1. 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.

2. Acid-Base Equilibria and Solubility Equilibria17
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

3. CH3COONa(aq) Na+(aq) + CH3COO–(aq) CH3COOH(aq) H+(aq) + CH3COO–(aq)
The Common Ion Effect
17.1
A system at equilibrium will shift in response to being stressed.
The addition of a reactant or a product can be an applied stress.
[H+] = [CH3COO–] = 1.34 x 10–3 M; pH = 2.87
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
0.10
Change in concentration (M) –x +x
Equilibrium concentration (M)
0.10 – x x
CH3COONa(aq) Na+(aq) + CH3COO–(aq)
H2O
addition
CH3COOH(aq) H+(aq) + CH3COO–(aq)
Equilibrium is driven toward reactant

4. Worked Example 17.1
Determine the pH at 25°C of a solution prepared by adding 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)
Initial concentration (M)
0.10
0.050
Change in concentration (M) –x +x
Equilibrium concentration (M)
0.10 – x x x

5. Worked Example 17.1 (cont.)
Solution These equilibrium concentrations are then substituted into the equilibrium expression to give
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 ( x) M ≈ M
Therefore, the equilibrium expression simplifies to
and x = 3.6×10-5 M. According to the equilibrium table, [H+] = x, so pH = –log(3.6×10-5) = 4.44.
(x)( x)
0.010 – x
1.8×10-5 =
(x)(0.050)
0.010
1.8×10-5 =

6. 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:
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 = We get the same pH either way.
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.34×10-3
5.134×10-2
Change in concentration (M) +y –y
Equilibrium concentration (M) y
1.34×10-3 – y
5.134×10-2 – y

7. CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Buffer Solutions
17.2
A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer.
Buffer solutions resist changes in pH.
acid
conjugate base
CH3COOH(aq) ⇌ H+(aq) CH3COO–(aq)
reacts with added base
reacts with added acid
CH3COOH(aq) + OH–(aq) CH3COO– (aq) + H2O(l)
CH3COOH(aq) CH3COO– (aq) + H+(aq)

8. Buffer Solutions
Calculate the pH of a buffer that contains 1.0 M acetic acid and 1.0 M sodium acetate.
x = [H+] = 1.8 x 10–5 M; pH = 4.74
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.0
Change in concentration (M)
Equilibrium concentration (M) –x +x +x
1.0 – x x
1.0 + x

9. CH3COO–(aq) + H+(aq) CH3COOH (aq)
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
1.0 mol
CH3COO–(aq) + H+(aq) CH3COOH (aq)
After H+ has been consumed:
0.9 mol
0 mol
1.1 mol
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.1
0.9
Change in concentration (M)
Equilibrium concentration (M) –x +x +x
1.1 – x x
0.9 + x

10. The original buffer had a pH = 4.74
Buffer Solutions
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
1.1
0.9
Change in concentration (M)
Equilibrium concentration (M) –x +x +x
1.1 – x x
0.9 + x
x = [H+] = 3.6 x 10–5 M;
pH = 4.66
The original buffer had a pH = 4.74

11. Buffer Solutions
The pH of a buffer solution can often be calculated with the Henderson-Hasselbalch equation.

12. CH3COOH(aq) + OH-(aq) ⇌ H2O(l) + CH3COO–(aq)
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 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 mol mol
After OH- consumed: mol mol mol

13. Worked Example 17.2 (cont.)
Solution These equilibrium concentrations are then substituted into the equilibrium expression to give
Thus, the pH of the buffer after addition of 0.10 mole of NaOH is 4.83.
1.10 M
0.90 M
pH = log
pH = 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.

14. Buffer Solutions
Buffers must have conjugate acid and base concentrations within a factor of 10.
The pH of a buffer cannot be more than one pH unit different than the pKa of the weak acid it contains.

15. Buffer Solutions
To make a buffer with a specific pH:
Pick a weak acid whose pKa is close to the desired pH.
Substitute the pH and pKa into the equation below to obtain the necessary [conjugate base]/[weak acid] ratio.

16. 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.
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
Strategy Select an acid with a pKa within one pH unit of 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.
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-]
[C6H5OH]
9.50 = log
[C6H5O-]
[C6H5OH]
9.50 – 9.89 = log = –0.39
[C6H5O-]
[C6H5OH]
= = 0.41

17. 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.

18. 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)

19. Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M HCl with M NaOH
Strong Acid-Strong Base Titrations

20. Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M HCl with M NaOH
Strong Acid-Strong Base Titrations

21. 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)

22. Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
Weak Acid-Strong Base Titrations

23. Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
The initial pH is determined by the ionization of acetic acid.
x = [H+] = 1.34 x 10–3 M; pH = 2.87
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Initial concentration (M)
0.10
Change in concentration (M)
Equilibrium concentration (M) –x +x +x
0.10 – x x x

24. Volume of OH– added (mL)
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
After the addition of base, some of the acetic acid has been converted to acetate ion:
The solution is a buffer and the Henderson-Hasselbalch equation can be used to calculate pH.
After mL of base has been added:
CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
Volume of OH– added (mL)
OH– added (mol)
CH3COOH remaining
CH3COO– produced pH
10.0
1.0
1.5
4.56

25. Volume of OH– added (mL)
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
At the equivalence point, all the acetic acid has been neutralized.
CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)
Volume of OH– added (mL)
OH– added (mol)
CH3COOH remaining
CH3COO– produced pH
25.0
2.5
0.0
8.72
Must use TOTAL VOLUME to calculate concentration.
CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)
Initial concentration (M)
0.050
Change in concentration (M) –x +x
Equilibrium concentration (M)
0.050 – x x

26. Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)
Initial concentration (M)
0.050
Change in concentration (M) –x +x
Equilibrium concentration (M)
0.050 – x x
Use Ka x Kb = Kw to get Kb for the acetate ion; Kb = 5.6 x 10–10
x = [OH–] = 5.3 x 10–6 ; pOH = 5.28; pH = 8.72

27. Volume of OH– added (mL) Volume of OH– added (mL)
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
After the equivalence point, all the acetic acid has been neutralized, nothing is left to neutralize the added strong base.
Volume of OH– added (mL)
OH– added (mol)
CH3COOH remaining
CH3COO– produced pH
25.0
2.5
0.0
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

28. Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH
Acid-Base Titrations
Titration of 25.0 mL of M acetic acid with M NaOH
Weak Acid-Strong Base Titrations

29. OH–(aq) + CH3COOH (aq) ⇌ CH3COO–(aq) + H2O(l)
Worked Example 17.4
Calculate the pH in the titration of 50.0 mL of M acetic acid by 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 =

30. 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 = 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 mmol acetate ion in solution.
Upon addition of OH-: 6.00 mol mol 0 mol
CH3COOH(aq) + OH-(aq) ⇌ H2O(l) + CH3COO–(aq)
After OH- consumed: mol mol mol
pH = pKa + log = 4.74 – 0.18 = 4.56
2.40
3.60

31. 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 mmol:
(volume)(0.240 mmol/mL) = 6.00 mmol
volume = = 25.0 mL
Therefore, the equivalence point occurs when 25.0 mL of base has been added, making the total volume 50.0 mL mL = 75.0 mL. The concentration of acetate ion at the equivalence point is therefore
= M
6.00 mmol
0.240 mmol/mL
6.00 mmol CH3COO-
75.0 mL

32. CH3COO-(aq) + H2O(l) ⇌ OH-(aq) + CH3COOH(aq)
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):
Using the equilibrium expression and assuming that x is small enough to be neglected,
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 = – 5.17 = 8.83.
CH3COO-(aq) + H2O(l) ⇌ OH-(aq) + CH3COOH(aq)
Initial concentration (M)
0.0800
Change in concentration (M) –x +x
Equilibrium concentration (M)
– x x
[CH3COOH][OH-]
[CH3COO-]
(x)(x)
– x x2
0.0800
Kb = = ≈
= 1.8×10-5
x = = 6.7×10-6 M

33. Titration of 25.0 mL of 0.100 M NH3 with 0.100 M HCl
Acid-Base Titrations
Titration of 25.0 mL of M NH3 with M HCl
Strong Acid-Weak Base Titrations

34. NH3(aq) + H+(aq) ⇌ NH4+(aq)
Worked Example 17.5
Calculate the pH at the equivalence point when 25.0 mL of M NH3 is titrated with 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 M HCl that contains 2.50 mmol is
(volume)(0.100 mmol/mL) = 2.50 mmol
volume = = 25.0 mL
2.50 mmol
0.100 mmol/mL

35. NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Worked Example 17.5 (cont.)
Solution It takes 25.0 mL of titrant to reach the equivalence point, so the total solution is = 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) = M. We must use this concentration as the starting concentration of ammonium ion in our equilibrium table.
The equilibrium expression is
Think About It In the titration of a weak base with a strong acid, the species in solution at the equivalence point is the conjugate acid. Therefore, we should expect an acidic pH. Once all the NH3 has been converted to NH4+, there is no longer anything in the solution to consume the added acid. Thus, the pH after the equivalence point depends on the number of millimoles of H+ added and not consumed divided by the new total volume.
NH4+(aq) H2O(l) ⇌ NH3(aq) H3O+(aq)
Initial concentration (M)
0.0500
Change in concentration (M) –x +x
Equilibrium concentration (M)
– x x
[NH3][H+]
[NH4+]
(x)(x)
– x x2
0.0500
Ka = = ≈
= 5.6×10-10
x = = 5.3×10-6 M = [H+]
pH = –log(5.3×10-6) = 5.28

36. HIn(aq) ⇌ H+(aq) + In–(aq)
Acid-Base Titrations
The equivalence point in a titration can be visualized with the use of an acid-base indicator.
The endpoint of a titration is the
point at which the color of the
indicator changes.
HIn(aq) ⇌ H+(aq) In–(aq)

37. Acid-Base Titrations

38. Acid-Base Titrations

39. 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.

40. 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.
(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.
Think About It If we don’t select an appropriate indicator, the endpoint (color change) will not coincide with the equivalence point.

41. Solubility Equilibria
17.4
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
Aluminum hydroxide
Al(OH)3(s) ⇌ Al3+(aq) + 3OH–(aq)
1.8 x 10–33
Calcium fluoride
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

42. 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.

43. 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)
Change in concentration (M) +s
Equilibrium concentration (M) s
Ksp = [Ag+][Br–] = 7.7 x 10–13
7.7 x 10–13 = (s)(s)
s = 8.8 x 10–7 M

45. 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 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× The molar mass of Cu(OH)2 is g/mol.
Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH-(aq)
Initial concentration (M)
Change in concentration (M) +s
+2s
Equilibrium concentration (M) s 2s

46. Worked Example 17.7 (cont.) Solution Therefore,
2.2×10-20 = (s)(2s)2 = 4s3
s = = 1.8×10-7 M
The molar solubility of Cu(OH)2 is 1.8×10-7 M. Multiplying by its molar mass gives
1.8×10-7 mol L
97.57 g Cu(OH)2
1 mol Cu(OH)2
solubility of Cu(OH)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).

47. 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 many of the Ksp values in Table 17.4). In fact, sulfates are listed as soluble compounds in Table 9.2, but calcium sulfate is listed as an insoluble exception. Remember that the term insoluble really refers to compounds that are slightly soluble, and that different sources may differ with regard to how soluble a compound must be to be considered soluble.
Solution The molar mass of CaSO4 is g/mol. The molar solubility of CaSO4 is
The equation and the equilibrium expression for the dissociation of CaSO4 are
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
0.67 g CaSO4
1 L
1 mol CaSO4
136.2 g CaSO4
molar solubility of CaSO4 = ×
s = 4.9×10-3 mol/L

48. Solubility Equilibria
For the dissociation of an ionic solid in water, the following conditions may exist:
The solution is unsaturated
The solution is saturated
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

49. Worked Example 17.9
Predict whether a precipitate will form when each of the following is added to 650 mL of M K2SO4: (a) 250 mL of 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× Concentrations of the constituent ions of BaSO4 are:
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.
[Ba2+] =
250 mL × M
650 mL mL
= M
[SO42-] =
650 mL × M
650 mL mL
= M

50. 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:
Using these concentrations in the equilibrium expression, [Ag+]2[SO42-], gives a reaction quotient of (0.032)2(0.0063) = 6.5×10-6, which is less than the Ksp of Ag2SO4 (1.5×10-5). Therefore, Ag2SO4 will not precipitate.
(c) SrSO4 might form and its Ksp = 3.8×10-7. Concentrations of the constituent ions of Ag2SO4 are:
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.
[Ag+] =
175 mL × 0.15 M
650 mL mL
= M
[SO42-] =
650 mL × M
650 mL mL
= M
Think About It Students sometimes have difficulty deciding what compound might precipitate. Begin by writing down the constituent ions in the two solutions before they are combined. Consider the two possible combinations; the cation from the first solution and the anion from the second, or vice versa. You can consult the information in Table 9.2 and 9.3 to determine whether one of the combinations is insoluble. Also keep in mind that only an insoluble salt will have a tabulated Ksp value.
[Sr2+] =
325 mL × 0.25 M
650 mL mL
= M
[SO42-] =
650 mL × M
650 mL mL
= M

51. Factors Affecting Solubility
17.5
Several factors exist that affect the solubility of ionic compounds:
Common ion effect pH
Formation of complex ions

52. AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
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)
Initial concentration (M)
6.5×10-3
Change in concentration (M) +s
Equilibrium concentration (M)
6.5× s s

53. Worked Example (cont.)
Solution Substituting these concentrations into the equilibrium expression gives
1.6×10-10 = (6.5× s)(s)
We expect s to be very small, so
6.5× s ≈ 6.5×10-3
and
1.6×10-10 = (6.5×10-3)(s)
Thus
Therefore, the molar solubility of AgCl in 6.5×10-3 M AgNO3 is 2.5×10-8 M.
s =
1.6×10-10
6.5×10-3
= 2.5×10-8 M
Think About It The molar solubility of AgCl in water is = 1.3×10-5 M. The presence of 6.5×10-3 M AgNO3 reduces the solubility of AgCl by a factor of ~500.

54. Factors Affecting Solubility
pH:
Mg(OH)2(s) ⇌ Mg2+ (aq) OH–(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 = – 3.55 = 10.45
In a solution with a pH of less than 10.45, the solubility of Mg(OH)2 increases.

55. Factors Affecting Solubility
pH:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq)
2H+(aq) + 2OH–(aq) → 2H2O(l)
Overall:
Mg(OH)2(s) + 2H+(aq) ⇌ 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.

56. 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–

57. 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)

58. Worked Example (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.

59. 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.

60. 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)

61. 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)

62. Factors Affecting Solubility

63. Factors Affecting Solubility

64. Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)
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× The equilibrium expression for the dissociation is
[Cd2+][CN-]4
[Cd(CN)42-]
1.4×10-17 =

65. Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)
Worked Example (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.
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 x ≈ 1.2), so the solution becomes
and x = 1.4×10-18 M.
Think About It When you assume that all the metal ion is consumed and converted to complex ion, it’s important to remember that some of the complexing agent (in this case, CN- ion) is consumed in the process. Don’t forget to adjust its concentration accordingly before entering it in the top row of the equilibrium table.
Cd(CN)42-(aq) ⇌ Cd2+(aq) + 4CN-(aq)
Initial concentration (M)
0.20
1.2
Change in concentration (M) -x +x
+4x
Equilibrium concentration (M) x x x
= 1.4×10-17
[Cd2+][CN-]4
[Cd(CN)42-]
x(1.2)4
0.20 =

66. Separation of Ions Using Differences in Solubility
17.6
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

67. Worked Example 17.13
Silver nitrate is added slowly to a solution that is M in Cl- ions and 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.

68. Worked Example (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
Ksp
[Br-]
[Ag+] =
7.7×10-13
0.020 =
= 3.9×10-11 M

69. Worked Example (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
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.
Ksp
[Cl-]
[Ag+] =
1.6×10-10
0.020 =
= 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.
Thus, by the time AgCl begins to precipitate, (9.6×10-5 M)÷(0.020 M) = , so less than 0.5 percent of the original bromide ion remains in the solution.
Ksp
[Ag+]
[Br-] =
7.7×10-13
8.0×10-9 =
= 9.6×10-5 M

70. 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.

71. 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