and that requirement is only
fulfilled by weak conjugate acid/base
pairs.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Buffers can be created by dissolving a
salt in its common ion acidic solution.
HC2H3O2/ NaC2H3O2
To better understand, consider the
following:
HX(aq)  H+(aq) + X-(aq)
Ka = [H+][X-]
[HX]
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Solving for [H+]:
Ka[HX] = [H+]
[X-]
Examining this we can see that the pH
will be dependent on two factors: the
value of Ka and the ratio of the
conjugate acid base pair.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Adding minute amounts of base will
alter the proportions of the acid/base
concentrations, but not enough to
affect the pH.
Fig. 17.2
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Two important characteristics of a
buffer are its capacity and its pH.
Buffer capacity is the amount of
acid or base a buffer can neutralize
before the pH begins to change to
an appreciable degree
The pH depends on the Ka of the
solution and the relative amounts of
acid/base pair
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Two important characteristics of a
buffer are its capacity and its pH.
The greater the amounts of
conjugate acid/base pair, the more
resistant the ratio of their
concentrations, and hence pH, is to
change.
Because a common ion is shared,
the same calculations can be used.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the pH of a
buffer composed of 0.12 M benzoic
acid and 0.20 M sodium benzoate.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the pH of a
buffer composed of 0.12 M benzoic
acid and 0.20 M sodium benzoate.
Appendix D - Ka = 6.3 x 10-5
HC7H5O2  C7H5O2- + H+
I
0.12
0.20
0
D
-x
+x
+x
E
0.12-x
0.20+x
x
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the pH of a
buffer composed of 0.12 M benzoic
acid and 0.20 M sodium benzoate.
Appendix D - Ka = 6.3 x 10-5
HC7H5O2  C7H5O2- + H+
Ka = [0.20+x][x]
* make
[0.12-x]
assumption
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the pH of a
buffer composed of 0.12 M benzoic
acid and 0.20 M sodium benzoate.
Appendix D - Ka = 6.3 x 10-5
HC7H5O2  C7H5O2- + H+
6.3 x 10-5 = [0.20][x]
[0.12]
x = 3.8 x 10-5
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the pH of a
buffer composed of 0.12 M benzoic
acid and 0.20 M sodium benzoate.
[H+] = 3.8 x 10-5
pH = 4.42
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
Buffered Solutions
Sample exercise: Calculate the
concentration of sodium benzoate that
must be present in a 0.20 M solution
of benzoic acid to produce a pH of
4.00.
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
35
Buffered Solutions
Sample exercise: Calculate the concentration
of sodium benzoate that must be present in
a 0.20 M solution of benzoic acid to produce
a pH of 4.00. (Ka = 6.3 x 10-5)
HC7H5O2  C7H5O2- + H+
I
0. 20
x
0
D
-1 x 10-4 +1 x 10-4 +1 x 10-4
E
0. 20
x
1 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
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Buffered Solutions
Sample exercise: Calculate the concentration
of sodium benzoate that must be present in
a 0.20 M solution of benzoic acid to produce
a pH of 4.00.
Ka = 6.3 x 10-5 = [C7H5O2-][H+]
[HC7H5O2]
= [x][1 x 10-4]
[0. 20]
x = 0.126 M C7H5O2Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
37
Buffered Solutions
Addition of a strong acid or base to a buffer:
reactions between strong acids and weak
bases proceed essentially to completion
as long as we do not exceed the buffering
capacity of the buffer, we can assume that
the strong acid or base will be completely
absorbed by the buffer
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
38
Buffered Solutions
Addition of a strong acid or base to a buffer:
to perform calculations
1. Consider the acid base neutralization
reaction to determine stoichiometric
proportions
2. Use Ka and the new concentrations to
calculate [H+]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
39
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
a) before any acid or base is added
b) after the addition of 0.015 mol of
HNO3
c) after the addition of 0.015 mol of
KOH
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
40
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
a) before any acid or base is added
HCNO  CNO- + H+
I
0.140
0.110 0
D
-x
+x
+x
E 0.140-x 0.110+x x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
41
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
a) before any acid or base is added
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.110][x]
[0.140]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
42
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
a) before any acid or base is added
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.110][x]
[0.140]
x = 4.5 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
43
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
a) pH = - log[H+]
= -log[4.5 x 10-4]
= 3.35
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
44
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) addition of 0.015 mol of HNO3
HCNO  CNO- + H+
I 0.140+0.015 0.110-0.015 x
D
-x
+x
+x
E 0.155-x
0.095+x
x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
45
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) addition of 0.015 mol of HNO3
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.095][x]
[0.155]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
46
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) addition of 0.015 mol of HNO3
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.095][x]
[0.155]
x = 5.7 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
47
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) addition of 0.015 mol of HNO3
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.095][x]
[0.155]
x = 5.7 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
48
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) pH = - log[H+]
= -log[5.7 x 10-4]
= 3.24
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
49
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
c) addition of 0.015 mol of KOH
HCNO  CNO- + H+
I 0.140-0.015 0.110+0.015 0
D
-x
+x
+x
E 0.125-x
0.125+x
x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
50
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
c) addition of 0.015 mol of KOH
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.125][x]
[0.125]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
51
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
c) addition of 0.015 mol of KOH
HCNO  CNO- + H+
Ka = 3.5 x 10-4 = [0.125][x]
[0.125]
x = 3.5 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
52
Buffered Solutions
Consider a 1.00 L buffer made by adding
0.140 mol cyanic acid, HCNO, and 0.110
mol potassium cyanate, KCNO, to sufficient
water. Calculate the pH of the buffer
b) pH = - log[H+]
= -log[3.5 x 10-4]
= 3.46
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
53
Acid Base Titrations
In an acid base titration, a solution containing
a known concentration of base is slowly
added to an acid.
Indicators can be used to signal the
equivalence point of a titration
A pH meter can be used to monitor the the
progress of a reaction producing a pH
titration curve, a graph of the pH as a
function of the volume of base added
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
54
Acid Base Titrations
The shape of a titration curve makes it
possible to determine the equivalence point
in the titration.
The titration curve produced when a strong
base is added to a strong acid looks like an
elongated S, adding an acid to a base would
produce an upside down elongated S
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Acid Base Titrations
Strong Base added to Strong Acid
Fig 17.6
Copyright T. L. Heise 2001 - 2002
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CHE 116 Prof. T.L. Heise
56
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
57
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Base
Acid
0.100 M * 0.025 L
0.100 M * 0.02490 L
2.50 x 10-3 mol
2.49 x 10-3
**more moles
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
58
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Base
0.100 M * 0.025 L
2.50 x 10-3 mol
1.00 x 10-5 mol
OH**more moles
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
59
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Get to concentration!!
1.00 x 10-5 mol OH0.04990 L
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
60
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Get to concentration!!
1.00 x 10-5 mol OH- = 2.00 x 10-4 M OH0.04990 L
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
61
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Use pOH formula!!
pOH = -log[OH-]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
62
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Use pOH formula!!
pOH = -log[OH-]
= -log[2.00 x 10-4]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
63
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Use pOH formula!!
pOH = -log[OH-]
= -log[2.00 x 10-4]
= 3.7
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
64
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Remember
pOH + pH = 14
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
65
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Remember
pOH + pH = 14
14 - 3.7 = pH
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
66
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
a) 24.90 mL
Remember
pOH + pH = 14
14 - 3.7 = pH
10.3 = pH
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
67
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
b) 25.10 mL
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
68
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mL
Base
Acid
0.100 M * 0.025 L
0.100 M * 0.02510 L
2.50 x 10-3 mol
2.51 x 10-3
**more moles
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
69
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mL
Acid
1.00 x 10-5 mol H+
0.100 M * 0.02510 L
0.0501 L
2.51 x 10-3
**more moles
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70
Acid Base Titrations
Sample exercise: Calculate the pH when the
following quantities of 0.100 M HNO3 have
been added to 25.00 mL of 0.100 M KOH.
c) 25.10 mL
Acid
2.00 x 10-4 M H+
0.100 M * 0.02510 L
pH = 3.70
2.51 x 10-3
**more moles
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71
Acid Base Titrations
Optimally an indicator turns colors at the
equivalence point in a titration, however, in
practice this is not necessary.
The pH changes very rapidly near the
equivalence point, and in this region a
single drop could change the pH greatly.
An indicator beginning and ending its
color change anywhere on the rapid rise
portion of the graph will work.
Copyright T. L. Heise 2001 - 2002
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72
Acid Base Titrations
Weak Acid Strong Base titrations: the curve
for a weak acid by a strong base is very
similar in shape
Fig 17.9
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73
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.050
M NaOH to 40.0 mL of 0.0250 M benzoic
acid (HC7H5O2, Ka = 6.3 x 10-5).
Copyright T. L. Heise 2001 - 2002
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74
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.050
M NaOH to 40.0 mL of 0.0250 M benzoic
acid (HC7H5O2, Ka = 6.3 x 10-5).
HC7H5O2 + OH-  C7H5O2- + H2O
I
1 x 10-3 5 x 10-4
0
D
-5 x 10-4 -5 x 10-4 +5 x 10-4
E
5 x10-4
0
5x 10-4
Copyright T. L. Heise 2001 - 2002
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75
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.050
M NaOH to 40.0 mL of 0.0250 M benzoic
acid (HC7H5O2, Ka = 6.3 x 10-5).
HC7H5O2 + OH-  C7H5O2- + H2O
E
5 x10-4 mol 0
5x 10-4 mol
0.050 L
0.050 L
0.01 M
0.01 M
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76
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.050
M NaOH to 40.0 mL of 0.0250 M benzoic
acid (HC7H5O2, Ka = 6.3 x 10-5).
HC7H5O2 +  C7H5O2- + H+
Ka = 6.3 x 10-5 = [0.01][x]
[0.01]
x = 6.3 x 10-5
pH = 4.20
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77
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.100
M HCl to 20.0 mL of 0.100 M NH3.
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78
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.100
M HCl to 20.0 mL of 0.100 M NH3.
H+ + NH3  NH4+
I 1 x 10-3 2 x 10-3
0
D -1 x 10-3 -1 x 10-3 +1 x 10-3
E 0
1 x 10-3
1 x 10-3
Copyright T. L. Heise 2001 - 2002
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79
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.100
M HCl to 20.0 mL of 0.100 M NH3.
NH4+  NH3 + H+
I
0.033
0.033
0
D
-x
+x
+x
E
0.033-x
0.033+x
x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
80
Acid Base Titrations
Sample exercise: Calculate the pH in the
solution formed by adding 10.0 mL of 0.100
M HCl to 20.0 mL of 0.100 M NH3.
NH4+  NH3 + H+
Ka = 5.5 x 10-10 = [x][0.033]
[0.033]
x = 5.5 x 10-10 M H+
pH = 9.26
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is
titrated with 0.050 M NaOH
Copyright T. L. Heise 2001 - 2002
81
CHE 116 Prof. T.L. Heise
82
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is
titrated with 0.050 M NaOH
HC7H5O2 + OH-  C7H5O2- + H2O
1 x 10-3 mol HC7H5O2 reacts with 1 x 10-3 mol
OH- to form 1 x 10-3 mol C7H5O2- in 60 mL
of solution, which is a weak base.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is
titrated with 0.050 M NaOH
C7H5O2- + H2O  HC7H5O2 + OHI
1.6 x 10-2
0
0
D
-x
+x
+x
E
1.6 x 10-2 -x
x
x
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83
CHE 116 Prof. T.L. Heise
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
a) 40.0 mL of 0.025 M benzoic acid is
titrated with 0.050 M NaOH
C7H5O2- + H2O  HC7H5O2 + OHKb = 1.6 x 10-10 = [x][x]
[1.6 x 10-2]
x = 1.6 x 10-6 M OH- pOH = 5.80
pH = 8.20
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84
CHE 116 Prof. T.L. Heise
85
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
b) 40.0 mL 0f 0.100 M NH3 is titrated
with 0.100 M HCl
H+ + NH3  NH4+
4 x 10-3 mol NH3 reacts with 4 x 10-3 mol H+
to form 4 x 10-3 mol NH4 in 80 mL of
solution, which is a weak acid.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
b) 40.0 mL 0f 0.100 M NH3 is titrated
with 0.100 M HCl
NH4+ 
NH3 + H+
I
0.050
0
0
D
-x
+x +x
E
0.050 -x
x
x
Copyright T. L. Heise 2001 - 2002
86
CHE 116 Prof. T.L. Heise
87
Acid Base Titrations
Sample exercise: Calculate the pH at the
equivalence point when
b) 40.0 mL 0f 0.100 M NH3 is titrated
with 0.100 M HCl
NH4+  NH3 + H+
Ka = 5.6 x 10-10 = [x][x]
[0.050]
x = 5.3 x 10-6 M H+
pH = 5.28
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88
Acid Base Titrations
The pH titration curves for weak acids and
strong base titrations differ from those of a
strong acid strong base titration in 3
noteworthy ways:
solutions of weak acids have higher intial
pH’s
the pH change at the rapid rise portion of
the curve is much shorter
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
89
Acid Base Titrations
The titration curve for a weak base and
strong acid is very similar to the strong base
with strong acid and follows the same 3
noteworthy differences proportionally
Fig. 17.12
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CHE 116 Prof. T.L. Heise
90
Acid Base Titrations
Polyprotic Acids: If the
acid has more than
one ionizable
proton, then the
titration curve has
more than one
equivalence point
CHE 116 Prof. T.L. Heise
91
Solubility Equilibia
The equilibria studied so far have involved
only acids and bases. They have also been
only homogeneous equilibria.
Another important type of equilibria exists:
The dissolution and precipitation of ionic
compounds
By considering solubility equilibria, we can
make quantitative predictions about the
amount of a given compound that will
dissolve Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
92
Solubility Equilibia
Solubility Product Constant:
Saturated solution - the solution is in
contact with undissolved solute. A particle
of solute dissolves and exactly the same rate
as a dissolved particle precipitates.
An equilibrium is established between the
dissolved and undissolved particles
BaSO4(s)  Ba2+(aq) + SO42-(aq)
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93
Solubility Equilibia
BaSO4(s)  Ba2+(aq) + SO42-(aq)
Keq = [Ba2+][SO42-]
[BaSO4]
however, this is a heterogeneous equilibrium
and solids are not included so…
Ksp = [Ba2+][SO42-]
*the solubility product is equal to the product
of the concentrations of the ions involved,
each raised to the power of its coefficients
Copyright T. L. Heise 2001 - 2002
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94
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
a) barium carbonate
Copyright T. L. Heise 2001 - 2002
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95
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
a) barium carbonate
Ba2+
BaCO3
CO32-
Copyright T. L. Heise 2001 - 2002
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96
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
a) barium carbonate
BaCO3  Ba2+ + CO32Ksp = 5.0 x 10-9 = [Ba2+][CO32-]
Copyright T. L. Heise 2001 - 2002
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97
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
b) silver sulfate
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
98
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
b) silver sulfate
Ag+
Ag2SO4
SO42-
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
99
Solubility Equilibia
Sample exercise: Give the solubility product
constant expressions and values of the Ksp
for the following compounds:
b) silver sulfate
Ag2SO4  2Ag+ + SO42Ksp = 1.5 x 10-5 = [Ag+ ]2[SO42-]
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
100
Solubility Equilibia
Please be careful to distinguish between
solubility and the solubility product
constant.
Solubility is the number of grams that
will dissolve in a given amount of solvent.
Solubility product constant is the
equilibrium constant for the equilibrium
between the ionic solid and the saturated
solution
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101
Solubility Equilibia
Sample exercise: A saturated solution of
AgCl in contact with undissolved solid is
prepared at 25°C. The concentration of Ag+
ions in the solution is found to be
1.35 x 10-5 M. Assuming that AgCl
dissociates completely in water and that
there are no other simultaneous equilibria
involving Ag+ and Cl- ion in the solution,
calculate Ksp for this compound.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
102
Solubility Equilibia
Sample exercise: A saturated solution of
AgCl in contact with undissolved solid is
prepared at 25°C. The concentration of Ag+
ions in the solution is found to be
1.35 x 10-5 M. Calculate Ksp for this
compound.
AgCl  Ag+ + Cl1.35 x 10-5 1.35 x 10-5
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
103
Solubility Equilibia
Sample exercise: A saturated solution of
AgCl in contact with undissolved solid is
prepared at 25°C. The concentration of Ag+
ions in the solution is found to be
1.35 x 10-5 M. Calculate Ksp for this
compound.
Ksp = [Ag+][Cl-] = [1.35 x 10-5][1.35 x 10-5]
Ksp = 1.82 x 10-10
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
104
Solubility Equilibia
Sample exercise: The Ksp for Cu(N3)2 is
6.3 x 10-10. What is the solubility of Cu(N3)2
in water in grams per liter?
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
105
Solubility Equilibia
Sample exercise: The Ksp for Cu(N3)2 is
6.3 x 10-10. What is the solubility of Cu(N3)2
in water in grams per liter?
Cu(N3)2  Cu2- + 2N3I
0
0
D
+x
+2x
E
x
2x
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
106
Solubility Equilibia
Sample exercise: The Ksp for Cu(N3)2 is
6.3 x 10-10. What is the solubility of Cu(N3)2
in water in grams per liter?
Cu(N3)2  Cu2- + 2N3Ksp = [Cu2+][N3-]2
6.3 x 10-10 = [x][2x]2
x = 5.4 x 10-4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
107
Solubility Equilibia
Sample exercise: The Ksp for Cu(N3)2 is
6.3 x 10-10. What is the solubility of Cu(N3)2
in water in grams per liter?
Cu(N3)2  Cu2- + 2N35.4 x 10-4 mol 147.588 g = 0.080 g/L
1L
1 mol
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
The solubility of a substance is affected by
temperature
presence of other solutes
 presence of common ion
 pH of solution
 presence of complexing agents
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108
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
109
Common Ion Effect: The presence of a
common ion reduces the amount the ionic
salt can dissolve shifting the equilibrium to
the left towards the ionic solid
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
110
Sample exercise: The value for Ksp for
manganese (II) hydroxide, Mn(OH)2, is
1.6 x 10-13. Calculate the molar solubility of
Mn(OH)2 in a solution that contains 0.020
M NaOH.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
111
Sample exercise: The value for Ksp for
manganese (II) hydroxide, Mn(OH)2, is
1.6 x 10-13. Calculate the molar solubility of
Mn(OH)2 in a solution that contains 0.020
M NaOH.
Mn(OH)2  Mn2+ + 2OHI
0
0.020
D
+x
+2x
E
x
2x+0.020
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
112
Sample exercise: The value for Ksp for
manganese (II) hydroxide, Mn(OH)2, is
1.6 x 10-13. Calculate the molar solubility of
Mn(OH)2 in a solution that contains 0.020
M NaOH.
Mn(OH)2  Mn2+ + 2OHKsp = 1.6 x 10-13 = [Mn2+][OH-]2
= [x][0.020 + 2x]2
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
113
Sample exercise: The value for Ksp for
manganese (II) hydroxide, Mn(OH)2, is
1.6 x 10-13. Calculate the molar solubility of
Mn(OH)2 in a solution that contains 0.020
M NaOH.
Ksp = 1.6 x 10-13 = [Mn2+][OH-]2
= [x][0.020 - 2x]2
= [x][0.020]2
x = 4.0 x 10-10
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
114
Solubility and pH
The solubility of any substance whose anion
is basic will be affected to some extent by
the pH of the solution.
Anions to be concerned about:
OHCO32PO43CNS2SO42Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
115
Sample exercise: Write the net ionic equation
for the reaction of the following copper (II)
compounds with acid:
a) CuS
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
116
Sample exercise: Write the net ionic equation
for the reaction of the following copper (II)
compounds with acid:
a) CuS
CuS + H+  Cu2+ + HS-
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
117
Sample exercise: Write the net ionic equation
for the reaction of the following copper (II)
compounds with acid:
b) Cu(N3)2
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
118
Sample exercise: Write the net ionic equation
for the reaction of the following copper (II)
compounds with acid:
b) Cu(N3)2
Cu(N3)2 + H+  Cu2+ + 2HN3
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
119
Formation of Complex Ions: a characteristic
property of metal ions is their ability to
accept an electron pair from water
molecules.
Other molecules than water can also donate
their electron pair to the metal ions to form
complex ions.
The stability of the complex ion depends upon
its size of the Keq for its formation
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
120
Sample exercise: Calculate [Cr3+] in
equilibrium with Cr(OH)4- when 0.010 mol
of Cr(NO3)3 is dissolved in a liter of solution
buffered at pH 10.0.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
121
Sample exercise: Calculate [Cr3+] in
equilibrium with Cr(OH)4- when 0.010 mol
of Cr(NO3)3 is dissolved in a liter of solution
buffered at pH 10.0.
Kf is 8 x 1029, with a value this large we can
assume that all the Cr(OH)4- dissolves to
form Cr3+ ions
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
122
Sample exercise: Calculate [Cr3+] in
equilibrium with Cr(OH)4- when 0.010 mol
of Cr(NO3)3 is dissolved in a liter of solution
buffered at pH 10.0.
Cr3+ + 4OH-  Cr(OH)4x
1x10-4
0.010
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
123
Sample exercise: Calculate [Cr3+] in
equilibrium with Cr(OH)4- when 0.010 mol
of Cr(NO3)3 is dissolved in a liter of solution
buffered at pH 10.0.
Cr3+ + 4OH-  Cr(OH)4Kf = 8 x 1029 = [Cr(OH)4-]
[Cr3+][OH-]4
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
124
Sample exercise: Calculate [Cr3+] in
equilibrium with Cr(OH)4- when 0.010 mol
of Cr(NO3)3 is dissolved in a liter of solution
buffered at pH 10.0.
Cr3+ + 4OH-  Cr(OH)4Kf = 8 x 1029 = [0.010]
[x][1x10-4]4
x = 1.25 x 10-16
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Factors that Affect
Solubility Equilibia
125
Amphoterism: Many metal hydroxides and
oxides that are insoluble in water will
dissolve in strong acids and bases. They will
do this because they themselves are capable
of behaving as an acid or a base.
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
126
Equilibrium can be achieved starting with the
substances on either side of a chemical
equation.
The use of the reaction quotient, Keq, to
determine the direction in which a reaction
must proceed is important
If Keq > Ksp, precipitation will occur
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
127
Equilibrium can be achieved starting with the
substances on either side of a chemical
equation.
The use of the reaction quotient, Keq, to
determine the direction in which a reaction
must proceed is important
If Keq = Ksp, equilibrium exists
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
128
Equilibrium can be achieved starting with the
substances on either side of a chemical
equation.
The use of the reaction quotient, Keq, to
determine the direction in which a reaction
must proceed is important
If Keq < Ksp, solid dissolves until Keq = Ksp
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
129
Sample exercise: Will a precipitate form
when 0.050 L of 2.0 x 10-2 M NaF is mixed
with 0.010 L of 1.0 x 10-2 M Ca(NO3)2?
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
130
Sample exercise: Will a precipitate form
when 0.050 L of 2.0 x 10-2 M NaF is mixed
with 0.010 L of 1.0 x 10-2 M Ca(NO3)2?
Possible reaction:
2NaF + Ca(NO3)2  CaF2 + 2NaNO3
final volume will be 0.060 L
sodium salts are very soluble, CaF2 has a Ksp
of 3.9 x 10-11
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
131
Sample exercise: Will a precipitate form
when 0.050 L of 2.0 x 10-2 M NaF is mixed
with 0.010 L of 1.0 x 10-2 M Ca(NO3)2?
Molarity of Ca+2:(0.010)(1.0 x 10-2) = 1.7 x 103
(0.060)
Molarity of F-:(0.050)(2.0 x 10-2) = 1.7 x 10-2
(0.060)
Keq = (1.7 x 10-3)(1.7 x 10-2) = 2.78 x 10-5
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Precipitation and
Separation of Ions
132
Sample exercise: Will a precipitate form
when 0.050 L of 2.0 x 10-2 M NaF is mixed
with 0.010 L of 1.0 x 10-2 M Ca(NO3)2?
Keq = (1.7 x 10-3)(1.7 x 10-2) = 2.78 x 10-5
Keq>Ksp so precipitation will
occur
Copyright T. L. Heise 2001 - 2002
CHE 116 Prof. T.L. Heise
Qualitative Analysis for
Metallic Elements
133
How can solubility equilibria and complex ion
formation be used to detect the presence of
particular metal ions in solution?
Fig. 17.21
Copyright T. L. Heise 2001 - 2002