Chapter Seventeen

1
Chapter Seventeen

• Additional Aspects of Aqueous Equilibria

2
The Common Ion Effect and Buffer Solutions

• Common ion effect - solutions in which the same
  ion is produced by two different compounds
• Buffer solutions - resist changes in pH when
  acids or bases are added to them
• due to common ion effect
• Two common kinds of buffer solutions
• solutions of a weak acid plus a soluble ionic
  salt of the weak acid
• solutions of a weak base plus a soluble ionic
  salt of the weak base

3
The Common Ion Effect and Buffer Solutions

• Weak Acids plus Salts of Weak Acids
• acetic acid CH3COOH
• sodium acetate NaCH3COO

4
The Common Ion Effect and Buffer Solutions

• Example Calculate the concentration of Hand the
  pH of a solution that is 0.15 M in acetic acid
  and 0.15 M in sodium acetate.

5
The Common Ion Effect and Buffer Solutions

• Example Calculate the concentration of Hand the
  pH of a solution that is 0.15 M in acetic acid
  and 0.15 M in sodium acetate.

6
The Common Ion Effect and Buffer Solutions

• Substitute these quantities into the ionization
  expression.

7
The Common Ion Effect and Buffer Solutions

• Apply the simplifying assumption

8
The Common Ion Effect and Buffer Solutions

• Compare the acidity of a pure acetic acid
  solution and the buffer we just described.

9
The Common Ion Effect and Buffer Solutions

• Compare the acidity of a pure acetic acid
  solution and the buffer we just described.
• H is 89 times greater in pure acetic acid than
  in buffer solution.

10
The Common Ion Effect and Buffer Solutions

• General expression for the ionization of a weak
  monoprotic acid.

11
The Common Ion Effect and Buffer Solutions

• Its ionization constant expression is

12
The Common Ion Effect and Buffer Solutions

• Solve the expression for H

13
The Common Ion Effect and Buffer Solutions

• Making the assumption that the concentrations of
  the weak acid and the salt are reasonable.
• The expression reduces to

14
The Common Ion Effect and Buffer Solutions

• The above relationship is valid for buffers
  containing a weak monoprotic acid and a soluble,
  ionic salt.
• The relationship changes if the salts cation is
  not univalent to

15
The Common Ion Effect and Buffer Solutions

• Simple rearrangement of this equation and
  application of algebra yields the
• Henderson-Hasselbach equation

16
Weak Bases plus Salts of Weak Bases

• Buffers that contain a weak base plus the salt of
  a weak base - for example - ammonia plus ammonium
  nitrate.

17
Weak Bases plus Salts of Weak Bases

• Buffers that contain a weak base plus the salt of
  a weak base - for example - ammonia plus ammonium
  nitrate.

18
Weak Bases plus Salts of Weak Bases

• Example 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.

19
Weak Bases plus Salts of Weak Bases

• Substitute these values into the ionization
  expression for ammonia and solve algebraically.

20
Weak Bases plus Salts of Weak Bases

• Lets compare the aqueous ammonia concentration
  to that of the buffer described above.

21
Weak Bases plus Salts of Weak Bases

• Lets compare the aqueous ammonia concentration
  to that of the buffer described above.
• The OH- in aqueous ammonia is 180 times greater
  than in the buffer.

22
Weak Bases plus Salts of Weak Bases

• Derive a general relationship for buffer
  solutions that contain a weak base plus a salt of
  a weak base.
• Ionization equation

23
Weak Bases plus Salts of Weak Bases

• Ionization expression
• general form

24
Weak Bases plus Salts of Weak Bases

• For salts that have univalent ions
• For salts that have divalent or trivalent ions

25
Weak Bases plus Salts of Weak Bases

• Simple rearrangement of this equation and
  application of algebra yields the
• Henderson-Hasselbach equation

26
Buffering Action

• Buffer solutions resist changes in pH.
• Example If 0.020 mole of HCl is added to 1.00
  liter of solution that is 0.100 M in aqueous
  ammonia and 0.200 M in ammonium chloride, how
  much does the pH change? Assume no volume change
  due to addition of the gaseous HCl.
• Calculate the pH of the original buffer solution

27
Buffering Action
28
Buffering Action

• Now we calculate the concentration of all species
  after the addition of HCl.
• HCl will react with some of the ammonia

29
Buffering Action

• Now that we have the concentrations of our salt
  and base, we can calculate the pH.

30
Buffering Action

• Calculate the change in pH.

31
Preparation of Buffer Solutions

• Example Calculate the concentration of H and
  the pH of the solution prepared by mixing 200 mL
  of 0.150 M acetic acid and 100 mL of 0.100 M
  sodium hydroxide solutions.
• Determine the amounts of acetic acid and sodium
  hydroxide (before reaction)

32
Preparation of Buffer Solutions

• Example Calculate the concentration of H and
  the pH of the solution prepared by mixing 200 mL
  of 0.150 M acetic acid and 100 mL of 0.100 M
  sodium hydroxide solutions.
• Determine the amounts of acetic acid and sodium
  hydroxide (before reaction)

33
Preparation of Buffer Solutions

• Sodium hydroxide and acetic acid react in a 11
  mole ratio.

34
Preparation of Buffer Solutions

• After the two solutions are mixed, the total
  volume is 300 mL (100 200), and the
  concentrations are

35
Preparation of Buffer Solutions

• Substitution into the ionization constant
  expression (or Henderson-Hasselbach equation)
  gives

36
Preparation of Buffer Solutions

• For biochemical situations, it is sometimes
  important to prepare a buffer solution of a given
  pH.
• Example Calculate the number of moles of solid
  ammonium chloride, NH4Cl, that must be used to
  prepare 1.00 L of a buffer solution that is 0.10
  M in aqueous ammonia, and that has a pH of 9.15.
• Because pH9.15

37
Preparation of Buffer Solutions

• For biochemical situations, it is sometimes
  important to prepare a buffer solution of a given
  pH.
• ExampleCalculate the number of moles of solid
  ammonium chloride, NH4Cl, that must be used to
  prepare 1.00 L of a buffer solution that is 0.10
  M in aqueous ammonia, and that has a pH of 9.15.
• Because pH9.15

38
Preparation of Buffer Solutions

• Appropriate equations and equilibria
  representations are

39
Preparation of Buffer Solutions

• Substitute into the ionization constant
  expression (or Henderson-Hasselbach equation) for
  aqueous ammonia

40
Acid-Base Indicators

• Equivalence point - point at which chemically
  equivalent amounts of acid and base have reacted
• End point - point at which chemical indicator
  changes color

41
Acid-Base Indicators

• Equilibrium constant expression for an indicator
  would be

42
Acid-Base Indicators

• Rearrange this expression to get a feeling for
  range over which indicator changes color.

43
Acid-Base Indicators

• Some Acid-Base Indicators

44
Strong Acid/Strong Base Titration Curves

• Plot pH vs. Volume of acid or base added in
  titration.
• Consider the titration of 100.0 mL of 0.100 M
  perchloric acid with 0.100 M potassium hydroxide.
• Plot pH vs. mL of KOH added
• 11 mole ratio

45
Strong Acid/Strong Base Titration Curves

• Before titration starts the pH of the HClO4
  solution is 1.00.
• Remember perchloric acid is a strong acid

46
Strong Acid/Strong Base Titration Curves

• After 20.0 mL of 0.100 M KOH has been added the
  pH is 1.17.

47
Strong Acid/Strong Base Titration Curves

• After 50.0 mL of 0.100 M KOH has been added the
  pH is 1.48.

48
Strong Acid/Strong Base Titration Curves

• After 90.0 mL of 0.100 M KOH has been added the
  pH is 2.28.

49
Strong Acid/Strong Base Titration Curves

• After 100.0 mL of 0.100 M KOH has been added the
  pH is 7.00.

50
Strong Acid/Strong Base Titration Curves

• We have calculated only a few points on the
  titration curve. Similar calculations for
  remainder of titration show clearly the shape of
  the titration curve.

51
Weak Acid/Strong Base Titration Curves

• Consider the titration of 100.0 mL of 0.100 M
  acetic acid, CH3 COOH, with 0.100 M KOH.
• react in a 11 mole ratio

52
Weak Acid/Strong Base Titration Curves

• Before the equivalence point is reached , both
  CH3COOH and K CH3COO are present in solution
  forming a buffer.

53
Weak Acid/Strong Base Titration Curves

• Determine the pH of the acetic acid solution
  before titration is begun.

54
Weak Acid/Strong Base Titration Curves

• Determine the pH of the acetic acid solution
  before titration is begun.

55
Weak Acid/Strong Base Titration Curves

• Determine the pH of the acetic acid solution
  before titration is begun.

56
Weak Acid/Strong Base Titration Curves

• After 20.0 mL of KOH solution has been added, the
  pH is

57
Weak Acid/Strong Base Titration Curves

• After 20.0 mL of KOH solution has been added, the
  pH is

58
Weak Acid/Strong Base Titration Curves

• After 20.0 mL of KOH solution has been added, the
  pH is

59
Weak Acid/Strong Base Titration Curves

• After 20.0 mL of KOH solution has been added, the
  pH is
• Similarly for all other cases before the
  equivalence point is reached.

60
Weak Acid/Strong Base Titration Curves

• At the equivalence point, the solution is 0.500 M
  in KCH3COO, the salt of a strong base and a weak
  acid which hydrolyzes to give a basic solution.
• Both processes make the solution basic
• Cannot have a pH7.00 at equivalence point.
• Let us calculate the pH at the equivalence point.

61
Weak Acid/Strong Base Titration Curves

• Set up the equilibrium reaction

62
Weak Acid/Strong Base Titration Curves

• Determine the concentration of the salt in
  solution.

63
Weak Acid/Strong Base Titration Curves

• Perform a hydrolysis calculation for the
  potassium acetate in solution.

64
Weak Acid/Strong Base Titration Curves

• After the equivalence point is reached, the pH is
  determined by the excess KOH - as in Strong
  Acid/Strong Base example.

65
Weak Acid/Strong Base Titration Curves

• After the equivalence point is reached, the pH is
  determined by the excess KOH - as in Strong
  Acid/Strong Base example.

66
Weak Acid/Strong Base Titration Curves

• We have calculated only a few points on the
  titration curve. Similar calculations for
  remainder of titration show clearly the shape of
  the titration curve.

67
Strong Acid/Weak BaseTitration Curves

• Titration curves for Strong Acid/Weak Bases look
  similar to Strong Base/Weak Acid but they are
  inverted.

68
Weak Acid/Weak BaseTitration Curves

• Titration curves have very short vertical
  sections.
• Solution is buffered both before and after the
  equivalence point.
• Visual indicators cannot be used.

69
Solubility Product Constants

• Silver chloride, AgCl,is rather insoluble in
  water.
• Careful experiments show that if solid AgCl is
  placed in pure water and vigorously stirred, a
  small amount of the AgCl dissolves in the water.

70
Solubility Product Constants

• The equilibrium constant expression for this
  dissolution is called a solubility product
  constant.
• Kspsolubility product constant
• Molar concentration of ions raised to their
  stoichiometric powers at equilibrium

71
Solubility Product Constants

• Solubility product constant for a compound is the
  product of the concentrations of the constituent
  ions, each raised to the power that corresponds
  to the number of ions in one formula unit of the
  compound.
• Consider the dissolution of silver sulfide in
  water.

72
Solubility Product Constants

• Its solubility product expression is

73
Solubility Product Constants

• The dissolution of solid calcium phosphate in
  water is represented as

74
Solubility Product Constants

• Its solubility product constant expression is

75
Solubility Product Constants

• In general, the dissolution of a slightly soluble
  compound and its solubility product expression
  are represented as
• Ksp has a fixed value for a given system at a
  given temperature

76
Solubility Product Constants

• The same rules apply for compounds that have more
  than two kinds of ions.
• An example is calcium ammonium phosphate.

77
Determination of Solubility Product Constants

• Example One liter of saturated silver chloride
  solution contains 0.00192 g of dissolved AgCl at
  25oC. Calculate the molar solubility of, and Ksp
  for, AgCl.
• Molar solubility can be calculated from the data

78
Determination of Solubility Product Constants

• The equation for the dissociation of silver
  chloride and its solubility product expression are

79
Determination of Solubility Product Constants

• Substitution into the solubility product
  expression gives

80
Uses of Solubility Product Constants

• We can use the solubility product constant to
  calculate the solubility of a compound at 25oC.
• Example Calculate the molar solubility of
  barium sulfate, BaSO4, in pure water and the
  concentration of barium and sulfate ions in
  saturated barium sulfate at 25oC.
• Ksp 1.1 x 10-10.

81
Uses of Solubility Product Constants

• Example Calculate the molar solubility of
  barium sulfate, BaSO4, in pure water and the
  concentration of barium and sulfate ions in
  saturated barium sulfate at 25oC. Ksp 1.1 x
  10-10.

82
Uses of Solubility Product Constants

• Substitute into solubility product expression and
  solve for x, giving the ion concentrations.

83
Uses of Solubility Product Constants

• Now we can calculate the mass of BaSO4 in 1.00 L
  of saturated solution.

84
The Reaction Quotient in Precipitation Reactions

• Use solubility product constants to calculate the
  concentration of ions in a solution and whether
  or not a precipitate will form.
• Example We mix 100 mL of 0.010 M potassium
  sulfate, K2SO4, and 100 mL of 0.10 M lead (II)
  nitrate, Pb(NO3)2 solutions. Will a precipitate
  form?

85
The Reaction Quotient in Precipitation Reactions

• Example We mix 100 mL of 0.010 M potassium
  sulfate, K2SO4, and 100 mL of 0.10 M lead (II)
  nitrate, Pb(NO3)2 solutions. Will a precipitate
  form?

86
The Reaction Quotient in Precipitation Reactions

• Calculate the Qsp for PbSO4.
• Solution volumes are additive.
• Concentrations of the important ions are

87
The Reaction Quotient in Precipitation Reactions

• Finally, we calculate Qsp for PbSO4.

88
Fractional Precipitation

• Fractional precipitation is a method of
  precipitating some ions from solution while
  leaving others in solution.
• Look at a solution that contains Cu, Ag, and
  Au
• We could precipitate them as chlorides

89
Fractional Precipitation

• Fractional precipitation is a method of
  precipitating some ions from solution while
  leaving others in solution.
• Look at a solution that contains Cu, Ag, and
  Au
• We could precipitate them as chlorides

90
Fractional Precipitation

• Example If solid sodium chloride is slowly added
  to a solution that is 0.010 M each in Cu, Ag,
  and Au ions, which compound precipitates first?
  Calculate the concentration of Cl- required to
  initiate precipitation of each of these metal I
  chlorides.

91
Fractional Precipitation

• Example If solid sodium chloride is slowly added
  to a solution that is 0.010 M each in Cu, Ag,
  and Au ions, which compound precipitates first?
  Calculate the concentration of Cl- required to
  initiate precipitation of each of these metal I
  chlorides.

92
Fractional Precipitation

• Example If solid sodium chloride is slowly added
  to a solution that is 0.010 M each in Cu, Ag,
  and Au ions, which compound precipitates first?
  Calculate the concentration of Cl- required to
  initiate precipitation of each of these metal I
  chlorides.

93
Fractional Precipitation

• Repeat the calculation for silver chloride.

94
Fractional Precipitation

• For copper (I) chloride to precipitate.

95
Fractional Precipitation

• We have calculated the Cl- required
• to precipitate AuCl, Cl- gt2.0 x 10-11 M
• to precipitate AgCl, Cl- gt1.8 x 10-8 M
• to precipitate CuCl, Cl- gt1.9 x 10-5 M
• We can calculate the amount of Au precipitated
  before Ag begins to precipitate, as well as the
  amounts of Au and Ag precipitated before Cu
  begins to precipitate.

96
Fractional Precipitation

• Example Calculate the percent of Au ions that
  precipitate before AgCl begins to precipitate.
• Use the Cl- from before to determine the Au
  remaining in solution just before AgCl begins to
  precipitate.

97
Fractional Precipitation

• Example Calculate the percent of Au ions that
  precipitate before AgCl begins to precipitate.
• Use the Cl- from before to determine the Au
  remaining in solution just before AgCl begins to
  precipitate.

98
Fractional Precipitation

• The percent of Au ions unprecipitated just
  before AgCl precipitates is

99
Fractional Precipitation

• The percent of Au ions unprecipitated just
  before AgCl precipitates is
• Therefore, 99.99989 of the Au ions precipitates
  before AgCl begins to precipitate.

100
Fractional Precipitation

• Similar calculations for the concentration of Ag
  ions unprecipitated before CuCl begins to
  precipitate gives

101
Fractional Precipitation

• The percent of Au ions unprecipitated just
  before AgCl precipitates is

102
Fractional Precipitation

• The percent of Au ions unprecipitated just
  before AgCl precipitates is
• Thus, 99.905 of the Ag ions precipitates before
  CuCl begins to precipitate.

103
Factors that Affect Solubility

• CaF2 (s) ? Ca2 (aq) 2F- (aq)
• Addition of a Common ion (F- from NaF)
• Solubility decreases
• Equilibrium shifts to left
• Changes in pH (H reacts with F-)
• Solubility increases (with increasing pH)
• Equilibrium shifts to right

104
Factors that Affect Solubility

• Ag(aq) 2NH3(aq) ? Ag(NH3)2 (aq)
• complex ion
• Formation of a complex ion
• Lewis Acid base chemistry
• Calculate Kf Formation Constant

105
Factors that Affect Solubility

• Ag(aq) 2NH3(aq) ? Ag(NH3)2 (aq)
• complex ion
• AgCl(s) ? Ag(aq) Cl-(aq)
• In formation of complex ion
• Removed Ag from the equilibrium
• Equilibrium shifts to right
• Favors dissolving AgCl

106
Synthesis Question

• Bufferin is a commercially prepared medicine that
  is literally a buffered aspirin. How could you
  buffer aspirin? Hint - what is aspirin?

107
Synthesis Question

• Aspirin is acetyl salicylic acid. So to buffer
  it all that would have to be added is the salt of
  acetyl salicylic acid.

108
Group Question

• Blood is slightly basic, having a pH of 7.35 to
  7.45. What chemical species causes our blood to
  be basic? How does our body regulate the pH of
  blood?

109
Synthesis Question

• Most kidney stones are made of calcium oxalate,
  Ca(O2CCO2). Patients who have their first kidney
  stones are given an extremely simple solution to
  stop further stone formation. They are told to
  drink six to eight glasses of water a day. How
  does this stop kidney stone formation?

110
Synthesis Question
111
Group Question

• The cavities that we get in our teeth are a
  result of the dissolving of the material our
  teeth are made of, calcium hydroxy apatite. How
  does using a fluoride based toothpaste decrease
  the occurrence of cavities?