Transcript Document

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 17
Acid-Base Equilibria and
Solubility Equilibria
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
17
Acid-Base Equilibria and Solubility Equilibria
17.1 The Common Ion Effect
17.2 Buffer Solutions
Calculating the pH of a Buffer
Preparing a Buffer Solution with as Specific pH
17.3 Acid-Base Titrations
Strong Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
Strong Acid-Weak base Titrations
Acid-Base Indicators
17.4 Solubility Equilibria
Solubility Product Expression and Ksp
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions
17.5 Factors Affecting Solubility
The Common Ion Effect
pH
Complex Ion Formation
17.6 Separation of Ions Using Differences in Solubility
Fractional Precipitation
Qualitative Analysis of Metal Ions in Solution
17.1
The Common Ion Effect
A system at equilibrium will shift in response to being stressed.
The addition of a reactant or a product can be an applied stress.
CH3COOH(aq)
⇌ H+(aq) + CH3COO–(aq)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
[H+] = [CH3COO–] = 1.34 x 10–3 M; pH = 2.87
CH3COONa(aq)
CH3COOH(aq)
H2O
Na+(aq) + CH3COO–(aq)
H+(aq) + CH3COO–(aq)
Equilibrium is driven toward reactant
addition
Worked Example 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.)
Strategy Construct a new equilibrium table to solve for the hydrogen ion
concentration. We use the stated concentration of acetic acid, 0.10 M, and
[H+] ≈ 0 M as the initial concentrations in the table.
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
0.10
0
0.050
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
0.050 + x
Initial concentration (M)
Worked Example 17.1 (cont.)
Solution These equilibrium concentrations are then substituted into the
equilibrium expression to give
(x)(0.050 + x)
-5
1.8×10 = 0.010 – x
Because we expect x to be very small (even smaller than 1.34×10-3 M–see
above), because the ionization of CH3COOH is suppressed by the presence of
CH3COO-, we assume
(0.10 – x) M ≈ 0.10 M
and
(0.050 + x) M ≈ 0.050 M
Therefore, the equilibrium expression simplifies to
(x)(0.050)
-5
1.8×10 = 0.010
and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so
pH = –log(3.6×10-5) = 4.44.
Worked Example 17.1 (cont.)
Think About It The equilibrium concentrations of CH3COOH, CH3COO-, and
H+ are the same regardless of whether we add sodium acetate to a solution of
acetic acid, add acetic acid to a solution of sodium acetate, or dissolve both
species at the same time. We could have constructed an equilibrium table starting
with the equilibrium concentrations in the 0.10 M acetic acid solution:
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
0.09866
1.34×10-3
5.134×10-2
Change in concentration (M)
+y
–y
–y
Equilibrium concentration (M)
0.09866 + y
1.34×10-3 – y
5.134×10-2 – y
Initial concentration (M)
In this case, the reaction proceeds to the left. (The acetic acid concentration
increases, and the concentrations of hydrogen and acetate ions decrease.) Solving
for y gives 1.304×10-3 M. [H+] = 1.34×10-3 – y = 3.6×10-5 M and pH = 4.44.
We get the same pH either way.
17.2
Buffer Solutions
A solution that contains a weak acid and its conjugate base (or a
weak base and its conjugate acid) is a buffer.
acid
conjugate base
CH3COOH(aq)
⇌ H+(aq)
reacts with
added base
CH3COOH(aq) + OH–(aq)
CH3COOH(aq)
Buffer solutions resist changes in pH.
+
CH3COO–(aq)
reacts with
added acid
CH3COO– (aq) + H2O(l)
CH3COO– (aq) + H+(aq)
Buffer Solutions
Calculate the pH of a buffer that contains 1.0 M acetic acid and 1.0
M sodium acetate.
CH3COOH(aq)
⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.0
0
1.0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
1.0 – x
x
1.0 + x
Ka 
 x  1.0  x 
1.0  x
 1.8  105
x = [H+] = 1.8 x 10–5 M; pH = 4.74
Buffer Solutions
Calculate the pH of the same buffer (1.0 M acetic acid and 1.0 M
sodium acetate) after the addition of 0.10 mol of HCl.
The added acid reacts with the conjugate base (acetate ion).
Upon addition of H+:
1.0 mol
0.1 mol
CH3COO–(aq) + H+(aq)
After H+ has been consumed:
0.9 mol
CH3COOH(aq)
0 mol
1.0 mol
CH3COOH (aq)
1.1 mol
⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.1
0
0.9
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
1.1 – x
x
0.9 + x
Buffer Solutions
CH3COOH(aq)
⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.1
0
0.9
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
1.1 – x
x
0.9 + x
Ka 
 x  0.9  x 
1.1  x
 1.8  10
5
x = [H+] = 3.6 x 10–5 M;
pH = 4.66
The original buffer
had a pH = 4.74
Buffer Solutions
The pH of a buffer solution can often be calculated with the
Henderson-Hasselbalch equation.
pH  pK a 
conjugate base 

log
 weak acid
Worked Example 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)
Write the starting amount of each species above the equation and the final amount
of each species below the equation. Use the final amounts as concentrations in
pH = pKa + log([A-]/[HA]).
Upon addition of OH-:
1.00 mol
0.10 mol
1.00 mol
CH3COOH(aq) + OH-(aq) ⇌ H2O(l) + CH3COO–(aq)
After OH- consumed:
0.90 mol
0 mol
1.10 mol
Worked Example 17.2 (cont.)
Solution These equilibrium concentrations are then substituted into the
equilibrium expression to give
1.10 M
pH = 4.74 + log
0.90 M
pH = 4.83
Thus, the pH of the buffer after addition of 0.10 mole of NaOH is 4.83.
Think About It Always do a “reality check” on a calculated pH. Although a
buffer does minimize the effect of added base, the pH does increase. If you find
that you’ve calculated a lower pH after the addition of a base, check for errors
like mixing up the weak acid and conjugate base concentrations or losing track of
a minus sign.
Buffer Solutions
Buffers must have conjugate acid and base concentrations within a
factor of 10.
conjugate base 

10 
 0.1
 weak acid
The pH of a buffer cannot be more than one pH unit different than
the pKa of the weak acid it contains.
Buffer Solutions
To make a buffer with a specific pH:
1) Pick a weak acid whose pKa is close to the desired pH.
2) Substitute the pH and pKa into the equation below to obtain the
necessary [conjugate base]/[weak acid] ratio.
 A  
pH = pK a  log
HA 
Worked Example 17.3
Select an appropriate weak acid from the table at right,
and describe how you would prepare a buffer with a pH
of 9.50.
Strategy Select an acid with a pKa within one pH unit
of 9.50. Use the pKa of the acid and pH = pKa + log([A]/[HA]) to calculate the necessary ratio of [conjugate
base]/[weak acid]. Select concentrations of the buffer
components that yield the calculated ratio.
Weak Acid
Ka
pKa
HF
7.1×10-4
3.15
HNO2
4.5×10-4
3.35
HCOOH
1.7×10-4
3.77
C6H5COOH
6.5×10-5
4.19
CH3COOH
1.8×10-5
4.74
HCN
4.9×10-10
9.31
C6H5OH
1.3×10-10
9.89
Solution Two of the acids listed have pKa values in the desired range:
hydrocyanic acid (HCN, pKa = 9.31) and phenol (C6H5OH, pKa = 9.89).
[C6H5O-]
9.50 = 9.89 + log
[C6H5OH]
[C6H5O-]
9.50 – 9.89 = log
= –0.39
[C6H5OH]
[C6H5O-]
-0.39 = 0.41
=
10
[C6H5OH]
Worked Example 17.3 (cont.)
Solution Therefore, the ratio of [C6H5O-] to [C6H5OH] must be 0.41 to 1. One
way to achieve this would be to dissolve 0.41 mole of C6H5ONa and 1.00 mole of
C6H5OH in 1.00 L of water.
Think About It There is an infinite number of combinations of [conjugate base]
and [weak acid] that will give the necessary ratio. Note that this pH could also be
achieved using HCN and a cyanide salt. For most purposes, it is best to use the
least toxic compounds available.
17.3
Acid-Base Titrations
Strong Acid-Strong Base Titrations
The reaction between the strong acid HCl and the strong base NaOH
can be represented by:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
or by the net ionic equation,
OH–(aq) + H+(aq) → H2O(l)
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M HCl with
0.100 M NaOH
Strong Acid-Strong Base Titrations
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M HCl with
0.100 M NaOH
Strong Acid-Strong Base Titrations
Acid-Base Titrations
Weak Acid-Strong Base Titrations:
Consider the neutralization between acetic acid and sodium
hydroxide:
CH3COOH(aq) + OH–(aq)
→ CH3COO– (aq) + H2O(l)
The acetate ion that results from this neutralization undergoes
hydrolysis:
CH3COO– (aq) + H2O(l)(aq) ⇌ CH3COOH(aq) + OH–(aq)
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Weak Acid-Strong Base Titrations
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Acid-Base Titrations
The initial pH is determined by the ionization of acetic acid.
CH3COOH(aq)
⇌ H+(aq) + CH3COO–(aq)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
x2
Ka 
 1.8  105
0.10  x
x = [H+] = 1.34 x 10–3 M; pH = 2.87
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
After the addition of base, some of the acetic acid has been
converted to acetate ion:
CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
Volume of OH– OH– added
added (mL)
(mol)
10.0
1.0
CH3COOH
remaining
CH3COO–
produced
pH
1.5
1.0
4.56
The solution is a buffer and the Henderson-Hasselbalch equation
can be used to calculate pH.
After 10.0 mL of base has been added:
 A  
pH = pK a  log
HA 
1.0 mmol

pH = 4.74  log
 4.56
1.5 mmol
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Acid-Base Titrations
At the equivalence point, all the acetic acid has been neutralized.
CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
Volume of OH– OH– added
added (mL)
(mol)
25.0
CH3COOH
remaining
CH3COO–
produced
pH
0.0
2.5
8.72
2.5
2.5 mmol
CH3COO  
 0.050 M
50.0 mL
Must use TOTAL VOLUME to
calculate concentration.
CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)
0.050
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.050 – x
x
x
Initial concentration (M)
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Acid-Base Titrations
CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)
0.050
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.050 – x
x
x
Initial concentration (M)
Use Ka x Kb = Kw to get Kb for the acetate ion; Kb = 5.6 x 10–10
OH  CH3COOH
x2
10
Kb 


5.6

10
0.050  x
CH3COO 
x = [OH–] = 5.3 x 10–6 ; pOH = 5.28; pH = 8.72
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Acid-Base Titrations
After the equivalence point, all the acetic acid has been neutralized, nothing is
left to neutralize the added strong base.
Volume of OH– OH– added
added (mL)
(mol)
25.0
2.5
CH3COOH
remaining
CH3COO–
produced
pH
0.0
2.5
8.72
Volume of
OH– added
(mL)
OH–
added
(mmol)
Excess
OH–
(mmol)
Total
Volume
(mL)
[OH– ]
(mol/L)
pOH
pH
30.0
3.0
0.5
55.0
0.0091
2.04
11.96
35.0
3.5
1.0
60.0
0.017
1.78
12.22
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M acetic
acid with 0.100 M NaOH
Weak Acid-Strong Base Titrations
Worked Example 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 between acetic acid and sodium hydroxide is
OH–(aq) + CH3COOH (aq) ⇌ CH3COO–(aq) + H2O(l)
Prior to the equivalence point [part (a)], the solution contains both acetic acid and
acetate ion, making the solution a buffer. We can solve part (a) using the
Henderson-Hasselbach equation. At the equivalence point [part (b)], all the acetic
acid has been neutralized and we have only acetate ion in solution. We must
determine the concentration of acetate ion and solve part (b) as an equilibrium
problem, using the Kb for acetate ion. After the equivalence point [part (c)], all the
acetic acid has been neutralized and there is nothing to consume the additional
added base. We must determine the concentration of excess hydroxide ion in
the solution and solve for pOH = -log[OH-] and pH + pOH = 14.00.
Worked Example 17.4 (cont.)
Solution Remember that M can be defined as either mol/L or mmol/mL. For
this type of problem, it simplifies the calculations to use millimoles rather than
moles. Ka for acetic acid is 1.8×10-5, so pKa = 4.74. Kb for acetate ion is
5.6×10-10.
(a) The solution originally contains (0.120 mmol/mL)(50.0 mL) = 6.00 mmol of
acetic acid. A 10.0-mL amount of base contains (0.240 mmol/mL)(10.0 mL) =
2.40 mmol of base. After the addition of 10.0 mL of base, 2.40 mmol of OH- has
neutralized 2.40 mmol of acetic acid, leaving 3.60 mmol of acetic acid and
2.40 mmol acetate ion in solution.
Upon addition of OH-:
6.00 mol
2.40 mol
0 mol
CH3COOH(aq) + OH-(aq) ⇌ H2O(l) + CH3COO–(aq)
After OH- consumed:
3.60 mol
0 mol
2.40 mol
pH = pKa + log
2.40
= 4.74 – 0.18 = 4.56
3.60
Worked Example 17.4 (cont.)
Solution (b) After the addition of 25.0 mL of base, the titration is at the
equivalent point. We calculate the pH using the concentration and the Kb of
acetate ion.
At the equivalence point, we have 6.0 mmol of acetate ion in the total volume. We
determine the total volume by calculating what volume of 0.24 M contains
6.0 mmol:
(volume)(0.240 mmol/mL) = 6.00 mmol
6.00 mmol
volume =
= 25.0 mL
0.240 mmol/mL
Therefore, the equivalence point occurs when 25.0 mL of base has been added,
making the total volume 50.0 mL + 25.0 mL = 75.0 mL. The concentration of
acetate ion at the equivalence point is therefore
6.00 mmol CH3COO= 0.0800 M
75.0 mL
Worked Example 17.4 (cont.)
Solution (b) We can construct an equilibrium table using this concentration and
solve for pH using the ionization constant for CH3COO- (Kb = 5.6×10-10):
CH3COO-(aq) + H2O(l) ⇌ OH-(aq) + CH3COOH(aq)
Initial concentration (M)
Change in concentration (M)
0.0800
0
0
–x
+x
+x
x
x
Equilibrium concentration (M) 0.0800 – x
Using the equilibrium expression and assuming that x is small enough to be
neglected,
[CH3COOH][OH-]
(x)(x)
x2
Kb =
= 1.8×10-5
=
≈
[CH3COO ]
0.0800 – x 0.0800
x = 4.481011 = 6.7×10-6 M
According to the equilibrium table, x = [OH-], so [OH-] = 6.7×10-6 M. At
equilibrium pOH = – log(6.7×10-6) = 5.17 and pH = 14.00 – 5.17 = 8.83.
Acid-Base Titrations
Titration of 25.0 mL of 0.100 M NH3 with
0.100 M HCl
Strong Acid-Weak Base Titrations
Worked Example 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
NH3(aq) + H+(aq) ⇌ NH4+(aq)
At the equivalence point, all the NH3 has been converted to NH4+. Therefore, we
must determine the concentration of NH4+ at the equivalence point and use the Ka
for NH4+ to solve for pH using an equilibrium table.
Solution 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. The
volume of 0.100 M HCl that contains 2.50 mmol is
(volume)(0.100 mmol/mL) = 2.50 mmol
volume =
2.50 mmol
= 25.0 mL
0.100 mmol/mL
Worked Example 17.5 (cont.)
Solution It takes 25.0 mL of titrant to reach the equivalence point, so the total
solution is 25.0 + 25.0 = 50.0 mL. At the equivalence point, all the NH3 originally
present has been converted to NH4+. The concentration of NH4+ is
(2.50 mmol)/(50.0 mL) = 0.0500 M. We must use this concentration as the
starting concentration of ammonium ion in our equilibrium table.
Think About It In the titration
of a weak
base with
a strong
acid, +
NH4+(aq)
+ H2O(l)
⇌ NH
3(aq) + H3O (aq)
the species in solution at the equivalence point is the conjugate acid.
Initial Therefore,
concentration
0 3 has
0
we (M)
should expect0.0500
an acidic pH. Once all the NH
been converted to NH4+, there is no longer anything in the solution
to consume
the added
point +x
Change
in concentration
(M)acid. Thus,
–x the pH after the equivalence
+x
depends on the number of millimoles of H+ added and not consumed
divided
by the new total
Equilibrium
concentration
(M) volume.
0.0500 – x
x
x
The equilibrium expression is
[NH3][H+]
(x)(x)
x2
Ka =
= 5.6×10-10
=
≈
+
[NH4 ]
0.0500 – x 0.0500
x = 2.8 1011 = 5.3×10-6 M = [H+]
pH = –log(5.3×10-6) = 5.28
Acid-Base Titrations
The equivalence point in a titration can be visualized with the use
of an acid-base indicator.
HIn(aq)
⇌
H+(aq)
The endpoint of a titration is the
point at which the color of the
indicator changes.
+ In–(aq)
Acid-Base Titrations
Acid-Base Titrations
Worked Example 17.6
Which indicator listed in Table 17.3 would you use for the acid-base titrations
shown in (a) Figure 17.3, (b) Figure 17.4, and (c) Figure 17.5?
Strategy Determine the pH range that corresponds to the steepest part of each
titration curve and select an indicator (or indicators) that changes color within that
range.
Solution (a) The titration curve at right is for
the titration of a strong acid with a strong base.
The steep part of the curve spans a pH range of
about 4 to 10.
Most of the indicators in Table 17.3, with the
exceptions of thymol blue, bromophenol blue,
and methyl orange, would work for the titration
of a strong acid with a strong base.
Worked Example 17.6 (cont.)
Solution (b) The figure at right shows the
titration of a weak acid with a strong base. The
steep part of the curve spans a pH range of about
7 to 10.
Cresol red and phenolphthalein are suitable
indicators.
Think About It If we don’t select an appropriate indicator, the
endpoint (color change) will not coincide with the equivalence point.
(c) The figure at right shows the titration of a
weak base with a strong acid. The steep part of
the curve spans a pH range of 7 to 3.
Bromophenol blue, methyl orange, methyl red,
and chlorophenol blue are all suitable indicators.
17.4
Solubility Equilibria
Quantitative predictions about how much of a given ionic compound
will dissolve in water is possible with the solubility product
constant, Ksp.
AgCl(s)
⇌ Ag+(aq)
+
Cl–(aq)
Ksp = [Ag+][Cl–]
Compound
Dissolution Equilibrium
Ksp
Al(OH)3(s) ⇌ Al3+(aq) + 3OH–(aq)
1.8 x 10–33
CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
4.0 x 10–11
Silver bromide
AgBr(s) ⇌ Ag+(aq) + Br–(aq)
7.7 x 10–13
Silver chloride
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
1.6 x 10–6
Zinc sulfide
ZnS(s) ⇌ Zn2+(aq) + S2–(aq)
3.0 x 10–23
Aluminum hydroxide
Calcium fluoride
Solubility Equilibria
Molar solubility is the number of moles of solute in 1 L of a
saturated solution (mol/L)
Solubility is the number of grams of solute in 1 L of a saturated
solution (g/L).
To calculate a compound’s molar solubility:
1) Construct an equilibrium table.
2) Fill in what is known.
3) Determine the unknowns.
Solubility Equilibria
The Ksp of silver bromide is 7.7 x 10–13. Calculate the molar
solubility.
AgBr(s) ⇌ Ag+(aq) + Br–(aq)
Initial concentration (M)
0
0
Change in concentration (M)
+s
+s
Equilibrium concentration (M)
s
s
Ksp = [Ag+][Br–] = 7.7 x 10–13
7.7 x 10–13 = (s)(s)
s = 8.8 x 10–7 M
8.8  107 mol AgBr
187.8 g

 1.7  104 g/L
1L
1 mol AgBr
Worked Example 17.7
Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in g/L.
Strategy Write the dissociation equation for Cu(OH)2, and look up its Ksp value
in Table 17.4. Solve for molar solubility using the equilibrium expression.
Convert molar solubility to solubility in g/L using the molar mass of Cu(OH)2.
Solution The equation for the dissociation of Cu(OH)2 is
Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq)
and the equilibrium expression is Ksp = [Cu2+][OH-]2. According to Table 17.4,
Ksp for Cu(OH)2 is 2.2×10-20. The molar mass of Cu(OH)2 is 97.57 g/mol.
Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq)
Initial concentration (M)
0
0
Change in concentration (M)
+s
+2s
Equilibrium concentration (M)
s
2s
Worked Example 17.7 (cont.)
Solution Therefore,
2.2×10-20 = (s)(2s)2 = 4s3
s=
3
2.2 1020
= 1.8×10-7 M
4
The molar solubility of Cu(OH)2 is 1.8×10-7 M. Multiplying by its molar mass
gives
1.8×10-7 mol
97.57 g Cu(OH)2
solubility of Cu(OH)2 =
× 1 mol Cu(OH)
L
2
= 1.7×10-5 g/L
Think About It Common errors arise in this type of problem when students
neglect to raise an entire term to the appropriate power. For example, (2s)2 is
equal to 4s2 (not 2s2).
Worked Example 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.
Strategy Convert solubility to molar solubility using the molar mass of CaSO4,
and substitute the molar solubility into the equilibrium expression to determine Ksp.
Think About It The Ksp for CaSO4 is relatively large (compared to
Solution
molar
ofinCaSO
136.2Ing/mol.
The molar
manyThe
of the
Kspmass
values
Table
17.4).
fact, sulfates
are solubility
listed as of
4 is
CaSO4soluble
is
compounds in Table 9.2, but calcium sulfate is listed as an
0.67that
g CaSO
1 mol CaSO
4
4 refers
insoluble
exception.
Remember
the
term
insoluble
really
molar solubility of CaSO4 =
×
1 L and that136.2
g CaSO
4
to compounds that are slightly soluble,
different
sources
-3 mol/L
s
=
4.9×10
may differ with regard to how soluble a compound must be to be
The equation
and the
equilibrium expression for the dissociation of CaSO4 are
considered
soluble.
CaSO4(s) ⇌ Ca2+(aq) + SO42-(aq)
and
Ksp = [Ca2+][SO42-]
substituting the molar solubility into the equilibrium expression gives
Ksp = (s)(s) = (4.9×10-3)2 = 2.4×10-5
Solubility Equilibria
For the dissociation of an ionic solid in water, the following
conditions may exist:
1) The solution is unsaturated
2) The solution is saturated
3) The solution is supersaturated
The following relationships are useful in making predictions on
when a precipitate might form.
Q < Ksp; no precipitate forms
Q = Ksp; no precipitate forms
Q > Ksp; a precipitate forms
Worked Example 17.9
Predict whether a precipitate will form when each of the following is added to
650 mL of 0.080 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.)
Strategy For each part, identify the compound that might precipitate and look
up its Ksp value in Table 17.4 or Appendix 4. Determine the concentration of each
compound’s constituent ions, and use them to determine the value of the reaction
quotient, Qsp; then compare each reaction quotient with the value of the
corresponding Ksp. If the reaction quotient is greater than Ksp, a precipitate will
form.
Solution (a) BaSO4 might form and its Ksp = 1.1×10-10. Concentrations of the
constituent ions of BaSO4 are:
[Ba2+] =
250 mL × 0.0040 M
= 0.0011 M
650 mL + 250 mL
[SO42-] =
650 mL × 0.0080 M
= 0.0058 M
650 mL + 250 mL
Using these concentrations in the equilibrium expression, [Ba2+][SO42-], gives a
reaction quotient of (0.0011)(0.0058) = 6.4×10-6, which is greater than the
Ksp of BaSO4 (1.1×10-10). Therefore, BaSO4 will precipitate.
Worked Example 17.9 (cont.)
Solution (b) Ag2SO4 might form and its Ksp = 1.5×10-5. Concentrations of the
constituent ions of Ag2SO4 are:
[Ag+] =
175 mL × 0.15 M
= 0.0032 M
650 mL + 175 mL
[SO42-] =
650 mL × 0.0080 M
= 0.0063 M
650 mL + 175 mL
Thinkconcentrations
About It Students
sometimes have
difficulty
deciding
what
+]2[SO 2Using these
in the equilibrium
expression,
[Ag
4 ], gives a
compound
might
precipitate.
Begin
by
writing
down
the
constituent
reaction quotient of (0.032)2(0.0063) = 6.5×10-6, which is less than the Ksp of
in the -5
two
solutions before
they
are
combined.
Consider the two
Ag2SOions
(1.5×10
).
Therefore,
Ag
SO
will
not
precipitate.
4
2
4
possible
combinations; the cation
from the first solution and the
anion
fromform
the second,
or vice
versa.-7You
can consult the
(c) SrSO
might
and
its
K
=
3.8×10
.
Concentrations
of the constituent
4
sp
information
in
Table
9.2
and
9.3
to
determine
whether
one
of the
ions of Ag2SO4 are:
combinations is insoluble. Also keep in mind that only an insoluble
salt 325
willmL
have
a tabulated
Ksp value.
× 0.25
M
650 mL × 0.0080 M
2+
2[Sr ] =
650 mL + 325 mL
= 0.083 M
[SO4 ] =
650 mL + 325 mL
= 0.0053 M
Using these concentrations in the equilibrium expression, [Sr2+][SO42-], gives a
reaction quotient of (0.083)(0.0053) = 4.4×10-4, which is greater than the Ksp of
SrSO4 (3.8×10-7). Therefore, SrSO4 will precipitate.
17.5
Factors Affecting Solubility
Several factors exist that affect the solubility of ionic compounds:
 Common ion effect
 pH
 Formation of complex ions
Worked Example 17.10
Calculate the molar solubility of silver chloride in a solution that is 6.5×10-3 M
in silver nitrate.
Strategy Silver nitrate is a strong electrolyte that dissociates completely in
water. Therefore, the concentration of Ag+ before any AgCl dissolves is 6.5×10-3
M. Use the equilibrium expression, the Ksp for AgCl, and an equilibrium table to
determine how much AgCl will dissolve.
Solution The dissolution equilibrium and the equilibrium expression are
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
6.5×10-3
0
Change in concentration (M)
+s
+s
Equilibrium concentration (M)
6.5×10-3 + s
s
Initial concentration (M)
Worked Example 17.10 (cont.)
Solution Substituting these concentrations into the equilibrium expression gives
1.6×10-10 = (6.5×10-3 + s)(s)
We expect s to be very small, so
6.5×10-3 + s ≈ 6.5×10-3
and
1.6×10-10 = (6.5×10-3)(s)
Thus
1.6×10-10
s=
= 2.5×10-8 M
-3
6.5×10
Therefore, the molar solubility of AgCl in 6.5×10-3 M AgNO3 is 2.5×10-8 M.
Think About It The molar solubility of AgCl in water is 1.6 1010 = 1.3×10-5
M. The presence of 6.5×10-3 M AgNO3 reduces the solubility of AgCl by a factor
of ~500.
Factors Affecting Solubility
pH:
Mg(OH)2(s)
⇌
Mg2+ (aq)
+ 2OH–(aq)
Ksp = [Mg2+][OH–]2 = 1.2 x 10–11
(s)(2s)2 = 4s3 = 1.2 x 10–11
s = 1.4 x 10–4 M
At equilibrium:
[OH–] = 2(1.4 x 10–4 M ) = 2.8 x 10–4 M
pOH = –log(2.8 x 10–4) = 3.55
pH = 14.00 – 3.55 = 10.45
In a solution with a pH of less than 10.45, the solubility of Mg(OH)2
increases.
Factors Affecting Solubility
pH:
Mg(OH)2(s)
2H+(aq) + 2OH–(aq)
Overall:
Mg(OH)2(s) + 2H+(aq)
⇌
Mg2+(aq) + 2OH–(aq)
→ 2H2O(l)
⇌
Mg2+(aq) + 2H2O(l)
If the pH of the medium were higher than 10.45, [OH–] would be
higher and the solubility of Mg(OH)2 would decrease because of the
common ion (OH-) effect.
Factors Affecting Solubility
pH:
BaF2(s) ⇌
Ba2+ (aq) + 2F–(aq)
2H+(aq) + 2F–(aq) → 2HF(aq)
Overall:
BaF2(s) + 2H+ (aq) ⇌
Ba2+ (aq) + 2HF(aq)
As the concentration of F – decreases, the concentration of Ba2+ must
increase so satisfy the equality:
Ksp = [Ba2+][F–]2
The solubilities of salts containing anions that do not hydrolyze are
unaffected by pH:
Cl– , Br – , NO3–
Worked Example 17.11
Which of the following compounds will be more soluble in acidic solution than in
water: (a) CuS, (b) AgCl, (c) PbSO4?
Strategy For each salt, write the dissociation equilibrium equation and
determine whether it produces an anion that will react with H+. Only an anion that
is the conjugate base of a weak acid will react with H+.
Solution The dissolution equilibrium and the equilibrium expression are
(a) CuS(s) ⇌ Cu2+(aq) + S2-(aq)
S2- is the conjugate base of the weak acid HS-. S2- reacts with H+ as follows:
S2-(aq) + H+(aq) ⇌ HS-(aq)
(b) AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Cl- is the conjugate base of the strong acid HCl. Cl- does not react with H+.
(c) PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)
SO42- is the conjugate base of the weak acid HSO4-. It reacts with H+ as follows:
SO42-(aq) + H+(aq) ⇌ HSO4-(aq)
Worked Example 17.11 (cont.)
Solution CuS and PbSO4 are more soluble in acid than in water. (AgCl is no
more or less soluble than in water.)
Think About It When a salt dissociates to give the conjugate base of a weak
acid, H+ ions in an acidic solution consume a product (base) of the dissolution.
This drives the equilibrium to the right (more solid dissolves) according to Le
Châtelier’s principle.
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.
A solution of CoCl2 is pink because of the
presence of Co(H2O)62+ ions.
Factors Affecting Solubility
Complex ion formation:
When HCl is added to a CoCl2 solution, there is
a color change from pink to blue.
This solution is blue because of the presence
of presence of CoCl42– ions.
Co2+(aq) + 4Cl– (aq) ⇌ CoCl42–(aq)
Factors Affecting Solubility
Complex ion formation:
Cu2+ (aq) + 2OH–(aq)
→ Cu(OH)2(s)
Cu(OH)2(s) + 4NH3(aq) ⇌ Cu(NH3)42+ (aq) + 2OH–(aq)
The formation of the Cu(NH3)42+ ion can be expressed as
Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+ (aq)
Cu NH 2 
3 4

  5.0  1013
Kf 
4
Cu2  NH3 
Factors Affecting Solubility
Factors Affecting Solubility
Worked Example 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 mol of Cd(NO3)2 is dissolved in a liter of
2.0 M sodium cyanide (NaCN).
Strategy Because formation constants are typically very large, we begin by
assuming that all the Cd2+ ion is consumed and converted to complex ion. We
then determine how much Cd2+ is produced by the subsequent dissociation of the
complex ion, a process for which the equilibrium constant is the reciprocal Kf.
Solution From Table 17.5, the formation constant (Kf) for the complex ion
Cd(CN)42- is 7.1×1016. The reverse process,
Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)
has an equilibrium constant of 1/Kf = 1.4×10-17. The equilibrium expression for
the dissociation is
[Cd2+][CN-]4
-17
1.4×10 =
[Cd(CN)42-]
Worked Example 17.12 (cont.)
Solution The formation of complex ion will consume some of the cyanide
originally present. Stoichiometry indicates that four CN- ions are required to react
with one Cd2+ ion. Therefore, the concentration of CN- that we enter in the top
row of the equilibrium table will be [2.0 M – 4(0.20 M)] = 1.2 M.
Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)
Initial concentration (M)
0 ion is 1.2
Think About It When you assume 0.20
that all the metal
consumed
converted to(M)
complex ion,
to remember
Change and
in concentration
-x it’s important
+x
+4x
that some of the complexing agent (in this case, CN ion) is
consumed
in the
process. Don’t
its concentration
Equilibrium
concentration
(M) forget
0.20to-adjust
x
x
1.2 + 4x
accordingly before entering it in the top row of the equilibrium table.
and, because the magnitude of K is so small, we can neglect x with respect to the
initial concentrations of Cd(CN)42- and CN- (0.20 – x ≈ 0.20 and 1.2 + 4x ≈ 1.2),
so the solution becomes
[Cd2+][CN-]4 x(1.2)4
=
= 1.4×102[Cd(CN)4 ]
0.20 17
and x = 1.4×10-18 M.
17.6
Separation of Ions Using Differences in Solubility
Some compounds can be separated based on fractional precipitation.
Fractional precipitation is the separation of mixture based upon the
components’ solubilities.
Compound
Ksp
AgCl
1.6 x 10–10
AgBr
7.7 x 10–13
AgI
8.3 x 10–17
Worked Example 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 Silver nitrate dissociates in solution to give Ag+ and NO3- ions.
Adding Ag+ ions in sufficient amount will cause the slightly soluble ionic
compounds AgCl and AgBr to precipitate from solution. Knowing the Ksp values
for AgCl and AgBr (and the concentrations of Cl- and Br- already in solution), we
can use the equilibrium expressions to calculate the maximum concentration of
Ag+ that can exist in solution without exceeding Ksp for each compound.
Worked Example 17.13 (cont.)
Solution The solubility equilibria, Ksp values, and equilibrium expressions for
AgCl and AgBr are
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ksp = 1.6×10-10 = [Ag+][Cl-]
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
Ksp = 7.7×10-13 = [Ag+][Br-]
Because the Ksp for AgBr is smaller (by a factor of more than 200), AgBr should
precipitate first; that is, it will require a lower concentration of added Ag+ to begin
precipitation. Therefore, we first solve for [Ag+] using the equilibrium expression
for AgBr to determine the minimum Ag+ concentration necessary to initiate
precipitation of AgBr. We then solve for [Ag+] again, using the equilibrium
expression for AgCl to determine the maximum Ag+ concentration that can exist
in the solution without initiating precipitation of AgCl.
Solving the AgBr equilibrium expression for Ag+ concentration, we have
[Ag+]
Ksp
7.7×10-13
=
=
= 3.9×10-11 M
[Br ]
0.020
Worked Example 17.13 (cont.)
Solution For AgBr to precipitate from solution, the silver ion concentration
must exceed 3.9×10-11 M. Solving the AgCl equilibrium expression for the Ag+
concentration, we have
Ksp
1.6×10-10
+
[Ag ] =
=
= 8.0×10-9 M
[Cl ]
0.020
For AgCl not to precipitate from solution, the silver ion concentration must stay
below 8.0×10-9 M. Therefore, to precipitate the Br- ions without precipitating the
Cl- from this solution, the Ag+ concentration must be greater than 3.9×10-11 M and
less than 8.0×10-9 M.
Think About It If we continue adding AgNO3 until the Ag+ concentration is
high enough to begin precipitation of AgCl, the concentration of Br- remaining in
solution can also be determined using the Ksp expression.
[Br-]
Ksp
7.7×10-13
=
=
= 9.6×10-5 M
+
-9
[Ag ]
8.0×10
Thus, by the time AgCl begins to precipitate, (9.6×10-5 M)÷(0.020 M) = 0.0048,
so less than 0.5 percent of the original bromide ion remains in the solution.
Separation of Ions Using Differences in Solubility
Qualitative analysis involves the
principle of selective precipitation
and can be used to identify the
types of ions present in a solution.
17
Key Concepts
Calculating the pH of a Buffer
Preparing a Buffer Solution with as Specific pH
Strong Acid-Strong Base Titrations
Weak Acid-Strong Base Titrations
Strong Acid-Weak base Titrations
Acid-Base Indicators
Solubility Product Expression and Ksp
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions
The Common Ion Effect
pH
Complex Ion Formation
Fractional Precipitation
Qualitative Analysis of Metal Ions in Solution