115 / 912
Chapter 2 107
Copyright © 2017 Pearson Education, Inc.
76.
a.
fraction present in the acidic form
=
amount in the acidic form
amount in the acidic form
+
amount in the basic form
=
3
HA
4
3
HA
4
+
3
A
-
4
Because there are two unknowns, we must define one in terms of the other.
By using the definition of the acid dissociation constant, we can define
3
A
-
4
in terms of
3
HA
4
,
3
H
+
4
,
and
K
a
,
so we have only one unknown.
K
a
=
3
H
+
4 3
A
-
4
3
HA
4
3
A
-
4
=
K
a
3
HA
4 3
H
+
4
Substituting the definition of
3
A
-
4
into the equation for the fraction present in the acidic form gives:
3
HA
4
3
HA
4
+
3
A
-
4
=
3
HA
4
3
HA
4
+
K
a
3
HA
4 3
H
+
4
=
1
1
+
K
a
3
H
+
4
=
3
H
+
4
3
H
+
4
+
K
a
Therefore, the percentage that is present in the acidic form is given by:
3
H
+
4
3
H
+
4
+
K
a
*
100
Because the
p
K
a
of the acid is given as 5.3, we know that
K
a
is
5.0
*
10
-
6
1
because p
K
a
=
-
log
K
a
2
.
Because the pH of the solution is given as 5.7, we know that
3
H
+
4
is
2.0
*
10
-
6
1
because pH
=
-
log
3
H
+
42
.
Substituting into the equation for the percentage present in the acidic form gives:
3
H
+
4
3
H
+
4
+
K
a
=
2.0
*
10
-
6
2.0
*
10
-
6
+
5.0
*
10
-
6
*
100
2.0
*
10
-
6
7.0
*
10
-
6
*
100
=
29%
b.
Fraction present in the acidic form
=
3
H
+
4
3
H
+
4
+
K
a
=
0.80
3
H
+
4
=
0.80
13
H
+
4
+
K
a
2
3
H
+
4
=
0.80
3
H
+
4
+
0.80
K
a
0.20
3
H
+
4
=
0.80
K
a
3
H
+
4
=
4
K
a
3
H
+
4
=
4
*
5.0
*
10
-
6
3
H
+
4
=
20
*
10
-
6
pH
=
4.7