1. Chapter 17 Acid-Base & Solubility Equilibria

2. 17.1
The Common Ion Effect
17.2
Buffer Solutions
17.3
Acid-Base Titrations (omitted)
17.4
Solubility Equilibria
17.5
Factors Affecting Solubility
17.6
Separation of Ions Using Differences in Solubility (omitted)

3. The Effect of Common Ion
17.1
A phenomenon known as the common ion effect states that:
When a compound containing an ion in common with an already dissolved substance (a weak electrolyte) is added to an aqueous solution at equilibrium, the equilibrium shifts to the left.
The ionization of the weak electrolyte is being suppressed by adding the common ion to the solution.

4. Le Châtelier's principle
17.1
The Common Ion Effect
A 1.0 L of 0.10 M solution of CH3COOH.
Adding mol of CH3COONa:
The result is that fewer H+ ions present with a higher pH.
CH3COO‒ is a common ion
Le Châtelier's principle

5. Equilibrium Calculation Involving Common Ion Effect
17.1
Equilibrium Calculation Involving Common Ion Effect
Before adding the CH3COO‒ ions:
1.0 L of 0.10 M
(M)
CH3COOH H+
CH3COO‒
Initial conc.
0.10
Change in conc. x
+ x
Equilibrium conc. x
x = 1.34  10-3 M
Ka < 10-3
[H+] = 1.34  10-3 M
pH = -log (1.34  10-3 ) = 2.87

6. Equilibrium Calculation Involving Common Ion Effect
17.1
Equilibrium Calculation Involving Common Ion Effect
After adding the CH3COO‒ (0.05 M) ions:
Method II
CH3COONa
(M)
CH3COOH H+
CH3COO‒
Initial conc.
0.10
0.050
Change in conc.
- x
+ x
Equilibrium conc. x x x
x = 3.6  10-5 M
[H+] = x = 3.6  10-5 M
pH = -log (3.6  10-5 ) = 4.44

7. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
17.1
The Common Ion Effect
Exercise:
Which of the following when dissolved in aqueous NH3 solution is (are) going to decrease the dissociation of NH3?
NH3(aq) H2O(l) NH4+(aq) OH-(aq)
(a) Ca(OH)2
(b) HNO2
(c) CH3COONa
(d) NH4NO3
(e) NH3  

8. 17.2
Buffer Solutions
A buffer solution is the one that resists the change in its pH when small amounts of either H+ or OH‒ ions are added.
Buffers are useful applications of the common ion effect.
Buffer solutions are important for:
Biological systems. (some enzymes can only function at a specific pH, pH of blood is always about 7.4, gastric “stomach” juices maintain a pH of about 1.5)
Chemical applications. (fermentation processes, dyes used in coloring fabrics, calibration for pH meters).

9. Buffer Solutions A buffer solution can be:
17.2
Buffer Solutions
A buffer solution can be:
a solution containing a weak acid and its conjugate base, or
CH3COOH (aq) CH3COO‒ (aq) + H+ (aq)
It is known as an acidic buffer solution and it maintains a pH value that is less than 7.
a solution containing a weak base and its conjugate acid.
NH3 (aq) + H2O(l) NH4+ (aq) + OH‒ (aq)
It is known as a basic buffer solution and it maintains a pH value that is greater than 7.
weak acid
conjugate base
weak base
conjugate acid

10. Henderson-Hasselbalch Equation
How to calculate pH of buffer?
Henderson-Hasselbalch Equation
For any buffer solution when a valid approximation is applied, its equilibrium expression is:
Ka =
[H+][A‒]
[HA]
From H-H equation, when the concentrations of the weak acid and its conjugate base in a buffer are equal, its pH = its pKa .
The slight change in the pH of the buffer is due to change in the concentrations of the weak acid and its conjugate base when small amounts of either H+ or OH‒ ions are added to the buffer.
HA is the weak acid
A‒ is the conjugate base
[H+] =
Ka [HA]
[A‒]
‒ log [H+] = ‒ log Ka + log
[HA]
[A‒]
pH = pKa log
[weak acid]
[conjugate base]

11. Requirement for a solution to be buffer?
A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added.
If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity.

12. 17.2 Therefore, we consider a solution a buffer, and can use
to calculate its pH, only if the following condition is met:
Consequently, the log term can only have values from –1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pKa of the weak acid it contains.
This is known as the range of the buffer, where pH = pKa 1.
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13. Exercise:
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.
Ka = 1.8 x 10-5 for acetic acid.
(Assume that the addition of sodium acetate does not change the volume of the solution.)
The solution will work as buffer or not?

14. pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74
pH = pKa log
[weak acid]
[conjugate base]
pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74
Log
[HA]
[A‒] =
[0.10]
[.050]
Log 0.5
-0.3
pH = 4.74 – 0.3 = 4.44

15. What happens when acid or base is added
into a buffer solution?
(1.00 M of acetic acid and 1.00 M of sodium acetate)
1.00 mol
1.10 mol

16. Consider what happens to the pH when we add 0.10 mole of HCl to
Pure water
The buffer solution prepared by adding 1.00 mole of sodium acetate to 1.0 L of 1.00 M acetic acid.
(Ka = 1.8 x 10-5 for acetic acid or pKa = 4.74).
(Assume that the addition of HCl causes no change in the volume of the solution.)

17. pH = 1 A change of only 0.08 pH units. How about the pH of pure water?OR
Since we are adding 0.1 M strong acid (HCl);
[acid] = [1 M M] = [1.1 M], [conjugate base] = [1 M M] = [0.9 M]
pH = pKa log
[weak acid]
[conjugate base]
pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74
log
[WA]
[CB] =
[1.10]
[0.90]
log 0.82
- 0.08
pH = 4.74 – 0.08 = 4.66
A change of only 0.08 pH units.
pH = 1
How about the pH of pure water?

18. Exercise:
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.)
(Ka = 1.8 x 10-5 for acetic acid or pKa = 4.74)

19. pH = 13 How about the pH of pure water? OR
1.00 mol
1.10 mol OR
Since we are adding 0.1 M strong base (NaOH);
[acid] = [1 M M] = [0.9 M], [conjugate base] = [1 M M] = [1.1 M]
pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74
log
[WA]
[CB] =
[0.90]
[1.10]
log 1.22
+ 0.08
pH = = 4.82
A change of only 0.08 pH units.
pH = 13
How about the pH of pure water?

20. Exercise:
Calculate the pH of 2.0 L of a buffer that is 1.0 mole in both acetic acid and sodium acetate after adding 0.15 M of Ca(OH)2. (pKa = 4.74 for acetic acid)

21. Concentration = pH = pKa + log (0.8) = 4.74 + log = 4.74 + 0.6 = 5.34
Mole
2 L
1.0 =
= 0.5 M
pH = pKa + log
[CH3COOH]
[CH3COO‒]
= log
(0.2)
(0.8)
= = 5.34

22. What we learned… Common Ion Effect
Buffer solution (properties, composition)
How to calculate pH of Buffer (Henderson-Hasselbalch Equation)
What happens when we add acid or base in buffer solutions

23. What we hope to learn today…
How to prepare a buffer with specific pH
Solubility and Solubility Product Expression (Ksp)
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions

24. How to Prepare a Buffer Solution with a Specific pH
17.2
How to Prepare a Buffer Solution with a Specific pH
Choose a weak acid whose pKa is close to the desired pH.
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25. 17.2
Substitute the pH and pKa values into the following equation to obtain the necessary ratio of [conjugate base]/[weak acid].
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26. How you would prepare a buffer with a pH of 9.50.? 17.3
Exercise:
How you would prepare a buffer with a pH of 9.50.?
17.3
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27. 17.3
One way to achieve this would be to dissolve 0.41 mol of C6H5ONa and 1.00 mol of C6H5OH in 1 L of water.
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28. Solubility Product Expression and Ksp
Calculations Involving Ksp and Solubility

29. Consider – soluble, less soluble, insoluble
17.4
Solubility Product Expression and Ksp
The solubility of ionic compounds is important in industry, medicine, and everyday life.
Consider – soluble, less soluble, insoluble
Solubility, solute, saturated solution, precipitate??
Equilibrium constant is represented by Ksp - called the solubility product constant.
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30. 17.4 There are two ways to express the solubility of a substance:
The molar solubility: number of moles of solute in 1 L of a saturated solution (mol/L).
The solubility: number of grams of solute in 1 L of a saturated solution (g/L).
Both of these refer to concentrations at a particular temperature (usually 25°C).
How to convert solubility into molar solubility or vice-versa?
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31. Solid
(precipitate)
Solvent
Solute
Saturated
Precipitate formation
Less saturated

32. All above compounds are very slightly soluble in water
All above compounds are very slightly soluble in water. None of them are soluble in water.
The smaller the Ksp value, the less soluble the compound. This is valid for compounds of similar formulas, such as comparing AgCl with CuBr, and CaF2 with Fe(OH)2 .

33. 17.4 Solubility Product Expression and Ksp
Ksp is equal to the concentrations of products over the concentrations of reactants, each raised to its coefficient from the balanced chemical equation.
Equilibrium constant is represented by Ksp - called the solubility product constant.
The smaller the Ksp, the less soluble the compound.
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34. 17.4 Calculations Involving Ksp and Solubility
To calculate molar solubility from Ksp :
The procedure is essentially identical to the procedure for solving weak acid or weak base equilibrium problems
Construct an equilibrium or ICE table.
Fill in what we know.
Figure out what we don’t know.
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35. Calculate the solubility of AgBr?
Exercise:
Calculate the solubility of AgBr?
The Ksp of silver bromide (AgBr) is 7.7 × 10‒13.
Let s be the molar solubility (in mol/L) of AgBr.
At equilibrium: [Ag+] = [Br‒] = s
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36. 17.4 Calculations Involving Ksp and Solubility
The equilibrium expression is
Therefore,
Is this solubility or molar solubility?
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37. Calculations Involving Ksp and Solubility
The molar solubility of AgBr is 8.8 × 10‒7 M.
Express this solubility in g/L by multiplying the molar solubility by the molar mass of AgBr:
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38. 17.4 Exercise: Solubility Equilibria 17.7
(Ksp for Cu(OH)2 = 2.2 × 10‒20) and the molar mass of Cu(OH)2 is g/mol.
The equation for the dissociation of Cu(OH)2 is
Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in g/L.
Calculations Involving Ksp and Solubility
Ksp = [Cu2+][OH‒]2
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39. Remember to raise an entire term to the appropriate power!
17.7
Therefore,
Ksp = [Cu2+][OH‒]2
Remember to raise an entire term to the appropriate power!
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40. The molar solubility of Cu(OH)2 is 1.8 × 10‒7 M.
17.7
The molar solubility of Cu(OH)2 is 1.8 × 10‒7 M.
Multiplying by its molar mass gives:
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41. Exercise:
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.
The molar mass of CaSO4 is g/mol.
The molar solubility of CaSO4 is
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42. 17.8
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43. Predicting Precipitation Reactions
To predict whether a precipitation will form or not, we calculate the reaction quotient (Q) for the possible precipitation for the initial state of mixing two solutions.
If Q < Ksp, no precipitation is going to form.
If Q > Ksp, precipitation is going to form.
If Q = Ksp,
the solution is saturated.

44. Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq)
Exercise:
Predict whether a precipitate will form when M Ag2SO4 is prepared?
Ag2SO4, Ksp = 1.5 × 10‒5
Ag2SO4(s) H2O(l) Ag+(aq) SO42-(aq)
0.001M
0.002M
Q =
[Ag+]2 [SO4-] =
[0.002]2 [0.001] 
Q < Ksp, (No precipitate)

45. Exercise:
Predict whether a precipitate will form when M AgNO3 is added to M K2SO4?
(Assume that the addition does not change the volume of the solution.)
Ksp values of Ag2SO4, Ksp = 1.5 × 10‒5

46. When we have 2 solutes… Q = [Ag+] [Cl‒] Q = [Ag+]2 [SO4‒]
AgNO3
(completely soluble)
Ag2SO4
(slightly soluble)
K2SO4
(completely soluble)
Q = [Ag+] [Cl‒]
Q = [Ag+]2 [SO4‒]
At the first stage, you should be able to determine which compound is soluble in water and which is very slightly soluble in water. You calculate Q for the latter compound and then compare it with the listed Ksp values.

47. AgNO3(s) + H2O(l) Ag+(aq) + NO3-(aq)
0.002 M
0.002 M
0.002 M
AgNO3(s) H2O(l) Ag+(aq) NO3-(aq)
0.1 M
0.2 M
0.1 M
K2SO4(s) H2O(l) K+(aq) SO42-(aq)
Ag+ reacts with SO42- to form slightly soluble Ag2SO4
Ag2SO4(s) H2O(l) Ag+(aq) SO42-(aq)
Q = [Ag+]2 [SO4‒] = [0.002]2[0.1] = 4  10-7
Ksp Ag2SO4 = 1.5 × 10‒5
Q < Ksp, (No precipitate)

48. Exercise:
17.9
Predict whether a precipitate will form when 250 mL of M BaCl2 is added to 650 mL of M K2SO4:
The Ksp values of BaSO4, Ksp = 1.1 × 10‒10
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49. Q = [Ba2+][SO42‒] = (0.0011)(0.0058) = 6.4 × 10‒6
Solution
17.9
Concentrations of the constituent ions of BaSO4 are:
Q = [Ba2+][SO42‒] = (0.0011)(0.0058) = 6.4 × 10‒6
Q > Ksp (1.1 × 10‒10)
Ksp BaSO4 = 1.1 × 10‒10
(BaSO4 will precipitate)
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50. Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0  mol/L at 25C.
Answer : 1.1  10-73

51. What we learned in last lecture…
How to prepare a buffer with specific pH
Solubility and Solubility Product Expression (Ksp)
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions

52. What we hope to learn today…
Factors affecting solubility

53. Factors Affecting Solubility
17.5
Factors Affecting Solubility
There are some factors that affect solubility. We will be discussing:
The common ion effect.
The pH.
Complex ion formation

54. Any factor that will shift the equilibrium left
(reverse) will decrease the solubility.
Any factor that will shift the equilibrium right
(forward) will increase the solubility.

55. Factors Affecting Solubility
17.5
The common ion effect.
Consider dissolving AgCl salt in pure water to get a saturated aqueous solution of AgCl.
The solubility of AgCl in water at 25C is 1.3  10-5 M.
AgCl salt
The solubility of AgCl in water can be calculated from its solubility product constant (Ksp = 1.6  10-10)
1.0 L Pure water
Cl‒
Ag+
Cl‒
Cl‒
Ag+
Ag+
Cl‒
Cl‒
Ag+
Ag+
Ag+
Cl‒
So how about if we dissolve AgCl in presence of other salts?
Cl‒
Ag+
AgCl (s)

56. Factors Affecting Solubility
17.5
Factors Affecting Solubility
The common ion effect.
The common ion effect is an example of Le Châtelier’s principle. The presence of a second salt (normally very soluble in water) that produces an ion common to a solubility equilibrium will reduce solubility.
AgCl (s) Ag+ (aq) + Cl‒ (aq)
AgNO3 solution
NaCl solution
What about NaNO3?

57. Factors Affecting Solubility
17.5
Factors Affecting Solubility
(less soluble)
(soluble)
Example:
Calculate the molar solubility of BaSO4 in M Na2SO4.
BaSO4 (s) Ba2+ (aq) + SO42‒ (aq) Ksp of BaSO4 = 1.1  10-10
(M)
BaSO4 (s)
Ba2+ (aq)
SO42‒ (aq)
Initial conc.
1  10-3
Change in conc.
+ s
Equilibrium conc. s
1  s
For comparison, the solubility in pure water is:
s = (1.1  10-10)1/2
= 1.0  10-5 M
Ksp = 1.1  = [Ba2+][SO42‒] = (s)(1  s)
1.1  ≈ (s)(1  10-3)
s = 1  10-7 M

58. In general, the common ion effect decreases the solubility of a substance.

59. Factors Affecting Solubility
17.5
Factors Affecting Solubility
The pH
Adding weak acids (HF, CH3COOH etc.)
Adding weak base (NH3)
Adding strong acids (HCl, HBr, HNO3, H2SO4)
Adding strong bases (NaOH, Ca(OH)2)

60. Exercise:
Which of the following compounds will be more soluble in acidic solution than in water:
CaF2
CuS
AgCl
HNO3
H2SO4
HSO4-  
Think about it: How solubility will change under basic solution?

61. Exercise:
Which of the following compounds will be more soluble in acidic solution than in water:
Cu(OH)2
CuCN
PbBr2
PbSO4
KClO4
KClO2    
Think about it: How solubility will change under basic solution?

62. Example:
Calculate the solution pH, above which the solubility of Ca(OH)2 will decrease (Ksp of Ca(OH)2 = 8.0  10-6 ).
pH = 12.41

63. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Ca(OH)2 (s) Ca2+ (aq) + 2OH‒ (aq) Ksp = 8.0  10-6
Ksp = 8.0  10-6 = [Ca2+][OH‒]2 = (s)(2s)2
s = 1.3  10-2 M
[OH‒] = 2(1.3  10-2 M) = 2.6  10-2 M
pOH= - log (2.6  10-2) = 1.59
Ca(OH)2 solubility decreases
14.00
pH = – 1.59 =
Ca(OH)2 solubility increases
1.00

64. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation.
A Complex ion is an ion that involves a central metal cation (mostly are transition metal ions) bonded to one or more ions or molecules ( ligands).
Tetra-ammine-copper(II) cation , Cu(NH3)42+ , is one example of complex ions.
- Complex ions exhibit colors when transition metal ions are contained at the central position.
- Different types of ligands give different color
Co(H2O)62+
CoCl42

65. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation.
Consider adding aqueous ammonia to a saturated AgCl solution.
AgCl dissolves, and Ag+ ions form Ag(NH3)2+ complex
NH3 (aq)
AgCl in water
Kf =

66. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation.
We measure the tendency of a metal ion to form a complex ion using the formation constant, Kf , (or stability constant).

67. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation.
Let’s write equilibrium equations for the previous experiment.
AgCl (s) Ag+ (aq) + Cl‒ (aq)
Ag+ (aq) + 2NH3 (aq) Ag(NH3)2+ (aq)
AgCl (s) + 2NH3 (aq) Ag(NH3)2+ (aq) + Cl‒ (aq)
K” = Ksp  Kf = 1.6   1.5  107 = 2.4  10-3
K” >> Ksp
In general, the formation of complex ions increases the solubility of a substance.
We can also apply Le Châtelier’s principle to understand!
Ksp = 1.6  10-10
Kf = 1.5  107
K ” = ?

68. Exercise:
Determine whether the solubility of Cu(OH)2 will increase or decrease when mixed with NaCN?
Ksp of Cu(OH)2 = 1.6  10-19
Kf Cu(CN)42- = 1 1025
Solubility will increase because CN- will react with Cu2+ making Cu(CN)42- and shifting the equilibrium to the right.

69. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation
Determine the molar conc. of free Cu2+ ion in solution when 0.10 mole of Cu(NO3)2 is dissolved in a L of 3.0 M NH3 to form Cu(NH3)42+. Kf for Cu(NH3)42+ = 5.0  1014
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70. Consider the following reaction
Solving an equilibrium problem involving complex ion formation is complicated by the magnitude of Kf and stoichiometry.
Consider the following reaction
Kf is so large that we cannot ignore “change in concentration (x)”.
A new approach is needed

71. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation
Assume that all the copper(II) ion is consumed to form the complex ion (since Kf is so large.)
Consider the equilibrium in terms of the reverse reaction; the dissociation of Cu(NH3)42+, for which the equilibrium constant is the reciprocal of Kf.
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72. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation
Because this K is so small, we can expect x to be insignificant.
Note that [NH3], which had been 3.0 M, has been diminished by 4 × 0.10 M due to the amount required to complex 0.10 mole of copper(II) ion.
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73. Factors Affecting Solubility
17.5
Factors Affecting Solubility
Complex Ion Formation
Neglecting x,
x = 4.4 × 10‒18 M
Because Kf is so large, the amount of copper that remains uncomplexed is extremely small.
Think about it! What is the conc. of NH3 and Cu(NH3)42+ at equilibrium??
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74. Kf for the complex ion Cd(CN)42‒ is 7.1 × 1016. The reverse process
17.12
In the presence of aqueous cyanide, cadmium(II) forms the complex ion Cd(CN)42‒. Determine the molar concentration of free (uncomplexed) cadmium(II) ion in solution when 0.20 mole of Cd(NO3)2 is dissolved in a liter of 2.0 M sodium cyanide (NaCN).
Setup
Kf for the complex ion Cd(CN)42‒ is 7.1 × The reverse process
has an equilibrium constant of 1/Kf = 1.4 × 10 ‒17
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75. The equilibrium expression for the dissociation is
17.12
The equilibrium expression for the dissociation is
Stoichiometry indicates that four CN‒ ions are required to react with one Cd2+ ion. Therefore, [CN‒] will be
[2.0 M ‒ 4(0.20 M)] = 1.2 M.
Solution
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76. 17.12
Neglecting x,
x = 1.4 × 10‒18 M
Don’t forget to adjust the concentration of the complexing agent before entering it in the equilibrium table.
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77. Review… Common Ion Effect
Buffer solution (properties, composition, required ratio for solution to be buffer)
How to calculate pH of Buffer (Henderson-Hasselbalch Equation)
What happens when we add acid or base in buffer solutions

78. How to prepare a buffer with specific pH
Solubility and Solubility Product Expression (Ksp)
Calculations Involving Ksp and Solubility
Predicting Precipitation Reactions
Factors affecting solubility (common ions, solution pH, complex ion formation)

79. Examples

80. Which of the following pairs will show common ion effect?
BaF2, Ba(NO3)2
NaCl, HCl
NH4OH,NH4Cl
NaOH, NaCl  
Which of the following will give buffer solutions
i) NH4Cl + NH3 solution ii) HCl + NaOH solution
iii) HA + NaA solution iv) NaHCO3 and Na2CO3 solution
v) NaHSO4/H2SO4
A) ii and iii only
B) i and iii only
C) iii only
D) i, iii iv and v only
E) i and ii only 

81. Which one of the following aqueous solutions, when mixed with an equal volume of 0.10 mol L-1 aqueous NH3, will produce a buffer solution?  a mol L-1 HCl  b mol L-1 NH4Cl
c mol L-1 NaOH  d mol L-1 NH4Cl   
Think about it – how will you calculate the pH?
How will calculate pH after adding acid or base?
Calculate the pH of the buffer 0.20 M Na2HPO4 / 0.30 M KH2PO4 [Ka (Phosphoric acid) = 6.2 × 10–8]
A) 3.07
B) 5.05
C) 9.01
D) 7.03
E) 2.06 
Think about it –How will calculate pH after adding acid or base?

82. Calculate the molar solubility and the solubility of SnS in g/L at 25°C. Ksp for SnS is 1.0  10-26.
Molar mass of SnS = g

83. Solubility Calculations
17.4
Solubility Calculations
Exercise:
Calculate the molar solubility and the solubility of SnS in g/L at 25°C. Ksp for SnS is 1.0 
SnS (s) Sn2+ (aq) + S2‒ (aq)
(M)
SnS
Sn2+
S2‒
Initial conc.
Change in conc.
+ x
Equilibrium conc. x
Molar solubility
Ksp = [Sn2+][S2‒] = (x)(x) = x2 = 1.0  10-26
s = 1.0  M
Molar solubility of SnS is then 1.0  mol/L
1 L
1.0  mol
1 mol g
Solubility = 
= 1.5  g/L

84. Calculate the molar solubility of CaF2 salt (Ksp = 4.0  10-11) in a 0.025 M NaF solution.
Answer : 6.4  10-8 mol/L

85. Ksp = [F‒]2 [Ca2+]= (0.025 + s)2 (s) = 4.0 × 10–11
CaF2 (s) Ca2+ (aq) + 2F‒ (aq)
(M)
CaF2
Ca2+
2F‒
Initial conc.
0.025
Change in conc.
+ s
+ 2s
Equilibrium conc. s s
Ksp = [F‒]2 [Ca2+]= ( s)2 (s) = 4.0 × 10–11
Because Ksp is small, we can ignore 2s
= (0.025)2 (s) = 4.0 × 10–11
s = 6.4  10-8 M
Since the Molar solubility of Ca2+ is = 6.4  10-8 M mol/L, the Molar solubility of CaF2 will same (6.4  10-8 M mol/L)

86. How many grams of CaCO3 will dissolve in 4
How many grams of CaCO3 will dissolve in 4.1 × 102 mL water containing 0.06 M Ca (NO3)2? [Ksp (Calcium Carbonate) = 8.7 × 10–9 at room temperature]
A) 6.0 × 10–6 g
B) 9.1 × 10–9 g
C) 5.1 × 10–5 g
D) 4.9 × 10–6 g
E) 3.6 × 10–7 g
Molar mass of CaCO3 = g

87. 1 mol 100.1 g Solubility = 1.45  10-7 mol/L  = 1.45  10-5 g/L 1 L
CaCO3
Ca2+
CO32‒
Initial conc.
0.06
Change in conc.
+ s
Equilibrium conc. s s
Ksp = [Ca2+][CO32‒] = ( s)(s) = 8.7 × 10–9
= (0.06) (s) = 8.7 × 10–9
s = 1.45  10-7 M
Molar solubility of CaCO3 is = 1.45  10-7 mol/L
1 mol
100.1 g
Solubility =
1.45  10-7 mol/L 
= 1.45  10-5 g/L
1 L
1.45  10-5 g
Solubility =
.410 L 
= 6.0  10-6 g

88. Examples on Solubility Equilibria
Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0  mol/L at 25C.
Answer : 1.1  10-73
Precipitation of bismuth sulfide

89. Predict whether a precipitate will form when 0
Predict whether a precipitate will form when M Ag2SO4 is prepared?
Ag2SO4, Ksp = 1.5 × 10‒5

90. Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq)
0.001M
0.002M
Q =
[Ag+]2 [SO4-] =
[0.002]2 [0.001] 
Ag2SO4, Ksp = 1.5 × 10‒5
Q < Ksp, (No precipitate)

91. Examples on Solubility Equilibria
A solution is prepared by adding mL of 4.00  10-3 M Ce(NO3)3 to mL of 2.00  10-2 M KIO3. Will Ce(IO3)3 solid (Ksp = 1.9  10-10) form from this solution?
Think about it – What will be the concentration of IO3-1,
if we use K2(IO3)2 ?

92. Ce(IO3)3 (s) Ce3+ (aq) + 3IO3‒ (aq)
Qsp = [Ce3+] [IO3-]3
M1V1 = M2V2
1050 mL
4.0  10-3 M  750 mL
2.857  10-3 M =
[Ce(NO3)3] =
1050 mL
2.0  10-2 M  300 mL
5.714  10-3 M =
[KIO3] =
Qsp = [Ce3+] [IO3-]3] = [2.857  10-3 M] [5.714  10-3 M]3
= 5.33  10-10
Qsp > Ksp(1.910-10), precipitate will form

93. Cu(OH)2(s) ⇄ Cu2+(aq) + 2OH - (aq)
Which will shift the following equilibrium in the left (reverse) direction.
Cu(OH)2(s) ⇄ Cu2+(aq) OH - (aq)
(i) addition of Cu(NO3)2.
(ii) addition of NaNO3.
(iii) due to increase in reaction pH.
(iv) decrease in reaction pH.
(v) addition of NH3 (Kf of Cu(NH3)42+ = 5.0 × 1013).
A) (iii), (iv)
B) (ii), (iii), (iv)
C) (i), (iii)
D) (i), (iv)
E) (i), (ii), (v) 