Chapter 16

Ionic Equilibria: Acids and Bases
1
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
A Review of Strong Electrolytes
The Autoionization of Water
The pH and pOH Scales
Ionization Constants for Weak Monoprotic Acids
and Bases
Polyprotic Acids
Solvolysis
Salts of Strong Bases and Strong Acids
2
Chapter Goals
8.
9.
10.
11.
Salts of Strong Bases and Weak Acids
Salts of Weak Bases and Strong Acids
Salts of Weak Bases and Weak Acids
Salts That Contain Small, Highly Charged
Cations
3
A Review of Strong Electrolytes

This chapter details the equilibria of weak acids
and bases.


Weak acids and bases ionize or dissociate
partially, much less than 100%.


We must distinguish weak acids and bases from strong
electrolytes.
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.
4
A Review of Strong Electrolytes

1
There are three classes of strong electrolytes.
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
5
A Review of Strong Electrolytes
100%
HNO3( )  H 2O (  )  H 3O

(aq)

3(aq)
 NO
or
100%
HNO3( )  H

(aq)

3(aq)
 NO
6
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)
7
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
8
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.
HNO3( )  H 2O(  ) 
 H 3O
100%
0.050 M

(aq)

3(aq)
 NO
0.050 M 0.050 M
9
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
H 2O
0.020M
 2 OH
(aq)
0.020M 20.020M 
0.040M
10
The Autoionization of Water

Pure water ionizes very slightly.

The concentration of the ionized water is less than onemillionth molar at room temperature.
11
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


12
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.

OH 
 1.0 x 10 1.0 x 10 
Kc  H 3O

+
-7
 1.0 x10
-7
14
13
The Autoionization of Water

This particular equilibrium constant is called the ionproduct 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
14
The Autoionization of Water

Example 18-3: Calculate the concentrations of
H3O+ and OH- in 0.050 M HCl.
HCl + H 2 O  H 3O +  Cl
0.050M



0.050M 0.050M

Thus the H 3O +  0.050M .
The H 3O + and K w will allow us to calculate [OH - ].
15
The Autoionization of Water

Use the [H3O+] and Kw to determine the [OH-].
You do it!
H O OH   1.0 10
1.0  10
1.0  10
OH   H O   5.0 10
OH   2.0 10 M

+
14
3
14

14
+
2
3

13
16
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!
The [H 3O  ] from HCl is 0.050M
 H O 
H 2O  H 2O 
OH
3

The [H 3O ] from K w is 2.0 10 M .
-13
The overall [H 3O  ]  0.050  2.0 10-13 M  0.050 M .
17
The Autoionization of Water

Now that we know the [H3O+] we can calculate the
[OH-].
You do it!
K w  [H 3O  ][OH-]  110-14
Since [H 3O  ]  0.050 M .
-14
1

10
[OH- ] 
[H 3O  ]
-14
1

10
[OH- ] 
[0.050 M ]
[OH- ]  2.0 10 13 M
18
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
+

19
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 .
20
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.

pH = -log H 3O

+

pH   log 3.0 10
pH  1.52
2

21
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!

pH  -log[H 3O ]
4.597  -log[H 3O  ]

log[H 3O ]  -4.597
[H 3O  ]  10-4.597
[H 3O  ]  2.53 10 5 M
22
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
23
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
24
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
25
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
3
pH  0

14
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
26
The pH and pOH scales
27
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+] ?
100%
HNO 3  H 2 O 
 H 3O   NO3-
0.020M
0.020 M 0.020 M
H O   2.0 10 M
pH  -log 2.0 10 M 

2
3
2
pH  1.70
28
The pH and pOH scales

Example 18-6: Calculate the [H3O+], pH, [OH-], and
pOH for a 0.020 M HNO3 solution.
  
1.0 10
1.0 10
OH   H O   2.0 10  5.0 10
pOH   log 5.0 10   12.30
K w  H 3O  OH   1.0 10 14
14

14

2
13
M
3
13
29
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
30
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
32
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
33
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
34
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.

H O CH COO 

 1.8 10

Ka
3

3
CH3COOH
5
for acetic acid
35
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
36
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
37
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
-
38
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
39
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!
+

HY  H + Y
H Y 


+
Ka
-
 HY
40
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.
H   Y   0.05(0.12M )  0.0060M
H   Y   6.0 10 M
+

+

3
HY   0.95(0.12M )  0.11M
41
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.
Ka

H Y 

Ka

6.0 10 6.0 10 



HY 
3
3
0.11
K a  3.3 10  4
42
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
H   10
H   1.110


2.97
3
M
43
Ionization Constants for Weak
Monoprotic Acids and Bases

Use the [H3O+] and the ionization reaction to
determine concentrations of all species.

HA

H+

A-

Equil. []'s 0.10 - 1.1  10-3

1.1  10-3
1.1  10-3
 0.10
44
Ionization Constants for Weak
Monoprotic Acids and Bases

Calculate the ionization constant from this
information.

H A  1.1 10 1.1 10 



Ka

HA 
-3
-3
0.10
K a  1.2  10 5
45
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.
CH3COOH  H 2 O 
 H 3O   CH3COO-

H O CH COO 

 1.8 10

Ka
3
-
3
CH3COOH
5
46
Ionization Constants for Weak
Monoprotic Acids and Bases

Next, combine the basic chemical concepts with
some algebra to solve the problem.
CH3COOH  H 2O 
 H 3O   CH3COOInitial []
0.15M
47
Ionization Constants for Weak
Monoprotic Acids and Bases

Next we combine the basic chemical concepts with
some algebra to solve the problem
CH3COOH  H 2O 
 H 3O   CH3COOInitial []
0.15M
Change
- xM
 xM
 xM
48
Ionization Constants for Weak
Monoprotic Acids and Bases

Next we combine the basic chemical concepts with
some algebra to solve the problem


CH3COOH  H 2 O  H 3O  CH3COO
Initial []
Change
0.15M
- xM
 xM
 xM
Equilibriu m [] ( 0.15-x)M
 xM
 xM
49
Ionization Constants for Weak
Monoprotic Acids and Bases

Substitute these algebraic quantities into the
ionization expression.
Ka

H





O
CH
COO
3
3
CH3COOH
 x  x 
0.15  x 

 1.8  10
5
50
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.
2
x
 1.8 10 5
0.15  x
x 2  0.15  x 1.8 10 5
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.
x2
 1.8 10 5
0.15  x
x 2  0.15  x 1.8 10 5
If
Ka

 10 3 then make this assumption .
x is small enough to ignore compared to [].
x  0.151.8 10
2
5
52
Ionization Constants for Weak
Monoprotic Acids and Bases

Complete the algebra and solve for the concentrations of the
species.
x 2  2.7 10 6

 
x  1.6 10 3 M  H 3O   CH 3COO
CH3COOH  0.15  1.6 10 M  0.15M

3
53
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.
 x  x 
0.15  X 
 1.8  10 5
5
x  1.8  10 x  2.7  10
2
a
x
b
b
6
0
c
b 2  4ac
2a
54
Ionization Constants for Weak
Monoprotic Acids and Bases
x

 1.8 10
5
  1.8 10 
5 2
21

 41  2.7 10 6

x  1.6 10 3 and - 1.6 10-3
55
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
% ionization
CH3COOHionized
=
100%
CH3COOHoriginal
3
1.6 10 M
% ionization 
100%  1.1%
0.15M
56
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!
57
Ionization Constants for Weak
Monoprotic Acids and Bases
Initial  
Change
 H O   CN HCN  H 2 O 
3
0.15 M
-xM
Equilibriu m 0.15 - x  M

H  CN 



Ka
HCN 
x 2  6.0 10 11

x x 
0.15  x
  
+xM
+xM
xM
xM
 4.0 10 10

x  7.7 10 6 M  H   CN 
HCN   0.15  x  M  0.15 M
58
Ionization Constants for Weak
Monoprotic Acids and Bases

The percent ionization of 0.15 M HCN solution is
calculated as in the previous example.
% ionization

HCN ionized
=
100%
HCN original
6
7.7 10 M
% ionization 
100%  0.0051%
0.15M
59
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
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.
60
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.
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!
Example 18-13: Calculate the concentrations of the
various species in 0.15 M aqueous ammonia.
NH3  H 2 O 
 NH4  OHInitial []
Change
0.15M
- xM
 xM
 xM
Equilibriu m [] ( 0.15-x)M
 xM
 xM
62
Ionization Constants for Weak
Monoprotic Acids
and
Bases
NH  H O 
 NH  OH
3
Initial []
Change

4
2
0.15M
- xM
 xM
Equilibriu m [] ( 0.15-x)M
 xM
Kb
NH OH  


4
-
NH3 
x x 
0.15  x 
-
 xM
 xM
 1.8 10 5
The simplifyin g assumption is valid.
x  0.15 thus 0.15 - x  0.15
x x   x x   1.8 105
0.15  x  0.15
x 2  (0.15)1.8 10 5
x 2  2.7 10 6 and x  1.6 10-3 M
NH   OH   1.6 10 M
NH   0.15  1.6 10 M  0.15M

4
-
-3
-3
3
63
Ionization Constants for Weak
Monoprotic Acids and Bases

The percent ionization for weak bases is calculated
exactly as for weak acids.
% ionization
NH3 ionized

100%
NH3 original
3
1.6 10 M

100%
0.15M
 1.1%
64
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!
65
Ionization Constants for Weak
Monoprotic Acids and Bases
pH = 11.37
From pH + pOH = 14.00, we can derive the pOH.
pOH = 14.00 - pH  14.00 - 11.37  2.63
OH   10  10  2.3 10
NH   2.3 10 M
-
 pOH

4
 2.63
3
M
3
66
Ionization Constants for Weak
Monoprotic Acids and Bases

Use the ionization equation and some algebra to get
the equilibrium concentration.


NH3  H 2O  NH4  OH
Initial[]
xM
Change
- 2.3 10-3


+ 2.3 10-3 + 2.3 10-3
Equilibriu m[] x - 2.3 10-3 M + 2.3 10-3 + 2.3 10-3
67
Ionization Constants for Weak
Monoprotic Acids and Bases

Substitute these values into the ionization constant
expression.
Kb

NH OH 

 1.8  10
1.8 10

4

NH3 
5
5

2.3  10 2.3 10 

x  2.3 10 
3
3
3
68
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.
2.3 10 
3 2
 1.8 10
5
x
x  0.30M NH3
69
Polyprotic Acids

Many weak acids contain two or more acidic hydrogens.


The calculation of equilibria for polyprotic acids is done
in a stepwise fashion.


1
2
3
Examples include H3PO4 and H3AsO4.
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
70
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
71
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
72
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
73
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.
74
Polyprotic Acids

1
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.



H 3AsO 4  H  H 2 AsO 4
0.100  x M
xM
xM
75
Polyprotic Acids
2
Substitute the algebraic quantities into the
expression for Ka1.

H H AsO 

 2.5 10

K a1
2

4
H 3AsO 4 
x x   2.5 104
K a1 
0.10  x 
4
x 2  2.5 10  4 x  2.5 10 5  0
In this case, the simplifyin g assumption does not apply.
76
Polyprotic Acids

Use the quadratic equation to solve for x, and obtain
both values of x.
x
4
 2.5 10 
2.5 10   412.5 10 
4 2
5
21
x  5.110 3 M and x  4.9 10 3 M
H   H AsO   xM  4.9 10

2

4
H3AsO 4   0.100  x M  0.095M
3
M
77
Polyprotic Acids
4
Next, write the equation for the second step
ionization and represent the concentrations.
H 2 AsO -4
 H + + HAsO 2
4
[ ]from 1st step (4.9 10-3 M )
algebraica lly (4.9 10-3  y ) M
yM
yM
78
Polyprotic Acids
5
Substitute the algebraic expressions into the second
step ionization expression.

H O  HAsO 
=
 5.6 10
H AsO 

4.9 10  y  y 
K =
4.9 10  y 

K a2
2
4
3
2
8

4
-3
a2
-3
For this step we can apply the assumption .
79
Polyprotic Acids

H O HAsO 
=
 5.6  10
H AsO 

4.9  10  y  y 
K =
4.9 10  y 

K a2
2
4
3
8

4
2
-3
a2
-3
In this case the assumption can be applied.
y  4.9  10 -3
Thus, 4.9  10 -3  y  4.9  10 -3
4.9 10  y   5.6 10
K =
4.9 10 
y  5.6  10 M  H   HAsO 
Note that H   H 
-3
a2
8
-3
8

2
4
2nd


1st
2nd
80
Polyprotic Acids
6
Finally, repeat the entire procedure for the third
ionization step.
HAsO 24


5.6 10 M  4.9 10
algebraic representa tions of [ ] changes 5.6 10  zM 
[ ]' s from 1st and 2 nd ionization s
-8
-8
H
-3

 5.6 10-8 M
zM

AsO 34zM
81
Polyprotic Acids
7.
Substitute the algebraic representations into the
third ionization expression.

H O AsO 
=
 3.0 10
HAsO 

4.9 10  5.6 10  z  z 
=
5.6 10  z 

K a3
3
4
3
13
2
4
3
K a3
8
8
The assumption can be applied, z  5.6 10 .
-8
82
Polyprotic Acids
4.9 10 z   3.0 10
5.6 10 
z   3.4 10 M  H   AsO 
3
13
8
18

3
4
3rd

Use Kw to calculate the [OH-] in the 0.100 M H3AsO4
solution.
H OH   1.0 10
1.0 10
1.0 10
OH   H   4.9 10
OH   2.0 10 M


14
14

14


3
12
83
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
84
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.
85
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
-

86
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
87
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
88
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.

HCl  H 2O   H 3O  Cl
~100%

89
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.
90
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 +  Fonly slightly
F- + H O + 

 HF + H O
3
2
nearly completely
91
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
92
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 2 O
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.
93
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?
94
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
-
95
Salts of Strong Bases and Weak Acids

The equilibrium constant for this reaction, called the
hydrolysis constant, is written as:
 HClOOH 
-
Kb =
ClO 
-
96
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
 HClOOH 
H 

=

ClO  H 
H OH 
HClO


=

1
H ClO 

-
Kb

-

Kb

-
-
97
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
98
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
99
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
100
Salts of Strong Bases and Weak Acids

1.
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, Ka=7.2 x 10-4.
F  H 2 O 
 HF  OH 
Kb 
 HFOH  
F 


Kw
Ka for HF
10
.  1014
11
Kb 

14
.

10
7.2  104
101
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!
 HCN + OHCN   H 2 O 
Kb




HCN OH 


CN 

Kw
K a for HCN
1.0 10 14
5
Kb 

2
.
5

10
10
4.0 10
102
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.

NaClO (s)   
 Na  ClO
~100% in H 2 O
0.10M


0.10M 0.10M
103
Salts of Strong Bases and Weak Acids

Set up the equation for the hydrolysis and the
algebraic representations of the equilibrium
concentrations.

ClO + H 2O  HClO + OH
Initial: 0.10 M
0M
0M
Change: - xM
+ xM + xM
-
At equil: 0.10 - x  M
xM
xM
104
Salts of Strong Bases and Weak Acids

Substitute the algebraic expressions into the
hydrolysis constant expression.
Kb

HClO  OH

ClO 


  2.9 10
7
x  x 

7
Kb 
 2.9  10
.  x
010
105
Salts of Strong Bases and Weak Acids

Substitute the algebraic expressions into the
hydrolysis constant expression.
The simplifyin g assumption can be made in this case.
x  0.10 and 0.10 - x  0.10
The equation reduces to x 2  2.9 10 8

 
Which becomes x  1.7 10  4 M  ClO   OH 

From the [OH - ] we get the pOH  3.77
From pH  pOH  14.00 we get the pH  10.23.
106
Salts of Strong Bases and Weak Acids

The percent hydrolysis for the hypochlorite ion may
be represented as:

ClO 
=
ClO 
-
% hydrolysis
hydrolyzed
original
100%
1.7 10-4 M
% hydrolysis =
100%  0.17%
0.10M
107
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
108
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?
109
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

110
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
111
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
-
112
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
113
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
114
Salts of Weak Bases and Strong Acids

1.
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:
Initial[]:
Change:
NH 4+  H 2O 
 NH 3  H 
0.10 M
+ xM + xM
- xM
+ xM + xM
Equilibrium[]:0.10  x  M
xM
xM
115
Salts of Weak Bases and Strong Acids
2.
Substitute the algebraic expressions into the
hydrolysis constant.
Ka

NH3  H  

 5.6 10 10
Ka 
NH 

4
x x 
0.10  x 
= 5.6 10-10
The assumption is applicable .
x  0.10 thus 0.10 - x  0.10
116
Salts of Weak Bases and Strong Acids
3.
Complete the algebra and determine the
concentrations and pH.
x x 
 5.6  10 10
0.10  x
x 2  5.6  10 11
x = 7.5  10-6 M
NH3   H    7.5 10-6 M
pH  5.12
117
Salts of Weak Bases and Strong Acids
4.
The percent hydrolysis of the ammonium ion in
0.10 M NH4Br solution is:

% hydrolysis =
NH 4+


hydrolized

NH 4+
original
 100%
7.5  10-6 M
% hydrolysis =
 100%
010
. M
% hydrolysis = 0.0075%
118
Salts of Weak Bases and Weak Acids
Salts made from weak acids and weak bases can
form neutral, acidic or basic aqueous solutions.


1.

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.
119
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
120
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
121
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!
122
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
123
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
124
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
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.
125
7
Salts of Weak Bases and Weak Acids
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.
3.


Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4
126
Salts of Weak Bases and Weak Acids

Both the cation, (CH3)3NH+, and the anion, F-,
hydrolyze.

CH
NH
F



CH
NH

F
 3 3 
 3 3 


H 2O~100%

127
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
128
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


Kw
K
F

Because the K for (CH ) NH ions is one order of
-
a for HF

a
314
3
+
magnitude larger
10
. than
 10the Kb for F- ions,H11+ ions are
K binexcess making
.solution
 10 acidic.
produced
the14
4 
7.2  10
129
Salts of Weak Bases and Weak Acids

1
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-.

A + H 2O  HA + OH
Kw
Kb 
Ka  HA 
130
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-.
+

BH + H 2O  B + H 3O
Kw
Ka 
B = weak base
K b  B
+
131
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.
132
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 sp 3 hybridized.
1s
2s 2 2p
2+
Be

Be(OH
)
2
4
aq


2


Be

1s 2s
 Be(OH 2 )4 
2


2p
form sp
3
hybrids
    
Be
4
xx 
xxxx
xx



e- pairs on coordinated water molecules
133
Hydrolysis of Small Highly-Charged Cations

In condensed form it is represented as:
 Be(OH 2 )4 
2


 H 2O   BeOH OH 2 3   H 3O
or, even more simply as:
Be
2+



 H 2O  Be(OH)  H
134
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+
135
Hydrolysis of Small Highly-Charged Cations

1.
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:



Be  H 2 O  Be(OH)  H
2
0.10  x M
xM
xM
136
Hydrolysis of Small Highly-Charged Cations
2.
Algebraic substitution of the expressions into the
hydrolysis constant:
x x 
0.10  x 
 1.0  10 5
The simplifyin g assumption applies.
x  0.10 and 0.10 - x  0.10
x x   1.0 105
0.10
x 2  1.0 10 6
x  1.0  10 3 M
H   Be 

2
3

1
.
0

10
M
hydrolyzed
pH  3.00
137
Hydrolysis of Small Highly-Charged Cations
3
Calculate the percent hydrolysis of Be2+.
Be 

% hydrolyzed =
 Be 
2+
hydrolyzed
2+
 100%
original
1.0  10
% hydrolyzed =
 100%  10%
.
0.10
-3
138
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.
139
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?
140
Synthesis Question
 
pH  6.5  H   10 6.5  3.2  10-7 M
CO 2  H 2 O  H 2 CO3


H CO  H  HCO
2
3
3

H HCO 

 4.2 10

Ka

3
H 2CO3 

3.2 10 3.2 10 

 4.2 10
7
Ka
7
H 2CO3 
7
7
H 2CO3   2.4 107 M  2.4 107 mol L
141
Synthesis Question
volume of water droplet  4 3  r 3  4 3  0.3 cm 
3
 0.11 cm 3 1L1000cm3   1.1  104 L
number of CO2 molecules  1.1  104 L2.4  107 mol L 
23

6.022

10
molecules
11
 2.6  10 mol 
mol

 1.6  1013 CO2 molecules
142



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?
143
End of Chapter 18

Weak aqueous acid-base mixtures are called buffers.
They are the subject of Chapter 19.
144