Chemistry part 26, Julia Burdge,2e (2009) ppsx

630
CHAPTER
15
Chemical Equilibrium
15.89
When
dissolved in water, glucose (corn sugar) and fructose
(f
ruit
sugar) exist
in
equilibrium as follows:
fructose
••
=='
glucose
A chemist prepared a
0.244 M fructose solution at 25°C.
At
equilibrium, it was found that its concentration
had
decreased to
0.113 M. (a) Calculate the equilibrium constant for the reaction.
(b)
At
equilibrium, what percentage
of
fructose was converted to
glucose?
15.90
At

room
temperature, solid iodine is in equilibrium with its vapor
through sublimation and deposition.
De
scribe
ho
w you would
use radioactive iodine,
in
either solid
or
vapor form, to show that
there is a dynamic equilibrium between these two phases.
15.91
At
1024°C, the pressure
of
oxygen gas from the decomposition
of
copper(II) oxide (CuO) is 0.49 atm:
4CuO(s).
' 2CU20(S) + 0
2(
g)
(a) What is
Kp
for the reaction? (b) Calculate the fraction
of
CuO
that will decompose

if
0.16
mole
of
it is placed in a 2.0-L flask
at
1024°C. (c)
What
would the fraction
be
if
a l.O-mole
sa
mple
of
CuO
were used? (
d)
What
is the smallest amount
of
CuO
(
in
moles) that would establish the equilibrium?
15.92 A mixture containing 3.9 moles
of
NO and 0.88 mole
of
CO

2
was
allowed to react in a flask at a certain temperature according to
the equation
15.93
15.94
NO(g
) + COi g)
+.
=~,
N0
2
(g) + CO(g)
At
equilibrium, 0.11 mole
of
CO
2
was present. Calculate the
equilibrium constant
Kc
of
this reaction.
The
equilibrium constant Kc for the reaction
is 54.3 at
430
°C.
At
the start

of
the reaction, there are 0.714
mole
of
H
2>
0.984
mole
of
1
2>
and 0.886 mole
of
HI
in a 2.40-L reaction
chamber. Calculate the concentrations
of
the gases at equilibrium.
When
heated, a gaseous compound A
di
ssociates as follows:
A(g
)
:;:::.
=:::!:' B(g) + C(g)
In an experiment, A was heated at a certain temperature until its
equilibrium pressure reached
0.14p, where P is the total pressure.
Calculate the equilibrium constant

Kp
of
this reaction.
15.95
When
a gas was heated under atmospheric conditions, its color
deepened. Heating above
150°C caused the color to fade, and
at
550°C the color was barely detectable.
Howe
ver, at 550°C,
the color was partially restored by increasing the pressure
of
the
system. Which
of
the following best fits the preceding description:
. (a) a mixture
of
hydrogen and bromine, (b) pure bromine, (c)
a mixture
of
nitrogen dioxide and dinitrogen tetroxide. (Hint:
Bromine has a reddish color, and nitrogen diox
ide
is a brown gas.
The
other gases are colorless.) Justify your choice.
15.96 Both

Mg
2+ and
Ca
2
+ are important biological
ion
s.
One
of
their
functions is to bind to the phosphate group
of
ATP molecules
or
amino acids
of
proteins.
For
Group
2A
metals
in
general, the
equilibrium constant for binding to the anions increases in the
order
Ba
2+
<
Sr
2+

< Ca
2+
<
Mg
2+
.
What
property
of
the Group
2A
metal cations might account for this trend?
•
15.97
The
equilibrium constant Kc for the following reaction is
l.2
at
375°C.
15.98
(a) What is the value
of
Kp
for this reaction?
(b)
What is the value
ofthe
equilibrium constant Kc
for
2NH

3
(g) • ' N
2
(g) +
3Hig)?
(c) What is Kc for i N
2
(g) +
~Hig)
• '
NH
3
(g)?
(d) What are
the values
of
Kpfor the reactions described in parts (b) and (c)?
In this chapter we learned that a catalyst has no effect on the
position
of
an equilibrium because it speeds up
both
the forward
and reverse rates to the same extent. To test this statement,
consider a situation in which an equilibrium
of
the type
2A(g)
:;:::.
=:::!:' B(g)

is establis
hed
inside a cylinder fitted with a weightless piston.
The piston is attached by a string to the cover
of
a box containing
a catalyst. When the piston moves upward (expanding against
atmospheric pressure), the cover is lifted and the catalyst is
exposed to the gases. When the piston moves downward, the box is
closed. Assume that the catalyst speeds up the forward reaction
(2A
' B) but does not affect the reverse process
(B
' 2A).
Suppose the catalyst is suddenly exposed to the equilibrium system
as
shown here. Describe what would happen subsequently. How
does this
"thought experiment" convince you that no such catalyst
can exist?
2A
•
'B
·,
String
~
Catalyst
~
_
.

_ ;;
15.99 A sealed glass bulb contains a mixture
of
N0
2
and N
2
0
4
gases.
Describe
what
happens to the following properties
of
the
gases when the bulb is heated from
20°C to 40°
C:
(a) color,
(b) pressure, (c) average molar mass, (d) degree
of
dissociation
(from N
2
0
4
to
N0
2
),

(e) density. Assume that volume remains
constant.
(Hint:
N0
2
is a brown gas; N
2
0
4
is colorless.)
15.100 At
20
°C, the vapor pressure
of
water is 0.0231 atm. Calculate
Kp
and Kc for the process
15.101 Industrially, sodium metal is obtained
by
electrolyzing
molten sodium chloride. The reaction at the cathode is
Na +
+ e - , Na. We might expect that potassium metal
would also be prepared by electrolyzing molten potassium
chloride. However, potassium metal is soluble in molten
potassium chloride and therefore is hard to recover. Furthermore,
potassium vaporizes readily at the operating temperature,
creating hazardous conditions. Instead, potassium is prepared by
the distillation
of

molten potassium chloride in the presence
of
sodium vapor at 892°
C:
Na(g) + KCI(l)
:;:.
=='
NaCl(l) +
K(g)
In view
of
the fact that potassium is a stronger reducing
agent than sodium, explain why this approach works. (The
boiling points
of
sodium and potassium are 892°C and 770°C,
respectively
.)
15.102 In the gas phase, nitrogen dioxide is actually a mixture
of
nitrogen dioxide
(N0
2
)
and dinitrogen tetroxide (N
2
0
4
).
If

the density
of
such a mixture is 2.3 gIL at
74
°C and 1.3 atm,
calculate the partial pressures
of
the gases and
Kp
for the
dissociation
of
N
2
0
4
.
15.103 A 2.50-mole sample
of
NOCI was initially
in
a 1.50-L reaction
chamber
at 400°C. After equilibrium was established, it was
found that
28.0 percent
of
the NOCI had dissociated:
15.104
2NOCl(g) +:.

:::=:='
2NO(g)
+ Cl
2
(g)
Calculate the equilibrium constant
Kc for the reaction.
About 75 percent
of
hydrogen for industrial use is produced by
the
steam-reforming process. This process is carried
out
in two
stages called primary and secondary reforming.
In
the primary
stage, a mixture
of
steam and methane at about
30
atm is heated
over a nickel catalyst at
800°C to give hydrogen and carbon
monoxide:
LlH
o = 206
kllmol
The
secondary stage is can'ied

out
at about 1000°C, in the
presence
of
air, to convert the remaining methane to hydrogen:
(a)
What
conditions
of
temperature and pressure would favor
the formation
of
products in both the primary and secondary
stages? (b)
The
equilibrium constant Kc for the primary stage is
18 at
800°C. (i) Calculate
Kp
for the reaction. (ii)
If
the partial
pressures
of
methane and
steam
were both
15
atm at the start,
what

are the pressures
of
all the gases at equilibrium?
15.105 Photosynthesis can
be
represented by
6C0
2
(g)
+
6H
2
0(l)
+:.
=='
C6
H,
20 6
(S
) +
60
2
(g)
LlW
= 2801 kJ/mol
Explain how the equilibrium would be affected
by
the following
changes: (a) partial pressure
of

CO
2
is increased, (b) O
2
is
removed from the mixture, (c) C6
H'
20 6 (glucose) is removed
from the mixture, (d) more water is added, (e) a catalyst is added,
(f) temperature is decreased.
15.106 Consider the decomposition
of
ammonium
chloride at a certain
temperature:
Calculate the equilibrium constant
Kp
if
the total pressure is
2.2 atm at that temperature.
15.107 Water
is
a very
weak
electrolyte that undergoes the following
ionization (called
autoionization):
kJ +
H
2

0(l)
• L J ' H (aq) +
OH
- (aq)
(a)
Ifk,
= 2.4 X
lO-
s
s- '
and L , = 1.3 X
lO"IM'
s,calculatethe
equilibrium constant K where K = [H+
][OH
-
]/[H
2
0].
(b) Calculate
the product [H+][OH-], [H+], and
[OH- ]. (Hint: Calculate the
concentration
of
liquid water using its density, 1.0 g/mL.)
15.108 Consider the following reaction, which takes place in a single
elementary step:
2A
+ B •
LI

• A2B
If
the equilibrium constant Kc is 12.6 at a certain temperature and
if
k,
= 5.1 X
10-
2
S-I,
calculate the value
of
L,.
QUESTIONS
AND
PROBLEMS
631
•
15.109
At
25°C, the equilibrium partial pressures
of
N0
2
and N
2
0
4
are
0.15
atm

and 0.20 atm, respectively.
If
the volume is doubled at
constant temperature, calculate the partial pressures
of
the gases
when a new equilibrium is established.
15.110 In 1899 the German chemist Ludwig
Mond
developed a process
for purifying nickel by converting it to the volatile nickel
tetracarbonyl
[Ni(CO)4] (b.p. =
42.2
°C):
15.111
Ni(s) +
4CO(g).
'
Ni(COMg)
(a) Describe how you can separate nickel and its solid impuritie
s.
(b)
How
would you recover nickel?
[LlH
f?
for Ni(CO)4
is
-602.9

kllmo!.]
Consider the equilibrium reaction described in Problem 15.30. A
quantity
of
2.
50
g
of
PCIs is placed in an evacuated 0.500-L flask
and heated to
250°C. (a) Calculate the pressure
of
PCIs, assuming
it does not dissociate. (b) Calculate the partial pressure
of
PCls at
equilibrium. (c) What
is
the total pressure at equilibrium? (d) What
is the degree
of
dissociation
of
PCl
s
? (The degree
of
dissociation is
given by the fraction
of

PCls that has undergone dissociation.)
15.112 Consider the equilibrium system
3A
• ' B. Sketch the
changes
in
the concentrations
of
A and B over time for the
following situations: (a) initially only A is present, (b) initially
only B is present, (c) initially both A and B are present (with A in
higher concentration). In each case,
assume
that the concentration
of
B is
higher
than that
of
A at equilibrium.
15.113 The vapor pressure
of
mercury is 0.0020 mrnHg at 26°C. (a)
Calculate
Kc and Kp for the process
Hg(l).
' Hg(g). (b) A
chemist breaks a thermometer and spills mercury onto the floor
of
a laboratory measuring 6.1 m long, 5.3 m wide, and 3.1 m high.

Calculate the mass
of
mercury (in grams) vaporized at equilibrium
and the concentration
of
mercury vapor (in
mg/m\
Does this
concentration exceed the safety
limit
of
0.05 mg/m
3
? (Ignore the
volume
of
furniture and other objects
in
the laboratory.)
15.114
At
25°C, a mixture
of
N0
2
and N
2
0
4
gases are in equilibrium in

a cylinder fitted with a movable piston.
The
concentrations are
[N0
2
]
= 0.0475 M and [N
2
0
4
] = 0.487
M.
The volume
of
the
gas mixture is halved by pushing down on the piston at constant
temperature. Calculate the concentrations
of
the gases when
equilibrium is reestablished. Will the color become darker or
lighter after the change?
[Hint: Kc for the dissociation
of
N
2
0
4
is
4.63
X

10-
3
. N
2
0i
g) is colorless, and
N0
2
(g) has a brown color.]
15.115 A student placed a few ice cubes in a drinking glass with water.
A few minutes later she noticed that
some
of
the ice cubes were
fused together. Explain what happened.
15.116 Consider the potential energy diagrams for two types
of
reactions
A
• • B.
In
each
ca
se, answer the following questions for
the system at equilibrium. (a) How would a catalyst affect the
forward and reverse rates
of
the reaction? (b)
How
would a

catalyst affect the energies
of
the reactant and product? (c) How
would an increase in temperature affect the equilibrium constant?
(
d)
If
the only effect
of
a catalyst is to lower the activation
energies for the forward and rev
er
se reactions, show that the
equilibrium constant remains unchanged
if
a catalyst
is
added to
the reacting mixture.
>
>
bJ)
OJ)

\
'-
'"
A
'"
B

c:
c:
'"
'"
A
~ ~
co
co


~
B
~
J::
c:
'"
'"
~
~
0
0
0.,
0.,
Reaction progress
Reaction progress
632
CHAPTER
15
Chemical
Equilibrium

15.117
The
equilibrium constant
Ke
for the reaction
is
0.83 at 375°
C.
A 14.6-g sample
of
ammonia
is placed in a
4.00-L fla
sk
and heated to 375°
C.
Calculate the concentrations
of
all the gases when equilibrium is reached.
15.118
The
dependence
of
the equilibrium constant
of
a reaction on
temperature is given by the
van't
Hoff
equation:

-tll/
0
In K =
RT
+ C
where C is a constant.
The
following table gives the equilibrium
constant
(Kp)
for the reaction at various temperatures.
2NO(g)
+ 0 2(g)
+=.
===='
2N0
2
(g)
138
5.12 0.436 0.0626 0.0130
T(K) 600
700
800 900 1000
Determine graphically the
6H
o for the reaction.
15.119 Cons
id
er the reaction between
N0

2
and N
2
0
4
in a closed container:
Initially,
1 mole
of
N
2
0
4
is present. At equilibrium, x
mole
of
N
2
0
4
has dissociated to form
N0
2
.
(a) Derive an expression
for
Kp
in terms
of
x and

P,
the total pressure. (b) How does the
expression in part (a) help you predict the shift in equilibrium
due to an increase in
P?
Doe
s yo
ur
prediction agree with
Le
Chiitelier's principle?
15.120 (a) Use the
van't
Hoff
equation in Problem 15.] 18 to derive the
following expression, which relates the equilibrium constants at
two different temperatures:
In KI =
t: H
o
(J
_
J )
K2 R T2 TI
How
does this equation support the prediction
ba
sed on
Le
Chiitelier's principle about the shift in equilibrium with

temperature? (b)
The
vapor pressures
of
water are 31.82
mmHg
at
30
°C and 92.51
mmHg
at 50°
C.
Calculate the molar heat
of
vaporization
of
water.
15.121
The
Kp for the reaction
S0
2C
I2(g)
:;:.
=='
S0
2(g) + CI
2
(g)
is 2

.05
at 648 K. A sample
of
S0
2Cl2 is placed in a container
and heated to 648
K,
while the total pressure is kept constant at
9.00 atm. Calculate the partial pressures
of
the gases at equilibrium.
15.122
The
"boat"
form and the "chair" form
of
cyclohexane (C
6
H
I2
)
interconvert
as
shown here:
>
•
Boat
Chair
In this representation, the H atoms are omitted and a C atom
is assumed to be at each intersection

of
two lines (bonds).
The
convers i on is
fi
rst order in each direction.
The
activation
energy for the chair
> boat conversion is
41
kJ/mol.
If
the
frequency factor is
1.0 X 10
12
S- I, what is
kl
at 298 K?
The
equilibrium constant Ke for the reaction is 9.83 X 10
3
at 298 K.
15.123 Consider the following reaction at a certain temperature
15
.124
A2 + B2
:;:.
=='

2AB
The
mixing
of
1 mole
of
A2 with 3 moles
of
B2 gives rise to x
mole
of
AB at equilibrium. The addition
of
2
more
moles
of
A2
produces another x mole
of
AB.
What
is the equilibrium constant
for the reaction?
Iodine is sparingly soluble
in
water but much more so in carbon
tetrachloride (CCI
4
) .

The
equilibrium constant, also called the
partition coefficient, for the distribution
of
12 between these two
phases
is 83 at 20°C. (a) A student adds 0.030 L
of
CC1
4
to 0.200 L
of
an aqueous solution containing 0.032 g
of
1
2
,
The
mixture at
20
°C is shaken, and the two phases are then allowed to separate.
Calculate the fraction
of
12 remaining
in
the aqueous phase.
(b)
The
student now repeats the extraction OfI2 with another
0.0

30
L
of
CCI
4
.
Calculate the fraction
of
the
12
from the original
solution that remains
in
the aqueous phase. (c)
Compare
the
result
in
part (b) with a single extraction using 0.060 L
of
CCI
4
.
Comment
on the difference.
PRE-PROFESSIONAL PRACTICE
EXAM
PROBLEMS:
PHYSICAL
AND

BIOLOGICAL SCIENCES
Lime
(CaO) is used to prevent
S0
2 from escaping from the s
moke
stacks
of
coal-burning power plants via the formation
of
solid
CaS04
.
2H
2
0
(gyp-
sum).
One
of
the important reactions in the overall process is the decom-
position
of
CaC0
3
:
CaC0
3
(s)
:;:.

==>
CaO(s) +
CO
2
(g)
which has an equilibrium constant (K
e)
of
3.0 X 10-
6
at
ns
o
c.
1.
Calculate the value
of
Kp
for the decomposition
of
CaC0
3
at 725°
C.
a) 3.0 X 10-
6
c) 2.0 X
10-
2
b) 2.5 X

10
-
4
d) 3.7 X
10-
8
2.
If
a 12.0-g sample
of
solid
CaC0
3
is placed in an evacuated vessel
at
ns
o
c,
what will the pressure
of
CO
2
be
when the system reaches
equilibrium?
3.
4.
it) 3.0 X 10-
6
atm

b) 3.3
X 10
5
atm
c) 1.7
X 10-
3
atm
d) 1.0 atrn
Which
of
the following actions will cause an increase in the pressure
of
CO
2
in
the vessel described in Question 2?
a)
Addition
of
He
gas
b) Addition
of
S02
gas
c) Addition
of
more
CaC0

3
solid
d) Increasing the volume
of
the vessel
If
a 12.0-g sample
of
solid
CaC0
3
is placed in a vessel at
nsoc
in
which the pressure
of
CO
2
is 2.5 atm,
wha
t mass
of
CaO
will form?
a)
6.n
g c) 6.00 g
b)
12.0 g d) None
ANSWERS

TO
IN-CHAPTER MATERIALS
63
3
ANSWERS TO IN-CHAPTER MATERIALS
lS.2B (a) Hz + Cl
z
• ' 2HCl, (b) H+ + F-
:;::.
==:'
HF
,
(c)
Cr
H
+
40H
- . ' Cr(OH
);
, (d) HC10 • ' H+ + ClO- ,
(e) H
Z
S0
3
•
' H+ +
HS0
3
,
(f)

2NO
+
Br
z . ' 2NOBr.
[HClt
I
lS.3A (a) Kc = , (b) Kc = ,
[SiCl
4
] [H2f [Hg2+]
[C1
- f
[
N'(CO)
]
[Z
2+] [NH+
][OW]
(c)K
c=
1
4,(d)K
c = n
.lS.3B
(a) Kc = 4 ,
[Cot [Fe
2
+] [
NH
3]

[
Aa
(N
H
3 )
~ ][Cl-]
(b) Kc = [Ag+][Cl- ], (c) Kc = [Ba
2
+
][F
- f , (d) Kc = b -2
[NH
3
]
•
lS.4A (a) 2.3 X 10
24
, (b) 6.6 X 10-
13
,
(c) 2.8 X
10-
15
lS.4B (a) 1.5 X
34 29 38 (P
eal
10 , (b) 1.3 X 10 , (c) 8.5 X 10 . lS.SA (a) Kp = ,
(Peoi (P
o
)

2 2
(P
NH
)
(b)
Kp
= P
eo
,
(c
) Kp = 3' lS.SB (a)
2NO
z
+ O
2
• '
2N
0
3
,
2 (P
N
)(P
H
)
2 2
(b)
CH
4
+ 2H

z
O.
' COZ + 4H
z
, (c) I
z
+ Hz • ' 2HI.
lS.6A
8.1
X
10-
6
lS.6B 0.104. 15.7 A Left. lS.7B Right. lS.8A
[H
z] =
[I
z
]
= 0.056 M; [HI] = 0.413 M. lS.8B [Hz] = [I
z
]
= 0.043 M; [HI] =
0.314 M. 15.9 [Br] = 8.8 X
10-
3
, [Brz] = 6.5 X 10-
2
lS.10A [PH] =
2
[PI]

= 0.37 atm,
PHI
= 2.75 atm. lS.10B
PH
= PI = 0.294 atm, P
HI
=
222
2.16
atm.1S.11
A (a) Right, (b) left, (c) right, (d)
left.1S.12A
(a) Right,
(b) right, (c) right.
lS.12B Right shift: remove HF; left shift: remove Hz;
no shift: reduce volume
of
container.
Answers
to
Checkpoints
15.2.1
e.
15.2.2 a. 15.3.1
d.
15.3.2
a.
15.3.3
c.
15.3.4

c.
15.4.1
d.
15.4.2 b.
15.5.1 a, c. 15.5.2 b, d,
e.
15.5.3 a, d,
e.
15.5.4
b.
Answers
to
Applying
What
You've
learned
a) [OD17X] . b) 3.05 X 10-
5
.
c)
[ODl7X-A77]
30.5.
[ODlA]
[OF2A]
[ODlA][OF
2
A][A77IAj'
d)
Qc = 3
.6

X 10-
5
;
therefore, the mixture is not at equilibrium. It
will have to proceed to the left (rev
er
se reaction) in order to achieve
equilibrium. e)
At
equilibrium,
[ODIA]
= 0.05 M, [OF2A] = 0.05 M,
[ODl
7X] = 0.95
M.
s
an
ases
16.1
Bf0nsted Acids and Bases
16.2
The
Acid-Base Properties
of
Water
16.3
The
pH
Scale
16.4

Strong Acids
and
Bases
•
Strong Acids
•
Strong Bases
16.5
Weak Acids
and
Acid
Ionization
Constants
•
The Ionization Constant,
Ka
•
Calculating
pH
from
Ka
•
Using
pH
to Determine
Ka
16.6
Weak Bases
and
Base

Ionization
Constants
·
•
The Ionization Constant,
Kb
•
Calculating
pH
from
Kb
•
Using
pH
to Determine
Kb
16.7
Conjugate Acid-Base Pairs
•
The Strength
of
a Conjugate
Acid
or Base
•
The Relationship Between
Ka
and
Kb
of

a Conjugate
Acid-Base Pair
16.8
Diprotic
and
Polyprotic
Acids
16.9
Molecular
Structure
and
Acid Strength
•
Hydrohalic
Acids
•
Oxoacids
•
Carboxylic
Acids
16.10 Acid-Base Properties
of
Salt Solutions
•
Basic Salt Solutions
•
Acidic Salt Solutions
•
Neutral Salt Solutions
•

Salts in
Which
Both the
Cation
and
the
Anion
Hydrolyze
16.11
Acid-Base Properties
of
Oxides
and
Hydroxides
•
Oxides
of
Metals
and
of
Nonmetals
•
Basic
and
Amphoteric
Hydroxides
16.12
Lewis Acids
and
Bases

An
Acid
Essential
to
Good
Health
Scurvy is a devastating disease caused by a deficiency
of
vitamin
C.
It
ha
s a host
of
terrible symptoms including extreme fatigue and weakness, hemorrhaging, and lo
ss
of
teeth due to inflamed gums. Untreated,
it
results in death, usually from pneumonia
or
other acute
infection-or
from heart failure. Most mammals make their own vitamin C,
but humans and other primates, as well as fruit bats and guinea pigs, must obtain
it
from
their diet. Since ancient times, people were at
risk
for contracting scurvy when their

diets did not include an adequate supply
of
fresh fruits and vegetables. One segment
of
the population historically deprived
of
an appropriate diet was sailors. During the
fourteenth and fifteenth centuries, as the development
of
ships and sails made voyages
on the open ocean possible, and returns to shore became less frequent, scurvy became
. . . .

. . . . . - - .
commonplace among sailors.
Although reports
of
cures for scurvy date back to the sixteenth century, it was not until
1747 that the effectiveness
of
citrus was proven with the experiments
of
Scottish naval
surgeon James Lind.
It
was another 40 years, though, before the British navy began
supplying citrus juice to its crews.
It
was the compulsory iss
ue

of
the juice
of
lemons or
limes that gave rise to the term
lime
y,
referring to a British sailor.
Vitamin C,
or
ascorbic acid (H
2
C
6
H
6
0
6
),
acts as an antioxidant, reacting with oxidizing
species such as the
·OH
radical, which can damage an organism's DNA.
It
is one
of
many acids that are important to human health.
•
Ascorbic acid
•

One
hundred of
Portuguese
explorer
Vasco
de
Gama
's
160
-
member
crew
died
from
scurvy
by
the
time
they
had
sailed
around
the
sou
thern tip
of
Africa
in
1497.
In

This Chapter, You Will Learn more about the properties
of
acids and bases and how those properties are
related to molecular structure.
Before you begin, you should review
•
The
list
of
strong acids and the list
of
strong bases
[~
.
Section
4.3,
Table
4.4]
• How to solve equilibrium problems
[
~
.
Section
15.4]
Acids
such as citric acid
and
ascorbic acid (v
itamin
C)

are
responsible for the
sour
tast
e
of
citrus fruits.
Media Player/
MPEG
Content
Chapter
in
Review
635
636 CHAPTER 16 Acids
and
Bases
Conjugate
Species
Base
CH
3
COOH
CH
3
COO
-
H
2
O

OH
-
NH3
NH2"
H
2
SO
4
HS0
4
Conjugate
Species
Acid
NH
3
NH
t
H
2
O
H3
0
+
OH-
H
2
O
H
2
NCONH

2
H
2
NCON
H!
(urea)
Think
About
It
A species does
not need to be what we think
of
as an acid
in
order for it to have a
conjugate bas
e.
For example,
we
would not
ref
er to the hydroxide
ion
(OH
- ) as an acid- but it
does have a conjugate base, the
oxide ion
(0
2
- ) .

Furthermore, a
species that can either lose or gain
a proton, such as
HC0
.1
, has b
ot
h
a conjugate base
(
CO
~-)
and a
conjugate acid (H
2
C0
3
).
Br¢nsted Acids and Bases
In
Chapter 4 we learned that a
Br
¢nsted acid is a substance that can donate a proton and a Br¢nsted
ba
se is a substance that can accept a proton
[
~~
Section
4.3]. In this chapter we extend our
di

s-
cussion
of
Br¢nsted acid-base theory to include conjugate acids and conjugate bases.
When
a Br¢nsted acid donates a proton, what remains
of
the acid is known as a conjugate
base.
For example, in the ionization
of
HCI in water,
HCI(aq) + H
2
0(l)
+=.
~.
H30 +(aq) +
Cl
- (aq)
acid con
jug
ate base
HCI donates a proton to water, producing the hydronium ion
CH30
+) and the chloride ion CCI-), which
is the conjugate base
of
HCl.
The

two species, HCI and Cl- , are known as a conjugate acid-base pair
or simply a conjugate pair. Table 16.1
li
s
ts
the conjugate bases
of
several familiar species.
Conversely, when a
Br¢n
sted base accepts a proton, the newly formed protonated species is
known as a
conjugate acid.
When
ammonia (NH3) ionizes in water,
NH
3(aq) + H
2
0(l)
+=.
~.
NH
t
Caq)
+
OH
- (aq)
base con
jug
ate acid

NH
3 accepts a proton from water to become the
ammonium
ion
(NH)t
.
The
ammonium
ion is the
conjugate acid
of
ammonia. Table 16.2 lists the conjugate acids
of
several common species.
Any reaction that we describe using
Br¢nsted acid-base theory involves an acid and a base.
The
acid donates the proton, and the base accepts it. Furthermore, the products
of
such a reaction
are always a conjugate base and a conjugate acid.
It
is useful to identify and label
each
species
in a
Br¢nsted acid-base reaction.
For
the ionization
of

HCl
in water, the species are labeled as
follows:
loses a proton
1
-
,
gains a proton
HCI(aq)
+
H2
0
(l)
•
H3
0
+(aq)
+
CI-(aq)
•
acid
base
r.

;.:.,
~d

~
conjugate
• 1

ell!
base
And for the ionization
of
NH
3 in water,
gains a proton
loses a proton
NH3(aq)
+
H
2
°(l)
NHt(aq)
•
+
OH-(aq)
•
base
acid
•
•
~
. r

,
conjugate
Ck
base
Sample Problems 16.1 and 16.2 let you practice identifying conjugate pairs and the species

in a
Br¢nsted acid-base reaction.
"
Sample
Problem
16.1
"."""
What
is
(a) the conju
ga
te base
of
H
N0
3
,
(b) the conjugate acid
of
0
2
- ,
(c) the conjugate base
of
HS0
4
,
and (d) the conjugate acid
of
HC0

.1
?
Strategy
To
find the conjugate base
of
a specie
s,
remove a proton from the formula. To find the
conjugate acid
of
a species, add a proton to the formula.
Setup
The word proton, in this context, refers to H+. Thus, the formula and the charge will both be
affected by the addition or removal
of
H+.
Solution
(a) NO
.1
(b)
OW
(c)
SO
~

(d) H
2
C0
3

SECTION
16.2 The Acid-Base Properties
of
Water
637
Practice Problem A
What
is (a) the conjugate acid
of
Cl0
4
,
(b) the conjugate acid
of
S2
- , (c) the
conjugate base
of
H
2
S, and (d) the conjugate base
of
H
2
C
2
0
4
?
Practice Problem B

HS0
3
is
the conjugate acid
of
what
species?
HS0
3
is the conjugate base
of
what species?
Label each
of
the species
in
the following equations as an acid,
base
, conjugate base,
or
conjugate
acid:
(a) HF(aq) +
NH
3
(aq).
• F- (aq) +
NH
r (aq)
(b)

CH
3
COO- (aq)
+ H
2
0(l).
•
CH
3
COOH(aq)
+
OW(aq)
Strategy
In each equation, the reactant that loses a proton is the acid and the reactant that gains a
proton is the base. Each product is the conjugate
of
one
of
the reactants. Two species that differ only
by a proton constitute a conjugate pair . .
Setup (a)
HF
loses a proton and becomes F- ;
NH
3 gains a proton and becomes
NH
r .
(b)
CH
3

COO
- gains a proton to become
CH
3
COOH; H
2
0 loses a proton to become
OH
Solution
(a) HF(aq) +
NH
3
(aq)
::0:.
=~.
F-(aq)
+
Nm(aq)
acid base conjugate conjugate
base acid
(b)
CH
3
COO- (aq) + H
2
0(l)
::o:.=~'
CH
3
COOH(aq) +

OH-(aq)
base acid conjugate acid conjugate
ba
se
Practice Problem A Identify and label the species in each reaction.
(a)
NHr(aq)
+ H
2
0(l).
•
NH
3(aq) + H30 +(aq)
(b)
CN
- (aq) + H
2
0(l).
• HCN(aq) +
OW(aq)
Practice Problem B (a) Write an equation in which
HS0
4
reacts (with water) to form its conjugate
base. (b) Write an equation in which
HS0
4
reacts (with water) to form its conjugate acid.
Checkpoint 16.1 Bnmsted Acids and Bases
16

.1.1
Which
of
the following pairs
of
species
are conjugate pairs? (Select all that
apply.)
a)
H
2
S and S2-
b)
NH2 and
NH
3
c)
O
2
and H
2
0
2
d)
HBr
and
Br
-
e)
HCI and

OH-
16.1.2 Which
of
the following species does
not have a conjugate base? (Select all
that apply.)
a)
HC
2
0
4
b)
OW
c) 0
2
-
d)
CO
~-
e) HCIO
The Acid-Base Properties
of
Water
Water is often referred to as the "universal solvent," because it is so common and so important to
life on Earth. In addition, most
of
the acid-base chemistry that you will encounter takes place in
aqueous solution. In this section, we take a closer look at water's ability to act as either a Br¢nsted
acid (as in the ionization
of

NH
3
)
or a Brlilnsted base (as in the ionization
of
Hel).
A species that
can behave either as a Br¢nsted acid or a Br¢nsted base is called
amphoteric.
Water is a very weak electrolyte, but it does undergo ionization to a small extent:
Thi
nk
About
It
In
a Br¢nsted
acid-base reaction, there
is
always
an acid
and
a base, and whether
a substance behaves as an acid
or
a base depends on what it is
combined with. Water, for example,
behaves as a base when combined
with
HCI but behaves as an acid
when combined with

NH
3
.
638
CHAPTER
16 Acids and
Bases
." t.
~

~.
Mu
ltimedia
Chemical
Equilibrium
equilibrium
(interactive).
Recall
that
in
a
heterogeneous
equilibr
iu
m
such
as
this,
liquids
and

solids
do
not
appear
in
the
equilibrium
express
i
on
[
~~
Section
15.3]
.
The
co
nst
an
t Kw is
so
metimes
referr
ed
to
as
the
ion-product constant.
Re
call

that we
disregard
the units when we
s
ub
stitute
conc
entrations into
an
equilib
riu
m
expression
[
~~
Section
15.5].
Bec
ause
the product of
H30+
and
OW
concentrations
is
a
constant,
we
cannot alter
the

conc
en
trations
independently.
Any
change
in o
ne
a
lso
affects
the
oth
er
.
Think
About
It
Remember
that equilibri
urn
constants are
temperature dependent. The value
of
Kw
is
1.0 X
10
-
14

only
at
25
°
C.
This
reaction is
known
as the autoionization o/water.
Because
we
can represent
the
aqueous pro-
ton as either H+
or
H30 +
[
~~
Section
4.3]
,
we
can also write the auto ionization
of
water as
H30 +(aq) +
OH-(aq)
+
•

acid
base
.
•
conjugate
acid
+
+
conjugate
base
where
one
water
molecule
acts as
an
acid and the other acts as a
base
.

As
·lildkatecn:'
y"
ih·e ·do
ubie
·arrov,i"iu· the· e·
ci.uatlOii
: the·
i-e
·actlon IS a

il
equiiibii"um:
The
equilib-
rium
expression for the autoionization
of
water is
or
Because
the autoionization
of
water is an
important
equilibrium
that
you
will
encounter
frequently
in
the study
of
acids and
base
s,
we
use the subscript w to indicate that the equilibrium constant is
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ,
that specifically for the autoionization

of
water.
It
is
important
to realize, though, that
Kw
is simply
a
Kc for a specific reaction. We will frequently replace the c in Kc expressions
with
a letter
or
a
series
of
letters
to
indicate the specific type
of
reaction to
which
the Kc refers.
For
example, Kc for
the ionization
of
a
weak
acid is called Ka,

and
Kc for the ionization
of
a
weak
base
is called K
b
.
In
Chapter
17
we
will
encounter
Ks
p,
where
"s
p" stands for "solubility product."
Each
specially
subscripted
K is simply a Kc for a specific type
of
reaction.
In
pure
water, autoionization is the only source
ofH

30 + and
OH-,
and the stoichiometry
of
the reaction tells us that their concentrations are equal.
At
25
°C, the concentrations
of
hydronium
and
hydroxide
ion
s in
pure
water
are [H30 +] =
[OH
- ] = 1.0 X
10
-
7
M.
Using
the equilibrium
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
express
ion
,
we

can
calculate the value
of
Kw at
25
°C as follows:
Kw = [H
3
0 +
][OH-]
= (1.0 X
10
-
7
)(1.0 X
10-
7
)
= 1.0 X
10
-
14
FUltherrnore, in any aqueous solution at
25
°
C,
the
product
of
H30 +

and
OH
- concentrations is
equal
to 1.0 X
10
-
14
.
Equation
16.1
Although
their
product
is a
constant
, the individual concentrations
of
hydronium
and
hydroxide
· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . +
can
be
influenced
by
the addition
of
an acid
or

a base.
The
relative amounts
of
H30 and
OH
-
determine
whether
a solution is neutral, acidic,
or
basic.
•
When
[H30 +] =
[OH
- ], the solution is neutral.
•
When
[H30 +] >
[OH
- ], the solution is acidic.
•
When
[H30 +] <
[OH
- ], the solution is basic.
Sample
Problem
16.3 shows how to use Equation 16.1.

Sample Problem 16.3 •
The concentration of hydronium ions in stomach acid
is
0.
10
M. Calculate the concentration of
hydroxide ions in stomach acid
at
25
°
C.
Strategy Use the value of Kw
to
determine
[OH
- ] when
[H
30 +] = 0.10
M.
Setup Kw =
[H
3
0 +
][OH
- ] =
1.0
X
10
-
14

at
25
°
C.
Rearranging Equation
16.1
to
solve for
[OH
- ],
[OH
- ] = 1.0 X
10
-
14
[H
3
0
+]
Solution
[OH
- ] = 1.0 X
10
-
14
= 1.0 X
10
-
13
M

0.10
Practice Problem A The concentration of hydroxide ions
in
the antacid milk of magnesia is 5.0 X
10
-
4
M. Calculate the concentration of hydronium ions
at
25
°
C.
Practice Problem B The value of Kw
at
normal body temperature
(37
°
C)
is
2.8
X
10
-
14
. Calculate
th
e concentration of hydroxide ions
in
stomach acid at body temperature. (
[H

30 +] = 0.10
M)
Checkpoint
16.2
The Acid-Base Properties
of
Water
16.2.1
Calculate
[OH-]
in a solution in which
[H30 +]
= 0.0012 M at 25°C.
a)
l.2
X
10-
3
M
b)
8.3 X 10-
17
M
c)
l.0
X 10-
14
M
d)
8.3 X 10-

12
M
e)
1.2XlO
II
M
The pH Scale
16.2.2 Calculate [H30 +] in a solution in
which
[OH-]
= 0.25 M at 25°C.
a)
4.0 X
10-
14
M
b)
1.0 X 10-
14
M
c)
2.5 X 10
13
M
d)
1.0 X 10-
7
M
e)
4.0 X

10-
7
M
The
acidity
of
an aqueous solution depends on the concentration
of
hydronium ions, [H30 +
].
This
concentration can range over many orders
of
magnitude, which can make reporting the numbers
cumbersome. To describe the acidity
of
a solution, rather than report the molar concentration
of
hydronium ions, we typically use the more convenient
pH
scale.
The
pH
of
a solution is defined as
SECTION 16.3 The pH Scale 639
the negative
base-l
0 logarithm '
of

the' hydi-oni'
um
'
{on
'coricenii-iti'on '
(In
'
motA.)".
··········

.
Equation
16
.2
converts
numbers
that
can
span
an
enormous
range
(-10
'
to
10-
14
)
to
numbers

generally
ranging
from
-1
to 14.
or
Equation 16.2
The
pH
of
a solution is a dimensionless quantity, so the units
of
concentration must
be
removed
from
[H30 +] before taking the logarithm. Because [H30 +] = [OH- ] = 1.0 X
10
-
7
M in pure
water at
25°C, the
pH
of
pure
water at 25°C is
-log
(1.0 X
10

-
7
)
= 7.00 "

.


A word about significant figu
res:
When
we
take
the
log
of a number with two
sign
ificant
figures,
we
report
the
result
to two
places
past
the
decimal
point.
Thus,

pH
7.00
has
two signifi
cant
fi
gures,
not
three.
Remember, too, that a solution in which [H30 +] = [OH- ] is neutral. At 25°C, therefore, a neutral
solution has
pH
7.00.
An
acidic solution, one in which [H30 +] >
[OH-]
,
ha
s
pH
< 7.00, whereas
a basic solution, in which
[H30 +] <
[OH-],
ha
s
pH
> 7.00. Table 16.3 shows the calculation
of
pH

for solutions ranging from 0.10 M to 1.0 X
10-
14
M.
In
the laboratory,
pH
is measured with a
pH
meter (Figure 16.1). Table 16.4 lists the
pH
val-
ues
of
a
number
of
common
fluids. Note that the
pH
of
body fluids varies greatly, depending
on
the
location and function
of
the fluid.
The
low
pH

(high acidity)
of
gastric
juices
is vital for digestion
of
food, whereas the higher
pH
of
blood is required to facilitate the transport
of
oxygen.
[H30+](M)
-log
[H30+]

0.10 - log (1.0 X
10-
1
)
0.010
-log
(1.0 X
10-
2
)
1.0 X 10 3
-log
(1.0 X
10-

3
)
1.0
X 10 4
-log
(1.0 X
10-
4
)
1.0 X
10
-
5
;;;''''I====~
l
=
o
=
g
=
(l
=
.0
=
X
=
I
=
0
~

5 )
1.0 X
10-
6
-log
(1.0 X 10-
6
)
1.0 X
10-
7
- log (1.0 X 10-
7
)
1.0 X
10-
8
-log
(1.0 X
10
-
8
)
1.0 X 10-
9
-log
(1.0 X
10-
9
)

1.0 X
10-
10
-log
(1.0 X 10-
1
°)
1.0 X 10-
11
-log
(1.0 X
10-
11
)
1.0 X
10-
12
-log
(1.0 X
10-
12
)
1.0 X
10-
13
- log (1.0 X
10-
13
)
1.0 X

10
-
14
-log
(1.0 X
10-
14
)
pH
1.00
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
Acidic
Neutral
Basic
Figure 16.1
ApHmeteris
commonly used in the laboratory
to determine the

pH
of
a solution.
Although many pH meters have a ran
ge
of
1 to 14,
pH
values can actually be
less than 1 and greater than 14.
640
CHAPTER
16
Acids
and
Bases
10'
is
the
inve
rse
function of
log.
(it
is
usually
the
second
function
on

the
same
key.)
You
must
be
comfortable performing
these
operations
on
your
calculator.
Fluid
pH
Fluid
pH
Stomach acid
l.0
Saliva
6.4-6.9
Lemon juice
2.0
Milk
6.5
Vinegar
3.0
Pure
water 7.0
Grapefruit
juice

3.2
Blood
7.35
-7.45
Orange
juice
3.5
Tears
7.4
Urine
4.8 - 7.5 Milk
of
magnesia
10.6
Rainwater (in clean air) 5.5
Household ammonia
11.5
A measured
pH
can be used to determine experimentally the concentration
of
hydronium
. . . . . . . . . . . . . . . . . . . . . . . . . . +
ion in solution. Solving Equation 16.2 for [H30 ] gives
Equation 16.3
Sample
Problems 16.4 and 16.5 illustrate calculations involving pH.
Sample Problem 16.4
Determine the
pH

of
a solution at 25°C in which the hydronium ion concentration is (a) 3.5 X
10
-
4
M, (b) 1.7 X
10
-
7
M,
and (c) 8.8 X
10-
11
M.
Strategy
Given [HP +], use Equation 16.2 to solve for pH.
Setup
(a) pH =
-log
(3
.5 X
10-
4
)
(b) pH = - log (1.7 X
10-
7
)
(c)
pH

=
-l
og
(8
.8 X
10-
11
)
Solution
(a)
pH
= 3.46
(b) pH = 6.77
(c) pH = 10.06
Think
About
It
When a hydronium ion concentration falls between two
"be
nchmark" concentrations in
Table 16.3, the pH falls between the two corresponding pH values. In part (c), for example, the hydronium
ion concentration
(8.8 X 10-
11
M)
is greater than 1.0 X
10
-
11
Mbut

less than 1.0 X 10-
10
M.
Therefore,
we expect the pH to be between
11.00 and 10.00.
1.
0 X 10-
10
8.8 X 10-
11
*
1.0 X
10
-
11
-lo
g (1.0 X 10-
10
)
- log
(8
.8 X 10-
11
)
-lo
g (1.0 X
10
-
11

)
*
[H
30 +] between t
wo
benchmark
va
lu
es
IpH between two benchmark values
pH
10.00
1O.06
t
11.00
Recog
ni
zing the benchmark concentrations and corresponding pH values is a good way to detennine
whether or not your calculated result is reasonable.
Practice Problem A Determine the pH
of
a so
lu
tion at 25°C in which the hydronium ion
co
ncentration is (a) 3.2 X 10-
9
M, (b) 4.0 X 10-
8
M, and (c) 5.6 X 10-

2
M.
Practice Problem B Detennine the pH
of
a solution at 25°C in which the hydronium ion
concentration is (a) 1.2
M, (b) 3.0 X
10
-
11
M, and (c) 8.6 X 10-
12
M.
Sample Problem 16.5
Calculate the h
yd
ronium ion concentration in a solution at 25°C in which the
pH
is (a) 4.76,
(b) 11.95, and (c)
8.01.
Strategy
Given pH, use Equation 16.3 to calculate [H30 +].
Setup
(a) [H3
0
+] = 10-
476
\
SECTION

16.3 The
pH
Scale 64 1
(
b)
[H3
0
+] = 10-
11
95
(c) [H3
0
+] =
10
-
80
1
Solution
(
a)
[H
30 +] = 1.7 X
10-
5
lvf
(
b)
[H
30 +] =
1.1

X
10-
12
lvf
(
c)
[H
30 +] =
9.8
X 10-
9
lvf
Think
About
It
If
you use the calculated hydronium ion
concentrations to recalculate pH, you will get numbers slightly different
from those given in the problem. In part (a), for example,
-log
(1.7 X
10
-
5
)
= 4.77.
The
small difference between this and 4.76 (the
pH
given in the problem)

is
due to a rounding error. Remember that a
concentration derived from a
pH
with two digits to the right
of
the
decimal point can have only two significant figures. Note also that
the benchmarks can be used equally well in this circumstance. A pH
between 4 and 5 corresponds to a hydronium ion concentration between
1
X
10
-
4
lvf and 1 X
10
-
5
lvf.
Practice Problem A Calculate the hydronium ion concentration in a solution at 25°C in which the
pH is (a) 9.90, (b) 1.45, and
(c
) 7.01.
Practice Problem B Calculate the hydronium ion concentration in a solution at 25°C
in
which the
pH
is (a) 2.11, (b) 11.59, and (c) 6.87.
ApOH

scale analogous to the
pH
scale can be defined using the negative base-lO logarithm
of
the hydroxide ion concentration
of
a solution,
[OH-].
pOH
= - log
[OH-]
Rearranging Equation 16.4 to solve for hydroxide ion concentration gives
Now consider again the
Kw equilibrium expression for water at 25°
C:
[H
3
0 +][OH- ] =
l.0
X 10-
14
Taking the negative logarithm
of
both sides,
we
obtain
Equation 16.4
Equation 16.5
-log
([H

3
0 +
][OH-])
= - log (
l.0
X 10-
14
)
-(log
[H30 +] + log [OH- ]) = 14.00
- log [H30 +] - log [OH- ] = 14.00
(-log
[H30 +]) +
(-log
[OH- ]) = 14.00
And from the definitions
of
pH
and pOH, we see that at 25°C
pH
+
pOH
= 14.00
Equation 16.6
Equation 16.6 provides another way to express the relationship between the hydronium ion con-
centration and the hydroxide ion concentration.
On the
pOH
scale, 7.00 is neutral, numbers greater
than

7.00 indicate that a solution is acidic, and numbers less than 7.00 indicate that a solution is
ba
sic. Table 16.5 lists
pOH
values for a range
of
hydroxide ion concentrations at
25
°
C.
[OH-]
(M)
0.10
1
X 10-
3
1 X 10-
5
1 X
10-
7
1 X 10 9
1 X
1O
-
1l
1 X 10-
13
pOH
l.00

3.00
5.00
7.00
9.00
1l.00
13.00
Basic
Neutral
Acidic
642
CHAPTER
16
Acids
and
Bases
Think
About
It
Remember
that
the
pOH
sca
le is, in essence, the
reverse
of
the
pH
scale.
On

the
pOH
sca
le,
numbers
below 7
indicate a bas
ic
solution, whereas
number
s above 7 indicate an acidic
so
lution.
The
pOH
benchmark
s
(abbreviated in Table 16.5)
work
the sa
me
way
the
pH
benchmark
s
do. In part (a), for example, a
hydroxide ion concentration
between 1 X
10-

4
M
and
1 X
10
-
5
M
conesponds
to a
pOH
between 4 and 5:
[OH
- ] (M)
pOH
1.0 X 10-
4
4.00
3.7 X 10-
5
*
4.43t
1.0 X
10
-
5
5.00
*[OH - ] between two benchmark values
tp
OH

between two benchmark values
Think
About
It
Use
the
benchmark
pOH
values to
determine
whether
these solutions
are rea
so
nable.
In
part (a), for
example
, the
pOH
between 4
and
5
corresponds to
[OH
- ] between 1 X
10
-
4
M

and
1 X
10-
5
M.
Sample Problems 16.6 and
16.
7 illustrate calculations involving pOH.
Determine
the
pOH
of
a solution at 25°C in which the
hydroxide
ion concentration is (a) 3.7 X
10
-
5
M,
(b) 4.1 X 10-
7
M, and (c) 8.3 X
10-
2
M.
Strategy
Given [OH-
],
use Equation 16.4 to calculate pOH.
Setup

(a)
pOH
=
-log
(3.7 X
10
-
5
)
(b)
pOH
=
-log
(4. 1 X
10
-
7
)
(c)
pOH
=
-log
(8.3 X
10
-
2
)
Solution
(a)
pOH

= 4.43
(b)
pOH
= 6.39
(c)
pOH
= 1.08
Practice Problem A
Determine
the
pOH
of
a
so
lution at 25°C in which the hydroxide ion
co
ncentration is (a) 5.7 X
10
-
12
M,
(b) 7.3 X
10
-
3
M,
and (c) 8.5 X 10-
6
M.
Practice Problem B

Determine
the
pOH
of
a solution at 25°C in which the hydroxide ion

_
co
_n_c_e_
nt
_ra_t_io_n_i_S_
ca
_)_2_.8
__
X
__
10
__
8_M
__
,C_b_)_9_.9
__
X_l_0
__
9
_M_,_a_n_d_
C
_
C)
_1_._0_X

__
l_0_
1
_
1
_M_.
____________
~I
Sample
Problem 16.7
Calculate the hydroxide ion concentration
in
a solution at 25°C
in
which the
pOH
is (a) 4.91,
Cb) 9.03, and (c) 10.55.
Strategy
Given
pOH
, use Equation 16.5 to calculate
[OH
- ].
Setup
(a)
[OH-]
=
10
-

491
(b) [OH- ] =
10
-
903
(c)
[OH-]
=
10
-
1055
Solution
(a)
[OW]
= 1.2 X
10
-
5
M
(b)
[OH-]
= 9.3 X
10
-
10
M
(c)
[OH
- ] = 2.8 X
10-

11
M
Practice Problem A Calculate the hydroxide ion concentration in a solution
at
25°C
in
which the
pOH
is
Ca)
13.02, (b) 5.14, and (c) 6.98.
Practice Problem B Calculate the hydroxide ion concentration
in
a solution
at
2YC
in which the
pOH
is Ca) 11.26, (b) 3.69, and (c) 1.60.
Bringing Chemistry
to
Life
Antacids and the
pH
Balance
in
Your Stomach
An average adult produces between 2 and 3 L of gastric juice daily. Gastric juice is
an
acidic

digestive fluid secreted
by
glands in the mucous membrane that lines the stomach.
It
contains
hydrochloric acid (Hel), among other substances. The pH
of
gastric juice is about 1.5, which
corresponds to a hydrochloric acid concentration
of
0.
03
M a concentration strong enough to
dissolve zinc metal!
The
inside lining
of
the stomach is made up
of
parietal cells, which are fused together to
form tight junctions.
The
interiors
of
the cells are protected from the surroundings by cell
mem
-
branes.
These
membranes allow water and neutral molecules to pass in and out

of
the stomach, but
they usually block the movement
of
ions such as H+,
Na
+, K+, and CI- .
The
H+ ions
come
from
carbonic acid (H
2
C0
3
) formed as a result
of
the hydration
of
CO
2
,
an end product
of
metabolism:
CO?(g) + H
2
0(l)
:;::.
===='

H
2
C0
3
(aq)
H
2
C0
3
(aq).
• H+(aq) +
HC0
3
(aq)
These reactions take place in the blood plasma bathing the cells in the mucosa.
By
a process
known as
active transport, H+ ions move across the membrane into the stomach interior. (Active
transport processes are aided by enzymes.) To maintain electrical balance, an equal number
of
CI-
ions also move from the blood
plasma
into the stomach. Once in the stomach, most
of
these ions
are prevented by cell membranes from diffusing
back
into the blood plasma.

The
purpose
of
t
he
highly acidic medium within the stomach is to digest food and to activate
certain digestive enzymes. Eating stimulates H+ ion secretion. A small fraction
of
these ions nor-
mally are reabsorbed by the mucosa, causing a number
of
tiny hemorrhages. About
half
a million
cells are shed by the lining every minute, and a healthy stomach is completely relined every few
days. However,
if
the acid content is excessively high, the constant influx
of
H+ ions through the
membrane back to the blood plasma can cause muscle contraction, pain, swelling, inflammation,
and bleeding.
One
way to temporarily reduce the H+ ions concentration in the stomach is to take an ant-
acid.
The
major function
of
antacids is to neutralize excess HCI in gastric juice.
The

table below
lists the active ingredients
of
some
popular antacids.
The
reactions by which these antacids neu-
tralize stomach acid are as follows:
NaHC0
3
(aq) + HCI(aq) -
-+.
NaCI(aq) + H
2
0(l)
+
CO
2
(g)
CaC0
3
(aq) + 2HCI(aq) • CaCI
2
(aq) + H
2
0(l)
+ CO
2
(g)
MgC0

3
(aq) + 2HCI(aq) • MgCI
2
(aq) + H
2
0(l)
+ CO
2
(g)
Mg(OH)z(s) + 2HCl(aq) • MgCI
2
(aq) +
2H
2
0 (l)
Al(OH)2NaC0
3(s) + 4HCl(aq) •
AlCl
3
(aq) + NaCl(aq) +
3H
2
0(l)
+ CO
2
(g)
Active Ingredients in Some Common Antacids
Commercial Name
Alka-Seltzer
Milk

of
magnesia
Rolaids
TUMS
Maalox
Active Ingredients
Aspirin, sodium bicarbonate, citric acid
Magnesium hydroxide
Dihydroxyaluminum
sodium carbonate
Calcium carbonate
Sodium
bicarbonate, magnesium carbonate
The
CO
2
released by most
of
these reactions increases gas pressure in the stomach, causing the
person to belch.
The
fizzing that takes place when Alka-Seltzer dissolves in water is caused by
carbon dioxide, which is released
by
the reaction between citric acid and
sodium
bicarbonate:
This effervescence helps to disperse the ingredients and enhances the palatability
of
the solution.

Checkpoint 16.3
The
pH
Scale
16.3.1
Determine the
pH
of
a solution at
25
°C
in which [H+] = 6.35 X 10-
8
M.
a)
7.65
b) 6.80
c) 7.20
d) 6.35
e) 8.00
16.3.2
Determine [H+] in a solution at 25°C
if
pH = 5.75.
a)
1.8 X
10
-
6
M

b)
5.6 X
10
-
9
M
c)
5.8 X 10-
6
M
d)
2.4 X
10
-
9
M
e)
1.0 X
10
-
6
M
SECTION 16.3 The pH Scale 643
644
CHAPTER
16 Aci
ds
and
Bases
16

.
3.3
Determine the pOH
of
a solution
at
25
°C in which [OH- ] = 4.65 X
10
-
3
M.
a)
11.67
b) 13.68
c)
0.32
d)
4.65
e)
2.33
Strong Acids and Bases
16
.3.4 Determine [OH- ]
in
a solution at 25°C
if
pH = 10.50.
a)
3.2 X

10-
11
M
b)
3.2
X
10-
4
M
c)
1.1
X 10-
2
M
d)
7.1
X
10-
8
M
e)
8.5 X 10-
7
M
Most
of
this
chapter
and
Chapter

17
deal
with
equilibrium
and
the
application
of
the
principles
of
equilibrium
to a variety
of
reaction
type
s.
In
the
context
of
our
discu
ssion
of
acids
and
ba
ses,
how-

ever,
it
is
necessary
to review
the
ionization
of
strong
acids
and
the
dissociation
of
strong bases.
· . . . . . . . . . . . . . . .
We
indicate that ionization of a
strong
acid
is
complete
by
using
a
single
arrow ( • )
instead
of the
double,

equilibri
um
arrow
(.
• )
in
the equation.
[
,
Mul
ti
media
Acids
and
Bases
- the
dissociat
i
on
of strong
and
weak
acids
(interactive).
These
reactions
generally
are
not
treated as equilibria

but
rather
as
processes
that
go
to completion.
This
makes
the
determination
of
pH
for
a solution
of
strong
acid
or
strong
base
relatively simple.
Strong Acids
There
are
many
different
acids,
but
as

we
learned
in
Chapter
4,
relatively
few
qualify
as
strong.
Strong
Acid
Hydrochloric
acid
Hydrobrorruc
acid
Hydroiodic
acid
Nitric
acid
Chloric
acid
Perchloric
acid
Ionization
Reaction
HCI(aq) + H
2
0(l)
• H30 +(aq) + CI (aq)

HBr(aq)
+ Hz
0(l)
• H
3
0 +(aq) +
Br-(aq)
HI(aq) + Hz
0(l)
• H
3
0 +(aq) + I - (aq)
HN0
3
(aq)
+ H
2
0(!)
• H
3
0 +(aq) +
N0
3
(aq)
HCI0
3
(aq) + H
2
0(l)
• H

3
0 +(aq) +
CI0
3
(aq)
HClOiaq)
+ H
2
0(l)
• H
3
0 + (aq) +
CI0
4
(aq)
• • • • • • • • • • • • •
H
2
S0
4
(aq) + H
2
0(!)
• H
3
0+(aq)
+
HS0
4
(aq)

Remember
that although sulfuric
acid
has
two
io
nizable
protons,
only
the fi
rst
ion
ization
is
complete
.
Sulfuric
acid
It
is a
good
idea
to comrrtit this
short
list
of
s
trong
acids
to memory.

Because
the
ionization
of
a
strong
acid
is
complete,
the
concentration
of
hydronium
ion
at
equilibrium
is
equal
to
the
s
tarting
concentration
of
the
strong
acid.
For
instance,
if

we
prepare
a
0.10 M solution
of
HCl
,
the
concentration
of
hydronium
ion
in
the
solution
is 0.10 M. All the
HCI
ionizes,
and
no
HCl
molecules
remain.
Thus,
at
equilibrium
(when
the
ionization
is

complete),
[HCI] = 0 M
and
[H
30 +] =
[Cl-]
= 0.10
M.
Therefore,
the
pH
of
the
solution
(at
25
°C)
is
pH
= -
log
(0.10) = 1.00
· . . . . . . . . , . . .

. . . . . . . . . . .




.

As
we
will
see
in
Section
16.5
, a solution of
equal
concentration but containing a
weak
acid
has
a hi
gher
pH.
This
is a very
low
pH,
which
is
consistent
with
a relatively
concentrated
solution
of
a
strong

acid.
Sample
Problems
16.8
and
16.9
let
you
practice
relating
the
concentration
of
a
strong
acid
to
the
pH
of
an
aqueous
solution.
Think
About
It
Again, note that when a hydronium ion concentration
fall
s
between two of the benchmark concentrations in Table

16.3,
the pH
falls
between
the two corresponding pH
val
ue
s. In part (b), for example, the hydronium
ion
concentration of
1.2
X
10
-
4
M
is
greater
than
1.0
X
10
-
4
M and
le
ss
than
1.0
X

10
-
3
M.
Therefore,
we
expect the pH
to
be
between 4.00
and
3.00.
1.0
X 10-
3
1.2 X
10-
4
*
1.0 X
10-
4
- log (1.0 X
10
-
3
)
-log
(1.2 X
10-

3
)
- log (1.0 X
10
-
4
)
*
[H
30 +] between two benchmark
va
lu
es
I
pH
between two benchmark values
pH
3.00
3.92t
4.00
Being comfortable with the benchmark hydronium ion concentrations and
the corresponding pH values will help you avoid some of the common errors
in pH calculations.
Calculate the pH of
an
aqueous solution at
25
°C that
is
(a) 0.035 M

in Hl,
(b)
1.2 X
10
-
4
Min
HN0
3
,
and (c) 6.7 X 10-
5
Min
HCl0
4
.
Strategy
Hl,
HN0
3
,
and
HCl0
4
are
all
strong acids, so the
concentration
of
hydronium ion

in
each solution is the same
as
the
stated concentration
of
the acid. Use Equation 16.2 to calculate
pH.
Setup
(a) [H30 +] = 0.035 M
(b)
[H
30 +] = 1.2 X 10-
4
M
(c)
[H
30 +] = 6.7 X 10-
5
M
Solution
(a) pH =
-log
(0.035) = 1.46
(b)
pH
=
-log
(1.2 X 10-
4

)
= 3.92
(c) pH
= - log (6.7 X 10-
5
)
= 4.17


•
SECTION
16.4
Strong Acids and
Bases
645
Practice Problem A Calculate the pH
of
an aqueous solution at 25°C that is (a) 0.081 M in HI,
(b) 8.2 X 10-
6
M in
HN0
3
, and (c) 5.4
X 10-
3
Min
HCI0
4
.

Practice Problem B Calculate the pH
of
an aqueous solution at 25°C that is (
a)
0.011
Min
HN0
3
,
(b) 3.5 X 10-
3
MinHBr,
and (c) 9.3 X
10-
10
Min
HCI.
Calculate the concentration
of
HCI in a solution at
25
°C that has pH (
a)
4.95, (b) 3.45, and (c) 2.78.
Strategy
Use Equation 16.3
to
convert from
pH
to

the molar concentration
of
hydronium ion.
In
a
strong acid solution, the molar concentration
of
hydronium ion is equal to the acid concentration.
Setup (a) [HCl] =
[HP
+] = 10-
49
5
(b) [HCI] =
[H3
0
+]
=
10-
345
(c) [HCI] =
[H3
0
+]
=
10-
278
Solution (a)
1.1
X 10-

5
M
(b) 3.5 X
10-
4
M
•
,

·
•
•
•
•
•
•
•
•
•
(c)1.7Xlo
-
3
M
··········································

.
!
•
I
I

,
Practice Problem A Calculate the concentration
of
HN0
3
in a solution at 25°C that has
pH
(a) 2.06,
(b) 1.77, and (c) 6.01.
Practice Problem B Calculate the concentration
of
HBr in a solution at 25°C that has
pH
(
a)
4.81 ,
(b) 1.82, and (c) 3.04.
I~ ~

•
Strong
Bases
The list
of
strong bases is also fairly short.
It
consists
of
the hydroxides
of

alkali metals (Group
1A) and the hydroxides
of
the heaviest alkaline earth metals (Group 2A). The dissociation
of
a
strong base is, for practical purposes, complete. Equations representing dissociations
of
the strong
bases are as follows:
Group
lA
hydroxides
LiOH(aq)
•
Li
+(aq) +
OH
(aq)
NaOH(aq)
•
Na
+(aq) +
OH-(aq)
KOH(aq)
• K +(aq) +
OH
- (aq)
RbOH(aq)
• Rb+(aq) +

OH-(aq)
CsOH(aq) •
Cs
+(aq) +
OH-(aq)
. . . . . , . . . . . . . . .

.
Group 2A hydroxides
Ca(OHh(aq)
•
Ca
2+
(aq) +
20H
(aq)
Sr(OHh(aq)
• Sr
2+
(aq) +
20H
- (aq)
Ba(OHhCaq) •
Ba
2+
(aq) +
20H
- (aq)
Again, because the reaction goes to completion, the pH
of

such a solution is relatively easy to cal-
culate. In the case
of
a Group 1A hydroxide, the hydroxide ion concentration is simply the starting
concentration
of
the strong base. In a solution that is 0.018 M in NaOH, for example,
[OH-]
=
0.018
M.
Its pH can be calculated in two ways. We can either use Equation 16.1 to determine
hydronium ion concentration,
[H
3
0 +
][OH-]
=
l.0
X
10-
14
[H 0 +] =
l.0
X 10-
14
=
l.0
X
10

-
14
= 5.56 X 10-
13
M
3 [OH ] 0.018
and then Equation 16.2 to determine pH:
pH
=
-log
(5.56 X 10-
13
M)
= 12.25
When
we take the
inverse
log
of a number with
two digits to the right of the
dec
i
mal
point, the
re
sult
has
two
sign
ificant fi

gures.
Th
i
nk
About It
As
pH
decreases,
acid concentration increases.
• •
oS
~
F -
•
. .
~
Multimedia
Acids
and
Bases
- ionization of a strong
base
and
weak
base
(interadive).
Reca
ll
that
Ca

(
OH
h
and
Sr
(
OHh
are
not
very
soluble
, but wh
at
does
dissolve
di
ssociates
completely
[
~
Section
4.3,
Table
4.4]
646 CHAPTER
16
Acids
and
Bases
Think

About
It
These are basic
pOH
values, which is what we
should expect for the solutions
described in the problem. Note
that while the solutions in parts
(a) and (b) have the same base
concentration, they do not have
the same hydroxide concentration
and therefore do
not
ha
ve the same
pOH.
or
we can calculate the
pOH
with Equation 16.3,
pOH
= - log (0.018) = 1.75
and use Equation 16.6 to convert to pH:
pH
+
pOH
= 14.00
pH = 14.00 - 1.75 = 12.25
Both methods give the same result.
In

the case
of
a Group
2A
metal hydroxide, we must be careful to account for the reaction
stoichiometry. For instance,
if
we prepare a solution that is 1.9 X 10-
4
M in barium hydroxide,
the concentration
of
hydroxide ion at equilibrium (after complete dissociation) is 2(1.9 X
10-
4
M)
or 3.8 X 10-
4
M twice the original concentration
of
Ba(OH)2' Once we have determined the
hydroxide ion concentration,
we
can determine pH as before:
[H
0 +] = 1.0 X
10
-
14
=

3 [OH ]
1.0 X 10-
14
= 2.63 X 10-
11
M
3.8 X 10-
4
and
or
pH
= - log (2.63 X 10-
11
M)
= 10.58
pOH
=
-log
(3
.8
X 10-
4
)
= 3.42
pH
+
pOH
= 14.00
pH = 14.00 - 3.42 = 10.58
Sample Problems 16.10 and 16.11 illustrate calculations involving hydroxide ion concentra-

tion,
pOH, and pH.
.
- -
_.
-
Sample Problem 16.10
Calculate the
pOH
of
the following aqueous solutions at 25°C: (a) 0.013 M LiOH,
(b)
0.013 M
Ba
(OH)2, (c) 9.2 X 10-
5
M KOH.
Strategy
L
iOH
,
Ba
(
OH
h,
and KOH are a
ll
strong bases. Use reaction stoichiometry to determine
hydroxide ion concentration and Equation 16.4 to determine
pOH.

Setup
(a)
The
hydroxide ion concentration is simply equal to the concentration
of
the base.
Therefore, [OH- ]
= [LiOH] = 0.013
M.
(b)
The
hydroxide ion concentration is twice that
of
the base:
Ba(OHhCaq)
-_.
Ba
2
+(aq) +
20H
- (aq)
Therefore,
[OW]
= 2 X [Ba(
OH
)2]
= 2(0.013
M)
= 0.026
M.

(c) The hydroxide ion concentration is equal to the concentration
of
the base. Therefore,
[OW]
= [KOH] = 9.2 X 10-
5
M.
Solution
(a) pOH = - log (0.013) = 1.89
(b)
pOH = - log (0.026) = 1.59
(c)
pOH
=
-lo
g (9.2 X
10-
5
)
= 4.04
Practice Problem A Calculate the pOH
of
the following aqueous solutions at 25°C:
(a) 0.15 M NaOH, (b) 8.4 X 10-
3
M RbOH, (c) 1.7 X 10-
5
M CsOH.
Practice Problem B Calculate the pOH
of

the following aqueous solutions at 25°C:
(a) 9
.5
X 10-
8
MNaOH,
(b) 6.1 X 10-
2
MLiOH,
(c) 6.1 X 10-
2
MBa(OHh.
Sample Problem 16.11
An aqueous solution
of
a strong base has
pH
8.15 at
25
°C. Calculate the original concentration
of
base in the solution (a)
if
the base is NaOH and (b)
if
the base is
Ba
(OH)2'
SECTION
16.5

Weak
A
ci
ds and Acid Ioniza
tion
Constants 647
Strategy
Use Equation 16.6 to convert from pH to pOH and Equation 16.5 to determine the
hydroxide ion concentration.
Consider the stoichiometry
of
dissociation in each case to determine the
concentration
of
the base itself.
Setup
pOH
= 14.00 - 8.15 = 5.85
(a)
The
dissociation
of
I mole
of
NaOH
produces 1 mole
of
OH
- . Therefore, the concentration
of

the
base is
equal to the concentration
of
hydroxide ion.
(b)
The
dissociation
of
1 mole
of
Ba
(
OH
h produces 2 moles
of
OH
- . Therefore, the concentration
of
the base is only one-half the concentration
of
hydroxide ion.
Solution
. . . . . . . . . . . . . . . . . . . . . . . .
[OH- ] = 10-
585
= 1.41 X 10-
6
M
(a)

[NaOH] =
[OW
] = 1.4 X 10-
6
M
(b)
[Ba(OHh] =
HOW]
= 7.1 X 10-
7
M
,
Practice Problem A
An
aqueous solution
of
a strong base has pH 8.98 at 25°
C.
Calculate the
concentration
of
base in the solution (a)
if
the base is LiOH and (b)
if
the base
is
Ba
(OH)2'
Practice Problem B An aqueous solution

of
a strong base has
pH
12.24 at 25°
C.
Calculate the
concentration
of
base in the solution (a)
if
the base is NaOH and (b)
if
the base is Ba(OH)2'
Checkpoint 16.4 Strong Acids and Bases
16.4.1
16.4.2
16.4.3
Calculate the
pH
of
a 0.075 M solution
of
perchloric acid (HCl0
4
)
at 25°
C.
a)
12.88
b) 7.75

c)
6.25
d)
1.12
e)
7.00
What
is
the concentration
of
HBr
in a
solution with
pH
5.89 at 25°
C?
a) 7.8 X 10-
9
M
b) 1.3 X 10-
6
M
c) 5.9 X 10-
14
M
d)
8.1
X 10-
7
M

e) 1.0 X
10-
7
M
What
is the pOH
of
a solution at
25
°C
that is 1.3 X
10-
3
Min
Ba(OHh ?
a)
2.89
b) 2.59
c)
3.19
d)
11.11
e) 11.41
16.4.4
16.4.5
16.4.6
What
is the concentration
of
KOH in a

solution at
25°C that has pOH 3.31?
a)
2.0 X
lO
-
II
M
b)
3.3
X
10
-
1
M
c)
3.3
X
10
-
7
M
d)
4.5
X 10-
4
M
e)
4.9
X 10-

4
M
What is the pH
of
a solution at 25°C
that is 0.0095
Min
LiOH?
a)
11.68
b)
2.02
c)
11.98
d)
1.72
e)
12.28
What
is the concentration
of
Ca(OH)2
in a solution at
25
°C
if
the
pH
is 9.01?
a)

1.0 X 10-
5
M
b)
5.1
X 10-
6
M
c)
2.0 X 10-
5
M
d)
9.8
X
10
-
9
M
e)
4.9
X
10
-
9
M
Weak Acids and Acid Ionization Constants
Most
acids
are

weak acids, w
hi
ch
ionize
only
to
a
limited
extent
in
water.
At
eq
u
ilibrium,
an
aq
u
eous
sol
u
tion
of
a
weak
acid
contains
a
mixture
of

aqueous
acid
mo
l
ecules,
hydronium
ions,
and
the
corresponding
conj
u
gate
base.
The
degree
to
which
a
weak
acid
ionizes
depends
on
the
concentration
of
the
acid
and

the
equilibrium constant
for
the
ionization
.
•
• •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

•
•
Re
mem
ber
to k
eep
an
add
it
io
n
al
sig
nifi
ca
nt
fi
gur
e or
tw
o until
the
e
nd
of t
he
pr
obl
em

- to
avoi
d
ro
un
d
i
ngerror
[
H~
Section
1.5].
Think
About
It Alternatively, we
could determine the hydroxide ion
concentration using Equation
16
.3,
[H
3
0
+] = 10-
81 5
= 7.1 X 10-
9
M
and Equation 16.1,
[OH-]
= 1.0 X 10-

14
7.1 X 10-
9
M
= 1.4 X 10-
6
M
Once [OH- ] is known, the solution
is the same as shown previousl
y.
648 CHAPTER 16 Acids
and
Bases
Remembe
r that H30+
and
H+
are
us
ed
in
te
r
cha
ng
eably.
Remember
that a Ka
is
a K

,.
The
sub
s
cr
ipt a
si
m
pl
y sta
nd
s for "
ac
id.
"
For
co
mpar
iso
n, the pH of a
0.1
M soluti
on
of
a
strong
aci
d s
uc
h

as
Hel or HN0
3
is
1
.0.
,
. .
The Ionization Constant,
Ka
Con
sider a weak monoprotic acid HA. Its ionization in water is represented by
or by
RA
(aq)
:;:::.
~.
H+(aq) + A
-(a
q)
The
equilibrium expression for this reaction is
[H
~
b
+
][
A-
]
K =

a
[RA]
or
. . . . . . . . .
where Ka is the equilibrium constant for the reaction. More specifically, Ka is called the acid
ionization constant.
Although all weak acids ionize less than 100 percent, they vary in strength.
Th
e magnitude
of
Ka indicates how strong a weak acid is. A large Ka value indicates a stronger
acid, whereas a small
Ka value indicates a weaker acid. For example, acetic acid (CH
3
COOH) and
hydrofluoric acid
(
HF
) are both weak acid
s,
but
HF
is the stronger acid
of
the two, as evidenced
by its larger
Ka value. Solutions of equal concentration
of
the two acids do not have the same pH.
The

pH
of
the
HF
solution is lower.
Solution
(at
25°C)
0.10MHF
0.10 M
CH
3
COOH
7.1 X 10-
4
1.
8 X
10-
5
pH
2.09
2.87
Table 16.6 lists a number
of
weak acids and their Ka values at 25°C in order
of
decreasing acid
strength.
Calculating
pH

from
Ka
Using the method outlined in Chapter 15, we can use acid concentration and Ka to determine the
equilibrium concentration
of
H30 +.
Fr
om [H30 +],
we
can determine pH. Suppose we want to
determine the
pH
of
a 0.50 M
HF
solution at 25°
C.
The
ionization
of
HF
is represented by
The
equilibrium expression for this reaction is
[
H
O+
][F
-]
K = 3 = 7 1 X 10-

4
a [HF] .
Name
of
Acid
Formula
Structure
Hydrofluoric acid
HF
H-F
Nitrous acid
HN0
2
O=N
-
O-H
0
II
Formic acid
HCOOH
H-C-O-H
0
/
"
I
Benzoic acid
C
6
H
5

COOH
C 0
\
,'
/,
o
I I
Acetic acid
CH
3
-C-O
H
Hydrocyanic acid
H-C
= N
Phenol
> 0
H
H
Ka
7.1 X 10-
4
4.5 X 10-
4
l.7
X 10-
4
6.5 X 10-
5
l.8

X 10-
5
4.9 X 10-
10
1.3 X 10-
10
SECTION 16.5
Weak
Acids
and
Acid
Ionization
Constants
649
;
We
construct an equilibrium table and enter the starting concentrations
of
all species in the equi-
librium expression.
HF(aq) + H
2
0(l)
:;:::.
===:!:'
H30 +(aq) + F- (aq)
Initial concentration (M): 0.50 o o
Change in concentration (M):
Equilibrium concentration (M):
,-

Using the reaction stoichiometry, we determine the changes in all species:
HF(aq) + H
2
0(l)
:;:::.
===:!:'
H30 +(aq) + F- (aq)
Initial concentration (M): 0.50
o o
Change in concentration (M):
-x
+x +x
Equilibrium concentration (M):
Finally, we express the equilibrium c0ncentration
of
each species in terms
of
x.
HF(aq) + H
2
0
(l
)
:;:::.
===:!:'
H30 +(aq) + F- (aq)
Initial concentration (M):
0.50 0 0
Change in concentration (M): - x
+x

+x
Equilibrium concentration (M):
0.50 - x
x x
These equilibrium concentrations are then entered into the equilibrium expression to give
Ka = (x) (x) = 7
.1
X
10-
4
0.50 - x
Rearranging this expression, we get
x
2
+ 7.1 X
1O
-
4
x - 3.55 X 10-
4
= 0
This is a quadratic equation, which we can solve using the quadratic formula given in Appendix
1.
In the case of a weak acid, however, often we can use a shortcut to simplify the calculation.
Because HF is
a weak acid, and weak acids ionize only
to
a slight extent, x must be small com-
pared to
0.50. Therefore, we can make the following approximation:

0.50 - x = 0.50
Now the equilibrium expression becomes
Rearranging, we get
x
1
x
2
-, ,
= = 7
.1
X 10-
4
0.50 - x 0.50
x
2
= (0.50)(7
.1
X 10-
4
)
= 3.55 X 10-
4
x =
~
3.55
X 10-
4
= 1.9 X 10-
2
M

Thus, we have solved for x without having to use the quadratic equation. At equilibrium we have
and the pH
of
the solution is
[HF]
= (0.50 - 0.019) M = 0.48 M
[H
30 +] = 0.019 M
[F- ]
= 0.019 M
pH =
-log
(0.019) = 1.72
This shortcut gives a good approximation as long
as
the magnitude
of
x is significantly smaller
than the initial acid concentration. As a rule, it is acceptable to use this shortcut
if
the calculated
value
of
x is le
ss
than 5 percent
of
the initial acid concentration. In this case, the approximation is
acceptable because
0.019 M X 100% = 3.8%

0.50M
-
Remember
that
soli
ds
an
d pu
re
li
quids
do
not
appear
in
th
e
eq
ui
li
bri
um
ex
pressio
n
[
~~
Section
15.3] .