16.2

In general
the components of the conjugate acidbase pair are on
opposite sides of the reaction arrow.
The base always has one fewer proton than the acid.
a.

The conjugate acidbase
pairs are (1) HCN (acid) and CN^{}
(conjugate base) and (2) CH_{3}COO^{}
(base) and CH_{3}COOH (conjugate acid).

b.

(1) HCO (acid)
and CO (conjugate base) and (2) HCO (base)
and H_{2}CO_{3} (conjugate acid).

c.

(1) H_{2}PO (acid)
and HPO (conjugate base) and (2) NH_{3}
(base) and NH (conjugate acid).

d.

(1) HClO (acid) and ClO^{}
(conjugate base) and (2) CH_{3}NH_{2} (base) and CH_{3}NH (conjugate
acid).

e.

(1) H_{2}O
(acid) and OH^{}
(conjugate base) and (2) CO (base)
and HCO (conjugate acid).



16.3

Tables 16.1 and 16.2 of the text list important Brønsted
acids and bases and their respective conjugates.
a.

both (why?)

b.

base

c.

acid

d.

base

e.

acid

f.

base

g.

base

h.

base

i.

acid

j.

acid



16.4

The conjugate
acid of any base is just the base with a proton added.
a.

H_{2}S

b.

H_{2}CO_{3}

c.

HCO

d.

H_{3}PO_{4}

e.

H_{2}PO

f.

HPO

g.

H_{2}SO_{4}

h.

HSO

i.

HSO



16.5

Recall that
the conjugate base of a Brønsted acid is the species that remains when one
proton has been removed from the acid.
a.

nitrite
ion: NO

b.

hydrogen sulfate ion (also called bisulfate
ion): HSO

c.

hydrogen sulfide ion (also called bisulfide
ion): HS^{}

d.

cyanide ion:
CN^{}

e.

formate ion:
HCOO^{}



16.6

a.

The Lewis structures of C_{2}HO and C_{2}O are
(The lone pairs on the oxygen atoms are not shown.)

b.

H^{+} and C_{2}H_{2}O_{4} can act
only as acids, C_{2}HO can act as both an acid and a base, and
C_{2}O can act only as a base.



16.7

The conjugate base of any acid is simply the acid minus
one proton.
a.

CH_{2}ClCOO^{}

b.

IO

c.

H_{2}PO

d.

HPO

e.

PO

f.

HSO

g.

SO

h.

IO

i.

SO

j.

NH_{3}

k.

HS^{}

l.

S^{2}^{}

m.

OCl^{}



16.10

The strength
of the HX
bond is the dominant factor in determining the strengths of binary
acids. As with the hydrogen halides
(see Section 16.9 of the text), the HX bond strength
decreases going down the column in Group 6A. The compound with the weakest HX
bond will be the strongest binary acid: H_{2}Se > H_{2}S
> H_{2}O.


16.11

All the
listed pairs are oxoacids that contain different central atoms whose
elements are in the same group of the periodic table and have the same
oxidation number. In this situation
the acid with the most electronegative central atom will be the strongest.
a.

H_{2}SO_{4} >
H_{2}SeO_{4}


b.

H_{3}PO_{4} >
H_{3}AsO_{4}




16.12

The CHCl_{2}COOH is a
stronger acid than CH_{2}ClCOOH.
Having two electronegative chlorine atoms compared to one, will draw
electron density away from the O–H group, making the OH bond more
polar. The hydrogen atom in CHCl_{2}COOH
is more easily ionized compared to the hydrogen atom in CH_{2}ClCOOH.


16.13

The conjugate bases are C_{6}H_{5}O^{} from phenol and
CH_{3}O^{}
from methanol. The C_{6}H_{5}O^{} is stabilized
by resonance:
The CH_{3}O^{} ion has no such
resonance stabilization. A more
stable conjugate base means an increase in the strength of the acid.


16.18

Strategy:

The
equilibrium concentrations of H^{+} and OH^{–} must
satisfy Equation 16.1, (25°C). Substitute the given concentrations and
solve for [OH^{–}]:

Solution:




16.19

Strategy:

The
equilibrium concentrations of H^{+} and OH^{–} must satisfy Equation
16.1, (25°C). Substitute
the given concentrations and solve for [H^{+}]:

Setup:


Solution:




16.20

Strategy:

The
equilibrium concentrations of H^{+} and OH^{–} must satisfy Equation
16.1, but with the appropriate equilibrium constant for 50°C:(50°C). Substitute the given concentrations and
solve for [OH^{–}]:

Solution:




16.21

Strategy:

Use
Equation 16.1 to calculate the [H_{3}O^{+}]
concentrations.
K_{w} =
[H_{3}O^{+}][OH^{–}]

Setup:

K_{w} = 5.13 × 10^{–13} at 100°C. Rearranging Equation 16.1 to
solve for [H_{3}O^{+}]
gives:

Solution:




16.24

[OH^{}] =
0.62 M


16.25

[H^{+}] =
1.4 ´
10^{3}
M


16.26

a.

Ba(OH)_{2} is ionic and
fully ionized in water. The
concentration of the hydroxide ion is 5.6 ´
10^{4}
M (Why? What is the
concentration of Ba^{2+}?) We find the hydrogen ion
concentration.
The pH is then:
pH =
log[H^{+}]
= log(1.8
´
10^{11}) =
10.74

b.

Nitric acid is a strong acid, so the concentration
of hydrogen ion is also 5.2 ´ 10^{4}
M. The pH is:
pH
= log[H^{+}]
= log(5.2
´
10^{4}) =
3.28



16.27

a.

HCl is a strong acid, so the
concentration of hydrogen ion is also 0.0010 M. (What is the concentration of chloride
ion?) We use the definition of pH.
pH =
log[H^{+}]
= log(0.0010) =
3.00

b.

KOH is an ionic compound
and completely dissociates into ions.
We first find the concentration of hydrogen ion.
The pH is then found from its
defining equation
pH
= log[H^{+}]
= log[1.3
´
10^{14}] =
13.89



16.28

Since pH = log[H^{+}], we write
[H^{+}] = 10^{pH}
a.

[H^{+}]
= 10^{5.20} =
6.3 ´ 10^{6}
M

b.

pH =
log[H^{+}]
= 16.00
log[H^{+}]
= 16.00
[H^{+}]
= 10^{16.00} =
1.0 ´ 10^{16}
M

c.

We are given the
concentration of OH^{} ions and asked to
calculate [H^{+}]. The relationship between [H^{+}] and
[OH^{}]
in water or an aqueous solution is given by the ionproduct of water, K_{w}
(Equation 16.1 of the text).
K_{w} =
1.0 ´
10^{14} =
[H^{+}][OH^{}]
Rearranging the equation to
solve for [H^{+}], we write
Since the [OH^{}] < 1 ´
10^{7}
M we expect the [H^{+}] to be greater than 1 ´
10^{7}
M.



16.29

Strategy:

Here we are given
the pH of a solution and asked to calculate [H^{+}]. Because pH is defined as
pH = log[H^{+}],
we can solve for [H^{+}] by taking the antilog of the pH; that
is, [H^{+}] = 10^{pH}.

Solution:

a.

From Equation 16.2 of the text:
pH = log[H^{+}] = 2.42
log[H^{+}] =
2.42
To calculate [H^{+}],
we need to take the antilog of 2.42.
[H^{+}]
= 10^{2.42} = 3.8 ´
10^{3} M
Check: Because the pH is between 2 and 3, we
can expect [H^{+}] to be between 1 ´
10^{2} M and 1 ´ 10^{3} M. Therefore, the answer is reasonable.

b.

pH = log[H^{+}] = 11.24
log[H^{+}] =
11.21
To calculate [H^{+}],
we need to take the antilog of 11.21.
[H^{+}] = 10^{11.21} =
6.2 ´ 10^{12} M
Check: Because the pH is between 11 and 12,
we can expect [H^{+}] to be between 1 ´
10^{11} M and 1 ´ 10^{12} M. Therefore, the answer is reasonable.

c.

pH = log[H^{+}] = 6.96
log[H^{+}] =
6.96
To calculate [H^{+}],
we need to take the antilog of 6.96.
[H^{+}] = 10^{6.96} =
1.1 ´ 10^{7} M

d.

pH = log[H^{+}] = 15.00
log[H^{+}] =
15.00
To calculate [H^{+}],
we need to take the antilog of 15.00.
[H^{+}]
= 10^{15.00} =
1.0 ´ 10^{15} M




16.30


16.31

pH

[H^{+}]

Solution is:

< 7

> 1.0 ´ 10^{7}
M

acid

> 7

< 1.0 ´ 10^{7}
M

basic

= 7

= 1.0 ´ 10^{7}
M

neutral






16.32

KOH is a
strong base and therefore ionizes completely. The OH^{} concentration
equals the KOH concentration, because there is a 1:1 mole ratio between KOH
and OH^{}.
[OH^{}] =
0.360 M
pOH = log[OH^{}] = 0.444


16.33

The pH can be found
by using Equation 16.6 of the text.
pH =
14.00 
pOH = 14.00  9.40 =
4.60
The hydrogen
ion concentration can be found as follows.
4.60 = log[H^{+}]
Taking the
antilog of both sides:
[H^{+}] = 2.5 ´ 10^{5}
M


16.34

Molarity
of the HCl solution is:
pH
=
log(0.762) =
0.118


16.35

We can
calculate the OH^{} concentration from the
pOH.
pOH =
14.00 
pH =
14.00 
10.00 = 4.00
[OH^{}] =
10^{pOH} =
1.0 ´
10^{4}
M
Since NaOH is a strong base, it ionizes completely. The OH^{} concentration
equals the initial concentration of NaOH.
[NaOH] =
1.0 ´
10^{4}
mol/L
So, we need to prepare 546 mL of 1.0 ´
10^{4}
M NaOH.
This is a dimensional analysis problem. We need to perform the following unit
conversions.
mol/L ® mol NaOH
® grams NaOH
546 mL =
0.546 L


16.38


16.39


16.40

a.

[HBr] = [H_{3}O^{+}]
= 10^{}^{0.12} =
0.76 M.

b.

[HBr] = [H_{3}O^{+}]
= 10^{}^{2.46} =
3.5 ´
10^{3}
M.

c.

[HBr]
= [H_{3}O^{+}] = 10^{}^{6.27}
= 5.4 ´
10^{7}
M.



16.41

Strategy:

HNO_{3}
is a strong acid. Therefore, the
concentration of HNO_{3} is equal to the concentration of
hydronium ion. We use Equation
16.3 to determine [H_{3}O^{+}]:

Solution:

a.

[HNO_{3}] = [H_{3}O^{+}]
= 10^{}^{4.21}
= 6.2 ´ 10^{5}
M.

b.

[HNO_{3}] = [H_{3}O^{+}]
= 10^{}^{3.55}
= 2.8 ´ 10^{4}
M.

c.

[HNO_{3}]
= [H_{3}O^{+}] = 10^{}^{0.98}
= 0.10 M.




16.42

a.

[OH^{}]
= [KOH] = 0.066 M.
pOH
= log
[OH^{}]
= 
log (0.066) = 1.18
pH
= 14.00 
pOH = 14.00 
1.18 = 12.82

b.

[OH^{}]
= [NaOH] = 5.43 M.
pOH
= log
[OH^{}]
= 
log (5.43) = 0.735
pH
= 14.00 
pOH = 14.00 
(0.735)
= 14.74

c.

[OH^{}]
= 2[Ba(OH)_{2}] = 2(0.74 M) = 1.48 M.
pOH
= log
[OH^{}]
= 
log (1.48) = –0.170
pH
= 14.00 
pOH = 14.00 
(–0.170) = 14.17



16.43

Strategy:

We
use Equation 16.4 to determine pOH and Equation 16.6 to determine
pH. The hydroxide ion
concentration of a monobasic strong base, such as LiOH and NaOH, is equal
to the base concentration. In the
case of a dibasic base such as Ba(OH)_{2}, the hydroxide
concentration is twice the base concentration.

Solution:

a.

[OH^{}] = [LiOH]
= 1.24 M.
pOH = log [OH^{}] =  log (1.24) = 0.093
pH = 14.00  pOH = 14.00  (0.093)=
14.09

b.

[OH^{}] = 2[Ba(OH)_{2}]
= 2(0.22 M) = 0.44 M.
pOH = log [OH^{}] =  log (0.44) = 0.36
pH = 14.00  pOH = 14.00  0.36 = 13.64

c.

[OH^{}] = [NaOH]
= 0.085 M.
pOH = log [OH^{}] =  log (0.085) = 1.07
pH = 14.00  pOH = 14.00  1.07 = 12.93




16.44

We calculate the pOH using Equation 16.6 and the
hydroxide ion concentration using Equation 16.5:
a.

pOH
= 14.00 
pH = 14.00 
9.78 = 4.22
[LiOH]
= [OH^{}]
= 10^{pOH}
= 10^{4.22}
= 6.0 ´ 10^{5}
M

b.

[OH^{}]
= 10^{pOH}
= 10^{4.22}
= 6.0 ´
10^{5}
M
[Ba(OH)_{2}]
= [OH^{}]==3.0 ´
10^{5}
M



16.45

Strategy:

We
use Equation 16.6 to determine pOH and Equation 16.5 to determine
hydroxide ion concentration. The
concentration of a monobasic strong base such as LiOH is equal to the
hydroxide concentration. In the case
of a dibasic base such as Ba(OH)_{2}, the base concentration is half
the hydroxide concentration.

Solution:

a.

pOH = 14.00  pH = 14.00  11.04 = 2.96
[KOH] = [OH^{}] = 10^{pOH} = 10^{2.96} = 1.1 ´ 10^{3} M

b.

[OH^{}]
= 10^{pOH} = 10^{2.96} = 1.1 ´ 10^{3} M
[Ba(OH)_{2}] = [OH^{}]==5.5 ´
10^{4} M




16.46

Strategy:

Construct an equilibrium table, and express the
equilibrium concentration of each species in terms of x.
Solve for x using the
approximation shortcut, and evaluate whether or not the approximation is
valid. If the shortcut is not
appropriate, use the quadratic equation given in Appendix 1 to solve for x.
Use Equation 16.2 to determine pH.

Setup:


+





Initial (M):

1.8 ´ 10^{5}



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

1.8 ´ 10^{5} – x



x

x

According to Table 16.8, K_{a} for carbonic acid is 4.2 × 10^{–7}.

Solution:

These equilibrium concentrations are then substituted
into the equilibrium expression to give
At a first approximation, assume that x is small compared to 1.8 ´
10^{5}
so that
1.8 ´ 10^{5}
– x ≈ 1.8 ´
10^{5}.
Note that this approximate
value of x is 15% of 1.8 ´
10^{5}, so it is probably not appropriate to
have assumed that 1.8 ´ 10^{5}
– x ≈ 1.8 ´
10^{5}.
Instead, attempt to solve for x without making any
approximations. Start by clearing
the denominator of equation (1) and then expanding the result to get:
This equation is a quadratic
equation of the form ax^{2} + bx + c =
0, the solutions of which are given by (see Appendix 1):
.
For this case, a = 1, b =
4.2 × 10^{–7}, and
c = –7.56×10^{–12}.
Substituting these values into the quadratic formula gives:
The negative solution is not physically
reasonable since there is no such thing as a negative concentration. The
solution is x = 2.5 ×10^{–6} M.
According to the equilibrium table, x = [H^{+}]. Therefore, [H^{+}] = 2.5
×10^{–6} M and
pH = –log(2.5 ×10^{–6}) = 5.6



16.47

Strategy:

HNO_{3} is a strong acid. Given pH, use Equation 16.3 to solve
for [H_{3}O^{+}].
[H_{3}O^{+}]
= 10^{–pH}
Use reaction stoichiometry to determine HNO_{3}
concentration.

Solution:

[H_{3}O^{+}]
= 10^{–4.65} = 2.24 × 10^{–5} M
The hydronium ion concentration is equal that of HNO_{3}
according to the balanced equation:
HNO_{3}(aq)
+ H_{2}O(aq) →
H_{3}O^{+}(aq)
+ NO(aq)
Therefore, [H_{3}O^{+}] = [HNO_{3}]
= 2.24 × 10^{–5} M



16.54

a.

strong acid

b.

weak acid

c.

strong acid (first stage of
ionization)

d.

weak acid

e.

weak acid

f.

weak acid

g.

strong acid

h.

weak acid

i.

weak acid



16.55


16.56

The maximum possible concentration of hydrogen
ion in a 0.10 M solution of HA is 0.10 M. This is the case if HA is a strong
acid. If HA is a weak acid, the
hydrogen ion concentration is less than 0.10 M. The pH corresponding to 0.10 M [H^{+}]
is 1.00. (Why three digits?) For a smaller [H^{+}] the pH is
larger than 1.00 (why?).
a.

false, the
pH is greater than 1.00

b.

false, they
are equal

c.

true

d.

false



16.57

Strategy:

Recall that a weak acid only partially ionizes in
water. We are given the initial
quantity of a weak acid (CH_{3}COOH) and asked to calculate the
pH. For this we will need to
calculate [H^{+}], so we follow the procedure outlined in Section
16.5 of the text.

Solution:

We set up a table for the dissociation.

C_{6}H_{5}COOH(aq)


H^{+}(aq)

+ C_{6}H_{5}COO^{}(aq)

Initial
(M):

0.10


0.00

0.00

Change
(M):

x


+x

+x

Equilibrium
(M):

(0.10  x)


x

x

Assuming that x is
small compared to 0.10, we neglect it in the denominator:
x =
2.55 ´
10^{3}
M = [H^{+}]
pH =
log(2.55
´
10^{3}) =
2.59
This problem could also have been solved using the
quadratic equation. The difference
in pH obtained using the approximation x »
0 is very small. (Using the
quadratic equation would have given pH = 2.60, a difference of less than
0.4%.)



16.58

We set up a table for the dissociation.

HF(aq)


H^{+}(aq)

+ F^{}(aq)

Initial
(M):

0.15


0.00

0.00

Change
(M):

x


+x

+x

Equilibrium
(M):

(0.15

x)


x

x

7.1
´
10^{4}
=
Assuming that x is
small compared to 0.15, we neglect it in the denominator:
7.1 ´ 10^{}^{4} =
x =
0.0103 M = [H^{+}]
pH = log(0.0103) = 1.99


16.59

We set up a table for the
dissociation.

HCN(aq)


H^{+}(aq)

+ CN^{}(aq)

Initial
(M):

0.095


0.00

0.00

Change
(M):

x


+x

+x

Equilibrium
(M):

(0.095  x)


x

x

4.9 ´ 10^{}^{10} =
Assuming that x is
small compared to 0.095, we neglect it in the denominator:
4.9 ´ 10^{}^{10} =
x = 6.82 ´ 10^{}^{6} M = [H^{+}]
pH = log(6.82 ´ 10^{}^{6}) = 5.17


16.60

We set up a table for the
dissociation.

C_{6}H_{5}OH(aq)


H^{+}(aq)

+
C_{6}H_{5}O^{}(aq)

Initial (M):

0.34


0.00

0.00

Change (M):

x


+x

+x

Equilibrium (M):

(0.34  x)


x

x

1.3 ´ 10^{}^{10} =
Assuming that x is
small compared to 0.34, we neglect it in the denominator:
1.3 ´ 10^{}^{10} =
x = 6.65 ´ 10^{}^{6} M = [H^{+}]
pH = log(6.65 ´ 10^{}^{6}) = 5.18


16.61

Strategy:

Formic acid is a weak
acid, and some of it will ionize in solution according to the equilibrium
HF(aq) + H_{2}O(l) F^{–}(aq) + H_{3}O^{+}(aq)
The
percent ionization of a weak acid is given by Equation 16.7:
,
where [HA]_{0} is the initial concentration of the acid
(that is, the concentration assuming there is no ionization).
Using the K_{a} expression above, set up a concentration
table and solve for [H^{+}]. Then, use this value to compute the
percent ionization.

Solution:

a.


HF(aq) +

H_{2}O(l)


F^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

0.016



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

0.016 – x



x

x

As a first approximation, assume that x is small
compared to 0.016 so that 0.016 – x » 0.016.
Note that this
approximate value of x is 10% of 0.016, so it is probably not
appropriate to have assumed that 0.016 – x » 0.016. Instead, attempt to solve for x without making
any approximations. Start by clearing the denominator of equation (1)
and then expanding the result to get:
This equation is a quadratic equation
of the form ax^{2} + bx + c = 0, the
solutions of which are given by (see Appendix 1):
.
For this case, a = 1, b
= 1.7×10^{–4}, and c = –2.72×10^{–6}.
Substituting these values into the quadratic formula gives:
The negative solution is not
physically reasonable since there is no such thing as a negative
concentration. The solution is x = 1.6×10^{–3} M.
Finally,
.
Think About It:

Did the result obtained using the quadratic formula differ
significantly from the approximate solution? What if the given K_{a}
value had three or more significant digits instead of two?


b.

Proceed in
the same manner as part (a). Note though that the initial concentration
is even less than it was in part (a), making it even more necessary to
avoid the approximate solution.

HF(aq) +

H_{2}O(l)


F^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

5.7×10^{–4}



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

5.7×10^{–4} –
x



x

x

The negative
solution is rejected, giving x = 6.3×10^{3} M. Thus,

c.

Proceed in the same manner as part (a). Note though that the
initial concentration is very large, and we may safely seek an
approximate solution.

HF(aq) +

H_{2}O(l)


F^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

1.75



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

1.75 – x



x

x





16.62

Strategy:

Phenol is a weak acid, and some of it will ionize in solution
according to the equilibrium
C_{6}H_{5}OH(aq)
+ H_{2}O(l) C_{6}H_{5}O^{–}(aq)
+ H_{3}O^{+}(aq)
The percent ionization of a weak acid is given by
Equation 16.7:
,
where [HA]_{0} is the
initial concentration of the acid (that is, the concentration assuming
there is no ionization). Using the K_{a} expression
above, set up a concentration table and solve for [H^{+}]. Then,
use this value to compute the percent ionization.

Solution:

a.


C_{6}H_{5}OH(aq) +

H_{2}O(l)


C_{6}H_{5}O^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

0.56



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

0.56 – x



x

x

Since the initial concentration is very large
compared to K_{a}, assume that 0.56 – x » 0.56. Then,

b.

Proceed as in part (a).

C_{6}H_{5}OH(aq) +

H_{2}O(l)


C_{6}H_{5}O^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

0.25



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

0.25 – x



x

x


c.

Proceed as in part (a).

C_{6}H_{5}OH(aq) +

H_{2}O(l)


C_{6}H_{5}O^{–}(aq)

+ H_{3}O^{+}(aq)

Initial (M):

1.8×10^{–6}



0

0

Change (M):

–x



+ x

+ x

Equilibrium (M):

1.8×10^{–6} – x



x

x





16.63

Strategy:

We are asked to compute K_{a} for the equilibrium
HA(aq) +
H_{2}O(l) A^{–}(aq) +
H_{3}O^{+}(aq)
We are given the percent
ionization of the acid, which is related to [H^{+}] and [HA]_{0}
by Equation 16.7:
or ,
where [HA]_{0} is the
initial concentration of the acid (that is, the concentration assuming
there is no ionization). Since the percent ionization (0.92%) is
very small, it is safe to assume that
[HA] »
[HA]_{0}. Using equation (2) above for [H^{+}], along
with the fact that [H^{+}] = [A–], substitute into equation (1)
and evaluate K_{a}.

Solution:




16.64

Strategy:

The relevant equilibrium for the weak acid ionization is
HA(aq) +
H_{2}O(l) A^{–}(aq) +
H_{3}O^{+}(aq)
where [H^{+}] = [A^{–}] (why?).
We are asked to find [HA]_{0}
when the percent ionization is 2.5%, and [HA]_{0} is related to
the percent ionization and [H^{+}] by Equation 16.7:
(2)
Since the percent ionization
is small (2.5%), assume that [HA] »
[HA]_{0}. Solve equation (2) for [H^{+}] and substitute
this value into equation (1), then solve the resulting equation for [HA]_{0}.

Solution:

Substituting [H^{+}]
for [A^{–}], for [H^{+}],
and [HA]_{0} for [HA] in equation (1), we get:



16.65

Strategy:

We use the pH to determine [H^{+}]_{eq},
and set up a table for the dissociation of the weak acid, filling in what
we know. The initial concentration
of weak acid is 0.19 M. The
initial concentrations of H^{+} and the anion are both zero; and
because they are both products of the ionization of the weak acid, their
equilibrium concentrations are equal: [H^{+}]_{eq} = [A^{}]_{eq}.

Solution:

[H^{+}] = 10^{}^{4.52}
= 3.02 ´ 10^{}^{5} M^{}
^{ }

HA(aq)


H^{+}(aq)

+ A^{–}(aq)

Initial
(M):

0.19


0

0

Change
(M):

3.02 ´ 10^{5}


+3.02 ´ 10^{5}

+3.02 ´ 10^{5}

Equilibrium
(M):

(0.19  3.02 ´ 10^{5})


3.02 ´ 10^{5}

3.02 ´
10^{5}

The
amount of HA ionized is very small compared to the initial concentration:
0.19

3.02 ´
10^{5}
»
0.19
Therefore,
= 4.8 ´
10^{9}



16.66

First we find the hydrogen ion concentration.
[H^{+}] = 10^{pH} =
10^{6.20} =
6.3 ´
10^{7}
M
If the concentration of [H^{+}]
is 6.3 ´
10^{7}
M, that means that 6.3 ´ 10^{7}
M of the weak acid, HA, ionized because of the 1:1 mole ratio
between HA and [H^{+}]^{
}.
Setting up a table:

HA(aq)


H^{+}(aq)

+ A^{–}(aq)

Initial
(M):

0.010


0

0

Change
(M):

–6.3
× 10^{–7}


+6.3
× 10^{–7}

+6.3
× 10^{–7}

Equilibrium
(M):

≈
0.010


6.3
× 10^{–7}

6.3
× 10^{–7}

Substituting into the acid ionization constant expression:
We have omitted the contribution to [H^{+}] due to water.


16.67

A pH of
3.26 corresponds to a [H^{+}]
of 5.5 ´
10^{4}
M. Let the original concentration
of formic acid be x. If the
concentration of [H^{+}]
is 5.5 ´
10^{4}
M, that means that 5.5 ´ 10^{4}
M of HCOOH ionized because of the 1:1 mole ratio between HCOOH and H^{+}.
^{ }
^{ }

HCOOH(aq)


H^{+}(aq)

+ HCOO^{}(aq)

Initial
(M):

x


0

0

Change
(M):

5.5
´
10^{4}


+5.5 ´
10^{4}

+5.5 ´
10^{4}

Equilibrium
(M):

x

(5.5 ´
10^{4})


5.5
´
10^{4}

5.5
´
10^{4}

Substitute K_{a} and
the equilibrium concentrations into the ionization constant expression to
solve for x.
x = [HCOOH] = 2.3 ´
10^{3}
M


16.68

A pH of
5.26 corresponds to a [H^{+}]
of 5.5 ´
10^{6}
M. Let the original
concentration of weak acid (HA) be x. If the concentration of [H^{+}] is 5.5 ´
10^{6}
M, that means that 5.5 ´ 10^{6}
M of HA ionized because of the 1:1 mole ratio between HA and H^{+}.
^{ }

HA(aq)


H^{+}(aq)

+ A^{}(aq)

Initial
(M):

x


0

0

Change
(M):

5.5
´
10^{6}


+5.5 ´
10^{6}

+5.5 ´
10^{6}

Equilibrium
(M):

x

(5.5 ´
10^{6})


5.5
´
10^{6}

5.5
´
10^{6}

Substitute K_{a} and
the equilibrium concentrations into the ionization constant expression to
solve for x.
x = [HA] = 6.4 ´
10^{6}
M


16.69

At 37°C,
K_{w} = 2.5 ´ 10^{14}. [H^{+}][OH^{}]
= 2.5 ´
10^{14}. Because in neutral water the
concentrations of hydronium and hydroxide are equal, we can solve for pH as
follows:
[H^{+}] = [OH^{}] = x
x^{2} = 2.5 ´ 10^{14}
pH = log(1.58 ´
10^{7})
= 6.80


16.72

a.

We construct the usual
table.






Initial (M):

0.10



0.00

0.00

Change (M):

–x



+ x

+ x

Equilibrium (M):

(0.10  x)



x

x

Assuming
(0.10 
x) »
0.10, we have:
x = 1.3 ´
10^{3}
M = [OH^{}]
pOH = log(1.3
´
10^{3}) =
2.89
pH = 14.00 
2.89 = 11.11

b.

By
following the identical procedure, we can show: pH = 8.96.



16.73

Strategy:

Weak bases only partially ionize in
water.
B(aq)
+ H_{2}O(l) BH^{+}(aq)
+ OH^{}(aq)
Note
that the concentration of the weak base given refers to the initial
concentration before ionization has started. The pH of the solution, on the other
hand, refers to the situation at equilibrium. To calculate K_{b}, we
need to know the concentrations of all three species, [B], [BH^{+}],
and [OH^{}]
at equilibrium. We ignore the
ionization of water as a source of OH^{}
ions.

Solution:

We proceed as follows.
Step 1: The major species in solution are B, OH^{}, and the conjugate acid BH^{+}.
Step 2: First,
we need to calculate the hydroxide ion concentration from the pH
value. Calculate the pOH from the
pH. Then, calculate the OH^{}
concentration from the pOH.
pOH =
14.00 
pH = 14.00 
10.66 = 3.34
pOH =
log[OH^{}]
pOH =
log[OH^{}]
Taking the antilog of both sides of the equation,
10^{pOH} =
[OH^{}]
[OH^{}] =
10^{3.34} =
4.57 ´
10^{4}
M
Step 3: If
the concentration of OH^{} is 4.57 ´
10^{4}
M at equilibrium, that must mean that 4.57 ´
10^{4}
M of the base ionized. We
summarize the changes.

B(aq) +

H_{2}O(l)


BH^{+}(aq)

+ OH^{}(aq)

Initial (M):

0.30



0

0

Change (M):

4.57 ´ 10^{4}



+4.57 ´ 10^{4}

+4.57 ´ 10^{4}

Equilibrium (M):

0.30  (4.57 ´ 10^{4})



4.57 ´ 10^{4}

4.57 ´ 10^{4}

Step 4: Substitute the equilibrium
concentrations into the ionization constant expression to solve for K_{b}.
K_{b}= 6.97 ´
10^{7}



16.74

A pH
of 11.22 corresponds to a [H^{+}] of 6.03 ´ 10^{12}
M and a [OH^{}] of 1.66 ´
10^{3}
M.
Setting
up a table:






Initial (M):

x



0.00

0.00

Change (M):

1.66
´
10^{3}



+1.66
´
10^{3}

+1.66
´
10^{3}

Equilibrium (M):

x

(1.66 ´
10^{3})



1.66
´
10^{3}

1.66
´
10^{3}

Assuming
1.66 ´
10^{3}
is small relative to x, then
x =
0.15 M = [NH_{3}]


16.75

We set up an equilibrium table to
solve for [OH^{}], and use pOH to get
pH. We use B and BH^{+} to
represent the weak base and its protonated form, respectively.






Initial (M):

0.61



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

0.61

x



x

x

Assuming
x is small relative to 0.61, then
x = 0.0096 M = [OH^{}]
pOH
= –log(0.0096) = 2.02
pH
= 14.00 
2.02 = 11.98


16.76

We use the pH given in the problem to
solve for pOH and then for [OH^{}]. We then set up an equilibrium table and
fill in what we know:
pOH = 14.00 
10.88 = 3.12
[OH^{}]
= 10^{3.12}
= 7.59 ´
10^{4}
M






Initial (M):

0.19



0.00

0.00

Change (M):

7.59
´
10^{4}



+7.59 ´
10^{4}

+7.59 ´
10^{4}

Equilibrium (M):

0.19

7.59 ´
10^{4}



7.59
´
10^{4}

7.59
´
10^{4}

K_{b}
3.0 ´ 10^{6}


16.77

a.

HA has the largest K_{a} value. Therefore, A^{}
has the smallest K_{b} value.

b.

HB has the smallest K_{a}
value. Therefore, B^{}
is the strongest base.



16.80

a.

C^{}
has the weakest conjugate acid (HC).

b.

In order of decreasing base strength: C^{}
> A^{} > B^{}.



16.81

Strategy:

We
calculate the K_{b} value for a conjugate base using
Equation 16.7.

Solution:

K_{b}(CN^{})
= 2.0 ´
10^{5}
^{ }
K_{b}(F^{})
= 1.4 ´
10^{11}
K_{b}(CH_{3}COO^{})
= 5.6 ´
10^{10}
K_{b}(HCO) = 2.4 ´
10^{8}



16.84

If we can
assume that the equilibrium concentration of hydrogen ion results only from
the first stage of ionization. In
the second stage this always leads to an expression of the type:
where c
represents the equilibrium hydrogen ion concentration found in the first
stage. If we can
assume (c ± y) »
c, and consequently
Think About It:

Is this
conclusion also true for the second stage ionization of a triprotic acid
like H_{3}PO_{4}?



16.85

The pH of a
0.040 M HCl solution (strong acid) is: pH = log(0.040) = 1.40. Follow the procedure for calculating the
pH of a diprotic acid to calculate the pH of the sulfuric acid solution.
Strategy:

Determining the pH of a diprotic acid in aqueous
solution is more involved than for a monoprotic acid. The first stage of ionization for H_{2}SO_{4}
goes to completion. We follow the
procedure for determining the pH of a strong acid for this stage. The conjugate base produced in the
first ionization (HSO_{4}^{})
is a weak acid. We follow the
procedure for determining the pH of a weak acid for this stage.

Solution:

We proceed
according to the following steps.
Step 1: H_{2}SO_{4}
is a strong acid. The first
ionization stage goes to completion.
The ionization of H_{2}SO_{4} is
H_{2}SO_{4}(aq) ® H^{+}(aq) + HSO (aq)
The concentrations of all the
species (H_{2}SO_{4}, H^{+}, and HSO_{4}^{})
before and after ionization can be represented as follows.





Initial
(M):

0.040


0

0

Change
(M):

0.040


+0.040

+0.040

Equilibrium
(M):

0


0.040

0.040

Step 2: Now,
consider the second stage of ionization.
HSO_{4}^{} is a weak acid. Set up a table showing the
concentrations for the second ionization stage. Let x be the change in
concentration. Note that the
initial concentration of H^{+} is 0.040 M from the first
ionization.





Initial
(M):

0.040


0.040

0

Change
(M):

x


+x

+x

Equilibrium
(M):

0.040  x


0.040 + x

x

Write the ionization constant expression for K_{a}. Then, solve for x. You can find the K_{a}
value in Table 16.8 of the text.
Since K_{a} is quite large, we cannot
make the assumptions that
0.040

x » 0.040 and 0.040 + x » 0.040
Therefore, we must solve a quadratic equation.
x^{2}
+ 0.053x  (5.2 ´
10^{4}) =
0
x =
8.5 ´
10^{3}
M or x =
0.062
M
The second solution is physically impossible because
you cannot have a negative concentration.
The first solution is the correct answer.
Step 3: Having
solved for x, we can calculate the H^{+} concentration at
equilibrium. We can then calculate
the pH from the H^{+} concentration.
[H^{+}] =
0.040 M + x
= [0.040 + (8.5 ´
10^{3})]M =
0.049 M
pH =
log(0.049) =
1.31

Think About It:

Without doing any calculations, could you have known
that the pH of the sulfuric acid would be lower (more acidic) than that
of the hydrochloric acid?



16.86

There is no H_{2}SO_{4} in the solution because
HSO has no
tendency to accept a proton to produce H_{2}SO_{4}. (Why?)
We are only concerned with the ionization:

HSO(aq)



+ SO(aq)

Initial
(M):

0.20


0.00

0.00

Change
(M):

x


+x

+x

Equilibrium
(M):

(0.20

x)


+x

+x

Solving
the quadratic equation:
x =
[H^{+}] = [SO] = 0.045 M
[HSO]
= (0.20 
0.045) M = 0.16 M


16.87

For the first
stage of ionization:




+ HCO (aq)

Initial
(M):

0.025


0.00

0.00

Change
(M):

x


+x

+x

Equilibrium
(M):

(0.025

x)


+x

+x

x =
1.0 ´
10^{4}
M
For the second ionization,





Initial
(M):

1.0
´
10^{4}


1.0
´
10^{4}

0.00

Change
(M):

y


+y

+y

Equilibrium
(M):

(1.0
´
10^{4})

y


(1.0
´
10^{4})
+ y

y

y =
4.8 ´
10^{11}
M
Since HCO_{3}^{} is a very weak
acid, there is little ionization at this stage. Therefore we have:
[H^{+}] =
[HCO_{3}^{}] =
1.0 ´ 10^{4}
M and [CO_{3}^{2}^{}] =
y = 4.8 ´ 10^{11}
M


16.88

Although phosphoric acid is polyprotic,
because its first and second ionization constants differ by significantly
more than a factor of 1000 (K_{a1}/K_{a2}
=120,000), we can determine pH from the first ionization alone. (The contributions of the second and
third ionizations to the hydronium ion concentration are negligible.) We set up an equilibrium table:




+
H_{2}PO (aq)

Initial (M):

0.25


0.00

0.00

Change (M):

x


+x

+x

Equilibrium (M):

(0.25  x)


+x

+x

We begin by
assuming that x is small enough to neglect in the denominator. This gives
x = = 0.00188
We then iterate
the solution until we get a constant value for x. For the first iteration, we substitute
the first value of x into the denominator and solve for x again:
x = 0.043
Substituting into the denominator again
gives the second iteration:
x = 0.039
Third iteration:
x = 0.040
Fourth iteration:
x = 0.040
The value of x converges at 0.040
so [H^{+}] = 0.040 M and pH = log(0.040) = 1.40.
Alternatively, we could have solved for
[H^{+}] using the quadratic equation.


16.89

Oxalic acid is diprotic. Because its first and second ionization
constants differ only by a factor of about 1000, we consider both the first
and second ionizations to determine [H^{+}]. We set up an equilibrium table:




+

Initial (M):

0.25


0.00

0.00

Change (M):

x


+x

+x

Equilibrium (M):

(0.25  x)


+x

+x

Using the
quadratic equation to solve for x we find
x = 0.099
For the second ionization:




+

Initial (M):

0.099


0.099

0.00

Change (M):

y


+y

+y

Equilibrium (M):

(0.099  y)


0.099+y

y

We neglect y
with respect to 0.099 and find
y = 6.1 ´ 10^{}^{5}
The
concentration of hydronium ion after the second ionization is therefore
0.099 + 6.1 ´ 10^{}^{5} » 0.099 M
and
pH = log(0.099) = 1.00
Note that the
second ionization did not contribute enough to the hydronium concentration
to change the value of pH.


16.90

1.

The two steps in the
ionization of a weak diprotic acid are:
H_{2}A(aq)
+ H_{2}O(l) H_{3}O^{+}(aq)
+ HA^{}(aq)
HA^{}(aq)
+ H_{2}O(l) H_{3}O^{+}(aq)
+ A^{2}^{}(aq)
The
diagram that represents a weak diprotic acid is (c). In this diagram, we only see the first
step of the ionization, because HA^{}
is a much weaker acid than H_{2}A.

2.

Both
(b) and (d) are chemically implausible situations. Because HA^{}
is a much weaker acid than H_{2}A, you would not see a higher
concentration of A^{2}^{} compared to HA^{}.



16.94

a.

The K^{+} cation does not
hydrolyze. The Br^{}
anion is the conjugate base of the strong acid HBr. Therefore, Br^{}
will not hydrolyze either, and the solution is neutral, pH »
7.

b.

Al^{3}^{+} is a small metal
cation with a high charge, which hydrolyzes to produce H^{+} ions. The NO
anion
does not hydrolyze. It is the
conjugate base of the strong acid, HNO_{3}. The solution will be
acidic, pH
< 7.

c.

The Ba^{2}^{+} cation does not
hydrolyze. The Cl^{}
anion is the conjugate base of the strong acid HCl. Therefore, Cl^{}
will not hydrolyze either, and the solution is neutral, pH »
7.

d.

Bi^{3}^{+} is a small metal
cation with a high charge, which hydrolyzes to produce H^{+} ions. The NO
anion
does not hydrolyze. It is the
conjugate base of the strong acid, HNO_{3}. The solution will be
acidic, pH
< 7.



16.95

Strategy:

In deciding whether a salt will undergo hydrolysis,
ask yourself the following questions:
Is the cation a highly charged metal ion or an ammonium ion? Is the anion the conjugate base of a
weak acid? If yes to either
question, then hydrolysis will occur.
In cases where both the cation and the anion react with water, the
pH of the solution will depend on the relative magnitudes of K_{a}
for the cation and K_{b} for the anion.

Solution:

We first break up the salt into its cation and anion
components and then examine the possible reaction of each ion with water.
a.

The Na^{+}
cation does not hydrolyze. The
Br^{} anion is the
conjugate base of the strong acid HBr.
Therefore, Br^{} will not hydrolyze
either, and the solution is neutral.

b.

The K^{+}
cation does not hydrolyze. The
SO anion
is the conjugate base of the weak acid HSO and
will hydrolyze to give HSO and OH^{}. The
solution will be basic.

c.

Both the NH and NO ions
will hydrolyze. NH is the
conjugate acid of the weak base NH_{3}, and NO is the
conjugate base of the weak acid HNO_{2}. Using ionization constants from
Tables 16.6 and 16.7, and Equation 16.7, we find that the K_{a}
of NH_{4}^{+}
(1.0 ´
10^{14}/1.8 ´ 10^{5} = 5.6 ´ 10^{10}) is greater than
the K_{b} of NO (1.0
´ 10^{14}/4.5 ´ 10^{4} = 2.2 ´ 10^{11}). Therefore, the solution will be acidic.

d.

Cr^{3}^{+} is a small metal cation with a high charge,
which hydrolyzes to produce H^{+} ions. The
NO anion
does not hydrolyze. It is the
conjugate base of the strong acid, HNO_{3}. The solution will be acidic.




16.96

There are two
possibilities: (i) MX is the salt of a strong acid and a
strong base so that neither the cation nor the anion react with water to
alter the pH and (ii) MX is the salt
of a weak acid and a weak base with K_{a} for the acid equal
to K_{b} for the base.
The hydrolysis of one would be exactly offset by the hydrolysis of
the other.


16.97

There is an
inverse relationship between acid strength and conjugate base
strength. As acid strength
decreases, the proton accepting power of the conjugate base increases. In general the weaker the acid, the stronger
the conjugate base. All three of the
potassium salts ionize completely to form the conjugate base of the
respective acid. The greater the pH,
the stronger the conjugate base, and therefore, the weaker the acid.
The order of increasing acid strength is HZ < HY
< HX.


16.98

The acid and
base reactions are:
acid: HPO(aq)
H^{+}(aq) + PO(aq)
base: HPO(aq) + H_{2}O(l)
H_{2}PO(aq) + OH^{}(aq)
K_{a}
for HPO is 4.2 ´
10^{13}. Note that HPO is the
conjugate base of H_{2}PO, so K_{b} is 1.6 ´
10^{7}. Comparing the two K's, we conclude
that the monohydrogen phosphate ion is a much stronger proton acceptor
(base) than a proton donor (acid).
The solution will be basic.


16.99

HCO H^{+} + CO K_{a} = 4.8 ´ 10^{11}
HCO + H_{2}O H_{2}CO_{3}
+ OH^{}
HCO has a
greater tendency to hydrolyze than to ionize (K_{b} > K_{a}). The solution will be basic (pH > 7).


16.100

The salt, sodium
acetate, completely dissociates upon dissolution, producing 0.36 M
[Na^{+}] and 0.36 M
[CH_{3}COO^{}] ions. The CH_{3}COO^{}
ions will undergo hydrolysis because they are a weak base.






Initial (M):

0.36



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

(0.36

x)



+x

+x

Assuming (0.36 
x) »
0.36, then
x = [OH^{}] = 1.4 ´ 10^{5}
pOH = log(1.4 ´
10^{5}) = 4.85
pH = 14.00  4.85 =
9.15


16.101

The salt ammonium chloride completely
ionizes upon dissolution, producing 0.42 M [NH] and 0.42 M [Cl^{}] ions. NH will
undergo hydrolysis because it is a weak acid (NH is the
conjugate acid of the weak base, NH_{3}).
Step 1: Express
the equilibrium concentrations of all species in terms of initial
concentrations and a single unknown x, that represents the change in
concentration. Let (x) be the
depletion in concentration (mol/L) of NH. From the
stoichiometry of the reaction, it follows that the increase in
concentration for both H_{3}O^{+} and NH_{3} must
be x. Complete a table that
lists the initial concentrations, the change in concentrations, and the
equilibrium concentrations.






Initial (M):

0.42



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

(0.42  x)



+x

+x

Step 2: You
can calculate the K_{a} value for NH from the K_{b}
value of NH_{3}. The
relationship is
K_{a} ´ K_{b} = K_{w}
or
Step 3: Write
the ionization constant expression in terms of the equilibrium
concentrations. Knowing the value of
the equilibrium constant (K_{a}), solve for x.
x = [H^{+}] =
1.5 ´ 10^{}^{5} M
pH = log(1.5 ´ 10^{}^{5}) = 4.82
Since NH_{4}Cl is the salt of a
weak base (aqueous ammonia) and a strong acid (HCl), we expect the solution
to be slightly acidic, which is confirmed by the calculation.


16.102

Aqueous NaF
contains the F^{} ion, which is the
conjugate base of a weak acid (HF)
We use Equation 16.7 to get K_{b} for the F^{}
ion:
K_{b}(F^{})
= 1.4 ´ 10^{11}
We construct
an equilibrium table to determine the concentration of hydroxide:

F^{}(aq) +

H_{2}O(l)


HF(aq)

+
OH^{–}(aq)

Initial (M):

0.082



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

0.082

x



x

x

Neglecting
x in the denominator gives
x
= = 1.07 ´
10^{6}
Therefore,
[OH^{}]
= 1.07 ´ 10^{6}
M, pOH = log(1.07
´
10^{6})
= 5.97
and
pH
= 14.00 
5.97 = 8.03


16.103

Strategy:

We are asked to determine the pH of a salt
solution. We must first determine
the identities of the ions in solution and decide which, if any, of the
ions will hydrolyze. In this case,
the ions in solution are C_{2}H_{5}NH and I^{}. The iodide ion, I^{},
is the anion of the strong acid, HI.
Therefore it will not hydrolyze.
The C_{2}H_{5}NH ion,
however, is the conjugate acid of the weak base C_{2}H_{5}NH_{2}
(ethylamine). It hydrolyzes to produce H^{+} and the weak base C_{2}H_{5}NH_{2}:
C_{2}H_{5}NH(aq)
+ H_{2}O(l) C_{2}H_{5}NH_{2}(aq) +
H_{3}O^{+}(aq)
In order to determine the pH, we must first determine
the concentration of hydronium ion produced by the hydrolysis. For this, we need the ionization
constant, K_{a}, of the C_{2}H_{5}NH ion.

Solution:

Using the K_{b}
value for ethylamine and Equation 16.7, we calculate K_{a}
for C_{2}H_{5}NH:
K_{a} =
We then
construct an equilibrium table and fill in what we know to find [H_{3}O^{+}]:






Initial (M):

0.91



0.00

0.00

Change (M):

x



+x

+x

Equilibrium (M):

(0.91  x)



x

x

Assuming that x is very small compared to 0.91,
we can write
and
x = [H_{3}O^{+}]
= 4.05 ´
10^{6}
M
pH = log(4.05
´
10^{6})
= 5.39



16.106

Metal ions with high oxidation numbers
are unstable. Consequently, these
metals tend to form covalent bonds (rather than ionic bonds) with
oxygen. Covalent metal oxides are
acidic while ionic metal oxides are basic.
The latter oxides contain the O^{2}^{}
ion which reacts with water as follows:
O^{2}^{} +
H_{2}O ® 2OH^{}


16.107

The most basic oxides occur with metal ions having the
lowest positive charges (or lowest oxidation numbers).
a.

Al_{2}O_{3}
< BaO < K_{2}O


b.

CrO_{3}
< Cr_{2}O_{3} < CrO




16.108

a.

2HCl(aq)
+ Zn(OH)_{2}(s) ® ZnCl_{2}(aq) + 2H_{2}O(l)

b.

2OH^{}(aq)
+ Zn(OH)_{2}(s) ® Zn(OH)(aq)



16.109

Al(OH)_{3}
is an amphoteric hydroxide. The
reaction is:
Al(OH)_{3}(s) + OH^{}(aq) ® Al(OH)(aq)
This is a Lewis acidbase reaction. Can you identify the acid and base?


16.110

According to Section 16.11, metal oxides change from basic to amphoteric to acidic moving from
left to right across a period. Metal
oxides become more basic moving from top to bottom within a group.


16.111

a.

The basic metallic oxides react with water to form
metal hydroxides:
Li_{2}O(s) + H_{2}O(l) →
2LiOH(aq)

b.

As in part (a) the basic metal oxide reacts to form a
metal hydroxide:
CaO(s) + H_{2}O(l) →
Ca(OH)_{2}(aq)

c.

The reaction between an acidic oxides and water is:
SO_{3}(g) + H_{2}O(l) →
H_{2}SO_{4}(aq)



16.114

a.

Lewis acid;
see the reaction with water shown in Section 16.12 of the text.

b.

Lewis base;
water combines with H^{+}
to form H_{3}O^{+}.

c.

Lewis base.

d.

Lewis acid; SO_{2} reacts with water to form H_{2}SO_{3}. Compare to CO_{2} above. Actually, SO_{2} can also act
as a Lewis base under some circumstances.

e.

Lewis base;
see the reaction with H^{+}
to form ammonium ion.

f.

Lewis base;
see the reaction with H^{+}
to form water.

g.

Lewis acid;
does H^{+} have
any electron pairs to donate?

h.

Lewis acid;
compare to the example of NH_{3} reacting with BF_{3}.



16.115

AlCl_{3} is a Lewis acid with
an incomplete octet of electrons and Cl^{}
is the Lewis base donating a pair of electrons.


16.116

a.

Both molecules have the same acceptor atom (boron) and both have
exactly the same structure (trigonal planar). Fluorine is more electronegative than
chlorine so we would predict based on electronegativity arguments that
boron trifluoride would have a greater affinity for unshared electron
pairs than boron trichloride.

b.

Since it has the
larger positive charge, iron(III) should be a stronger Lewis acid than
iron(II).



16.117

By definition
Brønsted acids are proton donors, therefore such compounds must contain at
least one hydrogen atom. In Problem
16.114, Lewis acids that do not contain hydrogen, and therefore are not
Brønsted acids, are CO_{2},
SO_{2}, and BCl_{3}.
Can you name others?


16.118

a.

Fe is the Lewis
acid; CO is the Lewis base.

b.

BCl_{3}
is the Lewis acid; NH_{3} is the Lewis base.

c.

Hg^{2+}
is the Lewis acid; I^{}
is the Lewis base.



16.119

a.

AlBr_{3}
is the Lewis acid; Br^{}
is the Lewis base.

b.

Cr is the Lewis
acid; CO is the Lewis base.

c.

Cu^{2+}
is the Lewis acid; CN^{}
is the Lewis base.



16.120

Loss of the first
proton from a polyprotic acid is always easier than the subsequent removal
of additional protons. The ease with
which a proton is lost (i.e., the strength of the acid) depends on the
stability of the anion that remains.
An anion with a single negative charge is more easily stabilized by
resonance than one with two negative charges.


16.121

A strong acid, such as HCl, will be
completely ionized, choice (b).
A weak acid will only ionize to a
lesser extent compared to a strong acid, choice (c).
A very weak acid will remain almost
exclusively as the acid molecule in solution. Choice (d) is the best choice.


16.122

The direction
should favor formation of F^{}(aq)
and H_{2}O(l).
Hydroxide ion is a stronger base than fluoride ion, and hydrofluoric
acid is a stronger acid than water.


16.123

