1. Acid-Base Equilibria and Solubility Equilibria
PowerPoint Lecture Presentation by J. David Robertson University of Missouri
Acid-Base Equilibria and Solubility Equilibria
Chapter 16
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. SOLUTION EQUILIBRIA HOMOGENOUS EQUILIBRIA HETEROGENOUS EQUILIBRIA
Involves dissolution & precipitation of slightly soluble subtances
Involves incomplete ionization of weak acid or weak base.
Solubility equilibria
Acid-base equilibria
Acid-base ionization
Salt hydrolisis
Buffer
Calculations using Ka or Kb
Solubility of ionic compound in water
Calculation using Ksp

3. Common Ion Effect Consider the ionization of weak acid HA:
HA + H2O  H3O+ + A–
If NaA salt is added to the solution, what is the effect of the addition to the acid ionization and pH of solution?
NaA  Na+ + A–
Which way does equilibrium shift? Does this agree with Le Chatelier’s principle?

4. A- is a “common ion”. Acid dissociation: HA + H2O  H3O+ + A–
Salt dissociation:
NaA  Na+ + A–
A– has 2 sources: from HA and NaA. Adding NaA increases [A-].
A- is a “common ion”.
In presence of NaA, the equilibrium will shift to the left.

5. ‘See-saw concept’:
Before adding NaA: HA + H2O  H3O+ + A–
Adding NaA, more A– ions:
HA + H2O  H3O+ + A– A–
Le Chatelier’s principle: The disturbance of an added reactant or product is relieved by shifting to the direction that consumes the added substance.
 To consumes the A–, equilibrium will be shifted to the left. Ionization of HA will be reduced.

6. CH3COONa (s) Na+ (aq) + CH3COO- (aq)
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance (the behaviour of solution in which the same ion is produced by two different compounds).
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).
common ion
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
16.2

7. Henderson-Hasselbalch equation
Consider mixture of salt NaA and weak acid HA.
NaA (s) Na+ (aq) + A- (aq)
Ka =
[H+][A-]
[HA]
HA (aq) H+ (aq) + A- (aq)
[H+] =
Ka [HA]
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
Henderson-Hasselbalch equation
-log [H+] = -log Ka + log
[A-]
[HA]
pH = pKa + log
[conjugate base]
[acid]
pH = pKa + log
[A-]
[HA]
pKa = -log Ka
16.2

8. Question #1: What is the pH of a solution containing 0
Question #1: What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Ka (HCOOH) = 1.7  10–4
Note: The solution is a mixture of weak acid, HCOOH and conjugate base (HCOO- from the salt)!
HCOOH (aq) H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
Initial (M)
Change (M)
Equilibrium (M) -x +x +x x x x
16.2

9. Method #1: Using assumption:
Assuming: x ≈ 0.52
0.30 – x ≈ 0.30
Checking assumption:
Assumption valid !!!
16.2

10. Method #2: using Henderson Hasselbalch equation:
*This method is recommended if the solution in question is a mixture of weak acid + salt or weak base + salt (buffer system).
16.2

11. Question #2: Calculate the pH of a solution containing 0
Question #2: Calculate the pH of a solution containing 0.50 M CH3COOH and 0.50 M CH3COONa. For CH3COOH, Ka = 1.8 × 10-5
Answer:
Salt dissociation:
CH3COONa → CH3COO– Na+
0.50M M M
Ionization of acid:
CH3COOH ⇄ CH3COO– H+
using Henderson Hasselbalch equation:

12. Question #3: Calculate the percent ionization of the acid and the pH of the solution that contains 0.500M HC2H3O2 (Ka = 1.8 x 10-5) and M NaC2H3O2.
Answer:
using Henderson Hasselbalch equation to calculate pH:
% ionization:

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)
16.3

14. Answer: Question #4: Which of the following are buffer systems?
(a) KF/HF
(b) KBr/HBr,
(c) Na2CO3/NaHCO3
Answer:
HF is a weak acid and F- is its conjugate base, therefore, it is a buffer system.
(b) HBr is a strong acid, so, it is not a buffer system.
(c) CO32- is a weak base and HCO3- is it conjugate acid, therefore, it is a buffer system.
16.3

15. Buffer Preparation
What if I need to make a buffer solution of known pH?
Select a buffer system with pKa of buffer acid close to the desired pH.
Solutions containing a weak acid and one of its ionic salts, are always less acidic than solutions containing the same concentration of the weak acid alone.
pH = pKa + log
[conjugate base]
[acid]
The equation above can be used to decide how to
prepare a buffer solution.

16. Question #5: Preparing a Buffer
You are given the following acids:
Methanoic acid, HCOOH: pKa = 3.75
Ethanoic acid, CH3COOH: pKa = 4.76
Explain how would you prepare a buffer solution with a pH of 5.70.
Answer:
This is an acidic buffer and ethanoic acid can be used as the pKa value is closer to the pH desired.
A suitable salt for ethanoic acid is sodium ethanoate, CH3COONa.
Next, use the Henderson Hasselbalch equation to determine the appropriate ratio of salt to acid to obtain the pH.

17. [conjugate base] [acid] [conjugate base] [acid] [conjugate base]
pH = pKa + log
[conjugate base]
[acid]
5.70 = log
[conjugate base]
[acid]
[conjugate base]
[acid]
[salt]
[acid] =
= 8.9
The buffer solution is prepared by dissolving 1 mol of ethanoic acid and 8.9 mol of sodium ethanoate in distilled water and make the solution up to 1 liter.

19. Calculation of pH before addition of NaOH:
Question #6: Adding base to a basic buffer system:
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. Kb = 1.8  10-5
What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
Answer:
Calculation of pH before addition of NaOH:
pOH = pKb + log
[NH4+]
[NH3]
pOH = log
[0.36]
[0.30]
pOH = 4.82
pH = – 4.82 = 9.18
16.3

20. Calculation of pH after addition of NaOH:
initial equation:
NH4+ (aq) H+ (aq) + NH3 (aq)
addition of NaOH:
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
0.36  80
1000
0.050  20
1000
0.30  80
1000
Initial
(mol) =
= 0.001
= 0.024
Change
(mol)
– 0.001
– 0.001
all NaOH will be used up
(*limiting reactant)
Equilibrium
(mol)
0.0278
0.025
16.3

21. Convert mol of each species to molarity:
Total volume = 80.0 mL mL = 100 mL
[NH3] =
0.025
0.10
[NH4+] =
0.0278
0.10
= 0.25 M
= M
using Henderson Hasselbalch equation:
pOH = pKb + log
[NH4+]
[NH3]
pOH = log
[0.278]
[0.25]
pOH = 4.79
pH = – 4.79 = 9.21
Before addition of NaOH: pH = 9.18
pH increases by only a small amount
After addition of NaOH: pH = 9.21
16.3

22. Question #7: Adding acid to a basic buffer system:
Calculate the pH of the 0.20 M NH3 / 0.20 M NH4Cl. For NH3, Kb = 1.8 × 10-5
What is the pH of the buffer after the addition of 10.0 mL 0.10 M HCl to 65.0 mL of the buffer?
Answer:
Calculation of pH before addition of HCl:
pOH = pKb + log
[NH4+]
[NH3]
pOH = log
[0.20]
pOH = 4.74
pH = – 4.74 = 9.26

23. Calculation of pH after addition of HCl:
initial equation:
H+ (aq) + NH3 (aq) NH4+ (aq)
addition of NaOH:
NH3 (aq) H+ (aq) NH4+ (aq)
0.20  65
1000
0.01  10
1000
0.20  65
1000
Initial
(mol)
= 0.013 =
= 0.013 – –
Change
(mol)
all HCl will be used up
(*limiting reactant)
Equilibrium
(mol)
0.0129
0.0131
16.3

24. Convert mol of each species to molarity:
Total volume = 65.0 mL mL = 75 mL
[NH3] =
0.0129
0.075
[NH4+] =
0.0131
0.075
= M
= M
using Henderson Hasselbalch equation:
pOH = pKb + log
[NH4+]
[NH3]
pOH = log
[0.175]
[0.172]
pOH = 4.75
pH = – 4.75 = 9.25
Before addition of HCl: pH = 9.26
pH decreases by only a small amount
After addition of HCl: pH = 9.25
16.3

25. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes colour at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7

26. Strong Acid-Strong Base Titrations (base added to acid):
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
OH- (aq) + H+ (aq) H2O (l)
*no hydrolysis of salt
16.4

28. Strong acid – strong base titration :
This reaction would require an indicator that changes color in pH range of about 7.
(1) Before any base is added, the acid dictates the pH.
(2) If some base is added, but before the equivalence point, the excess acid dictates the pH.
(3) At equivalence point, solution is neutral; there is equal amounts of acid and base.
(4) After the equivalence point, the excess base dictates the pH.
(4)
(3)
(2)
(1)

29. The curve for the acid titration of a base has the same general shape but is inverted.
Acid is in the conical flask, base is in the burette.
Base is in the conical flask, acid is in the burette.

30. Question #8: Calculating the pH of a Strong Acid-Strong Base Titration:
A titration between 0.20 M NaOH and 0.20 M HCl was conducted. Calculate the pH of the solution after:
Addition of 10.0 mL 0.20 M NaOH to 25.0 mL 0.20 M HCl;
Addition of equal amount : 25.0 ml 0.2 M NaOH to 25.0 mL 0.20 M HCl;
(c)Addition of 35.0 mL 0.20 M NaOH to 25.0 mL 0.20 M HCl.
Before equivalence point.
At equivalence point.
After equivalence point.

31. Addition of 10. 0 mL 0. 20 M NaOH to 25. 0 mL 0
Addition of 10.0 mL 0.20 M NaOH to 25.0 mL 0.20 M HCl (before equivalence point):
Moles of NaOH =
(0.20)(10.0)
1000
= mol
Moles of HCl =
(0.20)(25.0)
1000
= mol
NaOH HCl → NaCl H2O
Initial (mol)
Change (mol)
Equilibrium (mol)
0.002
0.005
– 0.002
– 0.002
0.003
[H+] =
0.003 mol
0.035 L
pH = - log = 1.07
= M

32. (b) Addition of equal amount : 25. 0 ml 0. 2 M NaOH to 25. 0 mL 0
(b) Addition of equal amount : 25.0 ml 0.2 M NaOH to 25.0 mL 0.20 M HCl (at equivalence point):
Complete neutralization reaction and the salt (NaCl) does not undergo hydrolysis.
At the equivalence point:
[H+] = [OH–] = 1.00  10-7
pH = 7.00

33. (c) Addition of 35. 0 mL 0. 20 M NaOH to 25. 0 mL 0
(c) Addition of 35.0 mL 0.20 M NaOH to 25.0 mL 0.20 M HCl (after equivalence point):
Moles of NaOH =
(0.20)(35.0)
1000
= mol
Moles of HCl =
(0.20)(25.0)
1000
= mol
NaOH HCl → NaCl H2O
Initial (mol)
Change (mol)
Equilibrium (mol)
0.007
0.005
– 0.005
– 0.005
0.002
[OH–] =
0.002 mol
0.060 L
pOH = – log = 1.48
= M
pH = – 1.48 = 12.52

34. Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)
16.4

35. At equivalence point, pH of solution is controlled
by hydrolysis of the anion of the acid:
Pr – + H2O ⇌ HPr + OH

36. Question #9: Calculating the pH of a Weak Acid-Strong Base Titration:
Calculate the pH during the titration of mL of M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of M NaOH :
(a) 0.00 mL (b) mL (c) mL.
Answer:
(a) 0.00 mL NaOH added:
HNO2 (aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)
Initial (M)
Change (M)
Equilibrium (M)
– x x x
0.250 – x x x

37. Assuming x << M:
Checking assumption:
Valid!!

38. (b) mL NaOH added:
HNO2 (aq) + OH–(aq) ⇌ H2O(aq) + NO2-–(aq)
0.250  20
1000
0.150  15
1000
Initial (mol)
= 0.005 =
Change (mol) – –
Equilibrium
(mol)
0.035
0.035
Equilibrium
(M)
= 0.064
= 0.079
pH = pKa + log
[NO2–]
[HNO2]
pH = log
0.064
0.079
= 3.26

39. (b) mL NaOH added:
HNO2 (aq) + OH–(aq) ⇌ H2O(aq) + NO2-–(aq)
0.250  20
1000
0.150  35
1000
Initial (mol)
= 0.005 =
Change (mol)
– 0.005
– 0.005
Equilibrium
(mol)
0.005
0.055
0.005
0.055
Equilibrium
(M)
= 0.091 =
*There is an excess of mol of NaOH which will control the pH.

40. Since all of the HNO2 has been neutralized, we only have to look at the concentration of hydroxide ion in the total volume of the solution to calculate the pH of the resultant solution.
Total volume = mL mL = mL
mol
0.055 L
[OH-] =
= M
pOH = – log = 2.34
pH = – 2.34 = 11.66

41. Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
H+ (aq) + NH3 (aq) NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)
16.4

42. At equivalence point, pH of solution is controlled
by hydrolysis of the cation of the base:
NH4+ + H2O ⇌ NH3 + H3O+

43. Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn]  10
Color of acid (HIn) predominates
 10
[HIn]
[In-]
Color of conjugate base (In-) predominates
16.5

44. Colors and Approximate pH Range of Some Common Acid-Base Indicators
Fig. 19.4

45. The titration curve of a strong acid with a strong base.  
Thymol blue
*Choose indicator that change colour along the steep portion of the curve
16.5

46. Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein
16.5

47. Solubility Equilibria – A Summary
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[PO33-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
No precipitate
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
Precipitate will form
16.6

48. Dissolving a salt...
A salt is an ionic compound - usually a metal cation bonded to a non-metal anion.
The dissolving of a salt is an example of equilibrium.
The cations and anions are attracted to each other in the salt.
They are also attracted to the water molecules.
The water molecules will start to pull out some of the ions from the salt crystal.

49. At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions.
However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation)
Eventually the rate of dissociation is equal to the rate of precipitation.
The solution is now “saturated”. It has reached equilibrium.

50. Solubility Equilibrium: Dissociation = Precipitation
In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker.
Concentration of the solution is constant.
The rate at which the salt is dissolving into solution equals the rate of precipitation.
Na+ and Cl - ions
surrounded by
water molecules
NaCl Crystal
Dissolving NaCl in water

51. The Basics of Solubility Equilibria
Dissolution of an ionic compound is an equilibrium process: CaF2 (s) ⇌ Ca2+ (aq) F– (aq)
Solubility Product, Ksp = [Ca2+][F-]2
The Solubility Product, Ksp of a sparingly soluble ionic compound is the product of the concentrations of its ion in a saturated solution of the compound, each raised to the power of its coefficient in the dissociation equation.
The Solubility Product, Ksp is an equilibrium constant, so it has only one value at a given temperature.
Remember, neither solids nor pure liquids (water) affect the equilibrium constant.

52. Question #10: Writing Ion-Product Expressions
Write the ion-product expression for:
silver bromide; (b) strontium phosphate;
(c) aluminum carbonate.
Answer:
(a) AgBr(s) ⇌ Ag+(aq) + Br -(aq)
Ksp = [Ag+] [Br -]
(b) Sr3(PO4)(s) ⇌ 3 Sr2+(aq) + 2 PO43-(aq)
Ksp = [Sr2+]3 [PO43-]2
(c) Al2(CO3)3 (s) ⇌ 2 Al3+(aq) + 3 CO32-(aq)
Ksp = [Al3+]2 [CO32-]3

53. 16.6

54. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. *Symbol used for molar solubility – s.
Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.
16.6

55. Question #11: Determining Ksp from Solubility (g/L)
Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It’s solubility is 5.8 x 10-6g/100mL water. What is the Ksp?
Answer:
Molar solubility of PbCrO4 :
= 1.79 x M PbCrO4 = s
PbCrO4 (s) ⇌ Pb2+(aq) + CrO42-(aq) s s s
Ksp = [Pb2+] [CrO42-] = (s)(s)
Ksp = (1.79 x 10 -7)2 = 3.20 x 10-14

56. Question #12: Determining Ksp from Solubility (g/L)
The solubility of zinc oxalate, ZnC2O4, is 7.9 × 10-3 M at 18 oC. Calculate the Ksp of zinc oxalate.
Answer:
Molar solubility of ZnC2O4 , s = 7.9 × 10-3 M
ZnC2O4 (s) ⇌ Zn2+(aq) + C2O42-(aq)
s s s
Ksp = [Zn2+] [C2O42-] = (s)(s)
Ksp = (7.9 x 10-3)2 = 6.24 x 10-5

57. b) The solubility of silver dichromate, Ag2Cr2O7, at 15 oC is 8
b) The solubility of silver dichromate, Ag2Cr2O7, at 15 oC is 8.3 × 10-3 g/100 mL solution. Calculate Ksp of silver dichromate (Molar mass = g/mol)
Answer:
Molar solubility of Ag2Cr2O7:
= 1.92 x M = s
Ag2Cr2O7 (s) ⇌ 2 Ag+(aq) + Cr2O72-(aq)
s s s
Ksp = [Ag+]2 [Cr2O72-] = (2s)2(s) = 4s3
Ksp = 4(1.92 x 10 -4)3 = 2.84 x 10-11

58. Question #13: Determining Solubility from Ksp
Lead chromate used to be used as the pigment for the yellow lines on roads, and is a very insoluble compound. Calculate the molar solubility of PbCrO4 in water if the Ksp is equal to 2.3 x
Answer:
PbCrO4 (s) ⇌ Pb2+(aq) + CrO42-(aq)
Ksp = [Pb2+] [CrO42-] = 2.3 x 10-13
Concentration (M) PbCrO4 ⇌ Pb CrO42-
Initial
Change s s
Equilibrium s s
s = 4.80 x 10-7M
Ksp = [Pb2+] [CrO42-] = (s)(s) = 2.3 x 10-13
Therefore, the molar solubility of PbCrO4 is 4.8 x M

59.  Question #14: 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 = 1.26 x 10-5
[Ag+] = 1.26 x 10-5 M
[Cl-] = 1.26 x 10-5 M
1.26 x 10-5 mol AgCl
1 L soln
g AgCl
1 mol AgCl x
Solubility of AgCl =
= 1.81 x 10-3 g/L
16.6

60. 16.6

61. Predicting Whether a Precipitate Will Form
Steps to be considered:
(a) First, see if the solutions will yield any insoluble ions;
(b) Then, calculate the concentrations, adding the two volumes together to get the total volume of the solution;
(c) Next, calculate the ion product constant, Qsp, and compare it to the solubility product constant to see if a precipitate will form.
Qsp < Ksp
Unsaturated solution, no precipitate.
Qsp = Ksp
Saturated solution, no precipitate.
Qsp > Ksp
Supersaturated solution,
precipitate will form.

62. Question #15: Predicting Whether a Precipitate Will Form–I
Will a precipitate form when L of a solution containing M barium nitrate is added to mL of a M solution of sodium chromate?
Answer:
Both Na2CrO4 and Ba(NO3)2 are soluble, so we will have Na+, CrO42-, Ba2+ and NO3- ions present in L of solution.
BaCrO4 would be insoluble. Therefore, calculate it’s ion-product constant and compare it to the solubility product constant, Ksp = 2.1 x

63. Calculate molarity for Ba2+ in the mixed solution:
Calculate molarity for CrO42- in the mixed solution:

64. BaCrO4 (s) ⇄ Ba2+(aq) + CrO42-(aq)
Qsp = [Ba2+][CrO42-]
Qsp = (0.0183)(0.067) = 1.23  10-3
Qsp >> Ksp ; therefore precipitate will form.

65. The ions present in solution are Na+, OH-, Ca2+, Cl-.
Question #16: If 2.00 mL of M NaOH are added to 1.00 L of M CaCl2, will a precipitate form?
Answer:
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 (Ca(OH)2) = 8.0 x 10-6
Q < Ksp
No precipitate will form.
16.6

66. Question #17:
If 20.0 mL of 0.10 M Ba(NO3)2 are added to 50.0 mL of 0.10 M Na2CO3, will BaCO3 precipitate?
Ksp (BaCO3) = 8.1  10-9
Answer: Qsp >> Ksp, therefore precipitate will form.
b) Does any solid Cu(OH)2 form when g KOH is dissolved in 1.0 L of 1.0 ×10-3 M Cu(NO3)2? (molar mass of KOH is 56.1 g/mol).
Ksp (Cu(OH)2) = 2.2  10-20
Answer: Qsp >> Ksp, therefore precipitate will form.

67. The Common Ion Effect and Solubility
The presence of a common ion decreases the solubility of the salt.
LeChatelier’s Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!
Question #18: What is the molar solubility of AgBr in
pure water ?
M NaBr?
Answer:
(a) in pure water:
AgBr (s) Ag+ (aq) + Br- (aq)
Ksp = (s)(s) = 7.7 x 10-13
s = 8.7 x 10-7
16.8

68. Answer:
(b) in M NaBr:
NaBr (s) Na+ (aq) + Br- (aq)
[Br-] = M
AgBr (s) Ag+ (aq) + Br- (aq)
[Ag+] = s
[Br-] = M + s
[Br-] ≈ M
Ksp = [Ag+][Br–] = (s)(0.0010) = 7.7  10–13
 s = 7.7  10–10
Compare: in pure water, s = 8.7  10-7
in NaBr solution, s = 7.7  10-10

69. The Effect of a Common Ion on Solubility
PbCrO4(s) ⇄ Pb2+(aq) + CrO42-(aq)
PbCrO4(s) ⇄ Pb2+(aq) + CrO42-(aq)
Adding CrO42– shifts the equilibrium to the left

70. Ag2CrO4 (s) ⇄ 2 Ag+(aq) + CrO42-(aq)
Question #19:Calculating the Effect of a Common Ion on Solubility
What is the solubility of silver chromate in M silver nitrate solution? Ksp = 2.6 x
Answer:
Ag2CrO4 (s) ⇄ 2 Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-]
Concentration (M): Ag2CrO4 (s) ⇄ 2 Ag+(aq) CrO42-(aq)
Initial
Change s s
Equilibrium s s
Ksp is small, therefore assume: M + 2s ≈ M
Ksp = 2.6 x = (0.0600)2(s)
s = 7.22 x M
Therefore, the solubility of silver chromate is 7.22 x M

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)
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Ksp = (s)(2s)2 = 4s3
s = 1.4 x 10-4 M
4s3 = 1.2 x 10-11
[OH-] = 2s = 2.8 x 10-4 M
pOH = pH = 10.45
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72. pH and Solubility Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq) pH = 10.45
Effect of adding acid (H+): the pH will be less than 10.45, as concentration of OH– will become low (the acid will react with OH– to produce H2O).
OH- (aq) + H+ (aq) H2O (l)
Equilibrium will be shifted to the right (to replace the lost OH–, and solubility of Mg(OH)2 increases.
Effect of adding base (OH–): the pH will be more than 10.45, as concentration of OH– is high (additional amount of OH–).
Equilibrium will be shifted to the left, and solubility of Mg(OH)2 decreases.
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73. Question #20: Predicting the Effect on Solubility of Adding Strong Acid
Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of:
(a) Iron(II) cyanide
(b) Potassium bromide
(c) Aluminum hydroxide
Answer:
Clue: Write the balanced dissolution equation and note the anion. Anions of weak acids react with H3O+ and shift the equilibrium position toward more dissolution. Strong acid’s anions do not react, so added acid has no effect.

74. Answer:
Fe(CN)2 (s) ⇄ Fe2+(aq) + 2 CN–
Since CN– is the anion of a weak acid, it will reacts with the added H3O+. Therefore, solubility of Fe(CN)2 will be increased.
(b) KBr(s) ⇄ K+(aq) + Br -(aq)
Br – is the anion of a strong acid, H3O+ will not react with it. Therefore, adding H3O+ will not affect the solubility.
(c) Al(OH)3 (s) ⇄ Al3+(aq) + 3 OH-(aq)
The OH- is the anion of water, a very weak acid, so it reacts with the added acid to produce water in a simple acid-base reaction. Equilibrium will be shifted to the right and more Al(OH)3 will dissolved.

75. Selective precipitation of ions.
An interesting application of solubility equilibria is the selective precipitation of ions, which is a useful way to analyze mixtures of ions in solution.
Suppose, for example, we had a solution that were 0.1 M in both Ag+ and Pb2+ ions. Is it possible to selectively precipitate out the Ag+ ions?
It is possible if we can add an anion to the solution that will form a solid with silver whose solubility is very low and at the same time form a solid with lead whose solubility is very high.

76. Question #21: Separating Ions by Selective Precipitation–I
A solution consists of 0.10 M AgNO3 and 0.15 M Cu(NO3)2. Calculate the [Ι -] that would separate the metal ions as their iodides.
Kspof AgΙ = 8.3 x 10-17; Kspof CuΙ = 1.0 x
Answer:
Since the two iodides have the same formula type (1:1),
compare their Ksp values. CuΙ is about 100,000 times more soluble than AgΙ. Therefore, AgΙ precipitates first, and we solve for [Ι-] that will give a saturated solution of AgΙ.

77. AgΙ(s) ⇄ Ag+(aq) + Ι -(aq) Ksp = [Ag+][Ι -]
CuΙ(s) ⇄ Cu+(aq) + Ι -(aq) Ksp = [Cu+][Ι -]
Needed: 8.3  M < [Ι─] < 6.7  M
Saturated solution of AgΙ will form when [Ι-] = 8.3  M, precipitate of AgΙ will be formed when [Ι-] > 8.3  M.
Precipitate of CuΙ will be formed when [Ι-] > 6.7  M.

78. AgBr (s) Ag+ (aq) + Br- (aq) Ksp = 7.7 x 10-13 Ksp = [Ag+][Br-]
Question #22: 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
[Br-]
1.6 x 10-10
0.020 =
= 8.0 x 10-9 M
 Needed: 3.9 x M < [Ag+] < 8.0 x 10-9 M
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