Chemistry part 27, Julia Burdge,2e (2009) potx

656
CHAPTER
16
Acids
and
Bases
Remember
that
the
hy
d
ron
iu
m
io
n
can
be
expressed
as
either H+
or
H
30
+
These
are
two
e
qu
iv

alent
ways
to
represent
the autoionization
of
water.
Think
About
It
Bec
ause the
conjugates
of
weak ac
id
s a
nd
ba
ses
have ionization constants, salts
containing these ions have an effect
on the
pH
of
a solution. In Section
16.10
we
will use the ionization
constants

of
conjugate acids and
conjugate
ba
ses to calculate
pH
for
solutions containing dissolved salts.
'
-' ' '
-~-~
-
•
•
As for any chemical equations, we
ca
n add these two equilibria and cancel identical terms:
_
CH
3
COO
II
(aq)
:;:::,
====:'
H+
(a
q) + _CH
3
COO (aq)

DA
"'1(-;;'a
q;;)
) + H)
O(l
) , '
S::
H3
COOII(aq)
+
OH
-
Caq)
The
sum is the autoionization
of
wate
r.
In
fact, this is the case for any weak acid and its conjugate
ba
se:
1M
:;:::,
====:'
H+ +
Pi"
+ K + H
2
0 , '

1M
+ OH
. . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

H
2
0
:;::,
=::z:'
H + +
OH
-
ar
for any weak base and its conjugate acid .
.B
+ H
2
0
:;::,
====:'
.wr
+ 0 H-
+
~
+ H
2
0.
'
J1
+ H3

0
+
. .

.




.

. . .

.
2H
2
0
:;::.
====:'
H30 + + OH-
Recall that when we add two equilibria, the equilibrium constant for the net reaction is the
product of the equilibrium constants for the individual equations
[
~~
Section
15
.3]. Thus, for any
conjugate acid-base pair,
Equation 16.7
Equation 16.7 gives

th
e quantitative basis for the reciprocal relationship between the strength
of
an
acid and that
of
its conjugate base
Co
r between the strength
of
a base and that
of
it
s conjugate acid).
Because
Kw
is
a constant, Kb
mu
st decrease
if
Ka increase
s,
and vice versa.
Sample Problem 16.16 shows how to determine ionization constants for conjugates.
Sample Problem 16.16
Determine (a) Kb
of
the acetate ion (CH
3

COO-
), (b) Ka
of
the methylammo
nium
ion
(CH
3
NH
~
)
,
(c) Kb
of
the fluoride ion (F
-),
and (d) Ka
of
t
he
anunonium ion
(NHt)
.
Strategy
Each species listed is eith
er
a conjugate base
or
a conjugate acid. Determine the identity
of

the acid corresponding to
eac
h conjugate base and the identity
of
the
ba
se corresponding to each
con
jug
ate
ac
id; then, consult Tables 16.6 and 16.7 for their ioni
za
tion constants.
Use
the tabulated
ionization constants and Equation 16.7 to calculate
eac
h indicated K value.
Setup (a) A Kb value is requested, indicating that the acetate ion is a conjugate
ba
se. To identify the
corresponding
Br
Sil
nsted acid, add a proton to
th
e formula to
get
CH

3
COOH
(acetic acid).
The
Ka
of
acetic acid (from Table 16.6) is 1.8 X 10
-'.
(b) A Ka value is requested, indicating that the
me
thy
lamm
onium ion
is
a conjugate acid. Determine
the identity
of
the corresponding
BrSiln
sted base
by
removing a proton from the formula to
get
CH
3
NH
2 (methylamine).
The
Kb
of

methylamine (from Table 16.7) is 4.4 X 10-
4
(c)
F-
is the conjugate base
of
HF; Ka = 7
.1
X
10-
4
(d) NHt is the conjugate acid
ofN
H
3
; Kb =
1.
8 X 10-
5
Solving Equation 16.7 separately for Ka and
Kb gives,
re
sp
ec
ti
vely,
and
Solution (a) Conjugate base
CH
3

COO
- : Kb = 1.0 X 10-
14
= 5.6 X
10
-
10
1.8 X
10-
5
(b) Conjugate acid
CH
3
NH
i : Ka = 1.0 X
10
-
14
=
2.3
X
10
-
11
4.4
X
10
-
4
(c) Conjugate base F- : Kb =

1.0 X
10
-
14
= 1.4 X 10-
11
7
.1
X
10
-
4
(d) Conjugate acid
NHt
: Ka =
1.
0 X
1O
-
1~
= 5.6 X 10-
10
1
.8
X
10
- )
SECTION 16.8
Diprotic
and Polyprotic Acids 657

,
I
Practice
Problem
A
Determine
(a)
Kb
of
the
benzoate
ion (C
6
H
s
COO- ), (b)
Kb
of
the
ascorbate
ion
I
(HC
6
H
6
0
6
),
and

(c)
Ka
of
the ethyl
ammonium
ion
(C
2
HsNH
~
).
Practice
Problem B
Determine
(a)
Kb
of
the
weak
base B
whose
conjugate acid
HB
+ has
Ka
. = 8.9 X
10-
4
and
(b)

Ka
of
the
weak
acid
HA
whose
conjugate
base
has
Kb
=
2.1
X 10-
8
I
~i
____________________________________________________________________
~EI
Checkpoint
16.7
Conjugate Acid-Base Pairs
16.7.1
Calculate the
Kb
of
the
cyanide
ion
16.7.2

Which
of
the anions listed is the
(CW).
strongest base? (See Table 16.6.)
a)
4.9
X 10-
10
a)
Ascorbate
ion
(HC6H60
(j
)
b)
2.0 X
lO-
s
b)
Benzoate
ion
(C
6
H
s
COO- )
c)
4.9
X

10-
24
c) Nitrite
ion
(N0
2
)
d)
1.0 X
10-
7
d)
Phenolate
ion (C
6
H
s
O-
)
e)
2.2
X
lO-
s
e)
Formate
ion
(HCOO-
)
Diprotic and Polyprotic Acids

Diprotic and polyprotic acids undergo successive ionizations, losing one proton at a time
[
~~
Sec
-

.
,.
. . . . . . . . . , . .


. . . . . . .
tion
4.3], and each ionization has a
Ka
associated with
it
Ionization constants for a diprotic acid
are designated
Ka
and
Ka
.
We
write a separate equilibrium expression for each ionization, and we
J 2
may need two or more equilibrium expressions to calculate the concentrations
of
species in solu-
tion at equilibrium. For carbonic acid (H

2
C0
3
),
for example, we write
HC0
3
(aq)
+.=='
H+(aq) +
CO~
-
(aq)
[H
+
][HC0
3
]
K
=

a
J
[H
2
C0
3
]
K =
a,

[H
+
][CO
~
-]
[HC0
3
]
Note that the conjugate base in the first ionization is the acid in the second ionization, Table 16,8
shows the ionization constants
of
several diprotic acids and one polyprotic acid, For a given acid,
the first ionization constant is much larger than the second ionization constant, and so
on, This
trend makes sense because it is easier to remove a proton from a neutral species than from one
that is negatively charged, and it is easier
to
remove a proton from a species with a single negative
charge than from one with a double negative charge,
Sample Problem 16.17 shows how to calculate equilibrium concentrations
of
all species in
solution for an aqueous solution
of
a diprotic acid.
Oxalic
acid (H
2
C
2

0
4
)
is a poisonous substance
used
mainly as a bleaching agent. Calculate the
concentrations
of
all species present at equilibrium
in
a 0.10 M solution
at
25°C.
Strategy Follow the
same
procedure
for
each ionization as for the determination
of
equilibrium
concentrations for a
monoprotic
acid.
The
conjugate
base
resulting
from
the
first ionization is the

acid
for
the second ionization,
and
its starting concentration is the equilibrium concentration
from
the
first ionization.
Setup
The
ionizations
of
oxalic
acid
and
the
corre
s
ponding
ionization constants are
H
2
C
2
0
4
(aq)
+.=='
H+(aq) +
HC

2
0
4
(aq)
HC
2
0
4
(aq)'
' H+(aq) +
C
2
0~-(aq)
K
aJ
= 6.5 X
10-
2
Ka =
6,1
X 10-
5
2
Construct
an equilibrium table for
each
ionization,
using
x as the
unknown

in
the first ionization
and
y as the unknown in the second ionization.
(Continued)
,
A triprotic acid h
as
Ka K
a
, and Ka .
I' 1 3
658
CHAPTER
16 Acids
and
Bases
Name
of
Acid
Formula
Structure
Ka
1
Ka
2
Ka
3
Sulfuric acid
Oxalic acid

Sulfurous acid
Ascorbic acid (vitamin
C)
Carbonic acid
Hydrosulfuric acid*
Phosphoric acid
•
H
2
SO
4
0
II
H-O
-
S-O-H
II
0
o 0
II
II
H-O-C-C-O
-H
o
II
H-O-S-O-H
H-O
O-H
"c
C/

H" / \
iC""-O'/C=O
CHOH
I
CH?OH
-
o
II
H-O-C-O-"H
H-S-H
o
II
H-O-P-O-H
I
o
I
H
Very large
6.5
X
10-
2
1.3 X
10-
2
8.0 X 10-
5
4.2 X
10
-

7
9.5
X
10
-
8
7.5 X
10-
3
1.3 x
10
-
2
6.1 X
10-
5
6.3
X
10-
8
1.6
X
10-
12
4.8
X
10-
11
1 X
10-

19
6.2 X
10-
8
4.2 X 10-
13
*Th e second
ionization
constant
of
H
2
5
is
very
low
and
difficult
to
measure. The
va
lue in this table
is
an
estimate.
H
2
C
2
0iaq)

:;:.=::!:' H+(aq) +
HC
2
0
4
(aq)
Initial concentration (
M):
0.10
o o
Change in concentration (M) :
-x
+x +x
Equilibrium concentration (
M):
0.10 - x
x
x
The equilibrium concentration
of
the hydrogen oxalate ion (HC
2
0
4
)
after the first ionization becomes
the starting concentration for the second ionization. Additionally, the equilibrium concentration
of
H+ is the starting concentration for the second ionization.
Initial concentration

(M):
x x o
Change in concentration (M): - y
+y +y
Equilibrium concentration (
M):
x - y
x+y
y
Solution
K =
[H
+][HC
2
0
4"
l
a,
[H
2
C
2
0
4
l .
2
6.5 X
10
-
2

= x
0.10 - x
Applying the approximation and neglecting x in the denominator
of
the expression gives
2
6.5 X
10
-
2
=
Ox
10
X
2
= 6.5 X
10
-
3
x = 8.1 X
10
-
2
M
SECTION
16.8
Diprotic
and
Po
l

yprotic
Acids 659
Testing the appr
ox
imation,
8.1
X 10-
2
M X 100% = 81%
O.lOM
Clearly the approximation is not valid, so we must solve the following quadratic equation:
X
Z
+
6.5
X lO-
z
x - 6.5 X 10-
3
= 0
The result is x = 0.054
M.
Thus, after the first io
ni
zation, the concentrations
of
species in solution are
[H+l =
0.054 M
[HC

2
0
4
l =
0.054M
[H
Z
C
2
0
4
l = (0.10 - 0.054) M = 0.046 M
Rewriting the equilibrium table for the second ionization, using the calculated value of
x,
gives the
following:
HC
Z
0
4
(aq)
+.
=::!:'
H+
(a
q) +
C20
~
-
(aq)

Initial concentration
(M):
Change in concentration
(M)
:
Equilibrium concentration (M):
0.054
0.054
-y
+y
0.054 - y
0.054 + Y
K =
[H+
][C
20~
-
l
a,
[HC
2
0
4
l
6
.1
X
10
- 5 =
_(

0_.0_S4
_+-= y) ,(y_)
0.054 - y
o
+y
y
Assuming that y is very small and applying the approximations 0.054 + Y = 0.054 and 0.054 - y =
0.054 gives
(
0.O
S4
)(y) -
= =-=,
'-'-
= =
6.
1 X 10 - ) M
0.054 y
We
must
test the approximation as follows to see if it is valid:
6.1
X
1O-
5
M X 100
'7l
=
011
'7l

0.054 M
O.
0
This time, because the ionization constant is much smaller, the approximation is valid. At
equilibrium, the
co
ncentrations
of
all species are
[H
2
C
2
0
4
l = 0.046 M
[HC
2
0
4
l = (0.054 - 6.1 X
10
-
5
)
M = 0.054 M
[H+l = (0.054 + 6
.1
X 10-
5

)
M = 0.054 M
[
Cz
O
~-
l
=
6.
1 X
10-
5
M
Practice Problem A Calculate the concentrations
of
H
Z
C
2
0
4
,
HC
2
0
4
,
C20
~-
,

and H+ ions in a
0.20 M oxalic acid solution at 2S°C.
Practice Problem B Calculate the
co
ncentrations
of
H
2
S0
4
,
HS0
4
,
SO
~-
,
and H+ ions in a 0.14 M
sulfuric acid solution at
25
°
C.
Checkpoint
16.8
Diprotic and Polyprotic Acids
16
.8.1
Calculate the equilibrium concentration
of
CO

~
-
in a 0.050 M solution
of
carbonic acid at
25
°
C.
a)
4.2
X
10-
7
M
b)
4.8
X
10-
11
M
c)
1.5
X
10-
4
M
d)
O.OSOM
e) 0.049 M
16

.
8.2
What
is the
pH
of
a 0.40 M solution
of
phosphoric acid at
25
°C?
a) 5.48
b) 1.26
c) 3.98
d) 12.74
e)
0.80
Think
About
It
Note that
the second ionization did not
contribute significantly to the H+
concentration. Therefore, we could
determine the
pH
of
this solution
by considering only the first
ionization. This is true in general

for polyprotic acids where
K
a
,
is at
least
1000 X K
a
, .
[It is necessary
to consider the second ionization
to determine the concentration
of
oxalate ion
(Cz
O
~-).
l
660
CHAPTER
16
Ac
ids
and
Bases
The
polarity of
the
H - X bo
nd

actually
decreases
from H - F to H
-I
,
large
ly
be
c
ause
F is the
most
electronegative
element.
This
would
suggest
tha
t HF would
be
the
str
o
nge
st
of
the
hydrohalic
acid
s.

Ba
s
ed
on
th
e
data
in
Table
16
.
9,
however,
bond
e
ntha
l
py
is
t
he
more
important factor
in
determining
the
stre
n
gths
of

these
acids.
Figure 16.2 Lewis structures
of
some common oxoacids.
Molecular Structure and Acid Strength
The
strength
of
an acid is
mea
sured by its tendency to ionize:
HX
• H+ + X-
Two factors influence the extent to which the acid undergoes ionization.
One is the strength
of
the
H-
X bond. The stronger the bond, the more difficult it is for the HX molecule to break up and
hence the weaker the acid. The other factor is the polarity
of
the H - X bond. The difference in the
electronegativities between H and X results in
a polar bond like
0+
0-
H-X
If
the bond is highly polarized (i.e.,

if
there is a large accumulation
of
positive and negative charges
on the H and X atoms, respectively),
HX
will tend to break up into H+ and X- ions. A high degree
of
polarity, therefore, gives rise to a stronger acid. In this section, we consider the roles
of
bond
stren
gt
h and bond polarity in determining the strength
of
an acid.
Hydrohalic Acids
The
halogens form a series
of
binary acids called the hydrohalic acids (HF, HCI, HBr, and HI).
Table 16.9 shows that of this series only
HF
is a weak acid
(Ka
= 7.1 X 10-
4
) .
The data in the
table indicate that the predominant factor in determining the strength

of
the hydrohalic acids is
.

bond strength.
HF
has the largest bond enthalpy, making its bond the most difficult to break. In
this series
of
binary acids, acid strength increases as bond strength decreases. The strength
of
the
acids increases as follows:
HF
< < HCI < HBr < HI
Oxoacids
An oxoacid,
as
we learned in Chapter 2, contains hydrogen, oxygen, and a central, nonmetal atom
[
~~
Section
2.7] . As the Lewis structures in Figure 16.2 show, oxoacids contain one
or
more
0-
H bonds.
If
the central atom is an electronegative element, or is in a high oxidation state, it will
attract electron

s,
causing the O- H bond to be more polar. This makes it easier for the hydrogen to
be lost as H+, making the acid stronge
r.
Acid
Strengtlis
Bond
Bond
Enthalpy
(kJ/mol)
Acid
Strength
H-F
562.8
Weak
H-CI
431.9
Strong
H-Br
366.1 Strong
H-I
298.3
Strong
••
•
H
-O-N
=O:
oo
•

• •
'0'
oo
II
oo
H-O-N
-O
:
••
• •
• •
'0 '
oo
II
oo
H-O-
C
-O-H
• • • •
Carbonic
ac
id
N itrous
ac
id
N
itri
c acid
• •
• •

'0'
'0 '
• •
'0'
oo
II
oo
H-O
-P
-O-H
oo
II
oo
H
-O
-S-O
- H
oo
II
oo
. . I .•
oo
II
oo
H-
O
-P-O
- H
'0'
• •

.0 .
• . I . •
I
• •
H
H
Ph
osphor
ou
s acid
Ph
osphoric acid
Sulfuric
ac
id
SECTION
16.9
Molecular
Structure
and
Acid
Strength
661
••
• •
H-O-CI:
•• •• ••
H-O-C1-0:
••
••

••
••
••
Hypochlorous acid (+ 1) Chlorous acid (+3)
••
:0:
••
:0:
• . I
H-O-CI-O:
. . I

H-O-Cl-O:
. . I

:0:
•• •• ••
• •
Chloric acid (+5)
Perchloric acid (+7)
To compare their strengths, it is convenient to divide the oxoacids into two groups:
1.
Oxoacids having different central atoms that are from the same group
of
the periodic table
and that have the same oxidation number.
Two examples are
• •
• •
'0'

• •
'0'
• •
••
I
••
• •
I
• •
H-O
Cl
0:
H-O
Br
0:
• • •
•• ••
• • • • • •
Within this group, acid strength increases with increasing electronegativity
of
the central
atom.
Cl
and
Br
have the
same
oxidation
number
in

these acids, +
S.
However, because
Cl
is
more
electronegative than Br, it attracts the electron pair it shares with oxygen (in the
CI-O-H
group) to a greater extent than
Br
does (in the corresponding
Br-O-H
group).
Consequently, the
O-H
bond
is more polar in chloric acid than in bromic acid and ionizes
more
readily.
The
relative acid strengths are
HCI0
3
>
HBr0
3
2.
Oxoacids having the same central atom but different numbers
of
oxygen atoms. Within this

•
group, acid strength increases with increasing oxidation
number
of
the central atom. Con-
sider the oxoacids
of
chlorine shown
in
Figure
16.3.
In
this series the ability
of
chlorine to
draw electrons away from the
OH
group (thus making the
O-H
bond
more
polar) increases
with the
number
of
electronegative 0 atoms attached to Cl. Thus,
HCI0
4
is the strongest
acid because it has the largest

number
of
oxygen atoms attached to Cl.
The
acid strength
decreases as follows:
HCIQ4
>
HCI0
3
>
HCI0
2
> HCIO
Sample
Problem
16.18 compares acid strengths based on molecular structure.
,
Predict the relative strengths
of
the
oxoacids in
each
of
the following groups: (a) HClO,
HErO,
and
HIO; (b) HNO}
and
RN0

2
.
Strategy
In
each group,
compare
the electronegativities or oxidation numbers
of
the
central
atoms
to
determine
which
O-H
bonds are the
most
polar.
The
more
polar
the
O-H
bond, the
more
readily
it
is
broken
and

the stronger the acid.
,
Setup
(a)
In
a
group
with different central atoms,
we
must
compare
electronegativitie
s,
The
electronegativities
of
the central atoms
in
this
group
decrease as follows:
Cl
>
Br
> L
(b)
These
two
acids have the
same

central atom but differ in the
number
of
attached
oxygen
atoms.
In a
group
such as this, the greater the
number
of
attached
oxygen
atoms,
the
higher
the oxidation
number
of
the central
atom
and the stronger the acid,
Solution
(a)
Acid
strength decreases as follows: HCIO >
HErO>
RIO,
. . . .

. .

. . .
· . . . .
.
. . . . . . , . . . .

.
(b)
RN0
3
is a stronger acid than
HN0
2
.
Practice
Problem
Indicate
which is the stronger acid: (a)
HBr03
or
HBr04;
(b) H2
Se04
or
H
2
S0
4
,

•
.'
,
,
,
,
•
,
Figure 16.3 Lewis structures
of
the oxoacids
of
chlorine.
The
oxidation
number
of
the
Cl
atom
is
shown
in
parentheses.
Note
that although
hypochlorous
acid is written as
HClO
,

the H
atom
is
bonded
to the ° atom.
As
the number of attached
oxygen
atoms
increases.
the
oxidat
i
on
number
of
the
central
atom
also
increases
[
~~
Section
4.4].
•
Another
way
to compare the strengths
of

these
two
is
to remember that HN0
3
is
one
of
the
seven
strong
acids,
HN0
2
is
not.
Think
About
It
Four
of
the strong
acids are oxoacids:
RN0
3
,
HCI0
4
,
HCI0

3
,
and
H
2
S0
4
,
662
CHAPTER
16 Acids
and
Bases
You
learned
in
Chapter 4
[
~~
Section
4.
1]
that
carboxylic
acid
formulas
are
often written
with the
ionizable

H atom
last,
in
order
to
keep
the functional group together.
You
should
recognize
the formulas for organic
acid
s written
either
way.
For
example
,
acetic
acid
may
be
written
as
HC
2
H
3
0
2

or
as
CH
3
COOH.
Recall
that a
salt
is
an
ionic compound formed
by
the
reaction
between
an
acid
and
a
base
[
~~
Section
4.3]
.
Salts
are
strong
electrolytes
that

dissociate
completely into
ions.
Carboxylic Acids
So far our discussion has focused on inorganic acids. A particularly important group
of
organic
acids is the
carboxylic acids, whose Lewis structures can be represented by
• •
"0'
II

R-C-O-H



where
' ids '
part
'
Of
the
'
acid
'
iiioiecuie
'
aiid
'

the
'
shaded
portioii
represents
'
fue
'
carijoxyi"
group
:
-COOH.
The
conjugate base
of
a carboxylic acid, called a carboxylate anion;
RCOO
- , can be repre-
sented by more than one resonance structure:
• •

-
U
'0'
• •

-
•
R C
0:

•
,
R C
0:

•
In terms
of
molecular orbital theory
[
~~
S
ecti
on
9.6],
we
attribute the stability
of
the anion to its
ability to spread out or delocalize the electron density over several atoms. The greater the extent
of
electron delocalization, the more stable the anion and the greater the tendency for the acid to
undergo ionization that is, the stronger the acid.
The
strength
of
carboxylic acids depends on the nature
of
the R group. Consider, for exam-
ple, acetic acid and chloroacetic acid:

• •
• •
H
'0'
CI
"0'
I
II

I
II

H
C C
0 H
H-C
C 0 H
I

I

H
H
Acetic acid
(Ka
=
l.8
x
10
-

5
)
Chloroacetic acid
(Ka
=
l.4
x
10-
3
)
The
presence
of
the electronegative CI atom in chloroacetic acid shifts the electron density toward
the R group, thereby making the
0-
H bond more polar. Consequently, there is a greater tendency
for chloroacetic acid to ionize:

• •

• •
:CI:
U
:CI:
U
I
II

I

II

H- C
C
O-H
,
H
C C
0:
+
H+
•
I

I

H
H
Chloroacetic acid is the stronger
of
the two acids.
•
Acid-Base Properties
of
Salt Solutions
In Section 16.7,
we
saw that the conjugate base
of
a weak acid acts as a weak Br\?lnsted base in

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
water. Consider a solution
of
the salt sodium fluoride (NaF). Because
NaF
is a strong electrolyte,
it
dissociates completely in water to give a solution
of
sodium cations (Na +) and fluoride anions
(F- ).
The
fluoride ion, which is the conjugate base
of
hydrofluoric acid, reacts with water to pro-
duce hydrofluoric acid and hydroxide ion:
F- (aq) + H
2
0(l)
:;:.
:=:z;
' HF(aq) +
OH
- (aq)
This is a specific example
of
salt hydrolysis, in whi
ch
ions produced by the dissociation
of

a salt
react with water to produce either hydroxide ions or hydronium ions thus impacting
pH
. Using
our
knowledge
of
how ions from a dissolved salt interact with water, we can determine (based on
the identity
of
the dissolved salt) whether a solution will be neutral, basic, or acidic. Note in the
preceding example that sodium ions (Na
+) do not hydrolyze and thus have no impact on the
pH
of
the solution.
Basic
Salt Solutions
Sodium fluoride is a salt that dissolves to give a basic solution. In general, an anion that is the
con
-
jugate base
of
a weak acid reacts with water to produce hydroxide ion. Other examples include the
acetate ion
(CH
3
COO
- ), the nitrite ion
(N0

2
),
the sulfite ion
(SO~
-
),
and the hydrogen carbonate
SECTION
16.10 Acid-Base Properties
of
Salt Solutions 663
.



.




.

. . .

. .
ion (HCO)). Each
of
these anions undergoes hydrolysis to produce the corresponding weak acid
and hydroxide ion:
A - (aq) + H

2
0(l)
:;::.
=::=:"
HA(aq)
+
OH
- (aq)
We
can therefore make the qualitative prediction that a solution
of
a salt in which the anion is
the conjugate base
of
a weak acid will be basic.
We
calculate the pH
of
a basic salt solution the
same way we calculate the pH
of
any weak base solution, using the
Kb
value for the anion. The
necessary
Kb
value is calculated u
si
ng the tabulated
Ka

value
of
the corresponding weak acid (see
Table 16.6).
Sample Problem 16.19 shows how to calculate the pH
of
a basic salt solution.
Calculate
the
pH
of
a
0.10
M solution
of
sodium
fluoride (NaF) at 25°C.
Strategy
A solution
of
NaF
contains
Na
+ ions and F- ions.
The
F- ion is the
conjugate
base
of
the

weak
acid, HF.
Use
the Ka value
for
HF
(7.1 X 10-
4
,
from Table 16.6)
and
Equation 16.7 to
determine
Kb
for F- :
1.0 X 10-
14
= 1.4 X 10-
11
7.1 X 10-
4
,
Then, solve this
pH
problem
like
any
equilibrium problem, using an equilibrium table.
Setup
It's

always a
good
idea to write the equation corresponding to
the
reaction that takes
plac
e
along
with
the equilibrium expression:
F-(aq)
+ H
zO(
I)
::;:.
==" HF(aq) +
OW(aq
)
Construct
an
equilibrium
table, and determine, in terms
of
the unkn
own
x,
the
equilibrium
concentrations
of

the
species
in
the
equilibrium
expression:
F- (aq) + HzO(l)
::;:.
==" HF(aq) +
OH
- (aq)
Initial concentration (M): 0.10
o o
Change
in
concentration
(M):
-x
+x
+x
Equilibrium concentration (M):
0.10
- x
x x
Solution
Substituting the equilibrium concentrations into the
eq
uilibrium expression and using the
shortcut to solve for
x,

we
get
z
1.4 X 10-
11
= X =
0.10
- x 0.10
x =
~(
1.4
X 10-
11
)(0.10) = 1.2 X 10-
6
M
According
to
our
equilibrium
table, x = [OH- j. In this case, the autoionization
of
water
makes a
significant contribution to the hydroxide
ion
concentration so the total concentration will
be
the
s

um
of
1.2 X 10-
6
M
(from
the ionization
of
F-)
and 1.0 X
10-
7
M
(from
the autoionization
of
water).
Therefore,
we
calculate
the
pOH
first as
pOH
=
-log
(1.2 X 10-
6
+ 1.0 X 10-
7

)
= 5.95
and then the
pH
,
pH
= 14.00 -
pOH
= 14.00 - 5.95 = 8.05
The
pH
of
a
0.10
M solution
of
NaF
at
25°C
is 8.05.
Practice
Problem
A
Determine
the
pH
of
a 0.15 M
so
lution

of
sod
ium
acetate
(CH
3
COONa)
at
25°C.
Practice
Problem
B
Determine
the concentration
of
a solution
of
sodium
fluoride (NaF)
that
has
pH
8.51
at
25
°C.
HC0
3"
has
an

ionizable
proton
and
can
also
act
as
a
Br0nsted
acid.
H
owever,
its
tendency
to
ac
ce
pt a proton
is
stronger than
its
tendency
to
donate a proton:
HC0
3
+ H
2
0
:;:.

=z·
H
2
C0
3
+
OW
Kb
= 10-
8
Remember
that for
any
conjugate
acid
-ba
se
pair
(
Equation
16.7):
Think
About
It
It'
s easy to mix
up
pH
and
pOH

in this type
of
problem. Always
make
a qualitative
prediction regarding the
pH
of
a
salt
solution first,
and
then
check
to
make
sure that
your
calculated
pH
agrees with
your
prediction.
In
this
case,
we
would
predict
a basic

pH
becau
se
the anion
in
the
salt
(F
- ) is
the
conjugate
base
of
a
weak
acid
(HF).
The
calculated
pH,
8.05, is
indeed
basic.
664
CHAPTER
16 Acids
and
Bases
Think
About

It In this case, we
would predict an acidic pH because
the cation in the salt
(NHt)
is the
conjugate acid
of
a weak base
(NH
3
).
The calculated pH
is
acidic.
•
Acidic Salt Solutions
When
the
cation
of
a
salt
is
the
conjugate
acid
of
a
weak
base,

a
solution
of
the
salt
will
be
acidic.
For
example
,
when
ammonium
chloride
dissolves
in
water,
it
dissociates
to
give
a
solution
of
ammonium
ion
s
and
chloride
ions:

The
ammonium
ion
is
the
conjugate
acid
of
the
weak
base
ammonia
(NH3)'
It
acts
as
a
weak
Br0n-
sted
acid,
reacting
with
water
to
produce
hydronium
ion:
We
would

therefore
predict
that
a
solution
containing
the
ammonium
ion
is acidic. To
calculate
the
pH,
we
must
deterllline
the
Ka
for
NHt
using
the
tabulated
Kb
value
for
NH
3
and
Equation

16.7.
Because
Cl
- is
the
weak
conjugate
base
ofthe
strong
acid
HCI,
Cl
-
does
not
hydrolyze
and
there-
fore
has
no
impact
on
the
pH
of
the
solution
.

Sample
Problem
16.20
shows
how
to
calculate
the
pH
of
an
acidic
salt
solution.
Sample Problem 16.20
·.
Calculate the pH
of
a 0.10 M
so
lution of ammonium chloride (NH
4
Cl) at
25°e.
Strategy
A solution
ofNH
4
Cl contains
NHt

cations and
Cl-
anions. The
NH
t ion is the conjugate
acid
of
the weak base NH
3
. Use the Kb value for NH3 (1.8 X
lO
-
s
from Table 16.7) and Equation
16.7 to determine
Ka
for
NH
t .
1.0 X 10-
14
= 5.6 X 10-
10
1.8 X
lO
-
s
Setup
Again, we write the balanced chemical equation and the equilibrium expression:
[NH

3
]
[H3
0
+]
K =
~ == ,: =.
a
[NH
t ]
Next, construct a table
to
determine the equilibrium concentrations of the species
in
the equilibrium
•
expre
ss
lOn:
NH
t(aq)
+ Hz
O(l)
::;:,
=~.
NH3(aq) + H3
0+(aq)
Initial concentration
(M)
: 0.10

o
o
Change
in
concentration (M):
+x
•
-rx
-x
x
Equilibrium concentration (M): 0.10 - x
x
,_._
Solution
Substituting the equilibrium concentrations into the equilibrium expression and using the
shortcut to solve for
x,
we get
2
5.6 X 10-
10
= X =
0.10
-x
0.10
x =
~(5
.6
X 10
10)(

0.10) = 7.5 X 10-
6
M
According to the equilibrium table, x =
[H
30 +]. The pH can be calculated
as
follows:
pH
= - log (7
.5
X 10-
6
)
= 5.12
The pH
of
a 0.10 M solution
of
ammonium chloride (at 25°C) is 5.12.
.
Practice Problem A Determine the pH
of
a 0.25 M solution
of
pyridinium nitrate (C
s
H
6
NN0

3
)
at
2Ye.
[Pyridinium nitrate dissociates
in
water to give pyridinium ions (CsH6N+), the conjugate acid
of
pyridinium (see Table 16.7), and nitrate ions
(N0
3
).]
Practice Problem B Determine the concentration
of
a solution
of
ammonium chloride (NH
4
Cl) that
has pH
5.37 at 25°
e.
The
metal
ion
in
a
dissolved
salt
can

also
react
with
water
to
produce
an
acidic
solution.
The
extent
of
hydrolysis
is
greatest
for
the
small
and
highly
charged
metal
y
ations
such
as
AI3+,
Cr
3
+,

Fe
3+
,
Bi
3
+,
and
Be2+.
For
example,
when
aluminum
chloride
dissolves
in
water,
each
AI
3+
ion
becomes
associated
with
six
water
molecules
(Figure
16.4).
SECTION 16.10 Acid-Base
Properties

of
Salt
Solutions
665

+
Al(OH)(H20)~
+
+
Consider one
of
the bonds that forms between the metal ion and an oxygen atom from one
of
the six water molecules
in
Al(H20)~+:
•
H
AI ' I
cf
l
~
H
The
positively charged
A1
3
+ ion draws electron density toward itself, increasing the polarity
of
the

0-
H bonds. Consequently, the H atoms have a greater tendency to ionize than those in water
molecules
not
associated with the Al3+ ion.
The
resulting ionization process can be written as
or
as
The
equilibrium constant for the metal cation hydrolysis is
given
by
[Al(OH)(H
2
0)
~
+
][H
+
]
-5
Ka
= =
1.3
X 10
[Al(H20)~
+
]
Al(OH)(H20)~+

can undergo further ionization:
and so on.
It
is generally sufficient, however, to take into account only the first stage
of
hydrolysis
when determining the
pH
of
a solution that contains metal ions.
Neutral Salt Solutions
The
extent
of
hydrolysis is greatest for the smallest and
mo
st highly charged metal ions because a
compact, highly charged ion is
more
effective
in
polarizing the
O-H
bond
and facilitating ioniza-
tion. This is
why
relatively large ions
of
low charge, including the metal cations

of
Groups
lA
and
. .

.

.

··
2
:.f
··
···························
2A
(the cations
of
the strong bases), do not undergo significant hydrolysis
(Be
is an exception).
Thus,
most
metal cations
of
Groups
lA
and
2A
do

not impact the
pH
of
a solution.
Similarly, anions that are conjugate bases
of
strong acids do not hydrolyze to any significant
degree. Consequently, a salt composed
of
the cation
of
a strong
base
and the anion
of
a strong acid,
such as NaCl, produces a neutral solution.
To summarize, the
pH
of
a salt solution can be predicted qualitatively
by
identifying the ions
in
solution and determining which
of
them,
if
any, undergoes significant hydrolysis.
Examples

A cation that will
make
a solution acidic is
•
The
conjugate acid
of
a
weak
base
• A small, highly charged metal ion (other than from Group
lA
or
2A)
An
anion that will
make
a solution basic is
•
The
conjugate
base
of
a
weak
acid
CN
- ,
N0
2

,
CH
3
COO-
A cation that will
not
affect the
pH
of
a solution is
• A Group
lA
or
heavy Group
2A
cation (except
Be
2
+)
An anion that will not affect the
pH
of
a solution is
•
The
conjugate base
of
a strong acid
Cl
- ,

N0
3
,
CI0
4
•
Figure 16.4 The six H
2
0
molecules surround the
AIH
ion
in
an
octahedral arrangement. The attraction
of
the small
AIH
ion for the lone pairs
on the oxygen atoms
is
so
great that
the
0-
H bonds
in
an
H
2

0 molecule
attached
to
the metal cation are
weakened, allowing the loss of a proton
(H
+)
to
an incoming H
2
0 molecule.
This hydrolysis of the metal cation
makes the
so
lution acidic.
The
metal
cations
of the strong
bases
are
those
ofthe
alkali
metals:
(Li
+,
Na
+,
K+,

Rb
+,
and
(s+)
and
those
of the
heavy
alkaline
earth
metals
(51"'+
and
Ba
2+)
.
666
CHAPTER
16
Acids
and
Bases
Think
About
It It's very important
that you be able to identify the ions
in solution correctly.
If
necessary,
review the formulas and charges

of
the common polyatomic ions
[
~
Sec
ti
on
2.7,
Table
2.8] .
Sample Problem 16.21 lets you practice predicting the pH
of
salt solutions.
Predict whether a 0.10 M solution
of
each
of
the following salts will be basic, acidic, or neutral:
(a) LiI, (b)
NH
4
N0
3
,
(c)
Sr(N0
3
h, (d)
KN0
2

,
(e) NaCN.
Strategy
Identify the ions present in each solution, and determine which,
if
any, will impact the
pH
of
the solution.
Setup
(a) Ions in solutio
n:
Li+ and
r.
Li+ is a Group
lA
cation; 1- is the conjugate base
of
the
strong acid
HI. Therefore, neither ion hydrolyzes
to
any significant degree.
(b) Ions
in solutio
n:
NH1 and
NO
) . NH1 is the conjugate acid
of

th
e weak base NH
3
;
N0
3"
is the
conjugate base
of
the strong acid
HN0
3
.
In this case, the cation will hydrolyze, making the
pH
acidic:
(c) Ions in solution:
Sr
2
+ and
N0
3"
.
Sr
2+
is a heavy Group 2A cation; NO) is the conjugate base
of
the strong acid
HN0
3

.
Neither ion hydrolyzes to any significant degree.
(d) Ions
in
solution: K+ and
N0
2
.
K+ is a Group I A cation;
N0
2
is the conjugate base
of
the weak
acid
HN0
2
.
In
this case, the anion hydrolyzes, thus making the
pH
basic:
N0
2
(aq) + H
2
0(l)
:;::.
=='
HN0

2
(aq) +
OW(aq)
(e) Ions in solution:
Na
+ and CN- .
Na
+
is
a Group 1A cation;
CN
- is the conjugate base
of
the weak
acid HCN. In this case, too, the anion hydrolyzes, thus making the pH
ba
sic:
CW(aq)
+ H
2
0(l)
:;:::.
==:!:'
HCN(aq) +
OH
- (aq)
Solution
(a) Neutral
(b) Acidic
(c) Neutral

(d) Basic
(e) Basic
Practice
Problem
A Predict whether a 0.10 M solution
of
each
of
the following salts will be basic,
acidic, or neutral: (a)
CH
3
COOLi, (b) CsHsNHCI, (c) KF, (d)
KN0
3
,
(e)
KCl0
4
.
Practice
Problem
B In addition
to
those given in Sample Problem 16.21 and Practice Problem
A, identify two salts that will dissolve to give (a) an acidic solution, (b) a basic solution, and (c) a
neutral solution.
•
Salts
in

Which Both the Cation and the Anion Hydrolyze
So far we have considered salts in which only one ion undergoes hydrolysis. In some salts, both
the cation and the anion hydrolyze. Whether a solution
of
such a salt is basic, acidic, or neutral
depends on the relative strengths
of
the weak acid and the weak base. Although the process
of
calculating the
pH
in these cases is more complex than in cases where only one ion hydrolyzes,
we can make qualitative predictions regarding
pH
using the values
of
Kb
(of the salt's anion) and
Ka
(of the salt's cation):
• When
Kb
> K
a
,
the solution is basic.
• When
Kb
<
Ka,

the solution is acidic.
• When
Kb
= K
a
,
the solution is neutral or nearly neutral.
The salt
NH
4
NO
z
, for example, dissociates in solution to give
NHt
(Ka
= 5.6 X
10-
1
°) and
NO
z
(Kb
= 2.2 X 10-
11
).
Because Ka for the ammonium ion is larger than
Kb
for the nitrite ion,
we would expect the
pH

of
an ammonium nitrite solution to be slightly acidic.
SECTION
16.
11
Acid-Base Properti
es
of
Oxides and H
ydrox
id
es
667
Checkpoint 16.10
Acid-Base Properties
of
Salt Solutions
16
.10.1 Calculate the
pH
of
a 0.075 M solution
of
NH
4
N0
3
at 25°C.
a) 5.19
b) 8.

81
c) 7.00
d)
2.93
e) 11.07
16
.
10
.2 Calculate the
pH
of
a 0.082 M solution
of
NaCN at 25°C.
a) 5.20
b) 8.80
c) 7.00
d)
2.89
e)
lLlI
16.10
.3 Which
of
the following salts will
produce a basic solution when dissolved
in water? (Select all that apply.)
a) Sodium hypochlorite (NaCIO)
b) Potassium fluoride (KF)
c) Lithium carbonate (Li

z
C0
3
)
d) Barium chloride (BaCl
z
)
e) Ammonium iodide
(NH
4
I)
16
.
10.4
Which
of
the following salts will produce
a neutral solution when dissolved in
water? (Select
all that apply.)
a)
Calcium chlorite [Ca(CIO
zh
]
b) Potassium iodide (KI)
c) Lithium nitrate
(LiN0
3
)
d) Barium cyanide [Ba(CN)z]

e) Ammonium iodide
(NH4I)
Acid-Base Properties
of
Oxides and Hydroxides
As we saw in Chapter 8, oxides can
be
classified as acidic, basic,
or
amphoteric. Thus,
our
discussion
of
acid-base reactions would
be
incomplete
if
we did not examine the properties
of
these compounds.
Oxides
of
Metals and Nonmetals
Figure 16.5 shows the formulas
of
a number
of
oxides
of
the main group elements in their highest oxi-

dation states. All alkali metal oxides and
all alkaline earth metal oxides except
BeO
are
ba
sic. Beryl-
lium oxide and several metallic oxides
in
Groups
3A
and
4A
are amphoteric. Nonmetallic oxides in
which the oxidation number
of
the main group el
ement
is high are acidic (e.g., N
1
0
s
,
S03, and
CI
2
0
7
),
but those in which the oxidation number
of

the main group element is low (e.g.,
CO
and
NO)
show no
measurable acidic properties.
No
nonmetallic oxides are known to have basic properties.
lA
1
Li
2
0
Na20
K
2
0
Rb
2
0
Cs
2
0
2A
2
BeO
3B
4B
MgO
3

4
CaO
SrO
BaO
Ba
sic oxide
Acidic oxide
Amphoteric oxide
5B 6B 7B 8B
I
I
5 6 7 8 9 10
Figure 16.5 Oxides
of
the main group elements in their highest oxidation states.
3A
4A
13
14
B
2
0
3
CO
2
1B
2B
Al
2
0

3
SiOz
11
12
Ga203
Ge02
In
Z
0
3
Sn02
TI
2
0
3
Pb0
2
SA
6A
15
16
N
2
0
S
P
4
0
10
S03

As
2
0
S
Se03
Sb
2
0
S
Te03
BizOs
Po0
3
7A
17
OF2
Clz07
Br207
12°7
Atz°7
8A
18
668
CHAPTER
16 Acids and
Bases
The
basic metallic oxides react with water to form metal hydroxides:
Na20(S)
+ H

2
0(l)
-_.
2NaOH(aq)
BaO(s) + H?O(I) •
Ba(OHMaq)
The
reactions between acidic oxides and water are as follows:
CO
2
(g) + H
2
0(I)
~.
=='
H
2
C0
3
(aq)
S0
3(g)
+ H
2
0(l).
• H
2
SOiaq)
N
2

0
S(g)
+ H
2
0(I).
•
2HN0
3
(aq)
P
4
0
IO
(g) +
6H
2
0(l).
•
4H
3
P0
4
(aq)
CI
2
0
7
(g) + H
2
0(l).

•
2HCIOiaq)
The
reaction between CO
2
and H
2
0 explains why pure water gradually becomes acidic when it is
exposed to air, which contains
CO
2
,
The
pH
of
rainwater exposed only to unpolluted air is about
5.5.
The
reaction between
S0
3 and H?O is largely responsible for acid rain.
Reactions
between
acidic oxides and bases and those between basic oxides and acids resem-
ble
normal acid-base reactions in that the products are a salt and water:
CO
2
(g) + 2NaOH(aq)
-_.

Na2C03(aq) + H
2
0(l)
BaO
(s) +
2HN0
3
(aq)
•
Ba(N0
3
Ma
q) + H
2
0(l)
Aluminum oxide (
Al
2
0
3
) is amphoteric. Depending on the reaction conditions, it can behave either
as an acidic oxide or as a basic oxide. For example,
Ah0
3 acts as a base with hydrochloric acid to
produce a salt (AICI
3
) and water:
and acts as an acid with sodium hydroxide:
Only a salt, sodium aluminum hydroxide [NaAI(OH)4, which contains the
Na

+ and AI(OH)4 ions]
is formed in the reaction with so
dium
hydroxide no water is produced. Nevertheless, the reaction
is still classified as an acid-base reaction because Al
2
0
3
neutralizes NaOH.
Some
transition metal oxides in which the metal has a high oxidation
number
act as acidic
oxides. Two examples are manganese(VII) oxide
(Mn
20
7)
and chromium(VI) oxide (Cr03), both
of
which react with water to produce acids:
pe
rman
ganic acid
Cr0
3
(S)
+ H
2
0(l)
-_.

H
2
Cr0
4(aq)
chromic acid
Basic
and Amphoteric Hydroxides
All the alkali and alkaline earth metal hydroxides, except
Be(OHh,
are basic. Be(OH)2, as well as
AI(0H)
3,
Sn
(OH)
2>
Pb(OH
)z,
Cr(OH)3, CU(OH)2,
Zn(OHh,
and
Cd(OH)2>
is amphoteric. All ampho-
teric hydroxides are insoluble, but beryllium hydroxide reacts with both acids and bases as follows:
Be(OHMs)
+
2H
+(aq)
-_.
Be
2

+(aq) +
2H
2
0(l)
Be(OH)is)
+
20H
- (aq) •
Be(OH)~-
(aq)
Aluminum
hydroxide reacts with both acids
and
bases in a similar fashion:
AI(OHMs)
+ 3H+(aq)
-_.
AI
3+
(aq) +
3H
2
0(l)
AI(OH
Ms)
+
OH
-(a
q) • AI(OH)4 (aq)
Lewis Acids

and
Bases
So
far
we
have discussed acid-base properties in terms
of
the Br0nsted theory.
For
example, a
Br0nsted
ba
se is a substance that must
be
able to accept protons.
By
this definition, both the
hydroxide ion and ammonia are bases:
+
+
• •
-
:O-H
••
H
:N-H
I
H
•
•

SECTION 16.12 Lewis Acids
and
Bases 669
••
H-O-H
• •
H
+
I
H-N-H
H
In
each
case, the atom to which the proton becomes attached possesses at least one unshared
pair
of
electrons. This characteristic property
of
OH-,
NH
3
, and other
Br0nsted bases suggests a more
general definition
of
acids and bases.
In 1932 G. N. Lewis defined what
we
now call a Lewis base as a substance that can donate a
pair

of
electrons. A Lewis acid is a substance that can accept a pair
of
electrons. In the protonation
of
ammonia, for example,
NH3
acts as a Lewis base because it donates a pair
of
electrons to the pro-
ton H+, which acts as a Lewis acid
by
accepting the pair
of
electrons. A Lewis acid-base reaction,
therefore, is one that involves the donation
of
a pair
of
electrons from one species to another.
The
significance
of
the Lewis concept is that
it
is more general than other definitions. Lewis
acid-base reactions include many reactions that do not involve
Br0nsted acids. Consider, for exam-
ple, the reaction between boron trifluoride
(BF

3)
and
ammonia
to form an adduct compound:
F H
F
H
I I
I
F B
+
:N
H
•
F B N H
I I
I I
F H
F
H
acid
base
The
B
atom
in
BF3
is sp2-hybridized
[
~~

Section
9.4].
The
vacant, un hybridized 2pz orbital
accepts the pair
of
electrons
from
NH
3
. Thus,
BF
3 functions as an acid according to the Lewis
definition, even though it does
not
contain an ionizable proton. A coordinate covalent
bond
[
~~
Section
8.8] is formed between the
Band
N atoms. In fact, every Lewis acid-base reaction results
in the formation
of
a coordinate covalent bond.
Boric acid is another Lewis acid containing boron. Boric acid (a
weak
acid used in eyewash)
is an oxoacia with the following structure:

H
I
·0·
• •
• •
I
••
H 0
B
0 H
• •
••
Boric acid does not ionize in water to produce H+. Instead, it produces H+ in solution
by
taking a
hydroxide ion away from water.
In this Lewis acid-base reaction, boric acid accepts a pair
of
electrons from the hydroxide ion that
is derived from the water molecule, leaving behind the hydrogen ion.
The
hydration
of
carbon dioxide to produce carbonic acid,
•
can
be
explained in
tenns
of

Lewis acid-base theory as well.
The
first step involves the donation
of
a lone
pair
on the 0 atom in H
2
0 to the C atom in
CO
2
.
An
orbital is vacated on the C atom to
accommodate the lone
pair
by relocation
of
the electron pair in one
of
the
C-O
pi bonds, chang-
ing the hybridization
of
the oxygen atom from
Sp2
to
Sp
3:

•

-
·0·
• •

+ I
H-O-C
I
II
H .0.
• •
As a result, H
2
0 is a Lewis
base
and CO
2
is a Lewis acid. Finally, a proton is transferred onto the
o atom bearing the negative charge to form H
2
C0
3
:

-

:0:

+ I

H-O-C
I
II
H .0.
• •
•
• •
H-O:
. . I
:O-C
I
II
H
.0.
• •
670
CHAPTER
16
Acids
and
Bases
An
electron-deficient
mo
l
ecu
le
is
one
with

le
ss
than a complete o
ct
et
around the
central
atom.
Think
About
It In Lewis acid-base
reactions, the acid is usually a
cation
or
an electron-deficient
molecule, whereas the base is an
anion
or
a molecule containing an
atom with lone pairs.
Other examples
of
Lewis acid-base reactions are
Ag
+(aq) +
2NH
3
(aq)
:;::,
~.

Ag(NH3)~(aq)
Cd
2
+(aq) +
4I-(aq)
, •
CdI~-(aq)
Ni(s) + 4CO(g) ' •
Ni(COMg)
The
hydration
of
metal ions is in itself a Lewis acid-base reaction. When copper(Il) sulfate (CUS04)
dissolves in water, each Cu
2
+ ion becomes associated with six water molecules as
Cu(H20)~+.
In this
case, Cu
2
+ acts as the acid, accepting electrons, whereas
H
2
0 acts as the base, donating electrons.
Sample
Problem 16.22 shows how to classify Lewis acids and bases.
Identify the Lewis acid and Lewis base in each
of
the following reactions:
(a) C

2
H
5
0C
2
H
5
+ AlCl
3
• • (C
2
H
5
)zOAlCI
3
(b)
Hg2
+(aq) +
4CW(aq)
' •
Hg(CN)~-(aq)
Strategy
Determine which species in each reaction accepts a pair
of
electrons (Lewis acid) and
which species donates a pair
of
electrons (Lewis base).
Setup
(a) It can

be
helpful to draw Lewis structures
of
the species involved:
(b) Metal ions act as Lewis acids, accepting electron pairs from anions
or
molecules with lone pairs.
• • • • •
•••
. . . .
·························
2

. .



.
Solution
(a)
The
AI
is
sp
-hybridized
in
AICl
3
with an empty 2pz orbital.
It

is electron deficient,
sharing only six electrons. Therefore, the
AI
atom has the capacity to gain two electrons to complete
its octet. This property makes AlCl
3
a Lewis acid. On the other hand, the lone pairs
on
the oxygen
atom in
C
2
H
5
0C
2
H
s
make the compound a Lewis bas
e:
(b)
Hg
2+
accepts four pairs
of
electrons from the
CN
- ions. Therefore, Hg2+ is the Lewis acid and
CN
- is the Lewis base.

Practice
Problem
Identify the Lewis acid and Lewis base in the following reaction:
Checkpoint 16.12 Lewis Acids and Bases
16.12.1 Which
of
the following cannot act as a
Lewis base? (Select all that apply.)
a)
NH
3
b)
OW
c) CH
4
d)
Fe
2+
e)
AI
3+
16.12.2 Which
of
the following is a Lewis acid
but not a
Bn
2jnsted acid? (Select all that
apply.)
a)
H

2
O
b)
BCl
3
c)
OW
d)
AI
3+
e)
NH
3
,
APPLYING
WHAT YOU'VE
LEARNED
67
1
Applying
What
You've Learned
Although it is a diprotic acid (Ka = 8.0 X
10
-
5
and K
a,
= 2.S X
10-

12
),
ascorbic acid
only loses one proton under biological conditions to give-H+ and the hydrogen ascorbate
•
IOn:
A person who takes vitamin C supplements but who has a sensitive stomach may
be
advised to take calcium ascorbate [Ca(HC
6
H
6
0
6
)2], which is less acidic than ascorbic
acid and contains the hydrogen ascorbate ion.
Problems:
a) Give the formulas for the conjugate acid and the conjugate base
of
the hydrogen
ascorbate ion.
[
~
Samp
le
Problem
16.
1]
•
b) Determine [H+] and

pH
for a solution of ascorbic acid prepared by dissolving a
1000-mg tablet in
SO.O
mL
of
water at
2S
°
C.
[
~
Sample
Problem
16.12]
c) Calculate [OH- ] in the solution in part (b).
[
~~
Sample
Problem
16.3]
d) Calculate the pOH
of
the solution in part (b).
[
~
Sample
Problem
16.6]
e) What concentration

of
HCI would have the same pH as the solution in part (b)?
[
~~
SampleProblem
16.9]
f) Detelmine
Kb
for the ascorbate ion
(C6H60~-).
[
~~
Sample
Problem
16.16]
g) Calculate the concentrations
of
all species in the solution in part (b).
[
~~
Sample
Problem
16.l7]
•
)
Calcium
Ascorbate
10()%
Pure
Buffered

Vitamin
C powder
Net
Wt.
8
oz.
(227
A
Dlata.,
Sb:JIpl.mpl
672
CHAPTER
16
Acids
and
Bases
CHAPTER SUMMARY
Section 16.1
• A
Br12lilsted
acid donates a proton; a
Br\<!nsted
ba
se accepts a proton.
•
When
a
Br\<!n
sted acid donates a proton, the anion that remains is a
conjugate

base.
•
When
a
Br\<!n
sted
base
accepts a proton, the resulting cation is a
conjugate acid.
•
The
combination
of
a
Br\<!nsted
acid and its conjugate base (or the
combination
of
a
Br\<!n
sted
ba
se
and its conjugate acid) is called a
conjugate pair.
Section 16.2
• Water is amphoteric, meaning it can act both as a
Br\<!nsted
acid and a
Br\<!n

sted base.
•
Pure
water undergoes auto ionization (to a very small extent), resulting
in concentrations
of
H+ and
OH
-
of
l.0
X
10-
7
Mat
25°
e.
• Kw is the eqUilibrium constant for the autoionization
of
water, also called
the
ion product constant. Kw = [H+] [OH- ] =
l.0
X
10-
14
at
2Ye.
Section 16.3
•

The
pH
scale measures acidity:
pH
= - log [H+].
•
pH
= 7.00 is neutral,
pH
<
7.00
is acidic, and
pH
> 7.00 is basic.
•
The
pOH
scale is analogous to the
pH
scale,
but
it
mea
sures basicity:
pOH
=
-log
[OW].
•
pOH

= 7.00 is neutral, pOH < 7.00
is
basic, and
pOH
> 7.00 is acidic.
•
pH
+
pOH
= 14.00 (at
2YC).
Section 16.4
• There are seven strong acids: HCI,
RBr
, HI,
RN0
3
,
HCI0
3
,
HCI0
4
,
and H
2
S0
4
.
• Strong acids ioni

ze
completely in aqueous solution. (Only the first
ionization
of
the diprotic acid H
2
S0
4
is complete.)
•
The
strong
ba
ses are the Group
IA
and the
hea
v
ie
st Group
2A
hydroxides:
LiOH
,
NaOH
,
KOH
,
RbOH
, CsOH,

Ca(OHh,
Sr
(
OH
h,
and
Ba(OH
)z.
Section 16.5
• A weak acid ionizes only partially. The acid ionization constant, K
a
,
is
the equilibrium constant that indicates to what extent a weak acid ionizes.
• We solve for the
pH
of
a solution
of
weak
acid using the concentration
of
the acid, the Ka value, and an equilibrium table.
• We can also determine the Ka
of
a
weak
acid
if
we

know the initial
acid concentration and the
pH
at equilibrium.
Section 16.6
• A weak
base
ionizes only partially. The
base
ionization constant,
K",
is
the equilibrium constant that indicates to what extent a weak base ionizes.
• We solve for the
pH
of
a solution
of
weak
ba
se using the concentration
of
the
base
, the
Kb
value, and an equilibrium table.
• We can also determine the
Kb
of

a
weak
ba
se
if
we know the
concentration and the pH at equilibrium.
Section 16.7
•
The
conjugate base
of
a strong acid is a weak conjugate base,
meaning that
it
does not react with water.
•
The
conjugate base
of
a
weak
acid is a strong conjugate base,
meaning that it acts as a weak
Br\<!nsted
base
in
water.
•
The

conjugate base
of
a strong base is a weak conjugate acid,
meaning that it does not react with water.
•
The
conjugate acid
of
a weak base is a strong conjugate acid,
meaning that it acts as a
weak
Br¢nsted acid in water.
•
For
any conjugate acid-base pair,
Ka
X
Kb
=
Kw.
Section 16.8
• Diprotic and polyprotic acids have more than one proton to donate.
They undergo stepwise ionizations.
Each
ionization has a
Ka
value
associated with it.
•
The

Ka values for stepwise ionizations
become
progressively smaller.
•
In
most cases, it is only necessary to consider the first ionization
of
an
acid to determine
pH
. To determine the concentrations
of
other species
at
eqUilibrium, it may
be
necessary to consider subsequent ionizations.
Section 16.9
•
The
strength
of
an acid is affected by molecular structure.
• Polar and weak bonds to the ionizable hydrogen lead to a stronger acid.
• Resonance stabilization
of
the conjugate
base
favors the ionization
process, resulting

in
a stronger acid.
Section 16.10
• Salts dissolve
in
water to give neutral, acidic,
or
basic solutions
depending on their constituent ions.
Salt hydrolysis is the reaction
of
an
ion with water to produce hydronium
or
hydroxide ions.
• Cations that are strong conjugate acids such as
NH
1
make
a solution
more acidic.
• Anions that are strong conjugate bases such as F- make a solution
more
basic. Anions that are conjugate bases
of
strong acids have
no
effect on pH.
• Small, highly charged metal ions hydrolyze to give acidic solutions.
Section 16.11

• Oxides
of
metals generally are basic; oxides
of
nonmetals generally
are acidic.
• Metal hydroxides may
be
basic
or
amphoteric.
Section 16.12
• Lewis theory provides
more
general definitions
of
acids and bases.
• A Lewis acid accepts a
pair
of
electrons; a Lewis base donates a pair
of
electrons.
• A Lewis acid is generally electron-poor and
need
not have a hydrogen
atom.
• A Lewis
ba
se

is an anion
or
a molecule with one
or
more
lone pairs
of
electrons.
KEyWORDS
Acid ionization constant
(Ka),
648
Amphoteric, 637
Autoionization
of
water, 638
Base ionization constant
(Kb)' 652
Conjugate acid, 636
KEY EQUATIONS
Conjugate base, 636
Conjugate pair, 636
Ion-product constant, 638
Lewis acid, 669
Lewis base, 669
•
16.2 pH = - log [H30 +]
16.3 [H3
0
+] =

lO
-
pH
16.4
pOH
= - log
[OW]
16.5 [OH- ] = lO-
poH
16.6
pH
+
pOH
= 14.
00
16.7 Ka X
Kb
= Kw
QUESTIONS AND PROBLEMS
Section 16.1: Br0nsted Acids and
Bases
Review Questions
16.1 Define Br(llnsted acids and bases. Give an example
of
a conjugate
pair in an acid-base reaction.
16.2 In order for a species to act as a
Br(llnsted base, an atom in the
species
must

possess a lone pair
of
electrons. Explain why this
is
so.
Problems
16.3 Classify each
of
the following species as a Br(llnsted acid
or
base,
or
both: (a) H
2
0,
(b)
OH
- , (c) H30 +, (d)
NH
3
, (e)
NH
1, (f)
NH
z
,
(g)
NO
;-
, (h)

co
j-
, (i)
HEr
, (j) HCN.
16.4 Identify the acid-base conjugate pairs in each
of
the following
reactions:
(a)
CH
3
COO
- +
HCN
• •
CH
3
COOH
+
CW
(b)
HCO
;-
+ HCO;- ( • H
Z
C0
3
+
co

j-
(c) H
Z
P0
4
+
NH
3 ( •
HPO
~
-
+
NH
1
(d) HCIO +
CH
3
NH
z
(
•
CH
3
NH
~
+
CIO-
(e)
co
j-

+ H
2
0 • •
HCO
;-
+
OW
16.5 Write the formulas
of
the conjugate bases
of
the following acids:
(a)
HNO
z
,
(b) H
2
S0
4
,
(c) HzS, (d)
HCN
, (e)
HCOOH
(formic
acid).
16.6 Write the formula for the conjugate acid
of
each

of
the following
bases: (a)
HS- , (b) HCO
;-
, (c)
CO
~
-
,
(d) H
2
P0
4
,
(e)
HPO
~
-
,
(f)
PO~
-
,
(g)
HS0
4
,
(h)
SO

~
-
,
(i) soj
16.7 Write the formula for the conjugate base
of
each
of
the following
acids: (a)
CH
2
CICOOH, (b)
HI0
4
,
(c) H
3
P0
4
,
(d) H
2
P0
4
,
(e)
HPO~-,
(f) H
2

S0
4
,
(g)
HS0
4
,
(h)
RI0
3
,
(i) HSO
;-
,
(j)
NH
1,
(k) HzS, (I)
HS-,
(m) HCIO.
QUESTIONS
AND
PROBLEMS
673
pH, 639
pOH, 641
Weak acid, 647
Weak base, 652
Salt hydrolysis, 662
Strong conjugate acid, 655

Strong conjugate base, 655
Weak conjugate acid, 655
Weak conjugate base, 654
or
16.8 Oxalic acid (H
Z
C
Z
0
4
)
has the following structure:
O=C-OH
I
O=C-OH
An oxalic acid solution contains the following species in varying
concentrations:
H
Z
C
Z
0
4
,
HC
Z
0
4
,
C

2
0
~
-,
and H+. (a) Draw
Le
w
is
structures
of
HC
Z
0
4
and
C
z
O~
-
.
(b) Which
of
the four species
can act only as acids, which can act only as bases, and which can
act as both acids and bases?
Section 16.2: The Acid-Base Properties
of
Water
Review
Questions

16.9 Write the equilibrium expression for the autoionization
of
wate
r.
16.10 Write an equation relating [H+] and
[OH-]
in solution at 25°C.
16.11
The
equilibrium constant for the autoionization
of
water
HzO(I) • • H+(aq) +
OH-(aq)
is
1.0 X 10-
14
at 25°C and 3.8 X
10-
14
at
40
°C. Is the forward
process endothermic
or
exothermic?
16.12 Define the term
amphoteric.
16.13 Compare the magnitudes
of

[H+] and
[OH-]
in aqueous solutions
that are acidic, basic, and neutral.
Section 16.3: The pH Scale
Review Questions
16.14 Define pH. Why do chemists normally choose to discuss the
acidity
of
a solution in terms
of
pH
rather than hydrogen ion
concentration [H+]?
674 CHAPTER 16 Acids
and
Bases
16.15
The
pH
of
a solution is 6.7.
From
this statement alone, can
you conclude that the solution is acidic?
If
not, what additional
information would you need?
Can
the

pH
of
a solution
be
zero
or
negative?
If
so, give examples to illustrate these values.
16.16 Define
pOH. Write the equation relating
pH
and pOH.
Problems
16.17 Calculate the concentration
of
OH-
ions in a 1.4 X
10-
3
M HCl
solution.
16.18 Calculate the concentration
of
H+ ions in a
0.62
M
NaOH
solution.
16.19 Calculate the

pH
of
each
of
the following solutions:
(a)
0.0010 M
HC1,
(b) 0.76 M KOH.
16.20
Calculate the
pH
of
each
of
the following solutions:
(a) 2.8
X
10-
4
M Ba(OH)z, (b) 5.2 X 10-
4
M
HN0
3
.
16.21 Calculate the hydrogen ion concentration in mol/L for solutions
with the following pH values: (a) 2.42, (b) 11.21, (c) 6.96,
(d)
15.00.

16.22 Calculate the hydrogen ion concentration in mol/L for each
of
the following solutions: (a) a solution
whose
pH
is 5.20, (b) a
solution whose
pH
is 16.00, (c) a solution whose hydroxide
concentration is 3.7
X 10-
9
M.
16.23 Complete the following table for a solution:
pH
[H
+] Solution is
<7
< 1.0 X 10 7 M
Neutral
16.24 Fill in the word
acidic, basic,
or
neutral for the following
solutions:
(a)
pOH
> 7; solution is
______
.

(b)
pOH
= 7; solution is
______
.
(c)
pOH
< 7; solution is
_______
.
16.25
The
pOH
of
a solution is 9.40 at 25°C. Calculate the hydrogen
ion concentration
of
the solution.
16.26 Calculate the
number
of
moles
of
KOH in 5.50
mL
of
a 0.360 M
KOH
solution.
What

is the
pOH
of
the solution at 25°
C?
16.27 How much
NaOH
(in grams) is needed to prepare 546
mL
of
solution with a
pH
of
10.00 at 25°C?
16.28 A solution is
made
by
dissolving 18.4 g
ofHCl
in enough water
to
make
662
mL
of
solution. Calculate the
pH
of
the solution at
25°C.

Section 16.4: Strong Acids
and
Bases
Review Questions
16.29 Without referring to the text, write the formulas
offour
strong
acids and four strong bases.
16.30
Which
of
the following statements are true regarding a 1.0 M
solution
of
a strong acid
HA
at 25°
C?
(Choose all that apply.)
(a) [A - ] > [H+]
(b)
The
pH
is 0.00.
(c) [H+] = 1.0 M
(d) [HA] = 1.0 M
16.31
Why
are ionizations
of

strong acids and strong bases generally
not treated
as
equilibria?
Problems
16.32 Calculate the
pH
of
an aqueous solution at 25°C that is
(a)
0.12
Min
HC1, (b) 2.4
Min
HN0
3
,
and (c) 3.2 X 10-
4
M
in
HCI0
4
.
16.33 Calculate the
pH
of
an aqueous solution at 25°C that is
(a)
1.02 M in HI, (b) 0.035 M in

HCI0
4
,
and (c) 1.5 X 10-
6
M
in HCl.
16.34 Calculate the concentration
of
HBr
in a solution at 25°C that has
a
pH
of
(a) 0.12, (b) 2.46, and (c) 6.27.
16.35 Calculate the concentration
of
HN0
3
in a solution at 25°C that
has a
pH
of
(a) 4.21, (b) 3.55, and (c) 0.98.
16.36 Calculate the
pOH
and
pH
of
the following aqueous solutions at

25°C: (a) 0.066 M KOH, (b) 5.43 M NaOH, (c) 0.74 M Ba(OH)z.
16.37 Calculate the
pOH
and
pH
of
the following aqueous solutions at
25°C: (a) 1.24 M LiOH, (b) 0.22 M Ba(OH)z, (c) 0.085 M NaOH.
16.38
An
aqueous solution
of
a strong base has a
pH
of
9.78 at 25°C.
Calculate the concentration
of
the base
if
the base is (a)
LiOH
and (b)
Ba(OHh-
16.39
An
aqueous solution
of
a strong base has a
pH

of
11.04 at 25°C.
Calculate the concentration
of
the base
if
the base is (a) KOH and
(b)
Ba(OH)z.
Section 16.
5:
Weak Acids
and
Acid Ionization Constants
Review Questions
16.40 Explain what is meant by the strength
of
an acid.
16.41
What
does the ionization constant tell us about the strength
of
an
acid?
16.42 List the factors on which the
Ka
of
a
weak
acid depends.

16.43 Why do we normally not quote
Ka
values for strong acids such as
HCl and
HN0
3
?
Why
is it necessary to specify temperature when
giving
Ka
values?
16.44 Which
of
the following solutions has the highest pH: (a) 0.40 M
HCOOH, (b) 0.40 M
HC10
4
,
(c) 0.40 M
CH
3
COOH?
16.45 Without referring to the text, write the formulas
of
four weak
acids.
Problems
16.46 Classify each
of

the following species as a weak
or
strong acid:
(a)
HN0
3
,
(b) HF, (c) H
2
S0
4
,
(d)
HS0
4
,
(e) H
2
C0
3
,
(f)
HC0
3
,
(g) HC1, (h) HCN, (i)
HN0
2
.
16.47 Classify each

of
the following species as a weak
or
strong base:
(a)
LiOH, (b)
CW,
(c) H
2
0,
(d)
CI0
4
,
(e)
NH
2
.
16.48 Which
of
the following statements are true for a 0.10 M solution
of
a
weak
acid HA? (Choose all that apply.)
(a)
The
pH is 1.00.
(b) [H+] »
[A

- ].
(c) [H+] =
[A
- l
(d)
The
pH
is less than
1.
16.49
The
Ka
for
benzoic
acid is 6.5 X
10-
5
.
Calculate
the
pH
of
a
0.10 M
aqueous
solution
of
benzoic
acid
at

25°C.
16.50
The
Ka
for
hydrofluoric acid is 7.1 X
10
-
4
.
Calculate
the
pH
of
a
0.15 M
aqueous
solution
of
hydrofluoric
acid
at
25
°C.
16.51
Calculate
the
pH
of
an

aqueous
solution
at
25°C
that
is
0.095
M
in
hydrocyanic
acid
(HCN). (Ka
for
hydrocyanic
acid =
4.9
X
10-
10
.)
16.52
Calculate
the
pH
of
an
aqueous
solution
at
25°C that

is
0.34
M
in
phenol
(C6H50H).
(Ka
for
phenol
=
1.3
X 10-
10
.)
16.53
Calculate
the
Ka
of
a
weak
acid
if
a 0.19 M
aqueous
solution
of
the
acid
has a

pH
of
4.52
at
25°C.
16.54
The
pH
of
an
aqueous
acid solution is
6.20
at
25°C.
Calculate
the
Ka
for
the acid.
The
initial
acid
concentration
is
0.010
M.
16.55
What
is the

original
molarity
of
a solution
of
formic acid
(HCOOH)
whose
pH
is
3.26
at 25°
C?
(Ka
for
formic acid =
1.7 X 10-
4
.)
16.56
16.57
What
is
the
original
molarity
of
a solution
of
a

weak
acid
who
se
Ka is
3.5
X
10
-
5
and
whose
pH
is
5.26
at
25°
C?
In
biological
and
medical
applications,
it
is often
necessary
to
study
the
autoionization

of
water
at
37°C
instead
of
25°C. Given
that
Kw
for
water
is 2.5 X
10
-
14
at
37
°
C,
calculate
the
pH
of
pure
water
at
this temperature.
Section 16.6:
Weak
Bases

and
Base
Ionization Constants
Review Questions
16.58
Compare
the
pH
values
for
0.10
M solutions
of
NaOH
and
of
NH3 to illustrate
the
difference
between
a s
trong
base
and
a
weak
base.
16.59
Which
of

the
following
has
a
higher
pH: (a) 1.0 M NH
3
,
(b)
0.20
M
NaOH
(Kb
for
NH3 = 1.8 X
1O
-
5
)?
Problems
16.60
Calculate
the
pH
for
each
of
the
following solutions
at

25°C:
(a)
0.10
M NH
3
,
(b)
0.050
M C5H5N (pyridine).
(Kb
for
pyridine
= 1.7 X
10
-
9
.)
16.61
The
pH
of
a
0.30
M solution
of
a
weak
ba
se is 10.66
at

25°C.
What
is the
Kb
of
the
base?
16.62
What
is the
original
molarity
of
an
aqueous
solution
of
ammonia
(NH
3
)
whose
pH
is 11.22
at
25°
C?
(Use
the
Kb

for
ammonia
provided
in
Problem
16.59.)
16.63
Calculate
the
pH
at
25°C
of
a 0.61 M
aqueous
solution
of
a
weak
base
B with a
Kb
of
1.5 X
10-
4
.
16.64
Determine
the

Kb
of
a
weak
base
if
a 0.19 M
aqueous
solution
of
the
base
at
25°C
has
a
pH
of
10.88.
16.65
The
following
diagrams
represent
aqueous
solutions
of
three
different
monoprotic

acids:
HA
,
HE,
and
HC.
(a
)
Which
conjugate
base
(A
-,
B - ,
or
C- ) has
the
smallest
Kb
value?
.
(b)
Which
anion
is
the
strongest
base?
The
water

molecule
s
have
been
omitted
for
clarity.
QUESTIONS
AND
PROBLEMS 675
=
HA,
HE,
or
HC
= A- ,
B-,
orC
-
HA
HE
HC
Section 16.7: Conjugate Acid-Base Pairs
Review Questions
16.66
Write
the
equation
relating
Ka

for
a
weak
acid
and
Kb
for
its
conjugate
base. Use NH3
and
its
conjugate
acid
NH
t to derive
the
relationship
between
Ka and K
b
.
16.67
From
the
relations
hip
Ka
Kb
=Kw,

what
can
you
deduce
about
the
relative strengths
of
a
weak
acid
and
its
conjugate
base?
Problems
16.68
The
following
diagram
s
represent
solutions
of
three s
alt
s
NaX
(X = A, B,
or

C). (a)
Which
X- has
the
weakest
conjugate
acid
?
(b)
Arrange
the
three X -
anion
s
in
order
of
decreasing
base
strength.
The
Na
+
ion
and
water
molecules
have
been
omitted

for
clarity.
= A- , B - ,
or
C- =
OH
-
__
=
HA,
HE,
or
HC
NaA
NaB
NaC
16.69
Calculate
Kb
for
each
of
the
following ions:
CN
- , F- ,
CH
3
COO
- ,

HCO
)"
.
Section 16.8: Diprotic and Polyprotic Acids
Review Questions
16.70
Write
all
the
s
pecie
s
(e
xcept
water)
that
are
present
in a
phosphoric
acid
solution.
Indicate
which
species
can
act
as a
Br!15nsted
acid

,
which
as a
Br!15nsted
base
, and
which
as both a
Br!15nsted
acid and a
Br!15n
s
ted
ba
se.
Problems
16.71 (1)
Which
of
the
following
diagrams
represents a solution
of
a
weak
diprotic
acid? (2)
Which
diagrams

represent
chemically
implausible
situations?
(The
hydrated
proton
is
shown
as a
hydronium
ion.
Water
molecules
are
omitted
for
clarity.)
(a)
(b)
(c)
(d)
16.72
The
first
and
second
ionization
constants
of

a diprotic acid H2A
are
K
a
,
and
Ka
2 at a certain temperature.
Under
what
conditions
will
[A
2
-]
= Ka?
,
676 CHAPTER
16
Acids
and
Bases
16.73 Compare the pH
of
a 0.040 M HCI solution with that
of
a
0.040 M H
2
S0

4
solution. (Hint: H
2
S0
4
is a strong acid;
Ka
for
HS0
4
= 1.3 X 10-
2
.)
16.74
What
are the concentrations
of
HS0
4
,
SO
~-,
and H+ in a 0.20 M
KHS0
4
solution? (Hint: H
2
S0
4
is a strong acid; Ka for

HS0
4
=
l.3
X
10
-
2
)
16.75 Calculate the concentrations
of
H+,
HC0
3
,
and
CO
~
-
in a
0.025 M H
2
C0
3
solution.
16.76 Calculate the
pH
at 25°C
of
a 0.25 M aqueous solution

of
phosphoric acid (H
3
P0
4
).
(Ka , K
a."
and Ka for phosphoric acid
are 7.5
X 10-
3
,6.25
X 10-
8
,
'and
-4
.8
X 10-
13
, respectively.)
16.77 Calculate the pH at
25
°C
of
a 0.25 M aqueous solution
of
oxalic
acid (H

2
C
2
0
4
).
(K
a and Ka for oxalic acid are 6.5 X 10-
2
_ , 2
and 6.1 X 10-
',
respectively.)
Section 16.9:
Molecular
Structure and Acid Strength
Review Questions
16
.78 List four factors that affect the strength
of
an acid.
16.79 How does the strength
of
an oxoacid depend on the
electronegativity and oxidation number
of
the central atom?
Problems
16.80 Predict the relative acid strengths
of

the following compounds:
H
2
0 , H
2
S, and H
2
Se.
16.81 Compare the strengths
of
the following pairs
of
acids: (a) H
2
S0
4
and
H2Se0
4,
(b) H
3
P0
4
and H
3
As0
4
·
16.82 Which
of

the following is the stronger acid:
CH
2
CICOOH or
CHCI
2
COOH? Explain your choice.
16.83 Consider the following compounds:
'> 0
H
CH
3
-O-H
Phenol
Methanol
Experimentally, phenol is found to
be
a stronger acid than methanoL
Explain this difference in terms
of
the structures
of
the conjugate
bases.
(Hint: A more stable conjugate base favors ionization. Only
one
of
the conjugate bases can be stabilized by resonance.)
Section 16.10: Acid-Base Properties
of

Salt Solutions
Review Questions
16.84 Define salt hydrolysis. Categorize salts according to how they
affect the pH
of
a solution.
16.85 Explain why small, highly charged metal ions are able to undergo
hydrolysis.
16.86
AlH is not a
Brs:;n
sted acid, but
Al
(
H20
)~+
is. Explain.
16.87 Specify which
of
the following salts will undergo hydrolysis: KF,
NaN0
3
,
NH
4
NO
z
,
MgS0
4

,
KCN, C
6
H
s
COONa, RbI,
Na
Z
C0
3
,
CaCI
2
, HCOOK.
Problems
16.88 Predict the pH
(>7,
<
7,
or = 7)
of
aqueous solutions containing
the following salts: (a) KEr,
(b) AlCN0
3)3,
(c) BaCI
2
,
(d) BiCN0
3h

16.89 Predict whether the following solutions are acidic, basic,
or
nearly neutral: (a) NaBr, (b) K
2
S0
3
,
(c)
~N02'
(d)
Cr(N0
3
h
16.90 A certain salt, MX (containing the M + and X- ions), is dissolved
in water, and the pH
of
the resulting solution is 7.0. What can you
say about the strengths
of
the acid and the base from which the
salt is derived?
16.91 In a certain experiment, a student finds that the pHs
of
0.10 M
solutions
of
three potassium salts KX,
KY,
and KZ are 7.0, 9.0,
and 11.0, respectively. Arrange the acids HX, HY, and

HZ
in
order
of
increasing acid strength.
16.92 Predict whether a solution containing the salt
K
2
HP0
4
will
be
acidic, neutral,
or
basic.
16.93 Predict the pH
(>7,
<7,
or = 7)
of
a
NaHC0
3
solution.
16.94 Calculate the
pH
of
a 0.36 M
CH
3

COONa
solution.
(Ka
for acetic
acid
= 1.8 X
lO
-
s
.)
16.95 Calculate the pH
of
a 0.42 M
NH
4
CI solution. (Kb for ammonia =
1.8 X lO-
s
.)
16.96 Calculate the pH
of
a 0.082 M
NaF
solution.
(Ka
for
HF
= 7.1 X
10-
4

.)
16.97 Calculate the
pH
of
a 0.91 M C2HsNH3I solution.
(Kb
for
C2HsNH2
= 5.6 X
10-
4
.)
Section 16.11: Acid-Base Properties
of
Oxides
and Hydroxides
Review Questions
16.98 Classify the following oxides as acidic, basic, amphoteric, or
neutral: (a)
CO
2
, (b) KzO, (c) CaO, (d) N
2
0
S
,
(e) CO, (f) NO,
(g)
Sn0
2,

(h)
S0
3,
(i)
Al
Z
0
3
, (j) BaO.
16.99 Write equations for the reactions between (a)
CO
2
and
NaOH(aq), (b)
Na20
and
HN0
3
(aq).
Problems
16.100 Explain why metal oxides tend to be basic
if
the oxidation
number
of
the metal is low and tend to be acidic
if
the oxidation
number
of

the metal is high. (Hint: Metallic compounds in which
the oxidation numbers
of
the metals are low are more ionic than
those in which the oxidation numbers
of
the metals are high.)
16.101 Arrange the oxides in each
of
the following groups in order
of
increasing basicity: (a) K
2
0,
Al
2
0
3
, BaO, (b) Cr03, cra, CrZ03'
16.102
Zn(OHh
is an amphoteric hydroxide. Write balanced ionic
equations to show its reaction with (a) HCI, (b)
NaOH
[the
product is
Zn(OH)~
-
].
16.103 AI(OH)3 is insoluble in water.

It
dissolves in concentrated NaOH
solution. Write a balanced ionic equation for this reaction.
What
type
of
reaction is this?
Section 16.12: Lewis Acids and
Bases
Review Questions
16.104 What are the Lewis definitions
of
an acid and a base? In what
way are they more general than the Brs:;nsted definitions?
16.105 In terms
of
orbitals and electron arrangements, what must be
present for a molecule or an ion to act as a Lewis acid (use H+
and
BF
3 as examples)?
What
must
be
present for a molecule or
ion to act as a Lewis base (use
OH
- and
NH3
as examples)?