1. Acid-Base Equilibria and Solubility Equilibria
Chapter 16

2. Acid-Base Equilibria and Solubility Equilibria
We will continue to study acid-base reactions
Buffers
Titrations
We will also look at properties of slightly soluble compunds and their ions in solution

3. Homogeneous vs. Heterogeneous
Homogeneous Solution Equilibria-
a solution that has the same composition throughout after equilibrium has been reached.
Heterogeneous Solution Equilibria-
a solution that after equilibrium has been reached, results in components in more than one phase.

4. The Common Ion Effect Acid-Base Solutions Common Ion
The Common Ion Effect- the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. pH
Instead of looking at solutions with one solute, we will be looking at solutions with two solutes.
We need to know that the two solutes have an ion in common. The result of this causes the equilibrium of the solution to shift.
Common ion effect- special case of le chatlier’s principle.

5. The Common Ion Effect
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
common ion
CH3COOH (aq) H+ (aq) + CH3COO- (aq)

6. Henderson-Hasselbach Equation
Relationship between pKa and Ka.
pKa = -log Ka
Henderson-Hasselbalch equation
pKa is calcualted the same way as pH. Taking this into consideration and solving for pH, we can create an equation that relates the pKa of a solute to the pH of that solute in solution.
Remember the stronger the acid, the larger the ka and the smaller the pKa.
pH = pKa + log
[conjugate base]
[acid]

7. The Henderson-Hasselbalch Equation
[H+] [Ac-]
HAc H+ + Ac-
Ka =
[HAc]
[Ac-]
take the -log on both sides
-log Ka = -log [H+] -log
[HAc]
[Ac-]
apply p(x) = -log(x)
pKa = pH log
[HAc]
and finally solve for pH…
Logxy = logx + logy
[Ac-]
[Proton acceptor]
pH =
pKa + log
= pKa + log
[HAc]
[Proton donor]

8. 1.0 = log [Acetate] 10x then *[Acetic acid] [Acetic acid]
Acetic acid has a pKa of How many ml of 0.1 M acetic acid and 0.1 M sodium acetate are required to prepare 1 liter of 0.1 M buffer with a pH of 5.8?
Substitute the values for the pKa and pH into the Henderson-Hasselbalch equation:
5.8 = log [Acetate]
[Acetic acid]
1.0 = log [Acetate] x then *[Acetic acid]
[Acetic acid]
on both sides
10 [Acetic acid] = [Acetate]
For each volume of acetic acid, 10 volumes of acetate must be added (total of 11 volumes).
Acetic acid needed: 1/11 x 1,000 ml = 91 ml
Acetate needed: /11 x 1,000 ml = 909 ml

9. Henderson-Hasselbach Equation
pH = pKa + log
[A- ]
[HA]
[HA] is the molar concentration of the undissociated weak acid,
[A⁻] is the molar concentration of this acid's conjugate base,
Ka and is the acid dissociation constant
pKa is calcualted the same way as pH. Taking this into consideration and solving for pH, we can create an equation that relates the pKa of a solute to the pH of that solute in solution.
Remember the stronger the acid, the larger the ka and the smaller the pKa.
pOH = pKb + log
[BH+]
[B]
[B] is the molar concentration of the undissociated weak base,
[BH+] is the molar concentration of this base's conjugate acid,
Kb and is the base dissociation constant

10. How does a buffer work? -
At the pKa, [HAc] = [Ac-] so the system is able to absorb the addition of HO- or H+. If we add HO- near the pH where [Hac] = [Ac-] (ie pH ~= pKa) then HAc can release H+ to offset the HO- added but the ratio of HAc to Ac- does not change much. If we add H+ then Ac- can absorb H+ to form HAc. Hence, the pH does not change much.

11. pH Calculations
What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
HCOOH pKa = 3.77
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00
0.52 -x
HCOOK—strong acid. Dissociates completely in water. +x +x x x x

12. pH Calculations [HCOO-] pH = pKa + log [HCOOH] Common ion effect
0.30 – x  0.30
x  0.52
pH = log
[0.52]
[0.30]
Remember to check assumption!!!
You can also use the h-h equation for bases too.
= 4.01

13. A buffer solution is a solution of:
A weak acid or a weak base and
The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
Add strong acid
H+ (aq) + CH3COO- (aq) CH3COOH (aq)
Add strong base
OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

14. Which of the following are buffer systems? (a) KF/HF
(b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) HF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution

15. Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.

16. Calculating pH Changes in Buffers
Before the reaction, since
mol CH3COOH = mol CH3COO−
pH = pKa = −log (1.8  10−5) = 4.74

17. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
CH3COOH (aq) + OH−(aq)  CH3COO−(aq) + H2O(l)
CH3COOH
CH3COO−
OH−
Before reaction
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol
New concentrations: mol/ 1 L = M CH3COOH
0.320 mol/ 1 L = M CH3COO-

18. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0. 280)
pH =
pH = 4.80

19. KH2PO4 H2PO4- + K+ and K2HPO4 HPO42- + 2 K+
To make a pH 7 buffer. You already have 0.1 M KH2PO4. What concentration of K2HPO4 do you need?
KH2PO H2PO K+ and K2HPO HPO K+
H2PO HPO H pKa = 6.86
[HPO42-]
pH = 7 = pKa + log
[H2PO4-]
7 = log(x / 0.1 M)
0.14 = log(x / 0.1 M)
= x / 0.1 M
x = M = [K2HPO4]

20. Another HH calculation
Make 200 ml of 0.1 M Na acetate (CH3COONa) buffer pH 5.1, starting with 5.0 M acetic acid (CH3COOH) and 1.0 M NaOH.
Strategy
1. Calculate the total amount of acetic acid needed.
2. Calculate the ratio of the two forms of acetate (A- and HA) that will exist when the pH is 5.1.
3. Use this ratio to calculate the % of acetate that will be in the A- form.
4. Assume that each NaOH will convert one HAc to an Ac-. Use this plus the % A- to calculate the amount of NaOH needed to convert the correct amount of HAc to Ac-.

21. (1) How much acetic acid is needed?
200 ml x 0.1 mol/l = 200 ml x 0.1 mmol/ml = 20 mmol
5.0 mol/l x x ml = 5.0 mmol/ml x x ml = 20 mmol
x = 4.0 ml of 5.0 M acetic acid are 20 mmol
(2) What is the ratio of Ac- to HAc at pH 5.1?
[Ac-]
pH =
pKa + log HH equation
[HAc]
= log[A-]/[HAc], thus [Ac-]/[HAc] = 2.19 / 1
(3) What fraction of total acetate is Ac- at pH 5.1?
[Ac-] [Ac-]
–––– = 2.19; –––––––––– = ––––––– = or 68.7%
[HAc] [HAc] + [Ac-]

22. (4) How much OH- is needed to obtain 68.7% Ac-?
Na+ + OH- + HAc  Na+ + Ac- + H2O
mmol NaOH = x 20 mmol = 13.7 mmol
1.0 mol/l x x ml = 1.0 mmol/ml x x ml = 13.7 mmol
x = 13.7 ml of 1.0 M NaOH
Final answer (Jeopardy…)
• 4.0 ml of 5.0 M acetic acid
• ml of 1.0 M NaOH
• Bring to final volume of 200 ml with water
(ie add about ml of H2O).

23. Calculate the pH of the 0. 30 M NH3/0. 36 M NH4Cl buffer system
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log
[NH3]
[NH4+]
pH = log
[0.30]
[0.36]
pKa = 9.25
= 9.17
start (moles)
0.029
0.001
0.024
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
end (moles)
0.028
0.0
0.025
final volume = 80.0 mL mL = 100 mL
[NH4+] =
0.028
0.10
[NH3] =
0.025
0.10
pH = log
[0.25]
[0.28]
= 9.20

24. The Common Ion Effect and Buffer Solutions
Example 16-1: Calculate the concentration of H+and the pH of a solution that is 0.15 M in CH3COOH and 0.15 M in CH3COONa.
This is another equilibrium problem with a starting concentration for both the acid and anion.

25. The Common Ion Effect and Buffer Solutions
Substitute the quantities determined in the previous relationship into the ionization expression.

26. The Common Ion Effect and Buffer Solutions
Apply the simplifying assumption to both the numerator and denominator.

27. Weak Bases plus Salts of Weak Bases
Example 16-2: Calculate the concentration of OH- and the pH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3.

28. Weak Bases plus Salts of Weak Bases
Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

29. HCl + CH3COO- CH3COOH + Cl-
HCl H+ + Cl-
HCl + CH3COO CH3COOH + Cl-

30. Equilibria of Acid-Base Buffer Systems
Calculating the Effect of Added H3O+ or OH- on Buffer pH
PROBLEM:
Calculate the pH:
(a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa
(b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a)
(c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a)
Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume changes.
PLAN:
We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system.
SOLUTION:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (M)
(a)
Initial
Change
Equilibrium
0.50 -
0.50
- x -
+ x
+ x
0.50-x -
0.50 +x x

31. Equilibria of Acid-Base Buffer Systems
Calculating the Effect of Added H3O+ and OH- on Buffer pH
continued (2 of 4)
[H3O+] = x
[CH3COOH]equil ≈ 0.50M
[CH3COO-]initial ≈ 0.50M
[H3O+][CH3COO-]
[CH3COOH]
[H3O+] = x = Ka
[CH3COO-]
[CH3COOH]
Ka =
= 1.8x10-5M
Check the assumption:
1.8x10-5/0.50 X 100 = 3.6x10-3 %
(b)
[OH-]added =
0.020 mol
1.0L soln
= 0.020M NaOH
CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)
Concentration (M)
Before addition
0.50 -
0.50 -
Addition -
0.020 - -
After addition
0.48
0.52 -

32. Equilibria of Acid-Base Buffer Systems
Calculating the Effect of Added H3O+ and OH- on Buffer pH
continued (3 of 4)
Set up a reaction table with the new values.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (M)
Initial
0.48 -
0.52
Change
- x -
+ x
+ x
Equilibrium
0.48 -x -
0.52 +x x
[H3O+] = 1.8x10-5
0.48
0.52
= 1.7x10-5
pH = 4.77
0.020 mol
1.0L soln
(c)
[H3O+]added =
= 0.020M H3O+
Before addition
0.50 -
0.50 -
Addition -
0.020 - -
After addition
0.48
0.52 -

33. Equilibria of Acid-Base Buffer Systems
Calculating the Effect of Added H3O+ and OH- on Buffer pH
continued (4 of 4)
Set up a reaction table with the new values.
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
Concentration (M)
Initial
0.52 -
0.48
Change
- x -
+ x
+ x
Equilibrium
0.52 -x -
0.48 +x x
[H3O+] = 1.8x10-5
0.48
0.52
= 2.0x10-5
pH = 4.70

34. 19.1 - Equilibria of Acid-Base Buffer Systems
The Henderson-Hasselbalch Equation
HA + H2O H3O+ + A-
Ka =
[H3O+] [A-]
[HA]
Ka [HA]
[A-]
[H3O+] =
[A-]
[HA]
- log[H3O+] = - log Ka + log
pH = pKa + log
[base]
[acid]

35. Equilibria of Acid-Base Buffer Systems
Buffer capacity is the ability to resist pH change.
The more concentrated the components of a buffer, the greater
the buffer capacity.
The pH of a buffer is distinct from its buffer capacity.
A buffer has the highest capacity when the component
concentrations are equal.
Buffer range is the pH range over which
the buffer acts effectively.
Buffers have a usable range within ± 1 pH unit of the pKa of
its acid component.

36. Buffer Capacity
Amount of H+ or OH- it can absorb without a significant change in pH.
For HA/A- system, buffer capacity depends on:
[HA] and [A-] (higher = higher)
[HA] ratio (closer to 1 = higher)
[A-]

37. Equilibria of Acid-Base Buffer Systems
The relation between buffer capacity and pH change.

38. Preparing a buffer 1. Choose and acid-conjugate base pair
2. Calculate the ratio of the buffer component pairs
3. Determine the buffer concentration
How would you prepare a benzoic acid/benzoate buffer with pH = 4.25, starting with 5.0 L of M sodium benzoate (C6H5COONa) solution and adding the acidic component?
Ka of benzoic acid (C6H5COOH) = 6.3 x 10-5

39. Equilibria of Acid-Base Buffer Systems
Preparing a Buffer
PROBLEM:
An environmental chemist needs a carbonate buffer of pH to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5L of freshly prepared 0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.
PLAN:
We know the Ka and the conjugate acid-base pair. Convert pH to [H3O+], find the number of moles of carbonate and convert to mass.
SOLUTION:
Ka =
[CO32-][H3O+]
[HCO3-]
HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq)
4.7x10-11 =
[CO32-](0.20)
1.0x10-10
pH = 10.00; [H3O+] = 1.0x10-10
[CO32-] = 0.094M
moles of Na2CO3 = (1.5L)(0.094mols/L)
= 0.14
0.14 moles
105.99g
mol
= 15 g Na2CO3

40. Titration (pH) Curves Plot pH of solution vs. amount of titrant added
Equivalence point:
Enough titrant added to react exactly with solution being analyzed.

41. Titration of Strong Acid with Strong Base

42. Weak Acid-Strong Base Titration
1. Stoichiometry
Reaction assumed to run to completion
2. Equilibrium
Use weak acid equilibrium to find pH

43. Weak Acid-Strong Base Titration

44. Differences

45. Differences and Ka

46. Titration Calculations
1. Solution of HA
2. Solution of HA and added base
3. Equivalent amounts of HA and added base
4. Excess base
A chemist titrates mL of M HBrO
(Ka = 2.3 x 10-9) with M NaOH. Find the pH:
(a) before any base is added
(b) when mL of NaOH is added
(c) at the equivalence point
(d) when the moles of OH- added are twice the moles of HBrO originally present?

47. Titration of Strong Base with Strong Acid

48. Weak Base-Strong Acid Titration

49. Acid-Base Indicator Indicates endpoint of a titration
Endpoint is not necessarily the equivalence point

51. Curve for a strong acid-strong base titration

52. Acid-base Titration Curves
Titration of 40.00mL of M HPr with M NaOH
Curve for a weak acid-strong base titration
pH = 8.80 at equivalence point
pKa of HPr = 4.89
methyl red
[HPr] = [Pr-]

53. Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl
pKa of NH4+ = 9.25
Curve for a weak base-strong acid titration
pH = 5.27 at equivalence point

54. Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH
Curve for the titration of a weak polyprotic acid (H2SO3).
pKa = 7.19
pKa = 1.85
Titration of 40.00mL of M H2SO3 with M NaOH

55. Acid-base Titration Curves
Calculating the pH During a Weak Acid-Strong Base Titration
PROBLEM:
Calculate the pH during the titration of mL of M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of M NaOH:
(a) 0.00mL
(b) mL
(c) mL
(d) mL
PLAN:
The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition.
SOLUTION:
(a) Find the starting pH using the methods of Chapter 18.
Ka = [Pr-][H3O+]/[HPr]
[Pr-] = x = [H3O+]
[Pr-] = x = [H3O+]
x = 1.1x10-3 ; pH = 2.96
(b)
HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)
Amount (mol)
Before addition - -
Addition - - -
After addition -

56. Acid-base Titration Curves
Calculating the pH During a Weak Acid-Strong Base Titration
continued
[H3O+] = 1.3x10-5
mol
mol
= 4.3x10-6M
pH = 5.37
(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be
( mol)
( L) + ( L)
= M
Ka x Kb = Kw
Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10
[H3O+] = Kw / = 1.6x10-9M
pH = 8.80
(d) 50.00mL of NaOH will produce an excess of OH-.
mol XS base = (0.1000M)( L L) = mol
[H3O+] = 1.0x10-14/ = 9.0x10-11M
M =
pH = 12.05

57. Ksp Solubility product constant
Shows the solubility of a given solid at a specific temperature

58. Solubility Constant, Ksp
For solids dissolving to form aqueous solutions.
For slightly soluble salts:
equilibrium between solid and component ions

59. Ksp Write the expression for Ksp for (a) CaSO4 (b) Cr2CO3 (c) Mg(OH)2
(d) As2S3

60. Solubility Equilibria
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
Ksp is the _solubility product constant_
MgF2 (s) Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
Ag2CO3 (s) Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
17.6

61. 17.6

62.  What is the solubility of silver chloride in g/L ?
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Initial (M)
Change (M)
Equilibrium (M)
0.00
0.00
Ksp = [Ag+][Cl-] +s +s
Ksp = s2
s = Ksp  s s
s = 1.3 x 10-5
[Ag+] = 1.3 x 10-5 M
[Cl-] = 1.3 x 10-5 M
1.3 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
Solubility of AgCl =
= 1.9 x 10-3 g/L
17.6

63. Determining Ksp from Solubility
PROBLEM:
(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?
PLAN:
Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.
SOLUTION:
PbSO4(s) Pb2+(aq) + SO42-(aq)
(a)
Ksp = [Pb2+][SO42-]
1000mL L
4.25x10-3g
100mL soln
303.3g PbSO4
mol PbSO4
= 1.40x10-4M PbSO4
Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =
1.96x10-8

64. Relationship Between Ksp and Solubility at 250C
No. of Ions
Formula
Cation:Anion
Ksp
Solubility (M) 2
MgCO3
1:1 s2
s = (Ksp)1/2 2
PbSO4
1:1 s2
s = (Ksp)1/2 2
BaCrO4
1:1 s2
s = (Ksp)1/2 3
Ca(OH)2
1:2
4s3
s = (Ksp/4)1/3 3
BaF2
1:2
4s3
s = (Ksp/4)1/3 4
Al(OH)3
1:3
27s4
s = (Ksp/27)1/4 3
Ca3(PO4)2
3:2
108s5
s = (Ksp/108)1/5
Ksp = [s][ns]n

65. Table 16.3 Relationship Between Ksp and Solubility at 250C
No. of Ions
Formula
Cation:Anion
Ksp
Solubility (M) 2
MgCO3
1:1
3.5 x 10-8
1.9 x 10-4 2
PbSO4
1:1
1.6 x 10-8
1.3 x 10-4 2
BaCrO4
1:1
2.1 x 10-10 ? 3
Ca(OH)2
1:2
5.5 x 10-6
1.2 x 10-2 3
BaF2
1:2
1.5 x 10-6 ? 3
Al(OH)3
1:3
1.8 x 10-13 ? 3
Ca3(PO4)2
3:2
1.2 x 10-26 ?
Ksp = [s][ns]n

66. 17.6

68. The ions present in solution are Na+, OH-, Ca2+, Cl-.
If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]0 2
= 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
17.6

69. AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13 Ksp = [Ag+][Br-]
What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br- and Cl- at a concentration of 0.02 M?
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
[Ag+] =
Ksp
[Br-]
7.7 x 10-13
0.020 =
= 3.9 x M
AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
[Ag+] =
Ksp
[Cl-]
1.6 x 10-10
0.020 =
= 8.0 x 10-9 M
3.9 x M < [Ag+] < 8.0 x 10-9 M
17.7

70. The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water and (b) M NaBr?
NaBr (s) Na+ (aq) + Br- (aq)
AgBr (s) Ag+ (aq) + Br- (aq)
[Br-] = M
Ksp = 7.7 x 10-13
AgBr (s) Ag+ (aq) + Br- (aq)
s2 = Ksp
[Ag+] = s
s = 8.8 x 10-7
[Br-] = s 
Ksp = x s
s = 7.7 x 10-10
17.8

71. pH and Solubility Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)
At pH less than 10.45
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Lower [OH-]
Ksp = (s)(2s)2 = 4s3
OH- (aq) + H+ (aq) H2O (l)
4s3 = 1.2 x 10-11
Increase solubility of Mg(OH)2
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
At pH greater than 10.45
pOH = pH = 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
17.9

72. Equilibria of Slightly Soluble Ionic Compounds
Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds
PROBLEM:
Write the ion-product expression for each of the following:
(a) Magnesium carbonate
(b) Iron (II) hydroxide
(c) Calcium phosphate
(d) Silver sulfide
PLAN:
Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d).
SOLUTION:
(a) MgCO3(s) Mg2+(aq) + CO32-(aq)
Ksp = [Mg2+][CO32-]
(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)
Ksp = [Fe2+][OH-] 2
(c) Ca3(PO4)2(s) Ca2+(aq) + 2PO43-(aq)
Ksp = [Ca2+]3[PO43-]2
(d) Ag2S(s) Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]
Ag2S(s) + H2O(l) Ag+(aq) + HS-(aq) + OH-(aq)

73. Equilibria of Slightly Soluble Ionic Compounds
Determining Ksp from Solubility
PROBLEM:
(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?
(b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2.
PLAN:
Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.
SOLUTION:
PbSO4(s) Pb2+(aq) + SO42-(aq)
(a)
Ksp = [Pb2+][SO42-]
1000mL L
4.25x10-3g
100mL soln
303.3g PbSO4
mol PbSO4
= 1.40x10-4M PbSO4
Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =
1.96x10-8

74. Equilibria of Slightly Soluble Ionic Compounds
Determining Ksp from Solubility
continued
(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2
0.64g
L soln
245.2g PbF2
mol PbF2
= 2.6x10-3 M
Ksp = (2.6x10-3)(5.2x10-3)2 =
7.0x10-8

75. Equilibria of Slightly Soluble Ionic Compounds
Solubility-Product Constants (Ksp)
of Selected Ionic Compounds at 250C
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
3 x 10-34
Cobalt (II) carbonate, CoCO3
1.0 x 10-10
Iron (II) hydroxide, Fe(OH)2
4.1 x 10-15
Lead (II) fluoride, PbF2
3.6 x 10-8
Lead (II) sulfate, PbSO4
1.6 x 10-8
Mercury (I) iodide, Hg2I2
4.7 x 10-29
Silver sulfide, Ag2S
8 x 10-48
Zinc iodate, Zn(IO3)2
3.9 x 10-6

76. Equilibria of Slightly Soluble Ionic Compounds
Determining Solubility from Ksp
PROBLEM:
Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.
PLAN:
Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table.
SOLUTION:
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Concentration (M)
Initial -
Change - +S
+ 2S
Equilibrium - S 2S
Ksp = (S)(2S)2
S =
= 1.2x10x-2M

77. Equilibria of Slightly Soluble Ionic Compounds
Relationship Between Ksp and Solubility at 250C
No. of Ions
Formula
Cation:Anion
Ksp
Solubility (M) 2
MgCO3
1:1
3.5 x 10-8
1.9 x 10-4 2
PbSO4
1:1
1.6 x 10-8
1.3 x 10-4 2
BaCrO4
1:1
2.1 x 10-10
1.4 x 10-5 3
Ca(OH)2
1:2
5.5 x 10-6
1.2 x 10-2 3
BaF2
1:2
1.5 x 10-6
7.2 x 10-3 3
CaF2
1:2
3.2 x 10-11
2.0 x 10-4 3
Ag2CrO4
2:1
2.6 x 10-12
8.7 x 10-5

78. Equilibria of Slightly Soluble Ionic Compounds
The effect of a common ion on solubility
PbCrO4(s) Pb2+(aq) + CrO42-(aq)
CrO42- added
PbCrO4(s) Pb2+(aq) + CrO42-(aq)

79. Equilibria of Slightly Soluble Ionic Compounds
Calculating the Effect of a Common Ion on Solubility
PROBLEM:
In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
PLAN:
Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved.
SOLUTION:
Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)
Concentration(M)
Initial -
0.10
Change - +S
+2S
Equilibrium - S 2S
Ksp = 6.5x10-6 = ( S)(2S)2 = (0.10)(2S)2
S << 0.10
Check the assumption:
S =
= 4.0x10-3
0.10M
4.0x10-3
x 100 =
4.0%

80. Equilibria of Slightly Soluble Ionic Compounds
Predicting the Effect on Solubility of Adding Strong Acid
PROBLEM:
Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:
(a) Lead (II) bromide
(b) Copper (II) hydroxide
(c) Iron (II) sulfide
PLAN:
Write dissolution equations and consider how strong acid would affect the anion component.
SOLUTION:
(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)
No effect.
(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)
OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility.
(c) FeS(s) Fe2+(aq) + S2-(aq)
S2- is the anion of a weak acid and will react with water to produce OH-.
FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)
Both weak acids serve to increase the solubility of FeS.

81. Equilibria of Slightly Soluble Ionic Compounds
Predicting Whether a Precipitate Will Form
PROBLEM:
A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?
PLAN:
Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.
SOLUTION:
CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11
mol Ca2+ = L(0.30mol/L) = 0.030mol
[Ca2+] = 0.030mol/0.300L = 0.10M
mol F- = L(0.060mol/L) = 0.012mol
[F-] = 0.012mol/0.300L = 0.040M
Q = [Ca2+][F-]2 =
(0.10)(0.040)2 = 1.6x10-4
Q is >> Ksp and the CaF2 WILL precipitate.

82. Ionic Equilibria in Chemical Analysis
Separating Ions by Selective Precipitation
PROBLEM:
A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.
PLAN:
Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility.
It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution.
SOLUTION:
Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10
Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20
[OH-] needed for a saturated Mg(OH)2 solution =
= 5.6x10-5M

83. Ionic Equilibria in Chemical Analysis
Separating Ions by Selective Precipitation
continued
Use the Ksp for Cu(OH)2 to find the amount of Cu remaining.
[Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 =
7.0x10-12M
Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.

84. Ionic Equilibria in Chemical Analysis
The general procedure for separating ions in qualitative analysis.
Add precipitating ion
Add precipitating ion
Centrifuge
Centrifuge

85. Ionic Equilibria in Chemical Analysis
A qualitative analysis scheme for separating cations into five ion groups.
Acidify to pH 0.5; add H2S
Centrifuge
Add NH3/NH4+ buffer(pH 8)
Centrifuge
Add (NH4)2HPO4
Centrifuge
Add
6M HCl
Centrifuge