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Special Topic I
125
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2.
a.
formic acid:
20 mL
*
0.10 M
=
2.0 mmol
sodium formate:
15 mL
*
0.50 M
=
7.5 mmol
p
pH log
HA
A
3.75 pH log
2.0
7.5
3.75 pH log 0.27
3.75 pH
a
K
= +
[ ]
= +
= +
= +
−
[ ]
( 0.57)
pH 4.32
−
=
b.
aniline:
10 mL
*
0.50 M
=
5.0 mmol
¡
3.5 mmol aniline
1
RNH
2
2
HCl:
15 mL
*
0.10 M
=
1.5 mmol
¡
1.5 mmol anilinium hydrochloride
(
)
RNH
+
3
p
pH log
HA
A
pH
pH
pH
a
K
= +
[ ]
= +
= +
= +
−
[ ]
.
log
.
.
.
log .
.
4 60
1 5
3 5
4 60
0 43
4 60
( . )
.
−
=
0 37
4 97
pH
c.
acetic acid:
15 mL
*
1.00 M
=
15 mmol
¡
10 mmol acetic acid
NaOH:
10 mL
*
0.50 M
=
5.0 mmol
¡
5.0 mmol sodium acetate
p
pH log
HA
A
4.76 pH log
10
5.0
4.76 pH log 2
4.76 pH 0.30
a
K
= +
[ ]
= +
= +
= +
−
[ ]
pH 4.46
=
3.
The ionized form is the basic form. Therefore, we need to use the equation that allows us to calculate the
fraction present in the basic form.
fraction of buffer in the basic form
=
+
=
×
×
+
×
=
×
+
−
−
−
−
K
K
a
a
6
6
6
6
H
5.89 10
(5.89 10
10.47 10
5.89 10
16.3
[ ]
) (
)
6 10
0.36
6
×
=
−