1. Chapter 17 Additional Aspects of Aqueous Equilibria

2. The Common Ion Effect

3. The common ion effect is an event that occurs when an ionic equilibrium shifts due to the addition of a solute that contains an ion that is the same as an existing ion in the equilibrium.

4. An Example of the Common-Ion Effect
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)

5. Example Problem The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8  10−4.
[H3O+] [F−]
[HF]
Ka =
= 6.8  10-4

6. The Common-Ion Effect HF(aq) + H2O(l) H3O+(aq) + F−(aq)
Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M
[H3O+], M
[F−], M
Initially
0.20
0.10
Change −x +x
At Equilibrium
0.20 − x  0.20
x  0.10 x

7. The Common-Ion Effect (0.10) (x) 6.8  10−4 = (0.20)
(0.20) (6.8  10−4)
(0.10)
= x
1.4  10−3 = x

8. The Common-Ion Effect Therefore, [F−] = x = 1.4  10−3
[H3O+] = x =  10−3 = 0.10 M
So, pH = −log (0.10)
pH = 1.00

9. Buffers

10. A buffer is an acid-base solution that resist changes in pH by neutralizing the effects of external acids and bases.
NH4Cl and NH3 is a buffer system.

11. Examples of five buffers.
Three solutions and two
powders.

12. SEM of a red blood cells
passing through a small artery
Blood is a suspension
and the serum of part
of blood is buffered at

13. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

14. Buffers
If acid is added, the F− reacts to form HF and water.

15. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

16. Derivation of the Henderson-Hasselbalch Equation

17. A specific equation exists for calculating the pH of a
buffer ..i.e. the Henderson-Hasselbalch (H-H) equation.

18. Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O
H3O+ + A−
[H3O+] [A−]
[HA]
Ka =

19. Buffer Calculations Rearranging slightly, this becomes
[HA]
Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + −log
pKa pH
acid

20. Buffer Calculations So pKa = pH − log [base] [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch equation.

21. Practice Problem: Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is
1.4  10−4.

22. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = −log (1.4  10−4) + log
(0.10)
(0.12)
pH = (−0.08)
pH = 3.77

23. pH Range
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH.

24. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.

25. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

26. Calculating pH Changes in Buffers
A buffer is made by adding mol HC2H3O2 and mol NaC2H3O2 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.

27. Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74

28. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
C2H3O2−
OH−
Before reaction
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol

29. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0. 200)
pH =
pH = 4.80

30. Acid-Base Titrations

31. Titration A lab method used to determine the concentration of unknown
solutions.

32. Titration For an acid-base titration the equivalence point is
the state where the moles of acid in the solution is equal to
the moles base in the solution.

33. Titration
For an acid-base titration the endpoint is when the indicator
changes color.

34. Strong Acid-Strong Base Titration

35. Titration of a Strong Acid with a Strong Base
From the start of the
titration to near the
equivalence point,
the pH goes up slowly.

36. Titration of a Strong Acid with a Strong Base
Just before and after the equivalence point, the pH increases rapidly.

37. Titration of a Strong Acid with a Strong Base
At the equivalence point,
moles acid = moles base

38. Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.

39. Weak Acid-Strong Base Titration

40. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
The pH at the equivalence point will be >7.
Phenolphthalein is commonly used as an indicator in these titrations.

41. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

42. Strong Acid-Weak Base Titration

43. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.

44. Titration of Polyprotic Acids

45. A polyprotic acid is a Bronsted acid with two or more ionizable protons. H2SO4 is the only polyprotic acid that is a strong acid.

46. Titrations of Polyprotic Acids
Polyprotic acids will have
An equivalence point for
each ionizable proton.

47. Solubility Product Constant (Ksp)

48. BaSO4(s) Ba2+(aq) + SO42−(aq)
The solubility product constant (Ksp) is an equilibrium
constant for a slightly soluble salt.
BaSO4(s)
Ba2+(aq) + SO42−(aq)

49. Ksp is used in calculating the solubility of a slightly
soluble salt.
Solubility is given as g/L which can be converted
to molar solubility mol/L (M).

50. Relationship Between Solubility and Ksp

51. Practice Exercise: Calculating Molar Solubility
Calculate the molar solubility of Al(OH)3 if the Ksp is
1.8 x

52. Factors Affecting Solubility

53. Three factors affect the solubility of a slightly soluble salt.
A Common Ion (solubility decreases)
(2) pH (solubility increases for a basic anion in an acidic
solution and for an acidic cation in a basic solution)
(3) Complex Ion Formation (solubility increases)

54. Effect of a Common Ion on Solubility
An ion that is already
dissolved in a solution
will shift the equilibrium to
the left which decreases the
solubility of the salt.

55. Practice Exercise: Calculating Molar Solubility in the
Presence of a Common Ion
a.Calculate the molar solubility of AgBr.
b. Calculate the molar solubility of AgBr in a M AgNO3 solution.
c. Calculate the molar solubility of AgBr in a 0.10 M NaBr solution. .

56. Effect of pH on Solubility

57. Explain how decreasing the pH increases the solubility of this precipitate.
Basic anions react with acid shifting the equilibrium to the right.

58. Explain how lower pH can cause sinkholes?
Limestone is
CaCO3.
Rainwater has a pH
~ 5-6

59. Effect of Complex Ions on Solubility
Recall metal ions are Lewis acids and can react with Lewis
bases to form complex ions.
The rxn is an equilibrium and is described by the
equilibrium constant Kf.

60. Effect of Complex Ions on Solubility
Write a chemical rxn and give an explanation that accounts for the dissolution of the solid.

61. Factors Affecting Solubility
Amphoterism
Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.

62. Predicting the Formation of a Precipitate
Q = Ksp, no change
Q < Ksp, more solid dissolves until Q = Ksp.
Q > Ksp, the salt will precipitate until Q = Ksp.

63. Qualitative Analysis of Ions by Selective Precipitation

64. What is selective precipitation (ppt) ?
Selective ppt is a qualitative analysis method where ions are separated and identified using a series of precipitation rxns.

65. General Flowchart for an Analysis
of an Unknown Aqueous Solution

66. Qualitative Analysis of Ions by Selective Precipitation

67. Practice Problems

68. Questions 1-3 refer to Equation I and Figure I
1. Which graph represents the change in solubility of BaCO3 if HBr is added to a saturated solution of BaCO3.
Figure: UNE7
Title:
Exercise 17.7
Caption:
Graphs representing the behavior of BaCO3 under different conditions.
Graph A
Graph B
Graph C

69. Questions 1-3 refer to Equation I and Figure I
2. Which graph represents the change in solubility of BaCO3 if K2SO4 is added to a saturated solution of BaCO3.
Figure: UNE7
Title:
Exercise 17.7
Caption:
Graphs representing the behavior of BaCO3 under different conditions.
Graph A
Graph B
Graph C

70. Questions 1-3 refer to Equation I and Figure I
3. Which graph represents the change in solubility of BaCO3 if Li2CO3 is added to a saturated solution of BaCO3.
Figure: UNE7
Title:
Exercise 17.7
Caption:
Graphs representing the behavior of BaCO3 under different conditions.
Graph A
Graph B
Graph C

71. The beaker below contains an aqueous solution of potassium
ions and barium ions. Recommend a qualitative analysis
strategy that will separate the ions.

72. Circle the salts that are more soluble in an acidic solution
than in pure water.
BiI CdS CaCO AgCN Ba3(PO4)2
Strategy
Write the chemical equation and determine if the conjugate
acid of the anion is weak.

73. Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

74. The beaker below contains an aqueous solution of chromium (III) ions
and iron (III) ions. Recommend a qualitative analysis strategy that
will separate the ions.