Ionic Equilibria: Acids and Bases

1
Chapter 16

• Ionic Equilibria Acids and Bases

2
Chapter Goals

1. A Review of Strong Electrolytes
2. The Autoionization of Water
3. The pH and pOH Scales
4. Ionization Constants for Weak Monoprotic Acids
   and Bases
5. Polyprotic Acids
6. Solvolysis
7. Salts of Strong Bases and Strong Acids

3
Chapter Goals

1. Salts of Strong Bases and Weak Acids
2. Salts of Weak Bases and Strong Acids
3. Salts of Weak Bases and Weak Acids
4. Salts That Contain Small, Highly Charged Cations

4
A Review of Strong Electrolytes

• This chapter details the equilibria of weak acids
  and bases.
• We must distinguish weak acids and bases from
  strong electrolytes.
• Weak acids and bases ionize or dissociate
  partially, much less than 100.
• In this chapter we will see that it is often less
  than 10!
• Strong electrolytes ionize or dissociate
  completely.
• Strong electrolytes approach 100 dissociation in
  aqueous solutions.

5
A Review of Strong Electrolytes

• There are three classes of strong electrolytes.
• Strong Water Soluble Acids
• Remember the list of strong acids from Chapter 4.

6
A Review of Strong Electrolytes
7
A Review of Strong Electrolytes

• Strong Water Soluble Bases
• The entire list of these bases was also
  introduced in Chapter 4.

8
A Review of Strong Electrolytes

• Most Water Soluble Salts
• The solubility guidelines from Chapter 4 will
  help you remember these salts.

9
A Review of Strong Electrolytes

• The calculation of ion concentrations in
  solutions of strong electrolytes is easy.
• Example 18-1 Calculate the concentrations of
  ions in 0.050 M nitric acid, HNO3.

10
A Review of Strong Electrolytes

• Example 18-2 Calculate the concentrations of
  ions in 0.020 M strontium hydroxide, Sr(OH)2,
  solution.
• You do it!

11
The Autoionization of Water

• Pure water ionizes very slightly.
• The concentration of the ionized water is less
  than one-millionth molar at room temperature.

12
The Autoionization of Water

• We can write the autoionization of water as a
  dissociation reaction similar to those previously
  done in this chapter.

• Because the activity of pure water is 1, the
  equilibrium constant for this reaction is

13
The Autoionization of Water

• Experimental measurements have determined that
  the concentration of each ion is 1.0 x 10-7 M at
  25oC.
• Note that this is at 25oC, not every temperature!
• We can determine the value of Kc from this
  information.

14
The Autoionization of Water

• This particular equilibrium constant is called
  the ion-product for water and given the symbol
  Kw.
• Kw is one of the recurring expressions for the
  remainder of this chapter and Chapters 19 and 20.

15
The Autoionization of Water

• Example 18-3 Calculate the concentrations of
  H3O and OH- in 0.050 M HCl.

16
The Autoionization of Water

• Use the H3O and Kw to determine the OH-.
• You do it!

17
The Autoionization of Water

• The increase in H3O from HCl shifts the
  equilibrium and decreases the OH-.
• Remember from Chapter 17, increasing the product
  concentration, H3O, causes the equilibrium to
  shift to the reactant side.
• This will decrease the OH- because it is a
  product!

18
The Autoionization of Water

• Now that we know the H3O we can calculate the
  OH-.
• You do it!

19
The pH and pOH scales

• A convenient way to express the acidity and
  basicity of a solution is the pH and pOH scales.
• The pH of an aqueous solution is defined as

20
The pH and pOH scales

• In general, a lower case p before a symbol is
  read as the negative logarithm of the symbol.
• Thus we can write the following notations.

21
The pH and pOH scales

• If either the H3O or OH- is known, the pH
  and pOH can be calculated.
• Example 18-4 Calculate the pH of a solution in
  which the H3O 0.030 M.

22
The pH and pOH scales

• Example 18-5 The pH of a solution is 4.597.
  What is the concentration of H3O?
• You do it!

23
The pH and pOH scales

• A convenient relationship between pH and pOH may
  be derived for all dilute aqueous solutions at
  250C.

• Taking the logarithm of both sides of this
  equation gives

24
The pH and pOH scales

• Multiplying both sides of this equation by -1
  gives

• Which can be rearranged to this form

25
The pH and pOH scales

• Remember these two expressions!!
• They are key to the next three chapters!

26
The pH and pOH scales

• The usual range for the pH scale is 0 to 14.

• And for pOH the scale is also 0 to 14 but
  inverted from pH.
• pH 0 has a pOH 14 and pH 14 has a pOH 0.

27
The pH and pOH scales
28
The pH and pOH scales

• Example 18-6 Calculate the H3O, pH, OH-,
  and pOH for a 0.020 M HNO3 solution.
• Is HNO3 a weak or strong acid?
• What is the H3O ?

29
The pH and pOH scales

• Example 18-6 Calculate the H3O, pH, OH-,
  and pOH for a 0.020 M HNO3 solution.

30
The pH and pOH scales

• To help develop familiarity with the pH and pOH
  scale we can look at a series of solutions in
  which H3O varies between 1.0 M and 1.0 x 10-14
  M.

H3O OH- pH pOH
1.0 M 1.0 x 10-14 M 0.00 14.00
1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00
1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00
2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30
1.0 x 10-14 M 1.0 M 14.00 0.00
31
The pH and pOH scales

• Example 18-7 Calculate the number of H3O and
  OH- ions in one liter of pure water at 250C.
• You do it!

32
Ionization Constants for Weak Monoprotic Acids
and Bases

• Lets look at the dissolution of acetic acid, a
  weak acid, in water as an example.
• The equation for the ionization of acetic acid is

• The equilibrium constant for this ionization is
  expressed as

33
Ionization Constants for Weak Monoprotic Acids
and Bases

• The water concentration in dilute aqueous
  solutions is very high.
• 1 L of water contains 55.5 moles of water.
• Thus in dilute aqueous solutions

34
Ionization Constants for Weak Monoprotic Acids
and Bases

• The water concentration is many orders of
  magnitude greater than the ion concentrations.
• Thus the water concentration is essentially that
  of pure water.
• Recall that the activity of pure water is 1.

35
Ionization Constants for Weak Monoprotic Acids
and Bases

• We can define a new equilibrium constant for weak
  acid equilibria that uses the previous
  definition.
• This equilibrium constant is called the acid
  ionization constant.
• The symbol for the ionization constant is Ka.

36
Ionization Constants for Weak Monoprotic Acids
and Bases

• In simplified form the dissociation equation and
  acid ionization expression are written as

37
Ionization Constants for Weak Monoprotic Acids
and Bases

• The ionization constant values for several acids
  are given below.
• Which acid is the strongest?

Acid Formula Ka value
Acetic CH3COOH 1.8 x 10-5
Nitrous HNO2 4.5 x 10-4
Hydrofluoric HF 7.2 x 10-4
Hypochlorous HClO 3.5 x 10-8
Hydrocyanic HCN 4.0 x 10-10
38
Ionization Constants for Weak Monoprotic Acids
and Bases

• From the above table we see that the order of
  increasing acid strength for these weak acids is

• The order of increasing base strength of the
  anions (conjugate bases) of these acids is

39
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-8 Write the equation for the
  ionization of the weak acid HCN and the
  expression for its ionization constant.

40
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-9 In a 0.12 M solution of a weak
  monoprotic acid, HY, the acid is 5.0 ionized.
  Calculate the ionization constant for the weak
  acid.
• You do it!

41
Ionization Constants for Weak Monoprotic Acids
and Bases

• Since the weak acid is 5.0 ionized, it is also
  95 unionized.
• Calculate the concentration of all species in
  solution.

42
Ionization Constants for Weak Monoprotic Acids
and Bases

• Use the concentrations that were just determined
  in the ionization constant expression to get the
  value of Ka.

43
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-10 The pH of a 0.10 M solution of a
  weak monoprotic acid, HA, is found to be 2.97.
  What is the value for its ionization constant?
• pH 2.97 so H 10-pH

44
Ionization Constants for Weak Monoprotic Acids
and Bases

• Use the H3O and the ionization reaction to
  determine concentrations of all species.

45
Ionization Constants for Weak Monoprotic Acids
and Bases

• Calculate the ionization constant from this
  information.

46
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-11 Calculate the concentrations of
  the various species in 0.15 M acetic acid,
  CH3COOH, solution.
• It is always a good idea to write down the
  ionization reaction and the ionization constant
  expression.

47
Ionization Constants for Weak Monoprotic Acids
and Bases

• Next, combine the basic chemical concepts with
  some algebra to solve the problem.

48
Ionization Constants for Weak Monoprotic Acids
and Bases

• Next we combine the basic chemical concepts with
  some algebra to solve the problem

49
Ionization Constants for Weak Monoprotic Acids
and Bases

• Next we combine the basic chemical concepts with
  some algebra to solve the problem

50
Ionization Constants for Weak Monoprotic Acids
and Bases

• Substitute these algebraic quantities into the
  ionization expression.

51
Ionization Constants for Weak Monoprotic Acids
and Bases

• Solve the algebraic equation, using a simplifying
  assumption that is appropriate for all weak acid
  and base ionizations.

52
Ionization Constants for Weak Monoprotic Acids
and Bases

• Solve the algebraic equation, using a simplifying
  assumption that is appropriate for all weak acid
  and base ionizations.

53
Ionization Constants for Weak Monoprotic Acids
and Bases

• Complete the algebra and solve for the
  concentrations of the species.

54
Ionization Constants for Weak Monoprotic Acids
and Bases

• Note that the properly applied simplifying
  assumption gives the same result as solving the
  quadratic equation does.

55
Ionization Constants for Weak Monoprotic Acids
and Bases
56
Ionization Constants for Weak Monoprotic Acids
and Bases

• Let us now calculate the percent ionization for
  the 0.15 M acetic acid. From Example 18-11, we
  know the concentration of CH3COOH that ionizes in
  this solution. The percent ionization of acetic
  acid is

57
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-12 Calculate the concentrations of
  the species in 0.15 M hydrocyanic acid, HCN,
  solution.
• Ka 4.0 x 10-10 for HCN
• You do it!

58
Ionization Constants for Weak Monoprotic Acids
and Bases
59
Ionization Constants for Weak Monoprotic Acids
and Bases

• The percent ionization of 0.15 M HCN solution is
  calculated as in the previous example.

60
Ionization Constants for Weak Monoprotic Acids
and Bases

• Lets look at the percent ionization of two weak
  acids as a function of their ionization
  constants. Examples 18-11 and 18-12 will suffice.

Solution Ka H pH ionization
0.15 M acetic acid 1.8 x 10-5 1.6 x 10-3 2.80 1.1
0.15 M HCN 4.0 x 10-10 7.7 x 10-6 5.11 0.0051

• Note that the H in 0.15 M acetic acid is 210
  times greater than for 0.15 M HCN.

61
Ionization Constants for Weak Monoprotic Acids
and Bases

• All of the calculations and understanding we have
  at present can be applied to weak acids and weak
  bases!
• One example of a weak base ionization is ammonia
  ionizing in water.

62
Ionization Constants for Weak Monoprotic Acids
and Bases

• All of the calculations and understanding we have
  at present can be applied to weak acids and weak
  bases!
• Example 18-13 Calculate the concentrations of
  the various species in 0.15 M aqueous ammonia.

63
Ionization Constants for Weak Monoprotic Acids
and Bases
64
Ionization Constants for Weak Monoprotic Acids
and Bases

• The percent ionization for weak bases is
  calculated exactly as for weak acids.

65
Ionization Constants for Weak Monoprotic Acids
and Bases

• Example 18-14 The pH of an aqueous ammonia
  solution is 11.37. Calculate the molarity
  (original concentration) of the aqueous ammonia
  solution.
• You do it!

66
Ionization Constants for Weak Monoprotic Acids
and Bases
67
Ionization Constants for Weak Monoprotic Acids
and Bases

• Use the ionization equation and some algebra to
  get the equilibrium concentration.

68
Ionization Constants for Weak Monoprotic Acids
and Bases

• Substitute these values into the ionization
  constant expression.

69
Ionization Constants for Weak Monoprotic Acids
and Bases

• Examination of the last equation suggests that
  our simplifying assumption can be applied.
• In other words (x-2.3x10-3) ? x.
• Making this assumption simplifies the calculation.

70
Polyprotic Acids

• Many weak acids contain two or more acidic
  hydrogens.
• Examples include H3PO4 and H3AsO4.
• The calculation of equilibria for polyprotic
  acids is done in a stepwise fashion.
• There is an ionization constant for each step.
• Consider arsenic acid, H3AsO4, which has three
  ionization constants.
• Ka1 2.5 x 10-4
• Ka2 5.6 x 10-8
• Ka3 3.0 x 10-13

71
Polyprotic Acids

• The first ionization step for arsenic acid is

72
Polyprotic Acids

• The second ionization step for arsenic acid is

73
Polyprotic Acids

• The third ionization step for arsenic acid is

74
Polyprotic Acids

• Notice that the ionization constants vary in the
  following fashion

• This is a general relationship.
• For weak polyprotic acids the Ka1 is always gt
  Ka2, etc.

75
Polyprotic Acids

• Example 18-15 Calculate the concentration of all
  species in 0.100 M arsenic acid, H3AsO4,
  solution.
• Write the first ionization step and represent the
  concentrations.
• Approach this problem exactly as previously done.

76
Polyprotic Acids

• Substitute the algebraic quantities into the
  expression for Ka1.

77
Polyprotic Acids

• Use the quadratic equation to solve for x, and
  obtain both values of x.

78
Polyprotic Acids

• Next, write the equation for the second step
  ionization and represent the concentrations.

79
Polyprotic Acids

• Substitute the algebraic expressions into the
  second step ionization expression.

80
Polyprotic Acids
81
Polyprotic Acids

• Finally, repeat the entire procedure for the
  third ionization step.

82
Polyprotic Acids

1. Substitute the algebraic representations into the
   third ionization expression.

83
Polyprotic Acids

• Use Kw to calculate the OH- in the 0.100 M
  H3AsO4 solution.

84
Polyprotic Acids

• A comparison of the various species in 0.100 M
  H3AsO4 solution follows.

Species Concentration
H3AsO4 0.095 M
H 0.0049 M
H2AsO4- 0.0049 M
HAsO42- 5.6 x 10-8 M
AsO43- 3.4 x 10-18 M
OH- 2.0 x 10-12 M
85
Solvolysis

• This reaction process is the most difficult
  concept in this chapter.
• Solvolysis is the reaction of a substance with
  the solvent in which it is dissolved.
• Hydrolysis refers to the reaction of a substance
  with water or its ions.
• Combination of the anion of a weak acid with H3O
  ions from water to form nonionized weak acid
  molecules.

86
Solvolysis

• Hydrolysis refers to the reaction of a substance
  with water or its ions.
• Hydrolysis is solvolysis in aqueous solutions.
• The combination of a weak acids anion with H3O
  ions, from water, to form nonionized weak acid
  molecules is a form of hydrolysis.

87
Solvolysis

• The reaction of the anion of a weak monoprotic
  acid with water is commonly represented as

88
Solvolysis

• Recall that at 25oC
• in neutral solutions
• H3O 1.0 x 10-7 M OH-
• in basic solutions
• H3O lt 1.0 x 10-7 M and OH- gt 1.0 x 10-7 M
  
• in acidic solutions
• OH- lt 1.0 x 10-7 M and H3O gt 1.0 x 10-7 M

89
Solvolysis

• Remember from BrØnsted-Lowry acid-base theory
• The conjugate base of a strong acid is a very
  weak base.
• The conjugate base of a weak acid is a stronger
  base.
• Hydrochloric acid, a typical strong acid, is
  essentially completely ionized in dilute aqueous
  solutions.

90
Solvolysis

• The conjugate base of HCl, the Cl- ion, is a very
  weak base.
• The chloride ion is such a weak base that it will
  not react with the hydronium ion.

• This fact is true for all strong acids and their
  anions.

91
Solvolysis

• HF, a weak acid, is only slightly ionized in
  dilute aqueous solutions.
• Its conjugate base, the F- ion, is a much
  stronger base than the Cl- ion.
• The F- ions combine with H3O ions to form
  nonionized HF.
• Two competing equilibria are established.

92
Solvolysis

• Dilute aqueous solutions of salts that contain no
  free acid or base come in four types
• Salts of Strong Bases and Strong Acids
• Salts of Strong Bases and Weak Acids
• Salts of Weak Bases and Strong Acids
• Salts of Weak Bases and Weak Acids

93
Salts of Strong Bases and Weak Acids

• Salts made from strong acids and strong soluble
  bases form neutral aqueous solutions.
• An example is potassium nitrate, KNO3, made from
  nitric acid and potassium hydroxide.

94
Salts of Strong Bases and Weak Acids

• Salts made from strong soluble bases and weak
  acids hydrolyze to form basic solutions.
• Anions of weak acids (strong conjugate bases)
  react with water to form hydroxide ions.
• An example is sodium hypochlorite, NaClO, made
  from sodium hydroxide and hypochlorous acid.

95
Salts of Strong Bases and Weak Acids

• We can combine these last two equations into one
  single equation that represents the total
  reaction.

96
Salts of Strong Bases and Weak Acids

• The equilibrium constant for this reaction,
  called the hydrolysis constant, is written as

97
Salts of Strong Bases and Weak Acids

• Algebraic manipulation of the previous expression
  give us a very useful form of the expression.
• Multiply the expression by one written as H/
  H.
• H/H 1

98
Salts of Strong Bases and Weak Acids

• Which can be rewritten as

99
Salts of Strong Bases and Weak Acids

• Which can be used to calculate the hydrolysis
  constant for the hypochlorite ion

100
Salts of Strong Bases and Weak Acids

• This same method can be applied to the anion of
  any weak monoprotic acid.

101
Salts of Strong Bases and Weak Acids

• Example 18-16 Calculate the hydrolysis constants
  for the following anions of weak acids.
• The fluoride ion, F-, the anion of hydrofluoric
  acid, HF. For HF, Ka7.2 x 10-4.

102
Salts of Strong Bases and Weak Acids

• The cyanide ion, CN-, the anion of hydrocyanic
  acid, HCN. For HCN, Ka 4.0 x 10-10.
• You do it!

103
Salts of Strong Bases and Weak Acids

• Example 18-17 Calculate OH-, pH and percent
  hydrolysis for the hypochlorite ion in 0.10 M
  sodium hypochlorite, NaClO, solution. Clorox,
  Purex, etc., are 5 sodium hypochlorite
  solutions.

104
Salts of Strong Bases and Weak Acids

• Set up the equation for the hydrolysis and the
  algebraic representations of the equilibrium
  concentrations.

105
Salts of Strong Bases and Weak Acids

• Substitute the algebraic expressions into the
  hydrolysis constant expression.

106
Salts of Strong Bases and Weak Acids

• Substitute the algebraic expressions into the
  hydrolysis constant expression.

107
Salts of Strong Bases and Weak Acids

• The percent hydrolysis for the hypochlorite ion
  may be represented as

108
Salts of Strong Bases and Weak Acids

• If a similar calculation is performed for 0.10 M
  NaF solution and the results from 0.10 M sodium
  fluoride and 0.10 M sodium hypochlorite compared,
  the following table can be constructed.

Solution Ka Kb OH- (M) pH hydrolysis
NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012
NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23 0.17
109
Salts of Weak Bases and Strong Acids

• Salts made from weak bases and strong acids form
  acidic aqueous solutions.
• An example is ammonium bromide, NH4Br, made from
  ammonia and hydrobromic acid.

110
Salts of Weak Bases and Strong Acids

• The reaction may be more simply represented as

111
Salts of Weak Bases and Strong Acids

• Or even more simply as

• The hydrolysis constant expression for this
  process is

112
Salts of Weak Bases and Strong Acids

• Multiplication of the hydrolysis constant
  expression by OH-/ OH- gives

113
Salts of Weak Bases and Strong Acids

• Which we recognize as

114
Salts of Weak Bases and Strong Acids

• In its simplest form for this hydrolysis

115
Salts of Weak Bases and Strong Acids

• Example 18-18 Calculate H, pH, and percent
  hydrolysis for the ammonium ion in 0.10 M
  ammonium bromide, NH4Br, solution.
• Write down the hydrolysis reaction and set up the
  table as we have done before

116
Salts of Weak Bases and Strong Acids

1. Substitute the algebraic expressions into the
   hydrolysis constant.

117
Salts of Weak Bases and Strong Acids

1. Complete the algebra and determine the
   concentrations and pH.

118
Salts of Weak Bases and Strong Acids

1. The percent hydrolysis of the ammonium ion in
   0.10 M NH4Br solution is

119
Salts of Weak Bases and Weak Acids

• Salts made from weak acids and weak bases can
  form neutral, acidic or basic aqueous solutions.
• The pH of the solution depends on the relative
  values of the ionization constant of the weak
  acids and bases.
• Salts of weak bases and weak acids for which
  parent Kbase Kacid make neutral solutions.
• An example is ammonium acetate, NH4CH3COO, made
  from aqueous ammonia, NH3,and acetic acid,
  CH3COOH.
• Ka for acetic acid Kb for ammonia 1.8 x 10-5.

120
Salts of Weak Bases and Weak Acids

• The ammonium ion hydrolyzes to produce H ions.
  Its hydrolysis constant is

121
Salts of Weak Bases and Weak Acids

• The acetate ion hydrolyzes to produce OH- ions.
  Its hydrolysis constant is

122
Salts of Weak Bases and Weak Acids

• Because the hydrolysis constants for both ions
  are equal, their aqueous solutions are neutral.
• Equal numbers of H and OH- ions are produced.

123
Salts of Weak Bases and Weak Acids

• Salts of weak bases and weak acids for which
  parent Kbase gt Kacid make basic solutions.
• An example is ammonium hypochlorite, NH4ClO, made
  from aqueous ammonia, NH3,and hypochlorous acid,
  HClO.
• Kb for NH3 1.8 x 10-5 gt Ka for HClO 3.5x10-8

124
Salts of Weak Bases and Weak Acids

• The ammonium ion hydrolyzes to produce H ions.
  Its hydrolysis constant is

125
Salts of Weak Bases and Weak Acids

• The hypochlorite ion hydrolyzes to produce OH-
  ions. Its hydrolysis constant is

• Because the Kb for ClO- ions is three orders of
  magnitude larger than the Ka for NH4 ions, OH-
  ions are produced in excess making the solution
  basic.

126
Salts of Weak Bases and Weak Acids

• Salts of weak bases and weak acids for which
  parent Kbase lt Kacid make acidic solutions.
• An example is trimethylammonium
  fluoride,(CH3)3NHF, made from trimethylamine,
  (CH3)3N,and hydrofluoric acid acid, HF.
• Kb for (CH3)3N 7.4 x 10-5 lt Ka for HF 7.2 x
  10-4

127
Salts of Weak Bases and Weak Acids

• Both the cation, (CH3)3NH, and the anion, F-,
  hydrolyze.

128
Salts of Weak Bases and Weak Acids

• The trimethylammonium ion hydrolyzes to produce
  H ions. Its hydrolysis constant is

129
Salts of Weak Bases and Weak Acids

• The fluoride ion hydrolyzes to produce OH- ions.
  Its hydrolysis constant is

• Because the Ka for (CH3)3NH ions is one order of
  magnitude larger than the Kb for F- ions, H ions
  are produced in excess making the solution
  acidic.

130
Salts of Weak Bases and Weak Acids

• Summary of the major points of hydrolysis up to
  now.
• The reactions of anions of weak monoprotic acids
  (from a salt) with water to form free molecular
  acids and OH-.

131
Salts of Weak Bases and Weak Acids

1. The reactions of anions of weak monoprotic acids
   (from a salt) with water to form free molecular
   acids and OH-.

132
Salts of Weak Bases and Weak Acids

• Aqueous solutions of salts of strong acids and
  strong bases are neutral.
• Aqueous solutions of salts of strong bases and
  weak acids are basic.
• Aqueous solutions of salts of weak bases and
  strong acids are acidic.
• Aqueous solutions of salts of weak bases and weak
  acids can be neutral, basic or acidic.
• The values of Ka and Kb determine the pH.

133
Hydrolysis of Small Highly-Charged Cations

• Cations of insoluble bases (metal hydroxides)
  become hydrated in solution.
• An example is a solution of Be(NO3)3.
• Be2 ions are thought to be tetrahydrated and sp3
  hybridized.

134
Hydrolysis of Small Highly-Charged Cations

• In condensed form it is represented as

• or, even more simply as

135
Hydrolysis of Small Highly-Charged Cations

• The hydrolysis constant expression for
  Be(OH2)42 and its value are

• or, more simply

136
Hydrolysis of Small Highly-Charged Cations

• Example 18-19 Calculate the pH and percent
  hydrolysis in 0.10 M aqueous Be(NO3)2 solution.
• The equation for the hydrolysis reaction and
  representations of concentrations of various
  species are

137
Hydrolysis of Small Highly-Charged Cations

1. Algebraic substitution of the expressions into
   the hydrolysis constant

138
Hydrolysis of Small Highly-Charged Cations

• Calculate the percent hydrolysis of Be2.

139
Hydrolysis of Small Highly-Charged Cations

• This table is a comparison of 0.10 M Be(NO3 )2
  solution and 0.10 M CH3COOH solution.

Solution H3O pH hydrolysis or ionization
0.10 M Be(NO3)2 1.0 x 10-3 M 3.00 1.0
0.10 M CH3COOH 1.3 x 10-3 M 2.89 1.3
Notice that the Be solution is almost as acidic
as the acetic acid solution.
140
Synthesis Question

• Rain water is slightly acidic because it absorbs
  carbon dioxide from the atmosphere as it falls
  from the clouds. (Acid rain is even more acidic
  because it absorbs acidic anhydride pollutants
  like NO2 and SO3 as it falls to earth.) If the
  pH of a stream is 6.5 and all of the acidity
  comes from CO2, how many CO2 molecules did a drop
  of rain having a diameter of 6.0 mm absorb in its
  fall to earth?

141
Synthesis Question
142
Synthesis Question
143
Group Question

• A common food preservative in citrus flavored
  drinks is sodium benzoate, the sodium salt of
  benzoic acid. How does this chemical compound
  behave in solution so that it preserves the
  flavor of citrus drinks?

144
End of Chapter 18

• Weak aqueous acid-base mixtures are called
  buffers. They are the subject of Chapter 19.