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128 Special Topic I
Copyright © 2017 Pearson Education, Inc.
10.
a.
x
x
y
y
mmol
50 mL
M
mmol of acetic acid
42.5 mmol
mL
M
28
=
=
=
=
0 85
42 5
1 50
.
.
.
.3 mL of 1.50 M acetic acid
mmol
mL
M
mmol of sodium ac
x
x
50
0 15
7 5
=
=
.
.
etate
7.5 mmol
mL
M
mL of
M sodium acetate
y
y
=
=
1 50
5 0
1 50
.
.
.
28.3 mL of 1.50 M acetic acid
5.0 mL of 1.50 M sodium acetate
16.7 mL
50.0 mL
of H
2
O
You will need to add 17 mL to the original solution, and the concentration of the buffer will be
1.28 M
1
150 mmol
>
117 mL
=
1.28 M
2
.
8.
acid:
100 mL
*
1.00 M
=
100 mmol HA
¡
90 mmol HA
NaOH:
10 mL
*
1.00 M
=
10 mmol HO
-
¡
10 mmol A
-
Therefore, 90% is in the acidic form.
For 40% to be in the acidic form, you need:
40 mmol HA
60 mmol A
-
You need to have 60 mmol rather than 10 mmol in the basic form. To get the additional 50 mmol in the basic
form, you need to add 50 mL of 1.0 M NaOH.
9.
fraction of buffer
in the basic form
A 0.15 M
HA 0.85 M
[ ]
−
[ ]
=
=
=
+
=
×
×
+
−
K
K
a
a
5
H
1.74 10
(1 10
10
[ ]
74
×
+
−
−
5
4
.
) ( .1 00
=
=
0.15
−
11.74
×
10
5
×
−
1 10
5
.74
)
=
×
×
−
5
1.74 10
(1 10
10
74
×
+
−
−
5
5
.
) (10.00
)
a.
3
acetic acid
4
=
0.85
M
3
sodium acetate
4
=
0.15
M
b.
3
acetic acid
4
=
1.00
M
3
NaOH
4
=
0.15
M
c.
3
sodium acetate
4
=
1.00
M
3
HCl
4
=
0.85
M