Chapter16

Chapter 16
Acid–Base
Equilibria

Copyright © Cengage Learning. All rights reserved.

16 | 1

Contents and Concepts
Solutions of a Weak Acid or Base
1. Acid−Ionization Equilibria
2. Polyprotic Acids
3. Base−Ionization Equilibria
4. Acid–Base Properties of Salt Solutions
Solutions of a Weak Acid or Base with Another
Solute
5. Common−Ion Effect
6. Buffers
7. Acid–Base Titration Curves

16 | 2

Learning Objectives
Solutions of a Weak Acid or Base
Acid−Ionization Equilibria
a. Write the chemical equation for a weak
acid undergoing acid ionization in aqueous
solution.
b. Define acid−ionization constant and
degree of ionization.
c. Determine Ka from the solution pH.
d. Calculate concentrations of species in a
weak acid solution using Ka (approximation
method).

16 | 3

1. Acid−Ionization Equilibria (cont.)
a. State the assumption that allows for using
approximations when solving problems.
b. Calculate concentrations of species in a
weak acid solution using Ka (quadratic
formula).
2. Polyprotic Acids
a. State the general trend in the ionization
constants of a polyprotic acid.
b. Calculate concentrations of species in a
solution of a diprotic acid.

16 | 4

3. Base−Ionization Equilibria

a. Write the chemical equation for a
weak base undergoing ionization in
aqueous solution.
b. Define base−ionization constant.
c. Calculate concentrations of species
in a weak base solution using Kb.


16 | 5

4. Acid–Base Properties of Salt Solutions
a. Write the hydrolysis reaction of an ion to
form an acidic solution.
b. Write the hydrolysis reaction of an ion to
form a basic solution.
c. Predict whether a salt solution is acidic,
basic, or neutral.
d. Obtain Ka from Kb or Kb from Ka.
e. Calculating concentrations of species in a
salt solution.


16 | 6

Solutions of a Weak Acid or Base with Another
Solute
5. Common−Ion Effect
a. Explain the common−ion effect.
b. Calculate the common−ion effect on acid
ionization (effect of a strong acid).
c. Calculate the common−ion effect on acid
ionization (effect of a conjugate base).


16 | 7

6. Buffers
a. Define buffer and buffer capacity.
b. Describe the pH change of a buffer
solution with the addition of acid or base.
c. Calculate the pH of a buffer from given
volumes of solution.
d. Calculate the pH of a buffer when a strong
acid or a strong base is added.
e. Learn the Henderson–Hasselbalch
equation.
f. State when the Henderson–Hasselbalch
equation can be applied.

16 | 8

7. Acid–Base Titration Curves
a. Define equivalence point.
b. Describe the curve for the titration of a
strong acid by a strong base.
c. Calculate the pH of a solution of a strong
acid and a strong base.
d. Describe the curve for the titration of a
weak acid by a strong base.
e. Calculate the pH at the equivalence point
in the titration of a weak acid by a strong
base.

16 | 9

7. Acid–Base Titration Curves (cont)
f. Describe the curve for the titration of a
weak base by a strong acid.
g. Calculate the pH of a solution at several
points of a titration of a weak base by a
strong acid.


16 | 10

The simplest acid–base equilibria are those in
which a weak acid or a weak base reacts with
water.
We can write an acid equilibrium reaction for the
generic acid, HA.
HA(aq) + H2O(l)  H3O+(aq) + A−(aq)


16 | 11

Acetic acid is a weak acid. It reacts with water as
follows:
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2−(aq)
Here A from the previous slide = C2H3O2−(aq)


16 | 12

The equilibrium constant for the reaction of a weak
acid with water is called the acid−ionization
constant (or acid−dissociation constant), Ka.

Ka

[H O ] [ A ]
=
3

+

−

[HA ]

Liquid water is not included in the equilibrium
constant expression.


16 | 13


16 | 14


16 | 15

Concept Check 16.1
You have prepared dilute solutions of equal molar
concentrations of HC2H3O2 (acetic acid), HNO2, HF,
and HCN. Rank the solutions from the highest pH to
the lowest pH. (Refer to Table 16.1.)
In order of decreasing Ka:
HF
HNO2
HC2H3O2
HCN

Ka
6.8 × 10−4
4.5 × 10−4
1.7 × 10−5
4.9 × 10−10


Highest pH

HCN
HC2H3O2
HNO2
HF
Lowest pH

16 | 16

Calculations with Ka
Given the value of Ka and the initial concentration
of HA, you can calculate the equilibrium
concentration of all species.
Given the value of Ka and the initial concentration
of HA, you can calculate the degree of ionization
and the percent ionization.
Given the pH of the final solution and the initial
concentration of HA, you can find the value of Ka
and the percent ionization.

16 | 17

We can be given pH, percent or degree ionization,
initial concentration, and Ka.
From pH, we can find [H3O+].
From percent or degree ionization, we can find Ka.
Using what is given, we can find the other
quantities.


16 | 18

?

Sore−throat medications sometimes
contain the weak acid phenol, HC6H5O.
A 0.10 M solution of phenol has a pH of
5.43 at 25°C.
a. What is the acid−ionization
constant, Ka, for phenol at 25°C?
b. What is the degree of ionization?


16 | 19

HC6H5O(aq) + H2O(l) 
Initial
Change
Equilibrium

H3O+(aq) + C6H5O−(aq)

0.10

0

0

–x

+x

+x

0.10 – x

x

x

We were told that pH = 5.43. That allows us to find
[H3O+] = 3.7 × 10−6 M = x = [C6H5O−].
Now we find [HC6H5O] = 0.10 – x = 0.10 M.

16 | 20

Finally, we write the expression for Ka and
substitute the concentrations we now know.
HC6H5O + H2O  H3O+ + C6H5O−
[H3O+] = [C6H5O−] = 3.7 × 10−6 M
[HC6H5O] = 0.10 M
H3 O + C 6 H5 O −
Ka =
[HC 6H5 O]

[

][

]

(3.7 × 10 −6 )2
Ka =
0.10
Ka = 1.4 × 10−10

16 | 21

The degree of ionization is the ratio of ionized
concentration to original concentration:
x
3.7 × 10−6
Degree of ionization =
=
0.10
0.10
Degree of ionization = 3.7 × 10−5
Percent ionization is the degree of ionization × 100%:
Percent ionization = 3.7 × 10−3% or 0.0037%

16 | 22

Simplifying Assumption for Acid and Base
Ionizations
The equilibrium concentration of the acid is most
often ([HA]0 – x).
If x is much, much less than [HA]0, we can assume
that subtracting x makes no difference to [HA]:
([HA]0 – x) = [HA]0
This is a valid assumption when the ratio of [HA] 0
to Ka is > 103. If it is not valid, you must use the
quadratic equation to solve the problem.

16 | 23

?

Para−hydroxybenzoic acid is used to
make certain dyes. What are the
concentrations of this acid, of hydronium
ion, and of para−hydroxybenzoate ion in
a 0.200 M aqueous solution at 25°C?
What is the pH of the solution and the
degree of ionization of the acid? The Ka
of this acid is 2.6 × 10−5.

We will use the generic formula HA for
para−hydroxybenzoic acid and the following
equilibrium:
HA + H2O  H3O+ + A−


16 | 24

HA(aq) +
Initial
Change
Equilibrium

H2O(l) 

H3O+(aq) + A−(aq)

0.200

0

0

–x

+x

+x

0.200 – x

x

x

[H O ] [ A ]
=
[HA ]
+

Ka

2.6 × 10

−

3

−5

x2
=
0.200 − x

0.200
> 1000, so 0.200 − x ≈ 0.200
−5
2.6 × 10


16 | 25

x2
Ka =
0.200
5.2 × 10 −6 = x 2
x = 2.3 × 10 −3 = [H3 O + ]

pH = − log [H3O+ ] = − log (2.3 × 10−3 )
pH = 2.64

16 | 26

Polyprotic Acids
A polyprotic acid has more than one acidic proton
—for example, H2SO4, H2SO3, H2CO3, H3PO4.
These acids have successive ionization reactions
with Ka1, Ka2, . . .
The next example illustrates how to do calculations
for a polyprotic acid.


16 | 27

?

Tartaric acid, H2C4H4O6, is a diprotic acid
used in food products.
What is the pH of a 0.10 M solution?
What is the concentration of the C 4H4O62−
ion in the same solution?
Ka1 = 9.2 × 10−4; Ka2 = 4.3 × 10−5.

First, we will use the first acid−ionization
equilibrium to find [H+] and [HC4H4O6−]. In these
calculations, we will use the generic formula H2A
for the acid.
Next, we will use the second acid−ionization
equilibrium to find [C4H4O62−].

16 | 28

H2A(aq) +
Initial

H3O+(aq)
+

HA−(aq)

0.10

0

0

–x

+x

+x

0.10 – x

x

x

Change
Equilibrium

K a1

H2O(l) 

[H O ] [HA ]
=
3

+

−

[H2 A ]

9.2 × 10

−4

2

x
=
0.10 - x

0.10
< 1000
−4
9.2 × 10
We cannot use the simplifying assumption .

16 | 29

9.2 × 10 −4 (0.10 − x ) = x 2
9.2 × 10 −5 − 9.2 × 10 −4 x = x 2
x 2 + 9.2 × 10−4 x − 9.2 × 10−5 = 0

a =1

b = 9.2 × 10

−4

c = − 9.2 × 10

−5

− b ± b 2 − 4ac
x=
2a
−(9.2 × 10−4 ) ± (9.2 × 10−4 )2 − 4(1)( −9.2 × 10−5 )
x=
2(1)
− (9.2 × 10 −4 ) ± (1.92 × 10 −2 )
−4
−3
x=
= −4.6 × 10 ± 9.6 × 10
2

16 | 30

x = 9.1× 10

−3

and

x = −1.0 × 10

−2

We eliminate the negative value because it is
physically impossible to have a negative
concentration.
At the end of the first acid ionization equilibrium,
the concentrations are
[H3O+] = 9.1 × 10−3 M
[HA−] = 9.1 × 10−3 M
[H2A] = 9.0 × 10−4 M
Now we use these for the second acid−ionization
equilibrium.

16 | 31

HA−(aq) +
Initial
Change
Equilibrium

Ka2

H2O(l) 

H3O+(aq) + A2−(aq)

0.0091

0.0091

0

–x

+x

+x

0.0091 – x

0.0091 + x

x

H3O+   A 2− 


=
HA − 



4.3 × 10

−5

(0.0091 + x ) x
=
(0.0091 − x )

We can assume that
0.0091 + x ≈ 0.0091 and 0.0091 − x ≈ 0.0091

16 | 32

4.3 × 10

−5

0.0091 x
=
0.0091

x = 4.3 × 10 −5 M
Our assumption was valid.

[C4H4O62− ] = 4.3 × 10−5 M
[HC4H4O6− ] = 9.1× 10−3 M
[H3O + ] = 9.1 × 10 −3 M
[H2C 4H4O6 ] = 9.0 × 10 −4 M

pH = −log (9.1× 10 −3 )

pH = 2.04
16 | 33

Base−Ionization Equilibrium
The simplest acid–base equilibria are those in
which a weak acid or a weak base reacts with
water.
We can write a base equilibrium reaction for the
generic base, B.
B(aq) + H2O(l)  HB+(aq) + OH−(aq)


16 | 34

Ammonia is a weak base. It reacts with water as
follows:
NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq)
Here the generic B from the previous slide is
NH3(aq).


16 | 35

The equilibrium constant for the reaction of a weak
base with water is called the base−ionization
constant, Kb.

Kb

[HB ][OH ]
=
+

−

[ B]

Liquid water is not included in the equilibrium
constant expression.


16 | 36


16 | 37

Writing Kb Reactions
The bases in Table 16.2 are nitrogen bases; that
is, the proton they accept adds to a nitrogen atom.
Next we’ll practice writing the Kb reactions.
Ammonium becomes ammonium ion:
NH3+ H2O → NH4+ + OH−
Ethylamine becomes ethyl ammonium ion:
C2H5NH2 + H2O → C2H5NH3+ + OH−

16 | 38

Dimethylamine becomes dimethylammonium ion:
(CH3)2NH2 + H2O → (CH3)2NH3+ + OH−
Pyridine becomes pyridinium ion:
C5H5N+ H2O → C5H5NH+ + OH−
Hydrazine becomes hydrazinium ion:
N2H4+ H2O → N2H5+ + OH−


16 | 39

We can be given pH, initial concentration, and Kb.
From the pH, we can find first [H3O+] and then
[OH−]. Using what is given, we can find the other
quantities.
We can also use a simplifying assumption: When
[B]0 / Kb > 103, the expression ([B]0 – x) = [B]0.


16 | 40

?

Aniline, C6H5NH, is used in the
manufacture of some perfumes.
What is the pH of a 0.035 M solution of
aniline at 25°C?
Kb= 4.2 × 10−10 at 25°C.

We will construct an ICE chart and solve for x.


16 | 41

C6H5H(aq) +
Initial
Change
Equilibrium

H2O(l) 

C6H5NH+(aq)
+

OH−(aq)

0.035

0

0

–x

+x

+x

0.035 – x

x

x

We are told that Kb = 4.2 × 10−10. That allows us to
substitute into the Kb expression to solve for x.


16 | 42

C6H5N + H2O

C6H5NH+ + OH−

[B] = 0.200 – x and [B+] = [OH−] = x

[C H NH ] [OH ]
=
+

Kb

0.200
4.2 × 10−10

6

−

5

[ C6H5N]

x2
−10
4.2 × 10 =
(0.200 − x )
> 1000, so we can assume that 0.200 − x = x
x2
4.2 × 10−10 =
0.200
x 2 = 8.4 × 10−11
x = 9.2 × 10−6 M


16 | 43

The question asks for the pH:

pH = 14.00 = pOH = 14.00 − log [OH− ]
pH = 14.00 − log (9.2 × 10−6 )
pH = 14.00 − 5.04
pH = 8.96


16 | 44

Salt Solutions
We will look at the cation and the anion separately,
and then combine the result to determine whether
the solution is acidic, basic, or neutral.


16 | 45

The conjugate acid of a strong base is very weak
and does not react with water. It is therefore
considered to be neutral.
Na+ + H2O → No Reaction (NR)
The conjugate base of a strong acid is very weak
and does not react with water. It is therefore
considered to be neutral.
Cl− + H2O → NR

16 | 46

The conjugate acid of a weak base reacts with
water to form an acidic solution:
NH4+ + H2O  NH3 + H3O+
The conjugate base of a weak acid reacts with
water to form a basic solution:
F− + H2O  HF + OH−


16 | 47

We classify each salt by examining its cation and
its anion, and then combining the result.

NaBr
Na+ is the conjugate acid of NaOH, a strong base.
It does not react with water, so it is neutral.
Br− is the conjugate base of HBr, a strong acid. It
does not react with water, so is neutral.
The cation is neutral; the anion is neutral.
NaBr is neutral.

16 | 48

NaC2H3O2
Na+ is the conjugate acid of NaOH, a strong base.
It does not react with water, so it is neutral.
C2H3O2− is the conjugate base of HC2H3O2, a weak
acid. It reacts with water to give a basic solution.
The cation is neutral; the anion is basic.
NaC2H3O2 is basic.

16 | 49

NH4Cl
NH4+ is the conjugate acid of NH3, a weak base. It
reacts with water to give an acidic solution.
Cl− is the conjugate base of HCl, a strong acid. It
does not react with water, so it is neutral.
The cation is acidic; the anion is neutral.
NH4Cl is acidic.

16 | 50

NH4F
NH4+ is the conjugate acid of NH3, a weak base. It
reacts with water to give an acidic solution.
F− is the conjugate base of HF, a weak acid. It does
not react with water, so it is basic.
The cation is acidic; the anion is basic. We need
more information in this case. We compare Ka for
HF to Kb for NH3.

16 | 51

If Ka > Kb, the solution is acidic. If Ka < Kb, the
solution is basic. If Ka = Kb, the solution is neutral.

K w 1.0 × 10 −14
+
For NH4 : K a =
=
= 5.6 × 10 −10
K b 1.8 × 10 −5
K w 1.0 × 10 −14
For F − : K b =
=
= 1.5 × 10 −11
K a 6.8 × 10 − 4
+

K a for NH 4 > K b for F −

NH4F is acidic.

16 | 52

?

Ammonium nitrate, NH4NO3, is
administered as an intravenous solution
to patients whose blood pH has
deviated from the normal value of 7.40.
Would this substance be used for
acidosis (blood pH < 7.40) or alkalosis
(blood pH > 7.40)?


16 | 53

NH4+ is the conjugate acid of a NH3, a weak base.
NH4+ is acidic.
NO3− is the conjugate base of HNO3, a strong acid.
NO3− is neutral.
NH4NO3 is acidic, so it could be used for alkalosis.


16 | 54

The hydrolysis equilibrium constant can be used in
problems to determine the pH of a salt solution. To
use the hydrolysis equilibrium, we need to
compute the K value for it.


16 | 55

We can determine K for the reaction of the ion with
water (hydrolysis reaction).
NH4+ + H2O <==> NH3 + H3O+
is the sum of
NH4+ + OH− <==> NH3 + H2O K1 = 1/Kb
2H2O <==> H3O+ + OH− K2 = Kw
Overall reaction
NH4+ + H2O <==> NH3 + H3O+
Kw
K a = K1 K 2 =
Kb

16 | 56

?

What is Kb for the F− ion, the ion added
to the public water supply to protect
teeth?
For HF, Ka = 6.8 × 10−4.


16 | 57

F− + H2O <==> HF + OH−
Is the sum of
H3O+ + F− <==> HF + H2O;
2H2O <==> H3O+ + OH−;

K1 = 1/Ka
K2 = Kw

Overall reaction:
F− + H2O <==> HF + OH−
K w 1.0 × 10 −14
K b = K1 K 2 =
=
-10
K a 6.8 × 10

For F- : Kb = 1.5 × 10−5

16 | 58

?

Household bleach is a 5% solution of
sodium hypochlorite, NaClO. This
corresponds to a molar concentration
of about 0.70 M NaClO.
What is the [OH−] and the pH of the
solution?
For HClO, Ka = 3.5 × 10−8.


16 | 59

ClO−(aq) +
Initial
Change
Equilibrium

H2O(l) 

HClO(aq) +

OH−(aq)

0.70

0

0

–x

+x

+x

0.70 – x

x

x

We are told that Ka = 3.5 × 10−8. That means that Kb=
2.9 × 10−7.
This allows us to substitute into the Kb expression
to solve for x.

16 | 60

ClO− + H2O  HClO + OH−
[H3O+] = [A−] = x and [HA] = 0.200 – x

0.70
2.9 × 10−7

HClO] OH− 
[


Kb =
ClO− 


2
x
−7
2.9 × 10 =
(0.70 − x )
> 1000, so we can assume that 070 − x = 0.70.
x2
2.9 × 10−7 =
0.70
x 2 = 2.0 × 10 -7
x = 4.5 × 10 -4 M


16 | 61

The question asks for [OH−] and the pH:

[OH− ] = x = 4.5 × 10−4 M
pH = 14.00 = pOH = 14.00 − log [OH− ]
−4

pH = 14.00 − log (4.5 × 10 )
pH = 14.00 − 3.35
pH = 10.65

16 | 62

Common−Ion Effect
The common−ion effect is the shift in an ionic
equilibrium caused by the addition of a solute that
takes part in the equilibrium.


16 | 63

Buffers
A buffer solution is characterized by the ability to
resist changes in pH when limited amounts of acid
or base are added to it.
A buffer is made by combining a weak acid with its
conjugate base or a weak base with its conjugate
acid.


16 | 64

?

What is the pH of a buffer made by
mixing 1.00 L of 0.020 M benzoic acid,
HC7H5O2, with 3.00 L of 0.060 M sodium
benzoate, NaC7H5O2? Ka for benzoic
acid is 6.3 × 10−5.

This problem involves dilution first. Once we know
the concentrations of benzoic acid and benzoate
ion, we can use the acid equilibrium to solve for x.
We will use HBz to represent benzoic acid and Bz −
to represent benzoate ion.

16 | 65

(0.020 M)(1.0 L)
[HBz] =
= 0.0050 M
(1.0 + 3.0) L
(0.060 M)(3.0 L)
[Bz ] =
= 0.045 M
(1.0 + 3.0) L
−

HBz(aq) + H2O(l) 
Initial
Change
Equilibrium

H3O+(aq) + Bz−(aq)

0.0050

0

0.045

–x

+x

+x

0.0050 – x

x

0.045 + x


16 | 66

H3O+  Bz− 


Ka = 
[ HBz]
6.3 × 10

−5

x (0.045 + x )
=
(0.0050 − x )

We can assume that
0.045 + x ≈ 0.045 and
6.3 × 10

−5

0.0050 − x ≈ 0.0050

0.045 x
=
0.0050

x = 7.0 × 10 −6 M

16 | 67

[H3O + ] = x = 7.0 × 10 −6 M
+

−6

pH = − log [H3O ] = − log (7.0 × 10 )
pH = 5.15


16 | 68

?

Calculate the pH of a 0.10 M HF
solution to which sufficient sodium
fluoride has been added to make the
concentration 0.20 M NaF.
Ka for HF is 6.8 × 10−4.

We will use the acid equilibrium for HF.
NaF provides the conjugate base, F−, so
[F−] = 0.20 M.


16 | 69

[HF] = 0.10 M
HF(aq) + H2O(l) 
Initial

H3O+(aq) +

F−(aq)

0.10

0

0.20

–x

+x

+x

0.10 – x

x

0.20 + x

Change
Equilibrium

[F − ] = 0.20 M

x (0.20 + x )
[H3O ] [F ]
−4
6.8 × 10 =
Ka =
(0.10 - x )
[HF]
We can assume that
0.20 + x ≈ 0.20 and 0.10 − x ≈ 0.10
+

−


16 | 70

6.8 × 10

−4

0.20x
=
0.10

x = 3.4 × 10 −4 M

[H3O + ] = x = 3.4 × 10 −4 M

pH = − log [H3O + ] = − log (3.4 × 10 −4 )
pH = 3.47


16 | 71

Henderson—Hasselbalch Equation
Buffers at a specific pH can be prepared using the
Henderson−Hasselbalch Equation.
Buffers are prepared from a conjugate acid−base
pair in which the ionization is approximately equal
to the desired H3O+ concentration.


16 | 72

Consider a buffer made up of a weak acid HA and
its conjugate base A−:
The acid−ionization constant is

[H3O + ][A − ]
Ka =
[HA]
Rearranging to solve for [H3O+]

 [HA] 
[H3O ] = Ka  − ÷
 [A ] 
+


16 | 73

The preceding equation can be used to derive
an equation for the pH of the buffer.
First, we take the negative logarithm of both
sides of the equation.


[HA] 
− log[H3O ] = −log K a x − 

[A ] 


+

 [HA] 
−log[H3O ] = −log Ka − log  − ÷
 [A ] 
+


16 | 74

The pKa of a weak acid is defined in a manner
similar to pH and pOH
pKa = − log Ka
We can now express the equation as

 [HA] 
pH = pKa − log  − ÷
 [A ] 
This is equivalent to

 [A − ] 
pH = pK a + log 
 [HA] 




16 | 75

This equation is generally shown as:

[base]
pH = pK a + log
[acid]
This equation relates buffer pH for different
concentrations of conjugate acid and base and is
known as the Henderson−Hasselbalch equation.


16 | 76

?

What is the [H3O+] for a buffer solution
that is 0.250 M in acid and 0.600 M in
the corresponding salt if the weak acid
Ka = 5.80 x 10−7?

Use as example the following equation of a
conjugate acid−base pair.
First, we use the value of Ka to find pKa:
pKa = −log(5.80 x 10−7)
pKa = −(−6.237)
pKa = 6.237

16 | 77

[base]
pH = pK a + log
[acid]
pKa = 6.237
[base] = 0.600 M
[acid] = 0.250 M
Substituting:

[0.600]
pH = 6.237 + log
[0.250]
pH = 6.237 + log 2.40
pH = 6.237 + 0.380
pH = 6.617


16 | 78

Using the pH, we can now find [H3O+]:
[H3O+] = 10−pH
[H3O+] = 10−6.617
[H3O+] = 2.42 × 10−7 M


16 | 79

Acid–Base Titration
An acid–base titration is a procedure for
determining the amount of acid (or base) in a
solution by determining the volume of base (or
acid) of known concentration that will completely
react with it.


16 | 80

An acid–base titration curve is a plot of the pH of
a solution of acid (or base) against the volume of
added base (or acid).
Such curves are used to gain insight into the
titration process. You can use the titration curve to
choose an indicator that will show when the
titration is complete.


16 | 81

This titration plot shows the titration of a strong
acid with a strong base. Note that the pH at the
equivalence point is 7.0.


16 | 82

The equivalence point is the point in a titration
when a stoichiometric amount of reactant has
been added.
The indicator must change color near the pH at the
equivalence point.


16 | 83

When solving problems in which an acid reacts
with a base, we first need to determine the limiting
reactant. This requires that we use moles in the
calculation.
The next step is to review what remains at the end
of the reaction. In the case of a strong acid and a
strong base, the solution is neutral.
When a weak acid or weak base is involved, the
product is a salt. After finding the salt
concentration, we use a hydrolysis equilibrium to
find the pH.

16 | 84

?

Calculate the pOH and the pH of a
solution in which 10.0 mL of 0.100 M
HCl is added to 25.0 mL of 0.100 M
NaOH.

First we calculate the moles of HCl and NaOH.
Then we determine what remains after the
reaction.


16 | 85

The overall reaction is
HCl + NaOH → H2O + NaCl
Moles HCl = (0.100 M)(10.0 × 10−3 L)
= 1.00 × 10−3 mol
Moles NaOH = (0.100 M)(25.0 × 10−3 L)
= 2.50 × 10−3 mol
All of the HCl reacts, leaving 1.50 × 10−3 mol
NaOH.
The new volume is 10.0 mL + 25.0 mL = 35.0 mL.

16 | 86

1.50 × 10−3 mol
[OH− ] =
= 4.29 × 10−2 M
35.0 × 10−3 L
−

pH = 14.00 − pOH = 14.00 − ( −log[OH ])
pH = 14.00 − 1.368
pH = 12.632


16 | 87

?

Calculate the [OH−] and the pH at the
equivalence point for the titration of 500.
mL of 0.10 M propionic acid with 0.050 M
calcium hydroxide. (This can be used to
prepare a preservative for bread.)
Ka = 1.3 × 10−5

First we need to find the moles of propionic acid to
find the volume of calcium hydroxide. Then we will
examine the salt that is produced and use the
hydrolysis equilibrium to find the pH.
We will use HPr to represent propionic acid and Pr − to
represent propionate ion.

16 | 88

Moles propionic acid = (0.10 M)(500. × 10−3 L)
= 0.050 mol HPr
To find the moles of calcium hydroxide, we need
the balanced chemical equation:
2HPr + Ca(OH)2 → Ca(Pr)2 + 2H2O
Moles calcium hydroxide =

1 mol Ca(OH)2
0.050 mol HPr
= 0.025 mol Ca(OH)2
2 mol HPr

16 | 89

Now we find the volume of Ca(OH)2 required to
give 0.025 mol:
0.025 mol
Volume Ca(OH)2 =
= 0.50 L
mol
0.050
L
The new volume is 500. × 10−3 L + 0.50 L = 1.00 L

The concentration of Pr− is 2(0.025 M) = 0.050 M

0.050 mol
[Pr ] =
= 5.0 × 10−2 M = 0.050 M
1.0 L
−


16 | 90

The hydrolysis reaction is
Pr− + H2O  HPr + OH−

K w 1.00 × 10−14
−10
Kb =
=
= 7.7 × 10
−5
Ka
1.3 × 10
Pr−(aq) +
Initial
Change
Equilibrium

H2O(l) 

HPr(aq) + OH−(aq)

0.050

0

0

–x

+x

+x

0.050 – x

x

x


16 | 91

−

[HPr] [OH ]
Ka =
[Pr − ]

7.7 × 10−10

x2
=
(0.050 − x )

0.050
> 1000
−10
7.7 × 10
We can assume that 0.050 − x ≈ 0.050.
x2
−10
7.7 × 10 =
0.050
x 2 = 3.85 × 10 −11
x = 6.20 × 10 -6 M
[OH- ] = x = 6.20 × 10−6 M

pOH = 5.21

pH = 8.79
16 | 92

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