Chapter 16. Acid –Base Equilibria
.
.
1
Equilibria in Solutions of Weak Acids
The dissociation of a weak acid is an
equilibrium situation with an equilibrium
constant, called the acid dissociation
constant, Ka based on the equation
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)

H O A 


Ka

3
HA
2
Equilibria in Solutions of Weak Acids
The acid dissociation constant, Ka is
always based on the reaction of one mole
of the weak acid with water.
If you see the symbol Ka, it always refers to
a balanced equation of the form
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)
3
Problem
The pH of 0.10 mol/L HOCl is 4.23. Calculate
Ka for hypochlorous acid.
HOCl (aq) + H2O (l)  H3O+ (aq) + ClO- (aq)

H O ClO 


Ka

3
HOCl
4
Calculating Equilibrium Concentrations in Solutions of
Weak Acids
We can calculate equilibrium
concentrations of reactants and
products in weak acid dissociation
reactions with known values for Ka.
To do this, we will often use the ICE
table technique we saw in the last
chapter on equilibrium.
5
Calculating Equilibrium Concentrations in Solutions of
Weak Acids
We need to figure out what is an acid and
what is a base in our system.
For example, if we start with 0.10 mol/L
HCN, then HCN is an acid, and water is a
base.
HCN (aq) + H2O (l)  H3O+ (aq) + CN- (aq)
Ka = 4.9 x 10-10
6
Like our previous equilibrium problems, we then create a
table of the initial concentrations of all chemicals, the
change in their concentration, and their equilibrium
concentrations in terms of known and unknown values.
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HCN (aq)
0.10
-x
0.10 – x
+
H2O (l)
N/A
N/A
N/A


[H
O
][CN
]
10
3
K a  4.9x10 
[HCN]

H3O+ (aq) +
0.0
+x
+x
CN- (aq)
0.0
+x
+x
(x)(x)
so 4.9x10 
(0.10 x)
10
7
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HCN (aq)
0.10
-x
0.10 – x
K a  4.9 x 1010
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.0
+x
+x
CN- (aq)
0.0
+x
+x
[H3O ][CN ]
(x)(x)
10

so 4.9 x 10 
[HCN]
(0.10 x)
We can ALWAYS solve this equation using
the quadratic formula and get the right
answer, but it might be possible to do it
more simply.
8


[H
O
][CN
]
(x)(x)
10
3
K a  4.9 x 10 
so 4.9 x 1010 
[HCN]
(0.10  x)
Every time we do an weak acid
equilibrium problem, divide the initial
concentration of the acid by Ka.
For this example
0.10 / 4.9 x 10-10 = 2 x 108
9


[H
O
][CN
]
(x)(x)
10
3
K a  4.9 x 10 
so 4.9 x 1010 
[HCN]
(0.10  x)
0.10 / 4.9 x
-10
10
=2x
8
10
Since this value is greater than 100, we can
assume that the initial concentration of
the acid and the equilibrium
concentration of the acid are the same.
This assumption will lead to answers with
less than 5% error since this pre-check is
greater than 100.
10


[H
O
][CN
]
(x)(x)
K a  4.9 x 1010  3
so 4.9 x 1010 
[HCN]
(0.10  x)
The assumption we will make is that
x << [HCN]i so [HCN]eqm  [HCN]I
4.9 x 10-10 = x2 / 0.10
x2 = (4.9 x 10-10)(0.10)
x = 4.9 x 10-11
x = 7.0 x 10-6 mol/L
11


[H
O
][CN
]
(x)(x)
K a  4.9 x 1010  3
so 4.9 x 1010 
[HCN]
(0.10  x)
Based on the assumption we’ve
made, at equilibrium
x = [H3O+]eqm = [CN-]eqm
= 7.0 x 10-6 mol/L
(-ve value isn’t physically possible)
[HCN]eqm = 0.10 mol/L.
12
.
Any time we make an assumption,
we MUST check it.
We assumed x << [HCN]i
To check the assumption, we divide
x by [HCN]i and express it as a
percentage
x
7.0 x 10-6 M
-5

 7.0 x 10  0.007%
HCNi
0.10M
13
As long as the assumption check is
less than 5%, then
the assumption is valid!
If the assumption was not valid, we
would have to go back and use the
quadratic formula!
14
Remember!
H2O (l) + H2O (l)  H3O+ (aq) + OH- (aq)
is always taking place in water
whether or not we have added
an acid or base.
This reaction also contributes
H3O+ (aq) and OH- (aq)
to our system at equilibrium
15
Remember!
Since at 25 C
Kw = [H3O+] [OH-] = 1.0 x 10-14
it turns out that if our acid-base equilibrium
we’re interested in gives a pH value between
about 6.8 and 7.2 then the auto-dissociation
of water contributes a significant amount of
[H3O+] and [OH-] to our system and the real
pH would not be what we calculated in the
problem.
16
Problem
Acetic acid CH3COOH (or HAc) is the
solute that gives vinegar its characteristic
odour and sour taste. Calculate the pH and
the concentration of all species present in:
a) 1.00 mol/L CH3COOH
b) 0.00100 mol/L CH3COOH
17
Problem a)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
1.00
-x
1.00 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.0
+x
+x
+x
+x
[H3O ][CH3COO ]
(x)(x)
5
K a  1.8 x 10 
so 1.8 x 10 
[CH3COOH]
(1.00 x)
5
Let’s check the initial acid
concentration / Ka ratio.
1.00 / 1.8 x 10-5  55000
is larger than 100.
18
Problem a)


[H
O
][CH
COO
]
(x)(x)
5
5
3
3
K a  1.8 x 10 
so 1.8 x 10 
[CH3COOH]
(1.00 x)
We can probably assume that
x << [HAc]i so [HAc]eqm  [HAc]i
1.8 x 10-5 = x2 / 1.00
x2 = (1.8 x 10-5)(1.00)
x = 1.8 x 10-5
x = 4.2 x 10-3 mol/L
(but must be + value since x = [H3O+])
19
Problem a)
So at equilibrium,
[H3O+] = [CH3COO-] = 4.2 x 10-3 mol/L
[CH3COOH] = 1.00 mol/L.
x
4.2 x 10-3 M

 4.2 x 10-3  0.42%
CH3COOHi
1.00M
The assumption was valid and so
pH = - log [H3O+]
pH = - log 4.2 x 10-3
pH = 2.38
20
Problem b)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.0100
0.00100
-x
0.0100 – x
0.00100
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.0
+x
+x
+x
+x


[H
O
][CH
COO
]
(x)(x)
5
5
3
3
K a  1.8 x 10 
so 1.8 x 10 
[CH3COOH]
(0.00100 x)
Let’s check the initial acid
concentration / Ka ratio.
0.00100 / 1.8 x 10-5  56
is smaller than 100.
21
Problem b)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.0100
0.00100
-x
0.0100 – x
0.00100
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.0
+x
+x
+x
+x


[H
O
][CH
COO
]
(x)(x)
5
5
3
3
K a  1.8 x 10 
so 1.8 x 10 
[CH3COOH]
(0.00100 x)
We can probably CAN NOT assume that
x << [HAc]i so [HAc]eqm  [HAc]i
5
[1.8 x 10 ](0.00100  x)  x  0
2
5
8
0  x  1.8 x 10 x - 1.8 x 10
2
22
Problem b)
 (1.8 x 105 )  (1.8 x 105 ) 2  4(1)(-1.8x 108 )
 b  b 2  4ac
x
so x 
2a
2(1)
- 1.8 x 105  3.24 x 1010  7.2 x 108
- 1.8 x 105  3.24 x 1010  7.2 x 108
x
or x 
2
2

4
- 1.8 x 105  2.69 x 10
- 1.8 x 105  2.69 x 10 4
x
or x 
2
2
4
4
 2.87 x 10
2.51 x 10
x
or x 
2
2
so x  1.25 x 104 mol/L or x  -1.44 x 104 mol/L
23
Problem b)
Since [H3O+] = x we must use the
positive value, so
[H3O+] = [CH3COO-] = 1.3 x 10-4 mol/L
[CH3COOH]
= 0.00100 mol/L – 1.3 x 10-4 mol/L
= 0.00087 mol/L.
24
Problem b)
Let’s confirm that x << [HAc]i
IS NOT TRUE
-4
x
1.3 x 10 M

 0.13  13 %
CH3COOHi 0.00100M
pH = - log [H3O+]
pH = - log 4.2 x 10-4
pH = 3.38
25
Problem
A vitamin C tablet containing 250 mg of
ascorbic acid (C6H8O6; Ka = 8.0 x 10-5
is dissolved in a 250 mL glass of water
to give a solution where [C6H8O6] =
5.68 x 10-3 mol/L.
What is the pH of the solution?
26
Problem
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
C6H8O6 (aq) +
5.68 x 10-3
-x
5.68 x 10-3 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + C6H7O6- (aq)
0.0
0.0
+x
+x
+x
+x


[H
O
][C6 H7 O6 ]
(x)(x)
K a  8.0 x 105  3
so 8.0 x 105 
[C6 H8O6 ]
(5.68x 10-3  x)
Check the initial acid concentration / Ka ratio.
5.68 x 10-3/ 8.0 x 10-5  71
which is not larger than 100 so
5
3
8.0 x 10 (5.68x 10  x)  x  0
2
5
7
0  x  8.0 x 10 x - 4.54 x10
2
27
Problem
 (8.0 x 105 )  (8.0 x 105 ) 2  4(1)(-4.54 x 107 )
 b  b 2  4ac
x
so x 
2a
2(1)
- 8.0 x 105  6.4 x 109  (1.82 x 106 )
- 8.0 x 105  6.4x109  (1.82 x106 )
x
or x 
2
2
5
3
5
3
- 8.0 x 10  1.35 x 10
- 8.0 x 10  1.35 x 10
x
or x 
2
2
3
3
1.4 x 10
 1.27 x 10
x 3
or x 
2
2
4
4
so x  7.1x10 mol/L or x  6.3x10 mol/L
28
Problem
Since [H3O+] = x the answer must be the
positive value
[H3O+] = [C6H7O6-] = 6.3 x 10-4 mol/L
[C6H8O6] = (5.68 x 10-3 - 6.3 x 10-4) mol/L
= 5.05 x 10-3 mol/L
pH = - log [H3O+]
pH = - log 6.3 x 10-4
pH = 3.20
29
Degree of ionization
The pH of a solution of a weak acid like acetic
acid will depend on the initial concentration of
the weak acid and Ka. Therefore, we can
define a second measure of the strength of a
weak acid by looking of the
degree (or percent) ionization of the acid.
%ionization = [HA]ionized / [HA]initial x 100%
30
Percent ionization
In part a) of an earlier problem an acetic
acid solution with initial concentration of
1.00 mol/L at equilibrium had
[H3O+]eqm = [HA]ionized = 4.2 x 10-3 mol/L
%ionization = [HA]ionized / [HA]initial x 100%
%ionization = 4.2 x 10-3 mol/L / 1.00 mol/L x 100%
%ionized = 0.42%
31
Percent ionization
In part b) of an earlier problem an acetic
acid solution with initial concentration of
0.00100 mol/L at equilibrium had
[H3O+] = [HA]ionized = 1.3 x 10-4 mol/L
%ionization = [HA]ionized / [HA]initial x 100%
%ionization = 1.3 x 10-4 mol/L / 0.00100 mol/L x 100%
%ionization = 13%
32
Figure
33
Equilibria in Solutions of Weak Bases
The dissociation of a weak base is an
equilibrium situation with an equilibrium
constant, called the base dissociation
constant, Kb based on the equation
B (aq) + H2O (l)  BH+ (aq) + OH- (aq)


[BH ][OH ]
Kb 
[B]
34
Equilibria in Solutions of Weak Bases
The base dissociation constant, Kb is
always based on the reaction of one mole
of the weak base with water.
If you see the symbol Kb, it always refers to
a balanced equation of the form
B (aq) + H2O (l)  BH+ (aq) + OH- (aq)
35
Equilibria in Solutions of Weak Bases
Our approach to solving equilibria
problems involving bases is exactly the
same as for acids.
1. Set up the ICE table
2. Establish the equilibrium constant
expression
3. Make a simplifying assumption when
possible
4. Solve for x, and then for eq’m amounts
36
Problem
Strychnine (C21H22N2O2), a deadly poison
used for killing rodents, is a weak base
having Kb = 1.8 x 10-6. Calculate the pH if
[C21H22N2O2]initial = 4.8 x 10-4 mol/L
37
Problem
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
C21H22N2O2 (aq)
4.8 x 10-4
-x
4.8 x 10-4 – x
+ H2O (l)
N/A
N/A
N/A
 C21H23N2O2+ (aq) + OH- (aq)
0.0
0.0
+x
+x
+x
+x

[C21H 23 N 2O2 ][OH ]
(x)(x)
6
K b  1.8 x 10 
so 1.8 x 106 
[C21H 22 N 2O2 ]
(4.8 x 10-4  x)
Check the initial base concentration / Kb ratio
4.8 x 10-4 / 1.8 x 10-6  267
which is greater than 100
We are probably good to make a simplifying
assumption that x << [C21H22N2O2]i
38

[C 21H 23 N 2 O 2 ][OH  ]
(x)(x)
6
6
K b  1.8 x 10 
so 1.8 x 10 
[C 21H 22 N 2O 2 ]
(4.8 x 10-4  x)
The assumption we will make is that
x << [C21H22N2O2]i so
[C21H22N2O2]eqm  [C21H22N2O2]I
1.8 x 10-6 = x2 / 4.8 x 10-4
x2 = (1.8 x 10-6)(4.8 x 10-4)
x = 8.64 x 10-10
x = 2.94 x 10-5 mol/L
39
Problem
Since x = [OH-], the answer must be the positive
value,
x = [C21H23N2O2+] = [OH-] = 2.9 x 10-5 mol/L
[C21H22N2O2] = 4.8 x 10-4 mol/L – 2.9 x 10-5 mol/L
= 4.5 x 10-4 mol/L.
We should check the assumption!
x
2.9 x 10-5 M

 0.060  6.0%
-4
C21H22 N2O2 i 4.8 x 10 M
40
Problem
x
2.9 x 10-5 M

 0.060  6.0%
-4
C21H22 N2O2 i 4.8 x 10 M
In this case, the error is more
than 5%.
I will leave it to you to go back
and use the quadratic formula.
Compare the two answers
41
Problem
To continue towards the answer of the
problem AS IF the assumption WERE
VALID
pOH = - log [OH-]
pOH = - log 2.9 x 10-5
pOH = 4.54
pH + pOH = 14.00
pH = 14.00 - pOH
pH = 14.00 - (4.54)
pH = 9.46
42
Relation Between Ka and Kb
The strength of an acid in water is
expressed through Ka, while the strength
of a base can be expressed through Kb
Since Brønsted-Lowry acid-base reactions
involve conjugate acid-base pairs there
should be a connection between the
Ka value and the Kb value of a
conjugate acid-base pair.
43
Relation Between Ka and Kb
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)

H O A 


Ka

3
HA
A- (aq) + H2O (l)  OH- (aq) + HA (aq)

OH HA

A 

Kb

44
Since these reactions take place in the same
beaker at the same time let’s
add them together
HA (aq) + H2O (l) + A- (aq) + H2O (l)  H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq)
2 H2O (l)  H3O+ (aq) + OH- (aq)
45
HA (aq) + H2O (l) + A- (aq) + H2O (l)  H3O+ (aq) + A- (aq) + OH- (aq) + HA (aq)
2 H2O (l)  H3O+ (aq) + OH- (aq)
The sum of the reactions is the dissociation of
water reaction, which has the ion-product
constant for water
Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25 °C
Closer inspection shows us that
H O A  OH HA  H O OH   K

HA
A 

Ka x K b



3

3

w
 1.0x1014 at 25 C
46
As the strength of an acid
increases (larger Ka) the strength
of the conjugate base must
decrease (smaller Kb) because
their product must always be the
dissociation constant for water
Kw.
47
Strong acids always have very weak
conjugate bases. Strong bases always
have very weak conjugate acids.
Since Ka x Kb = Kw
then Ka = Kw / Kb
and Kb = Kw / Ka
48
Problem
a) – Piperidine (C5H11N) is an amine found
in black pepper. Find Kb for piperidine in
Appendix C, and then calculate Ka for the
C5H11NH+ cation.
Kb = 1.3 x 10-3
b) Find Ka for HOCl in Appendix C, and
then calculate Kb for OCl-.
Ka = 3.5 x 10-8
49
Acid-Base Properties of Salts
When acids and bases react with each other,
they form ionic compounds called salts.
Salts, when dissolved in water, can lead to acidic,
basic, or neutral solutions, depending on the
relative strengths of the acid and base we derive
them from.
Strong acid + Strong base  Neutral salt solution
Strong acid + Weak base  Acidic salt solution
Weak acid + Strong base  Basic salt solution
50
Salts that Yield Neutral Solutions
Strong acids and strong
bases react to form neutral salt
solutions. When the salt
dissociates in water, the cation
and anion do not appreciably
react with water to form H3O+
or OH-.
51
Salts that Yield Neutral Solutions
Strong base cations like the alkali
metal cations (Li+, Na+, K+) or
alkaline earth cations (Ca2+, Sr2+,
Ba2+, but NOT Be2+) and strong
acid anions such as Cl-, Br-, I-,
NO3-, and ClO4- will combine
together to give neutral salt
solutions with pH = 7.
52
Salts that Yield Neutral Solutions
Sodium chloride (NaCl) will dissociate
into Na+ and Cl- in water.
Cl- has no acidic or basic tendencies.
Cl- (aq) + H2O (l) ⇌ no reaction
Chloride ions DO NOT HAVE
hydrolysis reactions with water since
it is the “conjugate” of a strong acid,
which makes it very, very weak.
53
Salts that Yield Neutral Solutions
Na+ has no acidic or basic tendencies.
Na+ (aq) + H2O (l) ⇌ no reaction
Sodium ions DO NOT HAVE
hydrolysis reactions with water since
it is the “conjugate” of a strong base,
which makes it very, very weak.
54
Salts that Yield Acidic Solutions
The reaction of a strong acid with anions like
Cl-, Br-, I-, NO3-, and ClO4with a weak base will lead to an
acidic salt solution.
The solution is acidic because the anion
shows no acidic or basic tendencies, but
the cation does, as it is the conjugate acid
of a weak base.
55
Salts that Yield Acidic Solutions
Ammonium chloride (NH4Cl) will dissociate
into NH4+ and Cl- in water.
Cl- has no acidic or basic tendencies.
Cl- (aq) + H2O (l) ⇌ no reaction
Chloride ions DO NOT HAVE
hydrolysis reactions with water since
it is the “conjugate” of a strong acid,
which makes it very, very weak.
56
Salts that Yield Acidic Solutions
NH4+ has acidic tendencies.
That is:
NH4+ (aq) + H2O (l)⇌ NH3 (aq) + H3O+ (aq)
Ammonium ions hydrolyze in
water because it is the conjugate
acid of the weak base NH3, which
means ammonium is a weak
acid.
57
Salts that Yield Basic Solutions
The reaction of a strong base with cations
like Li+, Na+, K+, Ca2+, Sr2+, and Ba2+
with a weak acid will lead to an
basic salt solution.
The solution is acidic because the cation
shows no acidic or basic tendencies, but
the anion does, as it is the conjugate base
of a weak acid.
58
Salts that Yield Basic Solutions
Sodium fluoride (NaF) will dissociate into Na+
and F- in water.
Na+ (aq) + H2O (l) ⇌ no reaction
Sodium ions DO NOT HAVE
hydrolysis reactions with water since
it is the “conjugate” of a strong base,
which makes it very, very weak.
59
Salts that Yield Basic Solutions
F- has basic tendencies.
That is:
F- (aq) + H2O (l)⇌ HF (aq) + OH- (aq)
Fluoride ions hydrolyze in water
because it is the conjugate base
of the weak acid HF, which means
fluoride is a weak base.
60
Problem
Predict whether the following salt solution is neutral,
acidic, or basic and calculate the pH.
0.25 mol/L NH4Br – NH3 has a Kb value of 1.8 x 10-5
NH4+ (aq)
0.25
-x
0.25 – x
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
10
K a  5.56 x10

+
H2O (l)
N/A
N/A
N/A
[H3O ][NH3 ]

[NH4 ]

H3O+ (aq) +
0.0
+x
+x
10
so 5.56 x10
NH3 (aq)
0.0
+x
+x
(x)(x)

(0.25 x)
61
Problem
Initial acid [HA] / Ka ratio is
0.25 / 5.56 x 10-10  4.5 x 108
we can probably assume 0.25 >> x
5.56 x 10-10 = x2 / 0.25
x2 = (5.56 x 10-10)(0.25)
x2 = 1.39 x 10-10
x = 1.39 x 10-10
x = 1.18 x 10-5 mol/L
62
Problem
Negative answer not physically possible so
therefore, [H3O+] = 1.18 x 10-5 mol/L
x
NH 

4 i
1.2 x 10-5 M

 4.8 x 10-5  4.8 x 10-3 %
0.25M
Since we’ve shown the assumption is valid
pH = -log [H3O+] = - log 1.18 x 10-5 = 4.93.
63
Salts that Contain Acidic Cations and
Basic Anions
If a salt is composed of an
acidic cation
and a
basic anion,
the acidity or basicity of the salt
solution
depends on the relative
strengths of the acid and base.
64
Salts that Contain Acidic Cations
and Basic Anions
If the acid cation is
“stronger” than the base
anion, it “wins” and the salt
solution is acidic.
If the base anion is
“stronger” than the acid
cation, it “wins” and the salt
solution is basic.
65
Salts that Contain Acidic Cations
and Basic Anions
Ka > Kb
the acid cation is “stronger” and the salt solution is
acidic.
Ka < Kb
the base anion is “stronger” and the salt solution is
basic.
Ka  Kb
the salt solution is close to neutral.
66
Problem
Classify each of the following salts
as acidic, basic, or neutral:
a) KBr
b) NaNO2
c) NH4Br
d) NH4F
Ka for HF = 6.6 x 10-4
Kb for NH3 = 1.8 x 10-5
67
The Common-Ion Effect
Solutions consisting of both an acid and
its conjugate base are very important
because they are very resistant to
changes in pH. Such buffer solutions
regulate pH in a variety of biological
systems.
68
The Common-Ion Effect
Let’s consider a solution made of 0.10 moles of acetic
acid and 0.10 moles of sodium acetate with a total
volume of 1.00 L, making the initial [CH3COOH] =
[CH3COO-] = 0.10 mol/L.
First we must identify all potential acids and bases
in the system.
CH3COOH
acid
CH3COObase
Na+
neutral
H2O
acid
or base
69
Point of view of the acid
Our reaction will be
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
Ka = 1.8 x 10-5
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.10
-x
0.10 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.10
+x
+x
+x
0.10 + x
Note that the initial concentration of our product
CH3COO- is NOT ZERO!
70
[H 3O  ][CH 3COO ]
(x)(0.10  x)
5
Ka 
 1.8 x 10 
[CH 3COOH]
(0.10  x)
Let’s check the
initial acid concentration / Ka ratio.
0.10 / 1.8 x 10-5  5500
It’s probably safe to assume that
x << [HAc]i so [HAc]eqm  [HAc]I
and x << [Ac-]i so [Ac-]eqm  [Ac-]i
1.8 x 10-5 = x (0.10 + x) / (0.10 –x)
1.8 x 10-5 = x (0.10) / (0.10)
x = 1.8 x 10-5 mol/L
71
[H 3O  ][CH 3COO ]
(x)(0.10  x)
5
Ka 
 1.8 x 10 
[CH 3COOH]
(0.10  x)
At equilibrium,
[H3O+] = 1.8 x 10-5 mol/L
[CH3COO-] = 0.10 + 1.8 x 10-5 = 0.10 mol/L
[CH3COOH] = 0.10 - 1.8 x 10-5 = 0.10 mol/L
Assumption was valid! Check for yourself!
pH = - log [H3O+]
pH = - log 1.8 x 10-5
pH = 4.74
72
If we had started out with only 0.10 mol/L
acetic acid, the pH would be found from
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.10
-x
0.10 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.00
+x
+x
+x
+x
[H3O ][CH3COO ]
(x)(x)
5
Ka 
 1.8 x 10 
[CH3COOH]
(0.10 x)
73
The initial acid concentration / Ka will still be the
same, so we can assume
x << [HAc]i so [HAc]eqm  [HAc]i
1.8 x 10-5 = x2 / (0.10 –x)
1.8 x 10-5 = x2/ (0.10)
x = 1.8 x 10-6 mol/L
x = 1.3 x 10-3 mol/L (can’t be –ve)
pH = - log [H3O+]
pH = - log 1.3 x 10-3
pH = 2.89
74
Without the acetate ion the pH of
0.10 M acetic acid is 2.89.
With an equal concentration of acetate ion
present, the pH of
0.10 M acetic acid – 0.10 M acetate is 4.74
The acetate ion makes a large difference on
the equilibrium pH!
75
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
Adding the conjugate
base (a stress!) to the
equilibrium system of
an acid dissociation
shows the commonion effect, where the
addition of a common
ion causes the
equilibrium to shift.
This is an example of
Le Chatalier’s
Principle.
Addition of the weak
base to the acid
dissociation
76
Problem
Calculate the concentrations of all
species present, and the pH in a
solution that is 0.025 mol/L HCN and
0.010 mol/L NaCN.
(Ka of HCN = 4.9 x 10-10)
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HCN (aq) +
0.025
-x
0.025 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq)
0.0
+x
+x
+
CN- (aq)
0.010
+x
0.010 + x
77
Problem
[H3O ][CN ]
(x)(0.010 x)
10
Ka 
 4.9 x 10 
[HCN]
(0.025 x)
The initial base concentration / Ka ratio is
0.010 / 4.9 x 10-10  2 x 107
It’s probably safe to assume that
x << [HCN]i so [HCN]eqm  [HCN]I
and x << [CN-]i so [CN-]eqm  [CN-]i
78
Problem
4.9 x 10-10 = x (0.010 + x) / (0.025 –x)
4.9 x 10-10 = x (0.010) / (0.025)
x = 1.2 x 10-9 mol/L
So at equilibrium,
[H3O+] = 1.2 x 10-9 mol/L
[CN-] = 0.010 + 1.2 x 10-9 = 0.010 mol/L
[HCN] = 0.025 - 1.2 x 10-9 = 0.025 mol/L.
Assumption was valid! Check this for
yourself!
79
Problem
pH = - log [H3O+]
pH = - log 1.2 x 10-9
pH = 8.91
80
Buffer Solutions
Solutions that contain both a
weak acid and its conjugate
base are buffer solutions.
These solutions are resistant
to changes in pH.
81
Buffer Solutions
If more acid (H3O+) or base (OH-) is added
to the system, the system has enough of
the original acid and conjugate base
molecules in the solution to react with the
added acid or base, and so the new
equilibrium mixture will be
very close in composition to the original
equilibrium mixture.
82
Buffer solutions
A 0.10 molL-1 acetic acid – 0.10 molL-1
acetate mixture has a pH of 4.74 and is a buffer
solution!
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
(all in mol/L)
Initial
Conc. changes
Equil. conc.
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
0.10
N/A
0.0
0.10
-x
N/A
+x
+x
0.10 - x
N/A
+x
0.10 + x


[H3O ][CH3COO ]
K a  1.8 x 10 
[CH3COOH]
5
83
Buffer solutions
If we rearrange the Ka
expression to solve for
+
[H3O ]
[CH3COOH]
5 [CH3COOH]
[H3O ]  K a
 1.8 x 10

[CH3COO ]
[CH3COO ]

84
Buffer solutions
[CH3COOH]
5 [CH3COOH]
[H3O ]  K a
 1.8 x 10

[CH3COO ]
[CH3COO ]

Assume x << [HAc]i so [HAc]eqm  [HAc]I
and x << [Ac-]i so [Ac-]eqm  [Ac-]i,
and we should see
If [CH3COOH]i = [CH3COO-]i,
then [H3O+] = 1.8 x 10-5 M = Ka
and pH = pKa = 4.74
85
Buffer solutions
What happens if we add 0.01 mol of NaOH (strong
base) to 1.00 L of the acetic acid – acetate buffer
solution?
CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)
This reaction goes to completion and keeps
occurring until we run out of the limiting reagent OH(all in moles)
Initial
Change
Final (where x = 0.01
due to limiting OH-)
CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)
0.10
0.01
N/A
0.10
-x
-x
N/A
+x
0.10 – x
0.01 – x =
N/A
0.10 + x
= 0.09
0.00
= 0.11
New [CH3COOH] = 0.09 M and new [CH3COO-] = 0.11 M
86
Buffer solutions
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.09
-x
0.09 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.11
+x
+x
+x
0.11 + x
[H3O ][CH3COO ]
(x)(0.11 x)
5
Ka 
 1.8 x 10 
With the assumption
smaller
[CH3COOH] that x is much (0.09
 x) than
0.09
mol[CH
(anCOOH]
assumption we always
need to5
0.09

5
[H3Ocheck
]  K aafter 3calculations
 1.8 xare
10 done!), we
1.5 xfind
10 M

[CH COO ]
0.11
3
Note we’ve made the assumption that x << 0.09 M!
pH = - log [H3O+]
pH = - log 1.5 x 10-5
pH = 4.82
87
Buffer solutions
Adding 0.01 mol of OH- to 1.00 L of water
would have given us a pH of 12.0 because
there is no significant amount of acid in
water for the base to react with.
Our buffer solution resisted this change
in pH because there is a significant
amount of acid (acetic acid) for the
added base to react with.
88
Buffer solutions
What happens if we add 0.01 mol of HCl (strong acid)
to 1.00 L of the acetic acid – acetate buffer solution?
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
This reaction goes to completion and keeps
occurring until we run out of the limiting reagent H3O+
(all in moles)
Initial
Change
Final (where x = 0.01
due limiting H3O+)
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
0.10
0.01
N/A
0.10
-x
-x
N/A
+x
0.10 – x
0.01 – x =
N/A
0.10 + x
= 0.09
0.00
= 0.11
New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M
89
Buffer solutions
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.11
-x
0.11 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.09
+x
+x
+x
0.09 + x
[H3O ][CH3COO ]
(x)(0.09 x)
5
Ka 
 1.8 x 10 
With the assumption
that x is much smaller than
[CH3COOH]
(0.110.09
 x)mol
(an assumption we always need to check after
[CH3COOH]

5 0.11
5
are
done!),
we
find
[H3O ]  K a calculations

1.8
x
10

2.2
x
10
M

[CH3COO ]
0.09
Note we’ve made the assumption that x << 0.09 M!
pH = - log [H3O+]
pH = - log 2.2 x 10-5
pH = 4.66
90
Buffer solutions
Adding 0.01 mol of H3O+ to 1.00 L of
water would have given us a pH of 2.0
because there is no significant
amount of acid in water for the base
to react with.
Our buffer solution resisted this
change in pH because there is a
significant amount of base (acetate)
for the added acid to react with.
91
.
92
Buffer capacity
Buffer capacity is the measure of the ability
of a buffer to absorb acid or base without
significant change in pH.
Larger volumes of buffer solutions have a
larger buffer capacity than smaller
volumes with the same concentration.
Buffer solutions of higher concentrations
have a larger buffer capacity than a buffer
solution of the same volume with smaller
concentrations.
93
Problem
Calculate the pH of a 0.100 L buffer solution
that is 0.25 mol/L in HF and 0.50 mol/L in NaF.
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq) +
0.25
-x
0.25 – x

H2O (l)
N/A
N/A
N/A

H3O+ (aq)
0.0
+x
+x
+
F- (aq)
0.50
+x
0.50 + x

[H3O ][F ]
(x)(0.50 x)
4
Ka 
 3.5 x 10 
[HF]
(0.25 x)
With thexassumption
x is much
smaller
Assume
<< [HCN]i that
so [HCN]

[HCN]i
eqm
than 0.25 mol (an assumption we always need
-] so [CN-]
-]
and
x
<<
[CN

[CN
to check after calculations
areeqm
done!), wei find
i
94
Problem
[HF]
 4 [HF]
[H3O ]  K a   3.5 x 10

[F ]
[F ]
 4 0.25
4
 3.5 x 10
 1.75 x 10 M
0.50

pH = - log [H3O+]
pH = - log 1.75 x 10-4
pH = 3.76
95
Problem
a) What is the change in pH on addition of
0.002 mol of HNO3?
(all in moles)
Initial
Change
Final (where x = 0.002
due to limiting H3O+)
F- (aq)
0.050
-x
0.050 – x
= 0.048
+ H3O+ (aq) → H2O (l) +
0.002
N/A
-x
N/A
0.002 – x
N/A
= 0.00
HF (aq)
0.025
+x
0.025 + x
= 0.027
New [HF] = 0.27 M and new [F-] = 0.48 M
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq)
0.27
-x
0.27 – x
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.00
+x
+x
F- (aq)
0.48
+x
0.48 + x
[H3O ][F ]
(x)(0.48 x)
4
Ka 
 3.5 x 10 
[HF]
(0.27 x)
96
Problem
[HF]
 4 0.27
[H3O ]  K a   3.5 x 10
 1.97 x 104 M
[F ]
0.48

Notice we’ve made the assumption that
x << 0.27 M. We should check this!
pH = - log [H3O+]
pH = - log 1.97 x 10-4
pH = 3.71
97
Problem
b) What is the change in pH on addition of
0.004 mol of KOH?
(all in moles)
Initial
Change
Final (where x = 0.004
due to limiting OH-)
HF (aq)
0.025
-x
0.025 – x
= 0.021
+ OH- (aq) → H2O (l) +
0.004
N/A
-x
N/A
0.004 – x
N/A
= 0.00
F- (aq)
0.050
+x
0.050 + x
= 0.054
New [HF] = 0.21 M and new [F-] = 0.54 M
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq)
0.21
-x
0.21 – x
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.00
+x
+x
F- (aq)
0.54
+x
0.54 + x


[H
O
][F
]
(x)(0.54 x)
4
4
3
K a  3.5x10 
so 3.5x10 
[HF]
(0.21 x)
98
Problem
[HF]
 4 0.21
[H3O ]  K a   3.5 x 10
 1.36 x 104 M
[F ]
0.54

Notice we’ve made the assumption that
x << 0.21 M. We should check this!
pH = - log [H3O+]
pH = - log 1.36 x 10-4
pH = 3.87
99
The Henderson-Hasselbalch Equation
We’ve seen that, for buffer solutions
containing members of a conjugate acidbase pair, that
[acid]
[H3O ]  K a
[base]

 [acid] 
[acid]
  log K a  log
 log [H3O ]  log K a
[base]
 [base] 

pH = pKa + log [base] / [acid]
This is called the HendersonHasselbalch Equation.
100
The Henderson-Hasselbalch Equation
If we have a buffer solution of a conjugate
acid-base pair, then the pH of the solution
will be close to the pKa of the acid.
This pKa value is modified by the logarithm
of ratio of the concentrations of the base
and acid in the solution to give the actual
pH.
101
Problem
Use the Henderson-Hasselbalch Equation to
calculate the pH of a buffer solution prepared
by mixing equal volumes of 0.20 mol/L NaHCO3
and 0.10 mol/L Na2CO3.
We need the Ka and the concentrations of the
acid (HCO3-) and the base (CO32-).
Ka = 5.6 x 10-11
(we use the Ka for the second proton of H2CO3!).
102
Problem
NOTE: The concentrations we are given for
the acid and the base are the
concentrations
before the mixing of equal
volumes!
103
Problem
If we mix equal volumes, the total volume is
TWICE the volume for the original acid or base
solutions.
Since the number of moles of acid or base
DON’T CHANGE on mixing,
the initial concentrations we use will be
half the given values.
pH = pKa + log [base] / [acid]
pH = (-log 5.6 x 10-11) + log (0.05) / (0.10)
pH = 10.25 – 0.30
pH = 9.95
104