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Chapter 2 101
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f.
As the pH becomes more basic than 2.34, the COOH group will become more negatively charged.
As the pH becomes more acidic than 9.69, the
+
NH
3
group will become more positively charged.
Therefore, the amount of negative charge will be the same as the amount of positive charge at the pH
that is equidistant from the two
p
K
a
values.
2.34
+
9.69
2
=
6.02
44.
Solved in the text.
45.
a.
6.4 (two log units more acidic than the
p
K
a
)
b.
7.3 (when the pH is equal to the
p
K
a
)
c.
5.6 (one log unit more basic than the
p
K
a
)
46.
a. 1.
pH
=
4.9
When the
pH
=
p
K
a
,
half the compound is in its acidic form (with its proton)
and half is in its basic form (without its proton).
2.
pH
=
10.7
b. 1.
pH
7
6.9
Because the basic form is the form in which the compound is charged,
the pH needs to be more than two units more basic than the
p
K
a
value.
2.
pH
6
8.7
Because the acidic form is the form in which the compound is charged,
the pH needs to be more than two units more acidic than the
p
K
a
value.
47.
Solved in the text.
48.
a.
For the carboxylic acid to dissolve in water, it must be charged (in its basic form), so the pH will have
to be greater than 6.8. For the amine to dissolve in ether, it must be neutral (in its basic form); the pH
must be greater than 12.7 to have essentially all of it in the neutral form. Therefore, the pH of the water
layer must be greater than 12.7.
b.
To dissolve in ether, the carboxylic acid has to be neutral; the pH must be less than 2.8 to have essen-
tially all the carboxylic acid in the acidic (neutral) form. To dissolve in water, the amine has to be
charged; the pH must be less than 8.7 to have essentially all the amine in the acidic form. Therefore,
the pH of the water layer must be less than 2.8.
49.
a.
The basic form of the buffer removes a proton from the solution.
+
H
+
CH
3
COO
−
CH
3
COO
Η
b.
The acidic form of the buffer donates a proton to remove hydroxide ion from the solution.
+
HO
−
CH
3
COO
Η
+
H
2
O
CH
3
COO
−
50.
Solved in the text.
51.
a.
CH
3
HO
ZnCl
2
−
+
ZnCl
2
+
CH
3
OH
b.
FeBr
3
+
Br
−
Br FeBr
3
−
c.
AlCl
3
+
Cl
−
Cl AlCl
3
−