124 / 912
116 Special Topic I
Copyright © 2017 Pearson Education, Inc.
2. Weak Acid and Strong Base
A buffer solution can be prepared by mixing a solution of a weak acid with a solution of a strong base such as
NaOH. The NaOH reacts completely with the weak acid, thereby creating the conjugate base needed for the buffer
solution.
The pH of a solution prepared by mixing 10 mL of a 2.0 M solution of a weak acid with a
p
K
a
of 5.86 with 5.0 mL
of a 1.0 M solution of sodium hydroxide can be calculated as follows:
When the 20 mmol of HA and the 5.0 mmol of
HO
-
are mixed, the 5.0 mmol of strong base react with 5.0 mmol of
HA, with the result that 5.0 mmol of
A
-
is formed and 15 mmol
1
20 mmol
-
5.0 mmol
2
of HA is left unreacted.
10
2 0
20
5 0
1 0
5 0
mL M mmol HA
mL M mmol HO
×
=
×
=
.
.
.
−
.
15
5 0
mmol HA
mmol A
−
.
p
pH log
HA
A
5.86 pH log
15
5
5.86 pH log 3
5.86 pH 0.48
pH
a
K
= +
[ ]
= +
= +
= +
−
[ ]
=
5.38
3. Weak Base and Strong Acid
A buffer solution can be prepared by mixing a solution of a weak base with a strong acid such as HCl. The strong
acid reacts completely with the weak base, thereby forming the conjugate acid needed for the buffer solution.
The pH of an ethylammonium ion/ethylamine buffer
1
p
K
a
of CH
3
CH
2
N
+
H
3
=
11.0)
prepared by mixing 30 mL of
0.20 M ethylamine with 40 mL of 0.10 M HCl can be calculated as follows:
30
0 20
6 0
40
0 10
4
2
mL
M mmol RNH
mL
M mmol H
×
=
×
=
+
.
.
.
.
2 0
0
4 0
mmol RNH
mm
2
.
RN
+
H
3
Notice that 4.0 mmol
H
+
reacts with 4.0 mmol
RNH
2
,
forming 4.0 mmol
RN
+
H
3
, and 2.0 mmol of
RNH
2
is left
unreacted.
p
pH log
HA
A
pH log
pH log
pH
a
K
= +
= +
= +
= +
−
[ ]
[ ]
.
.
.
.
.
.
11 0
4 0
2 0
11 0
2 0
11 0
0.
.
30
10 7
pH
=