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108 Chapter 2
Copyright © 2017 Pearson Education, Inc.
77.
In each problem, we define
3
H
+
4
=
x
.
Then
3
A
-
4
is also
x
. In part
a
, because we have a 1.0 M solution,
3
HA
4
=
1.0
-
x
;
in part
b
, because we have a 0.1 M solution,
3
HA
4
=
0.1
-
x
.
a.
K
a
=
3
H
+
4 3
A
-
4
3
HA
4
1.74
*
10
-
5
=
x
2
1.0
-
x
1.74
*
10
-
5
=
x
2
x
=
4.16
*
10
-
3
pH
=
2.38
b.
K
a
=
3
H
+
4 3
A
-
4
3
HA
4
2.00
*
10
-
11
=
x
2
0.1
-
x
2.00
*
10
-
12
=
x
2
x
=
1.41
*
10
-
6
pH
=
5.85
c.
This question can be answered by plugging the given concentrations into the Henderson–Hasselbalch
equation.
p
K
a
=
pH
+
log
3
acid
4 3
base
4
3.76
=
pH
+
log
0.3
0.1
3.76
=
pH
+
log 3
3.76
=
pH
+
0.48
pH
=
3.76
-
0.48
=
3.28