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126 Special Topic I
Copyright © 2017 Pearson Education, Inc.
4.
fraction of buffer in the basic form
=
+
=
×
×
+
−
K
K
a
a
4
H
1.78 10
(1 10
10
[ ]
78
×
+
−
−
4
4
.
) ( .7 59
= ×
×
=
−
−
1 10
9.37 10
0.19
4
4
.78
)
3
sodium formate
4
=
0.19 M
3
formic acid
4
=
0.81 M
5.
From the Henderson–Hasselbalch equation, we see that the pH of the solution will be the same as the
p
K
a
of
the compound when the concentration of the compound in the acidic form is the same as the concentration
of the compound in the basic form.
p
pH log
HA
A
when HA A ],
p
pH
a
a
K
K
= +
[ ]
[ ]
=
=
−
−
[ ]
[
Therefore, in order to have a solution in which the pH will be the same as the
p
K
a
,
the number of mmol of
acid must equal the number of mmol of conjugate base.
Preparing a solution of
x
mmol of RCOOH and
1
>
2
x
mmol NaOH
gives a solution in which
3
RCOOH
4
=
3
RCOO
-
4
.
For example:
20 mL of 1.00 M RCOOH
=
20 mmol
10 mL of 1.00 M NaOH
=
10 mmol
This gives a solution that contains 10 mmol RCOOH and
10 mmol RCOO
-
.
The pH of this solution is the
p
K
a
of RCOOH.
6.
a.
mmol
mL
M
mmol
mL
M
30 mmol of aceticaci
0 mm
100
0 30
100
0 20
ol of sodiumacetate
30 mmol
mL
M
20 mmol
mL
M
30 mL
y
=
1 0
00
of 1.00 M acetic aci
x
x
x
x
2
d
=
=
=
=
.
.
y
y
=
=
2
0
.
.
d
10 mL of 2.00 M acetic acid
y
=
The buffer solution can be prepared by mixing:
30 mL of 1.00 M acetic acid
10 mL of 2.00 M sodium acetate
60 mL of water
100 mL
b.
Because the concentration of buffer in the acidic form (0.30 M) is greater than the concentration of
buffer in the basic form (0.20 M), the pH of the solution will be less than 4.76.