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Chapter 2 105
Copyright © 2017 Pearson Education, Inc.
70.
a.
CH
2
CHCOOH
because an
sp
2
carbon is more electronegative than an
sp
3
carbon
b.
H H
N
O
+
because an oxygen can withdraw electrons inductively
c.
HC CCOOH
because an
sp
carbon is more electronegative than an
sp
2
carbon
d.
N
+
H
because an
sp
2
nitrogen is more electronegative than an
sp
3
nitrogen
71.
a.
between 9 and 10
b.
between 0 and 1
c.
between 2 and 3
a.
9.5
b.
0.08
c.
2.8
72.
a.
The first
p
K
a
is lower than the
p
K
a
of acetic acid because the middle COOH group of citric acid has
additional oxygen-containing groups that acetic acid does not have that withdraw electrons inductively
and thereby stabilize the conjugate base.
b.
The third
p
K
a
is greater than the
p
K
a
of acetic acid because loss of the third proton puts a third negative
charge on the molecule. Increasing the number of charges on a species destablizes it.
73.
K
a
=
3
H
+
4 3
HO
-
4
3
H
2
O
4
Because
3
H
+
4
=
3
HO
-
4
,
both must be
1
*
10
-
7
M
K
a
=
1
1
*
10
-
7
21
1
*
10
-
7
2
55.5
K
a
=
1.80
*
10
-
16
p
K
a
=
-
log 1.80
*
10
-
16
p
K
a
=
15.7
The answer can also be obtained in the following way:
K
a
=
3
H
+
4 3
HO
-
4
3
H
2
O
4
K
a
3
H
2
O
4
=
3
H
+
4 3
HO
-
4
take the log of both sides
log
K
a
+
log
3
H
2
O
4
=
log
3
H
+
4
+
log
3
HO
-
4
multiply both sides by
-
1
-
log
K
a
-
log
3
H
2
O
4
=
-
log
3
H
+
4
-
log
3
HO
-
4
p
K
a
-
log
3
H
2
O
4
=
pH
+
pOH
p
K
a
-
log
3
H
2
O
4
=
14
p
K
a
=
14
+
log
3
H
2
O
4
p
K
a
=
14
+
log 55.5
p
K
a
=
14
+
1.7
p
K
a
=
15.7