18
Ionic Equilibria:
Acids and Bases
2
Chapter Goals
1.
2.
3.
4.
A Review of Strong Electrolytes
The Autoionization of Water
The pH and pOH Scales
Ionization Constants for Weak Monoprotic
Acids and Bases
5. Polyprotic Acids
6. Solvolysis
7. Salts of Strong Bases and Strong Acids
3
Chapter Goals
8. Salts of Strong Bases and Weak Acids
9. Salts of Weak Bases and Strong Acids
10.Salts of Weak Bases and Weak Acids
11.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
5
aqueous solutions.
A Review of Strong Electrolytes
• There are three classes of strong
electrolytes.
1 Strong Water Soluble Acids
Remember the list of strong acids from Chapter 4.
100%
HNO3( )  H 2O (  )  H 3O

(aq)

3(aq)
 NO
or
100%
HNO3( )  H

(aq)

3(aq)
 NO
6
A Review of Strong Electrolytes
100%
HNO3( )  H 2O (  )  H 3O

(aq)

3(aq)
 NO
or
100%
HNO3( )  H

(aq)

3(aq)
 NO
7
A Review of Strong Electrolytes
2 Strong Water Soluble Bases
The entire list of these bases was also introduced in
Chapter 4.
H 2 O 100%
KOH(s)   K
H 2 O 100%

(aq)
 OH
2
(aq)
Sr(OH) 2(s)   Sr
(aq)
 2 OH
(aq)
8
A Review of Strong Electrolytes
3 Most Water Soluble Salts
The solubility guidelines from Chapter 4 will help you
remember these salts.
H 2 O 100%
NaCl(s)   Na
H 2 O 100%

(aq)
 Cl
Ca(NO3 ) 2s    Ca
2
(aq)
(aq)

3(aq)
 2 NO
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.

(aq)
HNO3( )  H2O( ) 
 H3O
100%

3(aq)
 NO
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!
2
(aq)
Sr(OH)2(s) 
Sr
H2 O
 2 OH
(aq)
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.


H 2O( )  H 2O()  H3O(aq)  OH(aq)
• Because the activity of pure water is 1, the
equilibrium constant for this reaction is:

Kc  H 3O
+

OH


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.
K  H O OH
c

 
 1.0 x 10 1.0 x 10 
3
-7
 1.0 x10
-7
14
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.

K w  H 3O
+
OH 
 1.0 x10

14
15
The Autoionization of Water
Example 18-3: Calculate the concentrations of H3O+
and OH- in 0.050 M HCl.
HCl + H 2O  H3O+  Cl
16
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:

pH = -log H 3O
+

17
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.
 
pAg = -logAg 
pOH = -log OH
-
+
and so forth for other quantities .
18
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.
19
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!
20
The pH and pOH scales
• A convenient relationship between pH and pOH
may be derived for all dilute aqueous solutions at
250C.


[H 3O ][OH ]  1.0 10
14
• Taking the logarithm of both sides of this
equation gives:






log H 3O  log OH  14.00
21
The pH and pOH scales
• Multiplying both sides of this equation by -1
gives:



- log H 3O   log OH


  14.00
• Which can be rearranged to this form:
pH  pOH  14.00
22
The pH and pOH scales
• Remember these two expressions!!
– They are key to the next three chapters!
H O OH   1.0 10


14
3
pH  pOH  14.00
23
The pH and pOH scales
• The usual range for the pH scale is 0 to 14.
H O   1.0 M to H O   1.0 10


3
14
3
pH  0
M
pH  14.00
to
• 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.
OH   1.0 10

pOH  14.00
14



M up to OH  1.0M
pOH  0
24
The pH and pOH scales
25
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+] ?
26
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+]
1.0 M
[OH-]
1.0 x 10-14 M
pH
0.00
pOH
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
27
Ionization Constants for Weak
Monoprotic Acids and Bases
• Let’s look at the dissolution of acetic acid, a
weak acid, in water as an example.
• The equation for the ionization of acetic acid is:


CH3COOH  H 2O  H3O  CH3COO
• The equilibrium constant for this ionization is
expressed as:

H O CH COO 


Kc
3

3
CH3COOHH 2O
29
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:
H2O  55.5M
30
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.

H O CH COO 
K H O 


3
c
3
CH3COOH
2

H O CH COO 
K

3

3
CH3COOH
31
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.


5
3
3
a
3
K

H O CH COO 

 1.8 10
CH COOH
for acetic acid
32
Ionization Constants for Weak
Monoprotic Acids and Bases
• In simplified form the dissociation equation and
acid ionization expression are written as:


CH3COOH  H  CH3COO

H CH COO 

 1.8 10

Ka

3
CH3COOH
5
33
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
34
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:
HF > HNO2 > CH3COOH > HClO > HCN
• The order of increasing base strength of the
anions (conjugate bases) of these acids is:
-
2
-
-
F < NO < CH3COO < ClO < CN
-
35
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.


HCN  H  CN

H CN 

 4.0 x 10

Ka
HCN 
-
-10
36
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!
37
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
38
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.
39
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.
40
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
41
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!
42
Ionization Constants for Weak
Monoprotic Acids and Bases
The percent ionization of 0.15 M HCN solution is
calculated as in the previous example.
43
Ionization Constants for Weak
Monoprotic Acids and Bases
• Let’s 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
0.15 M
HCN
1.8 x 10-5
1.6 x 10-3
2.80
1.1
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.
44
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.
45
Ionization Constants for Weak
Monoprotic Acids and Bases
Example 18-13: Calculate the concentrations of
the various species in 0.15 M aqueous ammonia.
46
Ionization Constants for Weak
Monoprotic Acids and Bases
The percent ionization for weak bases is
calculated exactly as for weak acids.
47
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!
48
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.
1 Ka1 = 2.5 x 10-4
2 Ka2 = 5.6 x 10-8
3 Ka3 = 3.0 x 10-13
49
Polyprotic Acids
• The first ionization step for arsenic acid is:


H 3AsO 4  H  H 2 AsO 4

H H AsO 

 2.5 10

Ka1
2
H3AsO 4 

4
4
50
Polyprotic Acids
• The second ionization step for arsenic acid is:

2
H 2 AsO  H  HAsO 4
14

H HAsO 

 5.6 10
H AsO 

K a2
2
4
2
8
14
51
Polyprotic Acids
• The third ionization step for arsenic acid is:

3
HAsO  H  AsO 4
24

H AsO 

 3.0 10
HAsO 

K a3
3
4
24
13
52
Polyprotic Acids
• Notice that the ionization constants vary in the
following fashion:
K a1  K a2  K a3
• This is a general relationship.
– For weak polyprotic acids the Ka1 is always > Ka2,
etc.
53
Polyprotic Acids
Example 18-15: Calculate the concentration of all
species in 0.100 M arsenic acid, H3AsO4, solution.
54
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
55
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.
56
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 acid’s anion
with H3O+ ions, from water, to form
nonionized weak acid molecules is a form
of hydrolysis.
A  H 3O 
 HA  H 2O



recall H 2O + H 2O  H 3O  OH
-

57
Solvolysis
• The reaction of the anion of a weak monoprotic
acid with water is commonly represented as:

A  H 2 O  HA  OH
-
+
The removal of H 3O upsets
the water equilibriu m
58
Solvolysis
• Recall that at 25oC
• in neutral solutions:
[H3O+] = 1.0 x 10-7 M = [OH-]
• in basic solutions:
[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M
• in acidic solutions:
[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M
59
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.
~100%


HCl  H 2O   H 3O  Cl 60
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.
Cl   H 3O  No rxn. in dilute aqueous solutions
• This fact is true for all strong acids and their
anions.
61
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.
+
-
HF + H 2O 

H 3O  F
only slightly
F- + H O + 

 HF + H O
3
2
nearly completely
62
Solvolysis
•
1.
2.
3.
4.
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
63
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.
~100% in H O
+

2
KNO3 ( s )   

 K  NO3

H 2O  H 2O 

OH-  H 3O 
The ions that are in solution  KOH  HNO3
The KOH and HNO 3 are present in equal amounts.


There is no reaction t o upset H 3O + OH-

Thus the solution is neutral.
64
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.
100% in H 2 O
NaClO ( s ) ~

 Na   ClO 
HO + HO 

OH- + H O 
2
2
Notice ions in solution
3
 NaOH  HClO
Which is the stronger acid or base?
65
Salts of Strong Bases and
Weak Acids

Na ClO ( s)   Na  ClO
+

HO + HO 
 OH + H O
+
~100% in H 2O
-
2
2

-
3
ClO  H 3O  HClO  H 2O
-
• We can combine these last two equations into
one single equation that represents the total
reaction.


ClO  H 2O 
 HClO  OH
-
66
Salts of Strong Bases and
Weak Acids
• The equilibrium constant for this reaction, called
the hydrolysis constant, is written as:
 HClOOH 
-
Kb =
ClO 
-
67
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+].
Kb
 HClO



=

ClO  H 
H OH 
HClO


=

1
H ClO 
-+

+/H
H
=
1
OH
H

-

Kb

-
-
68
Salts of Strong Bases and Weak
Acids
• Which can be rewritten as:
Kb =
Kb =
 HClO
H ClO 

-
1
Ka for HClO
H OH 



-
1
 Kw
69
Salts of Strong Bases and
Weak Acids
• Which can be used to calculate the hydrolysis
constant for the hypochlorite ion:
Kb =
Kb =
Kb =
1
Ka for HClO
Kw
Ka for HClO

 Kw
1  10-14
=
3.5  10-8

HClO
OH


ClO 

  2.9  10
7
70
Salts of Strong Bases and
Weak Acids
• This same method can be applied to the anion
of any weak monoprotic acid.

A  H 2 O  HA  OH

Kb

HA OH
=
A 



KW
K a for HA
71
Salts of Strong Bases and
Weak Acids
Example 18-16: Calculate the hydrolysis constants
for the following anions of weak acids.
72
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.
73
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
74
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.
NH4 Brs   

H OH O

H 2 O ~100%
2
2
Ions in solution are
NH

4
 Br
-
OH-  H 3O 
 NH4OH
 HBr
Which is the stronger acid or base?
75
Salts of Weak Bases and
Strong Acids
The relatively strong acid, NH 4 ,
reacts with the OH- ion removing
it from solution leaving excess H 3O 
 NH  H O
NH 4  OH- 
3
2
generates excess H 3O 
• The reaction may be more simply represented
as:

NH  H 2O 


4
NH3  H 3O

76
Salts of Weak Bases and
Strong Acids
• Or even more simply as:


NH 
 NH3  H

4
• The hydrolysis constant expression for this
process is:
Ka

NH3  H 3O

 or K  NH  H 
NH 
NH 

4


3
a

4
77
Salts of Weak Bases and
Strong Acids
• Multiplication of the hydrolysis constant
expression by [OH-]/ [OH-] gives:
Ka

NH3  H 3O

  OH 
NH 
OH 


NH 
H O OH 


NH OH 
1

-

4
-

Ka
3

4
-
3
-
78
Salts of Weak Bases and
Strong Acids
• Which we recognize as:
1
Kw
Kw
Ka 


K b NH 3 
1
K b NH 3 
14
10
.  10
10
Ka 

5
.
6

10
5
18
.  10
79
Salts of Weak Bases and
Strong Acids
• In its simplest form for this hydrolysis:

NH 4
Ka 
 NH  H 


3
 NH 3  H


NH 4


  5.6  10
10
80
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.
81
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.
1. 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.
82
Salts of Weak Bases and Weak
Acids
• The ammonium ion hydrolyzes to produce H+
ions. Its hydrolysis constant is:
+
NH 4
Ka 


 NH 3  H
 NH 3  H

+
NH 4


  5.6  10
10
83
Salts of Weak Bases and Weak
Acids
• The acetate ion hydrolyzes to produce OH- ions.
Its hydrolysis constant is:


CH 3COO  H 2O  CH 3COOH  OH

Kb 
CH 3COOHOH
CH COO 


  5.6  10
10
3
84
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.
2 O ~100%
NH 4 CH 3COO H
 NH 4  CH 3COO

HO  HO 

OH-  H O 
2
2
Ions in solution are
3
 NH4OH
 CH 3COOH
A weak acid and base are formed in solution!
85
Salts of Weak Bases and Weak
Acids
2. Salts of weak bases and weak acids for
which parent Kbase > 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 > Ka for HClO = 3.5x10-8
86
Salts of Weak Bases and Weak
Acids
• The ammonium ion hydrolyzes to produce H+
ions. Its hydrolysis constant is:


NH  NH3  H

4
Ka

NH3  H

NH 

4

  5.6 10
10
87
Salts of Weak Bases and Weak
Acids
• The hypochlorite ion hydrolyzes to produce OHions. Its hydrolysis constant is:


ClO  H 2O  HClO  OH

K 
 HClOOH
ClO 

  2.9  10
7
• Because
b 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.
88
Salts of Weak Bases and Weak
Acids
3. Salts of weak bases and weak acids for
which parent Kbase < 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 < Ka for HF = 7.2 x
10-4
89
Salts of Weak Bases and Weak
Acids
• Both the cation, (CH3)3NH+, and the anion, F-,
hydrolyze.
CH3 3 NH

F  CH 3 3 NH  F

H 2O~100%

90
Salts of Weak Bases and Weak
Acids
• The trimethylammonium ion hydrolyzes to
produce H+ ions. Its hydrolysis constant is:


(CH 3 ) 3 NH  (CH 3 ) 3 N  H
+
Ka 
 
(CH ) NH  K

(
CH
)
N
H
 33 
Kw
+
3 3
b for ( CH 3 ) 3 N
14
10
.  10
10
Ka 
.  10
5  14
7.4  10
91
Salts of Weak Bases and Weak
Acids
• The fluoride ion hydrolyzes to produce OH- ions.
Its hydrolysis constant is:


F  H 2O  HF  OH

Kb 
 HFOH
F 
-


Kw
Ka for HF
+ ions is one order
• Because the Ka for (CH
)
NH
3
3
14
.  10
of magnitude 10
larger
than the Kb for F-ions,
H+
11
Kb 
.  10the
4  14
ions are produced
in
excess
making
7.2  10
solution acidic.
92
Salts of Weak Bases and Weak
Acids
• Summary of the major points of
hydrolysis up to now.
1 The reactions of anions of weak
monoprotic acids (from a salt) with water
to form free molecular acids and OH-.

A + H 2O  HA + OH
Kw
Kb 
Ka  HA 
-
93
Salts of Weak Bases and Weak
Acids
2. The reactions of anions of weak monoprotic
acids (from a salt) with water to form free
molecular acids and OH-.
+
+
2
3

BH + H O  B + H O
Kw
Ka 
B = weak base
K b  B
94
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.
95
The values of Ka and Kb determine the pH.
Hydrolysis of Small Highly-Charged
Cations
• Cations of insoluble bases (metal hydroxides)
become hydrated in solution.
– An example is a solution of Be(NO3)2.
– Be2+ ions are thought to be tetrahydrated and sp3
2
hybridized. Be2+  Be(OH )
 aq 
1s
Be2

2 4

1s 2s 2s 2p2p


Be




4
form sp hybrids
3
 Be(OH 2 )4 
2

xx 
xxxx
xx



e- pairs on coordinated water molecules
96
Hydrolysis of Small Highly-Charged
Cations
• In condensed form it is represented as:

2


Be(OH
)

H
O

Be
OH
OH

H
O






2 4
2
2 3
3
or, even more simply
as:
Be
2+



 H 2O  Be(OH)  H
97
Hydrolysis of Small Highly-Charged
Cations
• The hydrolysis constant expression for
[Be(OH2)4]2+ and its value are:
Be(OH ) (OH)  H O 


 10
.  10
Be(OH ) 

+
Ka
2 3
3
5
2
2 4
or, more simply
Be(OH)  H 


 10
.  10
Be 
+
Ka

5
2+
98
Hydrolysis of Small Highly-Charged
Cations
Example 18-19: Calculate the pH and percent
hydrolysis in 0.10 M aqueous Be(NO3)2 solution.
99
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
1.3 x 10-3 2.89
1.3%
CH3COOH
M
Notice that the Be solution is almost as acidic as the acetic
acid solution.
100
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?
101
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?
102
18
Ionic Equilibria:
Acids and Bases
103