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Special Topic I
127
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7.
Original solution
fraction of buffer in the basic form
A 0.87 M
HA 0.63 M
[ ]
−
[ ]
=
=
=
+
=
×
×
+
−
K
K
a
a
5
H
1.74 10
(1 10
10
[ ]
74
×
+
−
−
5
5
.
) ( .1 26
= ×
×
=
−
−
1 10
3.00 10
0.58
5
5
=
×
0.58 1.50 M 0.87 M
.74
)
Desired solution
fraction of buffer in the basic form
A 0.53 M
HA 0.97 M
[ ]
−
[ ]
=
=
=
+
=
×
×
+
−
K
K
a
a
5
H
1.74 10
(1 10
10
[ ]
74
×
+
−
−
5
5
.
) ( .3 16
= ×
×
=
−
−
1 10
4.90 10
0.35
5
5
=
×
0.35 1.50 M 0.53 M
.74
)
The original solution contains
87 mmol of A
-
1
100 mL
*
0.87 M
2
.
The desired solution with a
pH
=
4.50
must contain 53 mmol of
A
-
.
Therefore, 34 mmol of
A
-
1
87
-
53
=
34
2
must be converted to HA.
This can be done by adding 34 mmol of HCl to the original solution.
If you have a 1.00 M HCl solution, you will need to add 34 mL of it to the original solution in order to change
its pH from 4.90 to 4.50.
34
1 00
mmol
mL
M
34 mL
x
x
=
=
.
Note that after adding HCl to the original solution, it will no longer be a 1.50 M buffer; it will be more dilute
1
150 mmol
>
134 mL
=
1.12 M
2
.
The change in the concentration of the buffer solution will be less if a more concentrated solution of HCl is
used to change the pH. For example, if you have a 2.00 M HCl solution:
34
2 00
mmol
mL
M
17 mL
x
x
=
=
.