CHAPTER 2
Acids and Bases
1.
2.
a. 1. +NH4
2. HCl
3. H20
4. H30+
b. 1. -NH2
2. Br-
3. N03-
4. HO-
if the lone pairs are not shown:
a. CH30H
as an acid
CH30H + NH3
as a base
CH30H + HCl
CH30
-
+
+ NH4
+
CH30H
CH30H + CI-
I
H
b. NH3 as an acid
NH3 as a base
-NH2
NH3 + HONH3 + HBr
~
NH4
+
H20
+ Br-
if the lone pairs are shown:
a. CH30H as an acid
CH30H
as a base
CH3Q:-
CH3S?H + NH3
..
+ NH4
+
..
CH30H
CH3QH +H~l:
I
+ :Cl:-
H
b. NH3 as an acid
NH3 as a base
3.
NH3 + HO:NH3
:NH2
+
+ HBr:
NH4
a. The lower the pKa, the stronger the acid,so
acid.
+ H20:
+ :Br:
the compound with apKa
=
5.2 is the stronger
b. The greater the dissociation constant, the stronger the acid, so the compound with an acid
dissociation constant = 3.4 X 10-3 is the stronger acid.
4.
pKa = 4.82; therefore, K« = l.51 X 10-5
The greater the acid dissociation constant, the stronger the acid, so butyric acid (K; = 1.51 x 10-5)
is a weaker acid than vitamin C (K, = 6.76 x 10-5).
or
The lower the pKa, the stronger the acid, so butyric acid (pKa = 4.82) is a weaker acid than
vitamin C (pKa = 4.17).
13
14
Chapter 2
b. HC03 - + HCl
c.
6.
C032-
+ 2 HCl
~
A neutral solution has pH
a. basic
7.
~
a.
=
7. Solutions with pH < 7 are acidic; solutions with pH > 7 are basic.
b. acidic
CH3COO-
c. basic
is the stronger base.
Because HCOOH is the stronger acid, it has the weaker conjugate base.
b.
-NH2
is the stronger base.
Because H20 is the stronger acid, it has the weaker conjugate base.
c.
8.
H20 is the stronger base.
+
Because CH30H2 is the stronger acid, it has the weaker conjugate base.
The conjugate acids have the following relative strengths:
o
II
+
CH30H2
> CH3COH
+
> CH3NH3
> CH30H
> CH3NH2
The bases, therefore, have the following relative strengths:
+
9.
CH3NH2
pKa = 40
+
CH30H
~
CH3NH3
+
CH~O
.)
pKa = 15.5
The stronger acid of the two reactants will be the acid (that is, it is the one that will donate a
proton); the weaker acid will accept the proton.
Chapter 2
10.
15
Notice that in each case, the equilibrium goes away from the strong acid and toward the weak
acid.
0
0
a.
II
....---1000.
+ H2O
CH3COH
II
CH3CO-
+ H30+
pKa = -1.7
pKa =4.8
+OH
0
II
+
CH3COH
....---
pKa =-1.7
+
CH30H
II
-1000.
H30+
+
CH3COH
H2O
pKa = -6.1
-~
HO-
+
CH3O-
H2O
pKa = 15.7
pKa = 15.5
The pKa values are so close that there will be essentially no difference
in the equilibrium arrows.
+
CH30H
H3O-
....---1000.
pKa = -1.7
+
CH3NH2
+
CH30H
H
....•.
....---
HO-
CH3NH
+
....-
----"-
H30+
pKa =-1.7
H2O
+
+
CH3NH3
+
H2O
pKa = 10.7
....-
----"-
H2O
H30+
+
cr
+
HO-
pKa = -1.7
pKa =-7
+
....---1000.
H2O
pKa = 15.7
11.
+
pKa = 15.7
CH3NH2
NH3
H2O
pKa = -2.5
pKa =40
b. HCl
+
+NH4
pKa = 9.4
a. HBr is the stronger acid because bromine is larger than chlorine.
+
b. CH3CH2CH20H2
nitrogen.
is the stronger
acid because
oxygen
is more electronegative
than
c. The compound on the right (an alcohol) is a stronger acid than the compound on the left (an
amine) because oxygen is more electronegative than nitrogen.
12.
a. Because HF is the weakest acid, F- is the strongest base.
b. Because HI is the strongest acid, I- is the weakest base.
16
Chapter 2
13.
a. oxygen
As you saw in Problem 11, the size of an atom is more important than its electronegativity in
determining stability. So even though oxygen is more electronegative than sulfur, H2S is a stronger
acid than H20, and CH3SH is a stronger acid than CH30H. Because the sulfur atom is larger, the
electrons associated with the negatively charged sulfur are spread out over a greater volume, thereby
causing it to be a more stable base. The more stable the base, the stronger is its conjugate acid.
14.
a.
HO-;
if the atoms are the same, the negatively charged one is a stronger base than the
neutral one.
b.
NH3;
H30+ is a stronger acid than +NH4 because oxygen is more electronegative
nitrogen; the stronger the acid, the weaker its conjugate base.
than
o
15.
16.
17.
18.
II
c.
CH30-;
CH3COH
d.
CH30-;
a.
CH3COO-
b.
CH3CH2NH3
is a stronger acid than CH30H.
CH3SH is a stronger acid than CH30H because sulfur is larger than oxygen.
+
c.
H2O
e. +NH4
g.
N02-
d.
Br-
f.
h.
N03-
a. 1. neutral
2. neutral
3. charged
4. charged
5. charged
6. charged
b.
1.
2.
3.
4.
5.
6.
HC
N
charged
charged
charged
charged
neutral
neutral
c. 1. neutral
2. neutral
3. neutral
4. neutral
5. neutral
6. neutral
a.
Because the pH of the solution is greater than the pKa value of the carboxylic acid group, the group
will be in its basic form (without its proton).
Because the pH of the solution is less than the pKa value of the ammonium group, the group will
be in its acidic form (with its proton).
b.
No, because that would require a weaker acid (the +NH3 group) to lose a proton more
readily than a stronger acid (the COOH group).
a.
The basic form of the buffer (CH3COO-)
removes added H+ .
b. The acidic form of the buffer (CH3COOH) removes added HO-.
CH3COOH
+
HO-
~
CH3COO-
+
H20
Due to the rapid equilibrium, the added H+ or HO- readily reacts with the species (CH3COOCH3COOH) in the solution and thereby the effect on the solution's pH is minimized.
or
Chapter 2
19.
..
a. ZnCI2 + CH3QH
-----......
.....----
17
-
ZnCI2
1+
~
CH3QH
b. FeBr3 + ::8r:~.
~
-
FeBr3
I
:Br:
+ :CI:
c. AICI3
~
..
~
.....----
AICl3
I
-
:C]:
20.
a, b, c, and hare Brensted acids (protonating-donating
by donating a proton to it.
acids). Therefore,
d, e, f, and g are Lewis acids. They react with HO- by accepting
they react with
a pair of electrons
HO-
from it.
I
+ H2O
a. CH30-
21.
e. CH30H
b. NH3 + H2O
f.
HO-FeBr3
c. CH3NH2 + H2O
g.
HO-AIC13
d. HO-BF3
h.
CH3COO-
+ H2O
Ifthe pH of the solution is less than the pKa of the compound, the compound will be in its acidic
form (with its proton). If the pH of the solution is greater than the pKa of the compound, the
compound will be in its basic form (without its proton).
o
II
a. at pH = 3
at pH
=
CH3COH
+
b. at pH
°II _
6
CH3CO
at pH
°II
at pH = 10 CH3CO-
°II
at pH = 14 CH3CO0
22.
II
3
CH3CH2NH3
=6
CH3CH2NH3
=
+
CH3CH2NH3
at pH = 10
CF3CH20H
atpH=
14
CH3CH2NH2
atpH=
14
CF3CH20-
°II
b. CH3CH20H + -NH2
----..
-..-
CH3CH20-
c. CH3COH + CH3NH2
d.
CH3CH20H
+ CH30H
+ NH3
0
II
+ HCl
CF3CH20H
10
CH3CO-
0
=6
at pH
at pH=
+ CH3O-
CH3COH
CF3CH20H
+
----..
-..-
a.
c. at pH = 3
II
+
----..
-..-
CH3CO-
----..'
-..-
CH3CH20H2
+ CH3NH3
+
+ cr
18
Chapter 2
23.
a. Nitric acid is the strongest acid, because it has the largest Ka value.
b. Bicarbonate is the weakest acid, because it has the smallest K; value.
c. Bicarbonate has the strongest conjugate base (CO}-), because bicarbonate is the weakest acid.
Remember that the stronger base has the weaker conjugate acid.
24.
a. HO-, because H2S is the stronger acid, since S is larger than O.
-
b. CH3NH, because CH30H is the stronger acid, since 0 is more electronegative than N.
c. CH30-, because the oxygen is negatively charged.
d. CI-, because HBr is the stronger acid, since Br is larger than Cl.
25.
The nitrogen in the top left-hand corner is the most basic because it has the greatest electron
density (it is the most red).
26.
27.
As long as the pH is greater than the pKa value of the compound, the majority of the compound
will be in its basic form. Therefore, as long as the pH is greater than pH 10.4, more than 50% of
the amine will be in its basic (neutral, nonprotonated) form.
b. The electronegative chlorine substituent makes the carboxylic acid more acidic.
c. The closer the chlorine is to the acidic proton, the more it increases the acidity of the
carboxylic acid.
d. The electronegative (electron-withdrawing)
substituent makes the carboxylic acid more
acidic, because it stabilizes its conjugate base by withdrawing electrons from the oxygen
atom thereby decreasing its electron density.
29.
Substituting an H of the CH3 group with a more electronegative atom increases the acidity of the
carboxylic acid. As the electronegativity (electron-withdrawing
ability) of the subsitutent
increases, the acidity of the carboxylic acid increases.
5.75 x 10-13
>
4.90 x 10 -I
3
>
1.29
X
10-13
b. The greater the number of electron-withdrawing chlorine atoms, the greater will be the stability
of the base, so the stronger will be its conjugate acid.
Chapter 2
31.
The equilibrium favors reaction of the stronger acid and formation of the weaker acid. Therefore,
the first reaction favors formation of ethyne, and the second reaction favors formation of ammonia.
a. HC
CH
+ HO-
~
->0..
HC-C
b. HC . CH + -NH2
pKa = 15.7
---...
~
HC
C
+
NH3
pKa = 36
pKa = 25
c.
H2O
+
pKa= 25
32.
19
~H2 would. be a better base to use to remove a proton from ethyne because it would favor
formation of the desired product.
The reaction with the most favorable equilibrium
reactant acid and the weakest product acid.
constant is the one that has the strongest
a. CH30H is a stronger reactant acid (pKa = 15.5) than CH3CH20H (pKa = 15.9), and both
reactions form the same product acid (NH4). Therefore, the reaction of CH30H with NH3 has
the more favorable equilibrium constant.
.
b. Both reactions have the same reactant acid (CH3CH20H). The product acids are different:
+NH4 is a stronger product acid (pKa = 9.4) than CH3NH/ (pKa = 10.7). Therefore, the
reaction of CH3CH20H with CH3NH2 has the more favorable equilibrium constant.
33.
At a pH that is equal to the pKa value of an acidic compound, half the compound will be in the
acidic form (the form that has the proton) and half the compound will be in the basic form (the
form without the proton).
At pH values less than the pKa value, more of the compound will be in the acidic form than in the
basic form.
At pH values greater than the pKa value, more of the compound will be in the basic form than in
the acidic form.
The pKa value of ,carbonic acid is 6.1. Physiological pH (7.3) is greater than the pKa value. More
molecules of carbonic acid, therefore, will be present in the basic form (HCO; ) than in the acidic
form (H2C03). That means that the buffer system will be better at neutralizing excess acid.
34.
From the following equilibria you can see that a carboxylic acid is neutral when it is in its acidic
form (with its proton) and charged when it is in its basic form (without its proton). An amine is
charged when it is in its acidic form and neutral when it is in its basic form.
RCOOH
RCOO- +
H+
RNH2
H+
+
RNH3
+
Charged species will dissolve in water and neutral species will dissolve in ether.
In separating compounds you want essentially
form or its basic form. To obtain a 100:1 ratio
units lower than the pKa of the compound;
form:acidic form, the pH must be two pH units
all (100: 1) of each compound in either its' acidic
of acidic form:basic form, the pH must be two pH
and in order to obtain a 100: 1 ratio of basic
greater than the pKa of the compound.
20
Chapter 2
a. If both compounds are to dissolve in water, they both must be charged. Therefore, the
carboxylic acid must be in its basic form, and the amine must be in its acidic form. To
accomplish this, the pH will have to be at least two pH units greater than the pKa of the
carboxylic acid and at least two pH units less than the pKa of the ammonium ion. In other
words, it must be between pH 6.8 and pH 8.7.
b. 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 will have to be neutral
(in its basic form), so the pH will have to 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.
c. To dissolve in ether, the carboxylic acid will have to be neutral, so the pH will have to be less than
2.8 to have essentially all the carboxylic acid in the acidic (neutral) fOnTI.To dissolve in water, the
amine will have to be charged, so the pH will have to 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.
35.
Charged compounds will dissolve in water and uncharged compounds will dissolve in ether. The
acidic forms of carboxylic acids and alcohols are neutral and the basic forms are charged. The acidic
forms of amines are charged and the basic forms are neutral. Notice that one of the compounds does
not have a pKa value because it is not an acid (that is, it does not have a proton it can lose.)
6 (5 6 6 (5
COOH
OH
pKa=4.l7
pKa =4.60
Cl
pKa = 9.95
pK
a=
10.66
I ether
water at pH = 2.0
I water
I ether
layer
(5 (5
COOH
OH
CI
6 6 6
add ether
adjust pH of H20 to between 7 and 8
water layer
layer
ether layer
add H20 at pH
between 7 and 8
ether layer
water layer
coo-
OH
Cl
6 66
add ether, adjust pH of H20 to 12.7
water layer
0-
ether layer
Cl
6 6