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118 Special Topic I
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fraction present in the acidic form
HA
HA A
HA
HA
HA
a
=
[ ]
[ ]
+
=
[ ]
[ ]
+
−
[ ]
K
[ ]
=
+
=
+
+
+
+
+
[ ]
[ ]
[ ]
[ ]
H
1
1
H
H
H
a
a
K
K
To derive the equation for the fraction present in the basic form, we need to define [HA] in terms of
3
A
-
4
,
so we
can get rid of the [HA] term.
K
K
a
a
H A
HA
HA
H A
=
[ ]
[ ]
=
+ −
+ −
[ ] [ ]
[ ] [ ]
fraction present in the basic form
A
HA A
A
A
H
=
[ ]
+
=
+
−
−
−
−
+
[ ]
[ ]
[ ]
[ ]
[ ] [A
1
1
H
H
a
a
a
a
−
+
+
=
+
=
+
]
[ ]
[ ]
K
K
K
K
Preparing Buffer Solutions
The type of calculations just shown can be used to determine how to make a buffer solution.
For example, how can 100 mL of a 1.00 M buffer solution with a
pH
=
4.24
be prepared if you have 1.50 M
solutions of acetic acid
1
p
K
a
=
4.76;
K
a
=
1.74
*
10
-
5
2
,
sodium acetate, HCl, and NaOH?
First, we need to determine what fraction of the buffer is present in each form at
pH
=
4.24;
3
H
+
4
=
5.75
*
10
-
5
.
We start by calculating the fraction of the buffer present in the acidic form.
[ ]
[ ]
) (
)
H
H
5.75 10
(1.74 10
5.75 10
5.75 10
7.49
a
5
5
5
5
+
+
−
−
−
−
+
=
×
×
+
×
= ×
K
×
=
−
10
0.77
5
If a 1.00 M buffer solution is desired, the buffer must be 0.77 M in acetic acid and 0.23 M in sodium acetate.