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114 Special Topic I
Copyright © 2017 Pearson Education, Inc.
The pH of a 1.20 M solution of sodium acetate can be calculated as follows:
CH COO H O
CH COOH HO
HO CH COOH
CH COO
1.
3
2
3
w
a
3
3
−
−
−
−
+
+
=
[
]
K
K
[
]
[
]
00 10
1.74 10
( ) ( )
1.20
5.75 10
1.20
6.86 10
14
5
10
2
2
×
×
=
−
×
=
= ×
−
−
−
x x
x
x
x
−
−
−
−
= × =
= −
×
=
=
−
10
5
5
2.62 10
HO
pOH log 2.62 10
pOH 4.58
pH 14.00 4.5
x
[
]
8
pH 9.42
=
Notice that by setting up the equation equal to
K
w
>
K
a
,
we can avoid the introduction of a new term
1
K
b
2
,
because
K
w
>
K
a
=
K
b
.
Buffer Solutions
A buffer solution is a solution that maintains nearly constant pH in spite of the addition of small amounts of
H
+
or
HO
-
.
That is because a buffer solution contains both a weak acid and its conjugate base. The weak acid can give a
proton to any
HO
-
added to the solution, and the conjugate base can accept any proton that is added to the solution,
so the addition of small amounts of
HO
-
or
H
+
does not significantly change the pH of the solution.
A buffer can maintain nearly constant pH in a range of one pH unit on either side of the
p
K
a
of the conjugate acid.
For example, an acetic acid/sodium acetate mixture can be used as a buffer in the pH range 3.76–5.76 because
acetic acid has a
p
K
a
=
4.76;
methylammonium ion/methylamine can be used as a buffer in the pH range 9.7–11.7
because the methylammonium ion has a
p
K
a
=
10.7.
The pH of a buffer solution can be determined from the Henderson–Hasselbalch equation. This equation comes
directly from the expression defining the acid dissociation constant. Its derivation is found on pages 72–73 of
the text.
Henderson–Hasselbalch equation
p
pH log
HA
A
a
K
= +
[ ]
−
[ ]