Chapter 13
Applications of Aqueous Equilibria
13.1 Solutions of Acids or Bases Containing a
Common Ion
13.2 Buffered Solutions
13.3 Exact Treatment of Buffered Solutions (skip)
13.4 Buffer Capacity
13.5 Titrations and pH Curves
13.6 Acid-Base Indicators
13.7 Titration of Polyprotic Acids (skip)
13.8 Solubility Equilibria and the Solubility Product
13.9 Precipitation and Qualitative Analysis (skip)
13.10 Complex Ion Equilibria (skip)
4/8/2015
Zumdahl Chapter 8
1
The Common Ion Effect (1)
This applies to weak acids, weak bases and solubility of
salts when a common ion is added to the equilibrium
reaction. You can change the pH, but adding salt.
HF (aq)
H+ (aq) + F- (aq)
NaF (s) → Na+ (aq) + F– (aq)

Ka = 7.2 x 10-4
[H3O  ][F ]
Ka 
[HF]
When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s
principle). The pH increases. See example 13.1 in the text to see how
to quantitatively determine this effect, with an ICE box)
The Common Ion Effect (2)
If a solution and a salt to be dissolved in it have an
ion in common, then the solubility of the salt is
depressed relative to pure.
General equation
AB (s)
A+ (aq) + B- (aq)
If you add BH(aq), which dissociates into B- and H+, then the [A+]
decreases, and AB is driven out of solution.
Applications of Aqueous Equilibrium
Buffered Solutions
The Common Ion Effect on Buffering
Buffers: resist change in pH when acid or
base is added.
Buffer Solutions: contain a common ion
and are important in biochemical and
physiological processes
Organisms (and humans) have built-in
buffers to protect them against changes
in pH.
Blood: (pH 7.4)
Human blood is a
Death = 7.0 <pH > 7.8 = Death
buffered solution
Human blood is maintained by a combination of CO3-2, PO4-3 and protein buffers.
Unbuffered solution
A solution that is buffered
by acetic acid/acetate
How Do Buffers Work?
HA
H+ + A –
HA = generic acid
• Ka = [H+][A-]/[HA] =>
[H+] = Ka[HA]/[A-]
•If Ka is small (weak acid) then [H+] does not change
much when [HA] and [A-] change.
If [HA] and [A-] are large, and [HA]/[A-] ≈ 1,then small
additions of acid ([H+]) or base ([OH-]) don’t change the
ratio much.
Example: A solution of 0.5 M acetic acid plus 0.5 M acetate
Ka = 1.8x10-5
pKa = 4.7
HA
H+ + AKa = [H+][A-]/[HA]
pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH
[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = 4.74
i.e., pH=pKa
Example: A solution of 0.5 M acetic acid plus 0.5 M acetate
Ka = 1.8x10-5
pKa = 4.74
HA
H+ + AKa = [H+][A-]/[HA]
pH = pKa + log[A-]/[HA]
Use an ICE box to calculate the pH
[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0
=> => pH = 4.74
i.e., pH=pKa
Now add NaOH to 0.01 M
(in an unbuffered solution this would give a pOH of 2 and pH of 12)
Use HA + OH- → H2O + ARedo the ICE Box
[AH]ini = 0.49, [A-]ini = 0.51
=> => pH = 4.76
Buffer Calculation: Add Acid (#1) to a Buffered Solution
Acetic Acid: Ka = 1.8x10-5
pKa = 4.74
HA
H+ + AKa = [H+][A-]/[HA]
Case 1)
=>
-pKa = -pH + log[A-]/[HA]
pH = pKa + log[A-]/[HA]
=>
[CH3COOH]tot = [CH3COOH] + [CH3COO-] = 1.0M
pH = pKa => [CH3COOH] = [CH3COO-] = 0.5
Now add 0.01 M HCl (strong acid)
[CH3COOH] = 0.51
[CH3COO-] = 0.49
pH = pKa + log[A-]/[HA] = 4.74 + log(0.49/0.51)
= 4.74 – 0.02 = 4.72
ΔpH = -0.02
Buffer Calculation: Add Acid (#2) to a dilute Buffered Solution
Acetic Acid: Ka = 1.8x10-5
pKa = 4.74
HA
H+ + AKa = [H+][A-]/[HA]
Case 2)
=>
-pKa = -pH + log[A-]/[HA]
pH = pKa + log[A-]/[HA]
=>
[CH3COOH]tot = [CH3COOH] + [CH3COO-] = 0.10 M
pH = pKa => [CH3COOH] = [CH3COO-] = 0.05
Now add 0.01 M HCl (strong acid)
[CH3COOH] = 0.06
[CH3COO-] = 0.04
pH = pKa + log[A-]/[HA] = 4.74 + log(0.04/0.06)
= 4.74 – 1.5 = 3.54
ΔpH = -1.5
Adding Base to a Buffered Solution:
OH- ions do not accumulate but are replaced
by A- ions.
OH– + HA ⇌ A– + H2O
When the OH- is added, the concentrations of HA and Achange, but only by small amounts. Under these conditions the
[HA]/[A-] ratio and thus the [H+] stay virtually constant.
OH– + HA ⇌ A– + H2O
When protons are added
to a buffered solution,
the conjugate base (A–)
reacts
H+ + A– ⇌ HA
Characteristics of Buffered solutions
• Contain relatively large
concentrations of a weak
acid and its conjugate
base.
• When acid is added, it
reacts with the conjugate
base
• When base is added, it
reacts with the acid.
• pH is determined by the
ratio of the base and acid.
 [A  ] 

pH  pK a  log
 [HA] 
Buffer Capacity
• Buffer capacity is the amount of protons or
hydroxide ions that can be absorbed without a
significant change in pH.
• pH is determined by the ratio of [A–]/[HA] and
pKa
• Capacity is determined by the magnitudes of
[HA] and [A–].
[A ] 
pH  pKa  log

[HA]
Equivalence Point
Buffer Summary
Buffer Design: Add known amount of HA (weak acid) and
salt of HA (its conjugate base, A─)
[H+] or pH depends on Ka and the ratio of acid to salt or
[A─].
Thus if both conc. HA and A- are large then small
additions of acid or base don’t change the ratio much
4/8/2015
Zumdahl Chapter 8
16
How to Actually Make a Buffer (a buffered solution in a lab)
HA
H+ + A –
• Assume you want to make a 1L of of 1M buffer at pH 4.74. Use acetic acid,
because the pKa of acetic acid is 4.74.
•To make the Buffer:
• The easy way (this is how we actually do it in the lab): Weigh out the
appropriate amount of Na+Acetate- (70 g). Add it to a volumetric flask. Add
water to around 0.9 Liters. Add a stir bar and stir. Slowly add HCl, while
monitoring the pH with a pH meter. When the pH reaches the pKa (4.74),
stop. Take out the stir bar and add enough water to give 1L.
• The hard way. Start with two solutions. Solution A is 1M acetic acid. Solution
B is 1M Na+Acetate-. Add solution A to solution B. Then check the pH with a
pH meter. Adjust the pH with HCl or NaOH as required, to get the pH to 4.74
But this will change the concentration of A- and HA.
Acid-Base Titrations
Titration: A controlled addition of measured volumes of a
solution of known concentration (the Titrant) from a buret to a
second solution of unknown concentration under conditions in
which the solutes react cleanly (without side reactions),
completely, and rapidly.
A titration is complete when the second solute is fully
consumed
Completion is signaled by a change in some physical property,
such as the color of the reacting mixture or the color of an
indicator that has been added to it.
[NaOH]
Indicator
phenolphthalein
“X”
Zumdahl Chapter 8
Titrations and pH Curves
4/8/2015
Zumdahl Chapter 8
20
Titration of a weak acid with a
strong base.
The base reacts with the acid.
.
OH– + HA ⇌ A– + H2O
Strong acid
Zumdahl Chapter 8
21
Equivalence
point
determined
defined by the
Stoichiometry,
not by the pKa.
nbase = nacid
number of moles of acid = number of moles of base
4/8/2015
Zumdahl Chapter 8
22
For a variety of weak
acids:
The Equivalence point
occurs at the same
stoichiometric amount
of base added
The weaker the acid,
the greater the pH
value for the
equivalent point.
Similar for titration of weak bases with strong acids
4/8/2015
Zumdahl Chapter 8
24
Acid-Base Indicators
The indicator phenolphthalein is pink
in basic solution and colorless in acidic solution.
Indicators (Weak Acid Equilibria)
pH = -log10[H3O+]
4/8/2015
Zumdahl Chapter 8
26
Indicators
A soluble compound, generally an organic dye, that
changes its color noticeably over a fairly short
range of pH.
Typically, Indicators are a weak organic acid
that has a different color than its conjugate base.
HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)
Acid: HIn (aq)
Phenolphtalein
4/8/2015
Indicator
denoted by In
Conjugate base: In– (aq)
Zumdahl Chapter 8
27
Acid: HIn (aq)
Conjugate base: In- (aq)
HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)
[H3O+][OH-] = Kw
pH = -log10[H3O+]
28
Methyl Red
Bromothymol blue
Phenolphtalein
4/8/2015
Zumdahl Chapter 8
HIn(aq) + H O(l) ↔ H O+(aq) + In-(aq)
29
The useful pH ranges for several common
indicators
4/8/2015
+(aq) + In–(aq)
Zumdahl
8
HIn(aq) + H2O(l)
↔Chapter
H3O
30
Indicator Selection
• Want indicator color change and titration equivalence point
to be as close as possible
• Easier with a large pH change at the equivalence point
Strong acid
Weak acid
Could use either indicator
4/8/2015
Methyl red changes color to
early
Zumdahl Chapter 8
31
Solubility Product Ksp
Describes a chemical equilibrium in which an
excess solid salt is in equilibrium with a
saturated aqueous solution of its separated ions.
General equation
AB (s) ↔ A+ (aq) + B- (aq)
The solubility
expression controls
the amount of solid
that will dissolve
Ksp =
Ksp =
4/8/2015
Zumdahl Chapter 8
32
Ksp Values at 25°C for Common Ionic Solids
4/8/2015
Zumdahl Chapter 8
33
The Solubility of Ionic Solids
The Solubility Product
AgCl(s) ↔Ag+(aq) + Cl-(aq)
excess
Ksp =
The solid AgCl, which is in excess, is understood
to have a concentration of 1 mole per liter.
Ksp = 1.6  10-10 at 25oC
4/8/2015
Zumdahl Chapter 8
34
The Solubility of Ionic Solids
The Solubility Product
Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)
excess
Ksp =
Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)
excess
Ksp =
4/8/2015
Zumdahl Chapter 8
35
Solubility and Ksp
Determine the mass of lead(II) iodate dissolved in 2.50 L
of a saturated aqueous solution of Pb(IO3)2 at 25oC. The
Ksp of Pb(IO3)2 is 2.6  10-13.
Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)
Let “y” = molar solubility in mol/L
4/8/2015
Zumdahl Chapter 8
36
Determine the mass of lead(II) iodate dissolved in 2.50 L of a
saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of
Pb(IO3)2 is 2.6  10-13.
Pb(IO3)2(s) ↔
[Pb2+][IO3-]2 = Ksp
Pb2+(aq) + 2 IO3-(aq)
“y” = molar solubility
[Pb2+][IO3-]2 =
y = 4.0  10-5
Gram solubility of
 [Pb(IO3)2] = [Pb2+] = y = 4.0  10-5 mol L-1
 [IO3-] = 2y = 8.0  10-5 mol L-1
Lead (II) iodate
Molar Mass of lead (II) iodate
4/8/2015
Pb(IO
3)2 = 557g per mole
= (4.0  10-5 mol L-1)  (557 g mol-1)
= 0.0223 g L-1
Zumdahl Chapter 8
 2.50 L
37
The Solubility of Salts
Exercise 9-3
Solubility and Ksp
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its
mass solubility is 8.3 g L-1.
1 Ag2SO4 (s) ↔
4/8/2015
2 Ag+(aq) +
Zumdahl Chapter 8
1 SO42-(aq)
38
Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its
mass solubility is 8.3 g L-1.
1 Ag2SO4 (s) ↔
2 Ag+(aq) +
1 SO42-(aq)
[y] = (8.3 g Ag2SO4 L-1)  (1 mol Ag2SO4/311.8 g)
[y] = 2.66  10-2 mol Ag2SO4 L-1
[Ag+]2[SO42-] = Ksp
Ksp =
4/8/2015
Zumdahl Chapter 8
39
The Nature of Solubility Equilibria
Dissolution and precipitation are reverse of each other.
Dissolution (Solubility)
General reaction
X3Y2 (s)
↔ 3X+2 (aq) +
2Y-3 (aq)
[s]
Ksp =
s = molar solubility
expressed in moles per liter
4/8/2015
Zumdahl Chapter 8
40
Relative Solubilities
A salt’s Ksp value gives us information about its solubility.
Salt ↔ Cation + Anion
If the salts being compared produce the same number of ions, eg.,
AgI, CuI, CaSO4
s = [cation]
s = [anion]
Ksp = [cation] [anion] = s2
Salt molar solubility = s = (Ksp)1/2
Salt
Ksp
Solubility
AgI
CuI
CaSO4
1.5 x 10-16
5.0 x 10-12
6.1 x 10-5
1.2 x 10-8
2.2 x 10-6
7.8 x 10-3
Solubility CaSO4 > CuI > AgI
4/8/2015
Zumdahl Chapter 8
41
The Effects of pH on Solubility
Solubility of Hydroxides
Many solids dissolve more readily in more acidic
solutions
Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)
[Zn2+][OH-]2 = Ksp = 4.5  10-17
If pH decreases (or made more acidic), the [OH-] decreases. In
order to maintain Ksp the [Zn2+] must increase and
consequently more solid Zn(OH)2 dissolves.
Make more acidic:
2+
-]2
2+]][OH
22 = K
2+
[Zn
[Zn
[OH
]
[Zn ][OH ] = Kspspsp
--
4/8/2015
Zumdahl Chapter 8
Zinc hydroxide is more
soluble in acidic
solution than in pure
water.
42
The Effects of pH on Solubility
Estimate the molar solubility of Fe(OH)3 in a solution that is
buffered to a pH of 2.9. Lookup Ksp = 1.1x10-36
Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq)
(pH = 2.9 and) pOH = 11.1
[OH-] = 7.9  10-12 mol L-1
[Fe3+][OH-]3 = Ksp
[Fe3+] = Ksp/[OH-]3 = 1.1  10-36 / (7.9  10-12)3
[Fe3+] = [Fe(OH)3] = 2.2  10-3 mol L-1 answer
[OH-] = 3y
In pure water: [Fe3+] = y
y(3y)3 = 27y4 = Ksp = 1.1  10-36
y = 4.5  10-10 mol L-1 = [Fe3+] = [Fe(OH)3]=
[OH-] = 3y = 1.3  10-9 mol L-1
pOH = 8.87
(and pH = 5.13)
4/8/2015
In pure
water, Fe(OH)3 is 5 xZumdahl
10 6Chapter
less 8soluble than at pH = 2.943
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at
25oC. Determine the molar solubility of TlIO3 in
0.050 mol L-1 KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) +
IO3-(aq)
[Tl+] (mol L-1) [IO3-] (mol L-1)
Initial concentration
Change in concentration
Equilibrium concentration
[Tl+][IO3-] = Ksp
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
4/8/2015
Zumdahl Chapter 8
44
The Common Ion Effect
The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at 25oC.
Determine the molar solubility of TlIO3 in 0.050 mol L-1
KIO3 at 25oC.
TlIO3(s) ↔ Tl+(aq) +
[s]
[s]
IO3-(aq)
[s]
With common ion (from previous calculation)
s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility
What if no common ion is added? i.e., dissolve thallium iodate in pure
water
[Tl+][IO3-] = Ksp
s= 1.76x10-3 mol L-1 = molar solubility
With common ion, s = [TlIO3]= 6.2 × 10-5 mol L-1
4/8/2015
Zumdahl Chapter 8
45
Chapter 8
Applications of Aqueous Equilibria
8.1 Solutions of Acids or Bases Containing a
Common Ion
8.2 Buffered Solutions
8.3 Exact Treatment of Buffered Solutions (skip)
8.4 Buffer Capacity
8.5 Titrations and pH Curves
8.6 Acid-Base Indicators
8.7 Titration of Polyprotic Acids (skip)
8.8 Solubility Equilibria and the Solubility Product
8.9 Precipitation and Qualitative Analysis (skip)
8.10 Complex Ion Equilibria (skip)
4/8/2015
Zumdahl Chapter 8
46