Chemistry
Third Edition
Julia Burdge
Lecture PowerPoints
Chapter 17
Acid-Base Equilibria and
Solubility Equilibria
Copyright © 2012, The McGraw-Hill Compaies, Inc. Permission required for reproduction or display.
CHAPTER
17.1
17.2
17.3
17.4
17.5
17.6
17
Acid-Base Equilibria and
Solubility Equilibria
The Common Ion Effect
Buffer Solutions
Acid-Base Titrations
Solubility Equilibria
Factors Affecting Solubility
Separation of Ions Using Differences in Solubility
2
17.1
The Common Ion Effect
Topics
The Common Ion Effect
3
17.1
The Common Ion Effect
The Common Ion Effect
Let’s examine how the properties of a solution change when a
second solute is introduced.
Consider a liter of solution containing 0.10 mole of acetic
acid.
[H3O+] = [CH3COO‒] = 1.34 × 10‒3 M and pH = 2.87
4
17.1
The Common Ion Effect
The Common Ion Effect
Now consider what happens when we add 0.050 mole of
sodium acetate (CH3COONa) to the solution.
5
17.1
The Common Ion Effect
The Common Ion Effect
In general, when a compound containing an ion in common
with a dissolved substance is added to a solution at
equilibrium, the ionization equilibrium shifts to the left.
This phenomenon is known as the common ion effect.
The common ion can also be H3O+ or OH–.
6
SAMPLE PROBLEM
17.1
Determine the pH at 25°C of a solution prepared by adding
0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid.
(Assume that the addition of sodium acetate does not change
the volume of the solution.)
7
SAMPLE PROBLEM
17.1
Setup
We use the stated concentration of acetic acid, 0.10 M, and
[H3O+]  0 M as the initial concentrations in the ice table.
Solution
8
SAMPLE PROBLEM
17.1
x = 3.6 × 10‒5 M.
[H3O+] = x, so pH = ‒log (3.6 × 10‒5) = 4.44
The common ion effect has shifted the ionization of acetic
acid to the left.
9
17.2
Buffer Solutions
Topics
Calculating the pH of a Buffer
Preparing a Buffer Solution with as Specific pH
10
17.2
Buffer Solutions
Calculating the pH of a Buffer
A solution that contains a weak acid and its conjugate base
(or a weak base and its conjugate acid) is a buffer solution or
simply a buffer.
A buffer’s ability to convert a strong acid to a weak acid
minimizes the effect of the addition on the pH :
11
17.2
Buffer Solutions
Calculating the pH of a Buffer
A buffer’s ability to convert a strong base to a weak base
minimizes the effect of the addition on the pH :
12
17.2
Buffer Solutions
Calculating the pH of a Buffer
Suppose that we have 1 L of the acetic acid-sodium acetate
solution:
13
17.2
Buffer Solutions
Calculating the pH of a Buffer
14
17.1
The Common Ion Effect
Calculating the pH of a Buffer
Consider what happens when we add 0.10 mole of HCl to the
buffer.
(We assume that the addition of HCl causes no change in the
volume of the solution.)
15
17.1
The Common Ion Effect
Calculating the pH of a Buffer
A change of only 0.08 pH units.
16
17.1
The Common Ion Effect
Calculating the pH of a Buffer
Had we added 0.10 mol of HCl to 1 L of pure water, the pH
would have gone from 7.00 to 1.00!
17
17.2
Buffer Solutions
Calculating the pH of a Buffer
In the determination of the pH of a buffer, we always
neglect the small amount of weak acid that ionizes (x)
because ionization is suppressed by the presence of a
common ion.
Similarly, we ignore the hydrolysis of the acetate ion
because of the presence of acetic acid.
This enables us to derive an expression for determining the
pH of a buffer.
18
17.2
Buffer Solutions
Calculating the pH of a Buffer
19
17.2
Buffer Solutions
Calculating the pH of a Buffer
Henderson-Hasselbalch equation
20
SAMPLE PROBLEM
17.2
Starting with 1.00 L of a buffer that is 1.00 M in acetic acid
and 1.00 M in sodium acetate, calculate the pH after the
addition of 0.100 mole of NaOH. (Assume that the addition
does not change the volume of the solution.)
Strategy
Added base will react with the acetic acid component of the
buffer, converting OH‒ to CH3COO‒ :
CH3COOH(aq) + OH‒(aq) → H2O(l) + CH3COO‒(aq)
21
SAMPLE PROBLEM
17.2
Setup
Solution
Thus, the pH of the buffer after addition of 0.10 mol of NaOH
is 4.83.
22
17.2
Buffer Solutions
Preparing a Buffer Solution with as Specific pH
A solution is only a buffer if it has the capacity to resist pH
change when either an acid or a base is added.
If the concentrations of a weak acid and conjugate base differ
by more than a factor of 10, the solution does not have this
capacity.
23
17.2
Buffer Solutions
Preparing a Buffer Solution with as Specific pH
Therefore, we consider a solution a buffer, and can use
to calculate its pH, only if the following condition is met:
Consequently, the log term can only have values from –1 to 1,
and the pH of a buffer cannot be more than one pH unit
different from the pKa of the weak acid it contains.
24
17.2
Buffer Solutions
Preparing a Buffer Solution with as Specific pH
This is known as the range of the buffer, where pH = pKa 1.
This enables us to select the appropriate conjugate pair to
prepare a buffer with a specific, desired pH.
25
17.2
Buffer Solutions
Calculating the pH of a Buffer
To prepare a buffer with a specific, desired pH:
1. Choose a weak acid whose pKa is close to the desired pH.
26
17.2
Buffer Solutions
Calculating the pH of a Buffer
2. Substitute the pH and pKa values into the following
equation to obtain the necessary ratio of [conjugate
base]/[weak acid].
27
SAMPLE PROBLEM
17.3
Select an appropriate weak acid from the table, and describe
how you would prepare a buffer with a pH of 9.50.
28
SAMPLE PROBLEM
17.3
Setup
Two of the acids listed in the table have pKa values in the
desired range: hydrocyanic acid (HCN, pKa = 9.31) and phenol
(C6H5OH, pKa = 9.89).
29
SAMPLE PROBLEM
17.3
Solution
One way to achieve this would be to dissolve 0.41 mol of
C6H5ONa and 1.00 mol of C6H5OH in 1 L of water.
30
17.3
Acid‒Base Titrations
Topics
Strong Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
Strong Acid-Weak base Titrations
Acid-Base Indicators
31
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
The titrant is the solution that is added from the burette.
32
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Titration curve (pH as a function of volume titrant added)
of a strong acid-strong base titration.
33
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Consider the addition of a 0.100 M NaOH solution (from a
burette) to a vessel containing 25.0 mL of 0.100 M HCl.
34
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Calculate the pH of the solution at every stage of titration.
Prior to the addition of base:
Simply calculate the pH of a strong acid: ‒log [H3O+] .
Here, the pH of the acid is given by ‒log (0.100), or 1.00
35
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Before the equivalence point:
Consider the addition of 10.0 mL
of 0.100 M NaOH to 25.0 mL of
0.100 M HCl.
Note: Using millimoles instead of moles simplifies the
calculations. Remember that millimoles = M × mL
36
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Before the equivalence point:
The total volume of the solution is 35.0 mL. The number of
millimoles of NaOH in 10.0 mL is
37
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
The number of millimoles of HCl originally present is
The amount of HCl left after partial neutralization is
2.50 ‒ 1.00, or 1.50 mmol.
38
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
Next, determine [H3O+].
We have 1.50 mmol in 35.0 mL:
Thus [H3O+] = 0.0429 M, and the pH of the solution is
pH = ‒log (0.04289) = 1.37
39
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
At the equivalence point:
Consider the addition of 25.0 mL
of 0.100 M NaOH to 25.0 mL of
0.100 M HCl.
This is a complete neutralization reaction.
[H3O+] = [OH‒] = 1.00 × 10‒7 M and the pH of the solution is
7.000.
40
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
After the equivalence point:
Consider the addition of 35.0 mL of
0.100 M NaOH to 25.0 mL of
0.100 M HCl.
The total volume of the solution is now 60.0 mL. The number
of millimoles of NaOH added is
41
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
There are 2.50 mmol of HCl in 25.0 mL of solution.
2.50 mmol of NaOH have been consumed, and there are
3.5 ‒ 2.5 or 1.00 mmol of NaOH remaining.
The [NaOH] in 60.0 mL of solution is
42
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
[OH‒] = 0.0167 M and pOH = ‒log (0.0167) = 1.78.
The pH of the solution is 14.00 ‒ 1.78 or 12.22.
43
17.3
Acid‒Base Titrations
Strong Acid-Strong Base Titrations
44
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Titration curve of a
weak acid-strong base titration:
Strong-acid/strong-base
curve, for comparison)
45
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Consider the neutralization reaction between acetic acid (a
weak acid) and sodium hydroxide (a strong base):
46
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Prior to the addition of base:
The pH is determined by the
ionization of acetic acid.
Use concentration (0.10 M) and Ka (1.8 × 10‒5) to calculate the
[H3O+]:
47
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
48
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Solving for x,
gives [H3O+] = 1.34 × 10‒3 M and pH = 2.87
49
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Before the equivalence point:
After the first addition of base, some of the acetic acid has
been converted to acetate ion via the reaction
50
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
The solution is a buffer and we
use the Henderson-Hasselbalch equation to calculate the pH.
After the addition of 10.0 mL of base, the solution contains
1.5 mmol of acetic acid and 1.0 mmol of acetate ion
51
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
At the equivalence point:
All the acetic acid has been neutralized and we are left with
acetate ion in solution.
pH is determined by the concentration and the Kb of acetate
ion.
52
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
The equivalence point occurs when 25.0 mL of base has been
added, making the total volume 50.0 mL.
The 2.5 mmol of acetic acid has all been converted to acetate
ion.
53
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
The Kb for acetate ion is 5.6 × 10‒10.
54
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
Solving for x,
gives [OH‒] = 5.3 × 10‒6 M, pOH = 5.28, and pH = 8.72
55
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
After the equivalence point:
The curve for titration of a weak acid with a strong base is
identical to the curve for titration of a strong acid with a
strong base.
Because all the acetic acid has been consumed, there is
nothing in solution to consume the additional added OH‒, and
the pH levels off between 12 and 13.
56
17.3
Acid‒Base Titrations
Weak Acid-Strong Base Titrations
57
SAMPLE PROBLEM
17.4
Calculate the pH in the titration of 50.0 mL of 0.120 M acetic
acid by 0.240 M sodium hydroxide after the addition of (a)
10.0 mL of base, (b) 25.0 mL of base, and (c) 35.0 mL of base.
Strategy
The reaction is
58
SAMPLE PROBLEM
17.4
Setup
a) The solution originally contains (0.120 mmol/mL)(50.0 mL)
= 6.00 mmol of acetic acid.
59
SAMPLE PROBLEM
17.4
Solution
a) Prior to the equivalence point, the solution is a buffer. We
use the Henderson-Hasselbalch equation.
60
SAMPLE PROBLEM
17.4
b) After the addition of 25.0 mL of base, the titration is at
the equivalence point. Calculate the pH using the
concentration and the Kb of acetate ion.
61
SAMPLE PROBLEM
17.4
The total volume 50.0 mL + 25.0 mL = 75.0 mL. The
concentration of acetate ion is
62
SAMPLE PROBLEM
17.4
x = [OH‒], so [OH‒] = 6.7 × 10‒6 M
pOH = –log (6.7 × 10‒6) = 5.17 and pH = 14.00 ‒ 5.17 = 8.83.
63
SAMPLE PROBLEM
17.4
A 35.0 mL amount of the base contains (0.240 mol/mL)(35.0
mL) = 8.40 mmol of OH‒.
c) After the addition of 35.0 mL of base, the titration is past
the equivalence point and we solve for pH by determining
the concentration of excess hydroxide ion.
8.40 ‒ 6.00 = 2.40 mmol of excess OH‒
64
SAMPLE PROBLEM
17.4
The total volume is 50.0 + 35.0 = 85.0 mL.
[OH‒] = 2.40 mmol/85.0 mL = 0.0280 M
pOH = ‒log (0.0280) = 1.553
pH = 14.000 ‒ 1.553 = 12.447
65
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
66
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
Consider the titration of HCl, a strong acid, with NH3, a weak
base:
or
The pH at the equivalence point is less than 7 because the
ammonium ion acts as a weak Brønsted acid:
67
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
Before the equivalence point:
The pH is determined by the concentration and the Kb of
ammonia.
Consider the titration of 25.0 mL of 0.10 M NH3 with 0.10 M
HCl.
68
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
69
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
At the equivalence point:
The pH is calculated using the concentration and Ka of the
conjugate base of NH3, the NH4+ ion, and an equilibrium table.
70
17.3
Acid‒Base Titrations
Strong Acid-Weak base Titrations
After the equivalence point:
The curve for titration of a weak base with a strong acid is
identical to the curve for titration of a strong base with a
strong acid.
Because all the ammonia has been consumed, there is
nothing in solution to consume the additional added H3O+,
and the pH levels off between 1 and 2.
71
SAMPLE PROBLEM
17.5
Calculate the pH at the equivalence point when 25.0 mL of
0.100 M NH3 is titrated with 0.100 M HCl.
Strategy
The reaction between NH3 and HCl is
Setup
The solution originally contains (0.100 mmol/mL)(25.0 mL) =
2.50 mmol NH4+ . At the equivalence point, 2.50 mmol of HCl
has been added.
72
SAMPLE PROBLEM
17.5
The volume of 0.100 M HCl that contains 2.50 mmol is
It takes 25.0 mL of titrant to reach the equivalence point, so
the total solution volume is 25.0 + 25.0 = 50.0 mL.
At the equivalence point, all the NH3 originally present has
been converted to NH4+.
73
SAMPLE PROBLEM
17.5
The [NH4+] is (2.50 mmol)/(50.0 mL) = 0.0500 M.
Solution
74
SAMPLE PROBLEM
17.5
The equilibrium expression is
[H3O+] = x = 5.3 × 10‒6 M
At equilibrium, pH = ‒log (5.3 × 10‒6 ) = 5.28
75
17.3
Acid‒Base Titrations
Acid-Base Indicators
The equivalence point in a titration can be determined using
an acid-base indicator.
An acid-base indicator is usually a weak organic acid or base
for which the ionized and un-ionized forms are different
colors.
76
17.3
Acid‒Base Titrations
Acid-Base Indicators
HIn and its conjugate base, In‒, must have distinctly different
colors.
Acidic
Basic
Acidic solutions are the color of HIn.
Basic solutions are the color of In‒.
77
17.3
Acid‒Base Titrations
Acid-Base Indicators
The endpoint of a titration is the point at which the color of
the indicator changes.
There is a range of pH over which the color change occurs.
We select an indicator whose color change occurs over a pH
range that coincides with the steepest part of the titration
curve.
78
17.3
Acid‒Base Titrations
Acid-Base Indicators
The titration
curves for
hydrochloric acid
and acetic acid—
each being
titrated with
sodium hydroxide
79
17.3
Acid‒Base Titrations
Acid-Base Indicators
Phenolphthalein or methyl red are suitable indicators for the
titration of a strong acid with a strong base because both
endpoints coincide with the steepest part of the HCl-NaOH
titration curve.
Methyl red is not a suitable indicator for use in the titration of
acetic acid with sodium hydroxide. Its endpoint occurs
significantly before the equivalence point.
80
17.3
Acid‒Base Titrations
Acid-Base Indicators
81
SAMPLE PROBLEM
17.6
Which indicator (or indicators) would you use for the
following acid-base titrations (a) strong acid-strong base(b)
weak acid-strong base, and (c) strong acid-weak base?
Setup
The steep part of the curve spans
a) a pH range of about 4 to 10.
b) a pH range of about 7 to 10.
c) a pH range of about 7 to 3.
82
SAMPLE PROBLEM
17.6
83
SAMPLE PROBLEM
17.6
Solution
a) Most of the indicators, with the exceptions of thymol
blue, bromophenol blue, and methyl orange, would work.
b) Cresol red and phenolphthalein are suitable indicators.
c) Bromophenol blue, methyl orange, methyl red, and
chlorophenol blue are all suitable indicators.
84
17.4
Solubility Equilibria
Topics
Solubility Product Expression and Ksp
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions
85
17.4
Solubility Equilibria
Solubility Product Expression and Ksp
The solubility of ionic compounds is important in industry,
medicine, and everyday life.
The compounds described as “insoluble” in Chapter 4 are
actually very slightly soluble.
The Ksp is called the solubility product constant. It allows for
quantitative predictions about how much of a given ionic
compound will dissolve in water.
86
17.4
Solubility Equilibria
Solubility Product Expression and Ksp
Ksp is equal to the concentrations of products over the
concentrations of reactants, each raised to its coefficient from
the balanced chemical equation.
The smaller the Ksp, the less soluble the compound.
87
17.4
Solubility Equilibria
Calculations Involving Ksp and Solubility
There are two ways to express the solubility of a substance:
1. The molar solubility: number of moles of solute in 1 L of a
saturated solution (mol/L).
2. The solubility: number of grams of solute in 1 L of a
saturated solution (g/L).
Both of these refer to concentrations at a particular
temperature (usually 25°C).
88
17.4
Solubility Equilibria
Calculations Involving Ksp and Solubility
To calculate molar solubility from Ksp :
The procedure is essentially identical to the procedure for
solving weak acid or weak base equilibrium problems
1. Construct an equilibrium table.
2. Fill in what we know.
3. Figure out what we don’t know.
89
17.4
Solubility Equilibria
Calculations Involving Ksp and Solubility
The Ksp of silver bromide (AgBr) is 7.7 × 10‒13.
Let s be the molar solubility (in mol/L) of AgBr.
At equilibrium: [Ag+] = [Br‒] = s
90
17.4
Solubility Equilibria
Calculations Involving Ksp and Solubility
The equilibrium expression is
Therefore,
91
17.4
Solubility Equilibria
Calculations Involving Ksp and Solubility
The molar solubility of AgBr is 8.8 × 10‒7 M.
Express this solubility in g/L by multiplying the molar
solubility by the molar mass of AgBr:
92
SAMPLE PROBLEM
17.7
Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in
g/L.
Setup
The equation for the dissociation of Cu(OH)2 is
Ksp = [Cu2+][OH‒]2
Ksp for Cu(OH)2 = 2.2 × 10‒20.
The molar mass of Cu(OH)2 is 97.57 g/mol.
93
SAMPLE PROBLEM
17.7
Solution
Therefore,
Remember to raise an entire term to the appropriate power!
94
SAMPLE PROBLEM
17.7
The molar solubility of Cu(OH)2 is 1.8 × 10‒7 M.
Multiplying by its molar mass gives:
95
SAMPLE PROBLEM
17.8
The solubility of calcium sulfate (CaSO4) is measured
experimentally and found to be 0.67 g/L. Calculate the value
of Ksp for calcium sulfate.
Setup
The molar mass of CaSO4 is 136.2 g/mol.
The molar solubility of CaSO4 is
96
SAMPLE PROBLEM
17.8
Solution
This is a relatively large Ksp value.
97
17.4
Solubility Equilibria
Predicting Precipitation Reactions
We use the reaction quotient (Q) to predict when a
precipitate will form.
Q has the same form as Ksp except the concentrations of ions
are not equilibrium concentrations.
98
17.4
Solubility Equilibria
Predicting Precipitation Reactions
If we mix a solution containing Ag+ ions with one containing
Cl‒ ions, we write:
where “i” denotes initial concentrations
If Q ≤ Ksp no precipitate will form
If Q > Ksp AgCl will precipitate
99
SAMPLE PROBLEM
17.9
Predict whether a precipitate will form when each of the
following is added to 650 mL of 0.0080 M K2SO4:
(a) 250 mL of 0.0040 M BaCl2
(b) 175 mL of 0.15 M AgNO3
(c) 325 mL of 0.25 M Sr(NO3)2 (Assume volumes are additive.)
Setup
The compounds that might precipitate and their Ksp values are
a) BaSO4, Ksp = 1.1 × 10‒10
b) Ag2SO4, Ksp = 1.5 × 10‒5
c) SrSO4, Ksp = 3.8 × 10‒7
100
SAMPLE PROBLEM
17.9
Solution
(a) Concentrations of the constituent ions of BaSO4 are:
Q = [Ba2+][SO42‒] = (0.0011)(0.0058) = 6.4 × 10‒6
Q > Ksp (1.1 × 10‒10)
BaSO4 will precipitate.
101
SAMPLE PROBLEM
17.9
(b) Concentrations of the constituent ions of Ag2SO4 are:
Q = [Ag+]2[SO42‒] = (0.032)2(0.0063) = 6.5 × 10‒6
Q < Ksp (1.5 × 10‒5)
Ag2SO4 will not precipitate.
102
SAMPLE PROBLEM
17.9
(c) Concentrations of the constituent ions of SrSO4 are:
Q = [Sr2+][SO42‒] = (0.083)(0.0053) = 4.4 × 10‒4
Q > Ksp (3.8 × 10‒7)
SrSO4 will precipitate.
103
17.5
Factors Affecting Solubility
Topics
The Common Ion Effect
pH
Complex Ion Formation
104
17.5
Factors Affecting Solubility
The Common Ion Effect
In a saturated solution of AgCl, Ksp = [Ag+][Cl‒]
In a solution in which AgCl is the only solute, [Ag+] = [Cl‒].
s = 1.3 × 10‒5 M
105
17.5
Factors Affecting Solubility
The Common Ion Effect
Determine the solubility of AgCl in a solution already
containing a solute that has an ion in common with AgCl.
For example, consider dissolving AgCl in a 0.10 M solution of
AgNO3.
In this case,
[Ag+] ≠ [Cl‒]
[Ag+] will be equal to 0.10 M plus the concentration
contributed by AgCl.
106
17.5
Factors Affecting Solubility
The Common Ion Effect
The equilibrium expression is
Because we expect s to be very small,
s = 1.6 × 10‒9 M
AgCl is significantly less soluble in 0.10 M AgNO3 than in pure
water—due to the common ion effect.
107
SAMPLE PROBLEM
17.10
Calculate the molar solubility of silver chloride in a solution
that is 6.5 × 10‒3 M in silver nitrate.
Setup
The dissolution equilibrium and the equilibrium expression
are
Solution
108
SAMPLE PROBLEM
17.10
Because we expect s to be very small,
The presence of AgNO3 reduces the solubility of AgCl by a
factor of ~ 500!
109
17.5
Factors Affecting Solubility
pH
The solubility of a substance can also depend on the pH of the
solution.
Insoluble bases tend to dissolve in acidic solutions while
insoluble acids tend to dissolve in basic solutions.
110
17.5
Factors Affecting Solubility
pH
At equilibrium
In a solution with a pH of less than 10.45, the solubility of
Mg(OH)2 would increase.
111
17.5
Factors Affecting Solubility
pH
The effect of additional H3O+ ions are summarized as follows:
If pH > 10.45, [OH‒] would be higher and the solubility of
Mg(OH)2 would decrease because of the common ion (OH‒)
effect.
112
17.5
Factors Affecting Solubility
pH
The pH also influences the solubility of salts that contain a
basic anion. For example,
In an acidic medium, the high [H3O+] will shift the following
equilibrium to the left, consuming F‒:
113
17.5
Factors Affecting Solubility
pH
As [F‒] decreases, [Ba2+]must increase to satisfy the equality
Ksp = [Ba2+][F‒]2. Thus, more BaF2 dissolves.
The solubilities of salts containing anions that do not
hydrolyze, such as Cl‒, Br‒, and NO3‒, are unaffected by pH.
114
SAMPLE PROBLEM
17.11
Which of the following compounds will be more soluble in
acidic solution than in water: (a) CuS, (b) AgCl, (c) PbSO4?
Setup
a. S2‒ is the conjugate base of the weak acid HS‒. S2‒ reacts
with H3O+
b. Cl‒ is the conjugate base of the strong acid HCl. Cl‒ does
not react with H3O+.
115
SAMPLE PROBLEM
17.11
c. SO42‒ is the conjugate base of the weak acid HSO4‒ . It
reacts with H3O+ as follows:
A salt that produces an anion that reacts with H3O+ will be
more soluble in acid than in water.
Solution
CuS and PbSO4 are more soluble in acid than in water. (AgCl is
no more or less soluble in acid than in water.)
116
17.5
Factors Affecting Solubility
Complex Ion Formation
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
Transition metals have a particular tendency to form complex
ions.
For example, a solution of cobalt(II) chloride (CoCl2) is pink
because of the presence of the Co(H2O)62+ ions.
When HCl is added, the solution turns blue because the
complex ion CoCl42‒ forms:
117
17.5
Factors Affecting Solubility
Complex Ion Formation
118
17.5
Factors Affecting Solubility
Complex Ion Formation
CuSO4 (Center). After the addition of a few drops of
concentrated NH3, a light-blue precipitate of Cu(OH)2 is
formed. (Right) When more NH3 is added, the Cu(OH)2
precipitate dissolves to form the dark-blue complex ion
Cu(NH3)42+.
119
17.5
Factors Affecting Solubility
Complex Ion Formation
A measure of the tendency of a metal ion to form a particular
complex ion is given by the formation constant (Kf) (also
called the stability constant), which is the equilibrium
constant for the complex ion formation.
The larger Kf is, the more stable the complex ion.
120
17.5
Factors Affecting Solubility
Complex Ion Formation
The formation of the Cu(NH3)42+ ion can be expressed as
The large value of Kf indicates that the complex ion is very
stable in solution.
121
17.5
Factors Affecting Solubility
Complex Ion Formation
122
17.5
Factors Affecting Solubility
Complex Ion Formation
The dissolution of silver chloride is represented by the
equation
The sum of this equation and the one representing the
formation of Ag(NH3)2+ is
123
17.5
Factors Affecting Solubility
Complex Ion Formation
The corresponding equilibrium constant is
(1.6 × 10‒10)(1.5 × 107) = 2.4 × 10‒3.
This is significantly larger than the Ksp value, indicating that
much more AgCl will dissolve in the presence of aqueous
ammonia than in pure water.
In general, the effect of complex ion formation generally is to
increase the solubility of a substance.
124
17.5
Factors Affecting Solubility
Complex Ion Formation
Amphoteric hydroxides, can react with both acids and bases.
Examples are Al(OH)3, Pb(OH)2, Cr(OH)3, Zn(OH)2, and
Cd(OH)2.
For example,
125
17.5
Factors Affecting Solubility
Complex Ion Formation
Solving an equilibrium problem involving complex ion
formation is complicated by the magnitude of Kf and
stoichiometry.
Consider the following reaction
We cannot neglect x, and have to raise [NH3]to the fourth
power. This equation is not easily solved.
Another approach is needed.
126
17.5
Factors Affecting Solubility
Complex Ion Formation
1. Assume that all the copper(II) ion is consumed to form the
complex ion (since Kf is so large.)
2. Consider the equilibrium in terms of the reverse reaction;
the dissociation of Cu(NH3)42+, for which the equilibrium
constant is the reciprocal of Kf.
127
17.5
Factors Affecting Solubility
Complex Ion Formation
Because this K is so small, we can expect x to be insignificant.
Note that [NH3], which had been 3.0 M, has been diminished
by 4 × 0.10 M due to the amount required to complex 0.10
mole of copper(II) ion.
128
17.5
Factors Affecting Solubility
Complex Ion Formation
Neglecting x,
x = 4.4 × 10‒18 M
Because Kf is so large, the amount of copper that remains
uncomplexed is extremely small.
129
SAMPLE PROBLEM
17.12
In the presence of aqueous cyanide, cadmium(II) forms the
complex ion Cd(CN)42‒. Determine the molar concentration of
free (uncomplexed) cadmium(II) ion in solution when 0.20
mole of Cd(NO3)2 is dissolved in a liter of 2.0 M sodium
cyanide (NaCN).
Setup
Kf for the complex ion Cd(CN)42‒ is 7.1 × 1016. The reverse
process
has an equilibrium constant of 1/Kf = 1.4 × 10 ‒17
130
SAMPLE PROBLEM
17.12
The equilibrium expression for the dissociation is
Stoichiometry indicates that four CN‒ ions are required to
react with one Cd2+ ion. Therefore, [CN‒] will be
[2.0 M ‒ 4(0.20 M)] = 1.2 M.
Solution
131
SAMPLE PROBLEM
17.12
Neglecting x,
x = 1.4 × 10‒18 M
Don’t forget to adjust the concentration of the complexing
agent before entering it in the equilibrium table.
132
17.6
Separation of Ions Using Differences in
Solubility
Topics
Fractional Precipitation
Qualitative Analysis of Metal Ions in Solution
133
17.6
Separation of Ions Using Differences in
Solubility
Fractional Precipitation
One way to separate ions is to convert them to insoluble salts.
When a soluble compound such as silver nitrate is slowly
added to a solution that contains Cl‒, Br‒, and I‒ ions, AgI
precipitates first, followed by AgBr, and then AgCl. This
practice is known as fractional precipitation.
The solubility of the silver
halides decreases from
AgCl to AgI.
134
SAMPLE PROBLEM
17.13
Silver nitrate is added slowly to a solution that is 0.020 M in
Cl‒ ions and 0.020 M in Br‒ ions. Calculate the concentration
of Ag+ ions (in mol/L) required to initiate the precipitation of
AgBr without precipitating AgCl.
Strategy
Use the equilibrium expressions to calculate the maximum
[Ag+] that can exist without exceeding Ksp for each compound.
135
SAMPLE PROBLEM
17.13
Setup
AgBr should precipitate first.
Use the equilibrium expression for AgBr to determine the
minimum [Ag+] necessary to initiate precipitation of AgBr.
Solve for [Ag+] again, using the equilibrium expression for
AgCl to determine the maximum [Ag+] that can exist in the
solution without initiating the precipitation of AgCl.
136
SAMPLE PROBLEM
17.13
Solution
Solving the AgBr equilibrium expression for [Ag+],
For AgBr to precipitate from solution, [Ag+] must exceed
3.9 × 10‒11 M.
Solving the AgCl equilibrium expression for the [Ag+],
137
SAMPLE PROBLEM
17.13
For AgCl not to precipitate from solution, [Ag+] must stay
below 8.0 × 10‒9 M.
To precipitate Br‒ ions without precipitating Cl ‒ from this
solution,
3.9 × 10‒11 M < [Ag+] < 8.0 × 10‒9 M
138
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
The principle of selective precipitation can be used to identify
the types of ions present in a solution. This practice is called
qualitative analysis.
There are about 20 common cations that can be analyzed
readily in aqueous solution.
Analysis of these ions must be carried out systematically from
group 1 through group 5.
139
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
• Group 1 cations. When dilute HCl is added to the unknown
solution, only the Ag+, Hg22+, and Pb2+ ions precipitate as
insoluble chlorides.
• Group 2 cations. The chloride precipitates are removed by
filtration. Hydrogen sulfide is added to the unknown acidic
solution to produce metal sulfides:
140
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
Only the metal sulfides with the smallest Ksp values
precipitate under acidic conditions. These are Bi2S3, CdS, CuS,
and SnS. The solution is filtered to remove the insoluble
sulfides.
• Group 3 cations. NaOH is added to the solution. In a basic
solution, the more soluble sulfides (CoS, FeS, MnS, NiS, ZnS)
precipitate. The Al3+ and Cr3+ ions precipitate as hydroxides.
The solution is filtered again to remove the insoluble
sulfides and hydroxides.
141
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
• Group 4 cations. Sodium carbonate is added to the basic
solution to precipitate Ba2+, Ca2+, and Sr2+ ions as BaCO3,
CaCO3, and SrCO3. These precipitates are filtered.
• Group 5 cations. The only cations possibly remaining in
solution are Na+, K+, and NH4+. NH4+ ions can be
determined by adding sodium hydroxide:
142
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
Ammonia gas is detected by either its odor or a litmus test.
To confirm the presence of Na+
and K+ ions, a flame test is used in
which a piece of platinum wire is
dipped into the solution and held
over a flame.
Na+ ions emit a yellow flame,
whereas K+ ions emit a violet
flame.
143
17.6
Separation of Ions Using Differences in
Solubility
Qualitative Analysis of Metal Ions in Solution
A summary of the scheme for
separating metal ions:
144