Acid and Base Equilibria
Electrolytes
Strong
Conduct electricity
 Weak
Poor conductors of
electricity
 Nonelectrolytes
Do not conduct
electricity

Strong Electrolytes




Completely ionize or dissociate in dilute
aqueous solutions.
Strong acids
Strong soluble bases
Most soluble salts
HCl + H2O  H3O+ + ClKOH  K+ + OH-
Concentration of Ions

Calculated directly
from molarity of
strong electrolyte
initial
change
final
HCl

0.25 M
-0.25 M
0M

What is the
concentration of
hydrogen ion in a
0.25 M HCl solution?
H+ +
Cl-
0.25M 0.25M
0.25M 0.25M
Concentration of Ions

Calculated directly
from molarity of
strong electrolyte
initial
change
final

What is the
concentration of
hydroxide ion in a
0.25 M Ba(OH)2
solution?
Ba(OH)2  Ba2+ + 2OH0.25 M
-0.25 M
0.25M 2(0.25M)
0M
0.25M 0.50M
Weak Electrolyte




Slightly ionized in dilute aqueous
solutions.
Weak acids
Weak bases
A few covalent salts
Auto-Ionization of Water
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)
Kw = [H3
+
O ][OH ]
OH-
H3O+
H2O + H2O
H3O+ + OH
Auto-Ionization of Water
OH-
H3O+
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25oC
In a neutral solution [H3O+] = [OH-]
and so [H3O+] = [OH-] = 1.00 x 10-7 M
H2O + H2O
H3O+ + OH
Kw
Kw = [H3O+] [OH-]
14
5
1.00 x10
OH 
OH 



 1.00 x10
14
1.00 x10

5
1.00 x10
9
 1.00 x10 M
OH 

The Auto-Ionization of Water

Calculate the concentrations of H3O+ and
OH- in 0.050 M HCl.
HCl + H2 O  H3O+  Cl
0.050M

thus H3O
0.050M 0.050M
+
  0.050M
The Auto-Ionization of Water

Calculate the concentrations of H3O+ and
OH- in 0.050 M HCl.
HCl + H2 O  H3O+
0.050M

0.050M
Cl
0.050M
 
H O OH   1.0  10
1.0  10
1.0  10
OH   H O   5.0  10
OH   2.0  10 M
thus H3O+  0.050M
3
+

14
14

3

+
13
14
2
[H3O+], [OH-] and pH

A common way to express acidity and
basicity is with pH
pH = - log [H3
+
O ]
In a neutral solution,
[H3O+] = [OH-] = 1.00 x 10-7 at 25oC
pH = -log (1.00 x 10-7) = - (-7)
pH = 7.00

[H3O+], [OH-] and pH
What is the pH of the
1.00 x10 14  H   0.001
0.0010 M NaOH solution?  1.00 x1014
H   0.001
+
-11
[H3O ] = 1.0 x 10 M

H    1.00 x10 11 M
pH = - log (1.0 x 10-11) = 11.00
General conclusion —
Basic solution pH > 7
Neutral
pH = 7
Acidic solution pH < 7
[H3O+], [OH-] and pH
If the pH of Coke is 3.12, What is [H3O+]?
Because pH = - log [H3O+] then
log [H3O+] = - pH
Take antilog and get
+
[H3O ]
=
-pH
10
[H3O+] = 10-3.12
[H3O+] = 7.6 x 10-4 M
Other pX Scales
In general
pX = -log X
and so
pOH = - log [OH-]
Kw = [H3O+] [OH-] = 1.00 x 10-14
Take the -log of both sides
-log (10-14) = - log [H3O+] + (-log [OH-])
14 = pH + pOH
pKa Values

The Ka of acetic acid is 1.8 x 10-5, what
is the pKa?
pKa = -log Ka = -log (1.8 x 10-5) = 4.74
pH
+
[H ]
[OH ]
pH
0 1
2
3
4 5 6 7 8 9 10 11 12 13 14
The pH and pOH scales

Develop familiarity with pH scale by looking
at a series of solutions in which [H3O+] varies
between 1.0 M and 1.0 x 10-14 M.
[H3O+]
1.0 M
-3
1.0 x 10 M
1.0 x 10-7 M
2.0 x 10-12 M
-14
1.0 x 10 M
[OH-]
-14
1.0 x 10 M
-11
1.0 x 10 M
1.0 x 10-7 M
5.0 x 10-3 M
1.0 M
pH
0.00
3.00
7.00
11.70
14.00
pOH
14.00
11.00
7.00
2.30
0.00
The pH and pOH scales

Calculate [H3O+], pH, [OH-], and pOH for
0.020 M HNO3 solution.
100%

HNO 3  H 2 O  H 3O  NO
0.020 M
H O   2.0 10

3
3
0.020 M 0.020 M
2


M pH  -log 2.0  10  2 M  1.70
The pH and pOH scales

Calculate [H3O+], pH, [OH-], and pOH for
0.020 M HNO3 solution.
100%
HNO3  H2 O 

 H3O   NO-3
0.020M
0.020M 0.020M
H O   2.0  10 M pH  -log2.0  10 M   1.70
H O OH   1.0  10
1.0  10
1.0  10
OH   H O   2.0  10  5.0  10 M
pOH   log5.0  10   12.30

3

3
2

2
14
14

3
14

2
13
13
pH Examples
Ionization Constants for Weak
Monoprotic Acids


Let’s look at the dissolution of acetic acid, a
weak acid, in water as an example.
The equation for the ionization of acetic acid
is:


CH 3COOH  H 2 O  H 3O  CH 3COO
Kc

H O CH COO 

3

3
CH3COOH

Ionization Constants for Weak
Monoprotic Acids and Bases

Write the equation for the ionization of the
weak acid HCN and the expression for its
ionization constant.
HCN 
Ka
+
H
+
CN

H CN 

 4.0  10
+
HCN
-
10
Equilibria Involving
Weak Acids and Bases
Consider acetic acid
HOAc + H2O
Acid

H3O+
+
OAcConj. base
Equilibria Involving
Weak Acids and Bases
Consider acetic acid
HOAc + H2O

Acid
H3O+
+
OAcConj.
base
[H3O+ ][OAc- ]
Ka 
 1.8 x 10-5
[HOAc]
(K is designated Ka for ACID)
Because [H3O+] and [OAc-] are SMALL, Ka << 1.
Ka From Equilibrium
Concentrations

A solution of a weak acid has the
following concentrations at equilibrium.
What is the Ka? [HA] = 0.049 M; [H+]
= [A-] = 0.00084 M
HA + H2O H3O+ + AKa = [H3O+][A-]/[HA]
Ka = (0.00084)2/0.049
Ka = 1.4 x 10-5
Ka From Percent Ionization

In a 0.0100 M solution, acetic acid is
4.2% ionized. What is +the Ka?
-
HA + H2O H3O + A
initial
0.0100
change -0.00042
0.00042 0.00042
equil
0.00958
0.00042 0.00042
Ka = [H3O+][A-]/[HA]
Ka = (0.00042)2/0.00958
Ka = 1.8 x 10-5
Ka From pH
The pH of a 0.115 M weak acid is 1.92.
What is the Ka?
HA + H2O H3O+ + A[H3O+] = 10-1.92 = 0.012 M
initial
0.115
change -0.012
0.012
0.012
equil
0.103
0.012
0.012
Ka = [H3O+][A-]/[HA]
Ka = (0.012)2/0.103
Ka = 1.4 x 10-3

Weak Acid
You have 1.00 M HOAc. Calc. the equilibrium
concs. of HOAc, H3O+, OAc-, and the pH.
Step 1. Define equilibrium concs.
[HOAc]
Initial
Change
Equilib
[H3O+]
[OAc-]
Weak Acid
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc-,
and the pH.
Step 1. Define equilibrium concs.
[HOAc] [H3O+] [OAc-]
Initial
1.00
0
0
Change -x
+x
+x
Equilib
1.00-x
x
x

You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 2. Write Ka expression
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 2. Write Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 2. Write Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x

This is a quadratic. Solve using
quadratic formula or method of
approximations
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x

First assume x is very small because Ka
is so small.
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x

First assume x is very small because Ka
is so small. K  1.8 x 10-5 = x2
a
1.00
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka expression
+
2
[H
O
][OAc
]
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x

First assume x is very small because Ka
2
is so small.
x
-5
Ka  1.8 x 10

=
1.00
And so x = [H3O+] = [OAc-] = [Ka •
1.00]1/2
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka approximate
expression
2
x
Ka  1.8 x 10-5 =
1.00

x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka approximate
expression
2
x
Ka  1.8 x 10-5 =
1.00


x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
You have 1.00 M HOAc. Calc. the equilibrium concs. of
HOAc, H3O+, OAc-, and the pH.
Weak Acid

Step 3. Solve Ka approximate expression
2
x
Ka  1.8 x 10-5 =
1.00




x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3)
pH = 2.37
Percent Ionization
Calculate the percent ionization of a
0.10 M solution of acetic acid.
HA + H2O H3O+ + Ainitial
0.10
change -x
x
x
equil
0.10 - x
x
x
Ka = [H3O+][A-]/[HA] = x2/0.10 - x
1.8 x 10-5 = x2/0.10 - x = x2/0.10
x = 1.3 x 10-3

Calculations Based on
Ionization Constants


Calculate the concentrations of the
species in 0.15 M hydrocyanic acid,
HCN, solution.
Ka = 4.0 x 10-10 for HCN
You do it!
Calculations Based on
Ionization Constants

Note that the properly applied simplifying
assumption gives the same result as solving
the quadratic equation does.
x x   4.0  10 10
0.15  X 
x 2  4.0  10 10 x  6.0  10 11  0
a
x 
b
b
b 2  4ac
2a
c
Calculations Based on
Ionization Constants
 4.0  10   4.0  10 
x 
10 2
10
21
11



 4 1  6.0  10 
x  7.7  10 and - 7.7  10
6
-5
7.7 x 10-6 is the VALID chemical solution!
Calculations Based on
Ionization Constants

Let’s look at the percent ionization of two weak acids
as a function of their ionization constants.
Solution
Ka
[H+]
pH % ionization
-5
-3
0.15 M 1.8 x 10 1.6 x 10 2.80
1.1
CH3COOH
-10
-6
0.15 M 4.0 x 10 7.7 x 10 5.11
0.0051
HCN

Note that the [H+] in 0.15 M acetic acid is 210 times
greater than for 0.15 M HCN.
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O

NH4+ + OHKb = 1.8 x 10-5
Step 1. Define equilibrium concs.
[NH3]
[NH4+]
[OH-]
initial
0.010
0
0
change
-x
+x
+x
equilib
0.010 - x
x
x
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O

NH4+ + OHKb = 1.8 x 10-5
Step 1. Define equilibrium concs.
[NH3]
[NH4+]
[OH-]
initial
0.010
0
0
change
-x
+x
+x
equilib
0.010 - x
x
x
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
2
+ ][OH- ]
[NH
x
4
Kb  1.8 x 10-5 =
=
[NH3 ]
0.010 - x
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
-5
Kb  1.8 x 10
[NH4+ ][OH- ]
x2
=
=
[NH3 ]
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
2
+ ][OH- ]
[NH
x
4
Kb  1.8 x 10-5 =
=
[NH3 ]
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
and [NH3] = 0.010 - 4.2 x 10-4 = 0.010 M
Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
Example 18-15

The pH of household ammonia is 11.50.
What is its molarity?
+
-
NH3 + H2O NH4 + OH
pOH = 14 - 11.50 = 2.50
[OH-] = 10-2.50 = 0.0032 M
initial
x
change
- 0.0032
0.0032 0.0032
equil
x - 0.0032
0.0032 0.0032
Kb = [NH4+][OH-]/[NH3 ] = (0.0032 )2/x - 0.0032
1.8 x 10-5 = (0.0032 )2/x
x = 0.57 M
Polyprotic Acids

Many weak acids contain two or more acidic
hydrogens.



polyprotic acids ionize stepwise
ionization constant for each step
Consider arsenic acid, H3AsO4, which has
three ionization constants
1
2
3
K1=2.5 x 10-4
K2=5.6 x 10-8
K3=3.0 x 10-13
Polyprotic Acids

The first ionization step is



H 3AsO4  H  H 2AsO4
H 



K1

H 2AsO4
 H 3AsO4 
  2.5  10
4
Polyprotic Acids

The second ionization step is
H 2AsO4

2

 H  HAsO4





K2
2
H HAsO4
2H 2AsO4

  5.6  10
8
Polyprotic Acids

The third ionization step is
2HAsO4

3

 H  AsO4





K3
3
H AsO4
2HAsO4

  3.0  10
13
Polyprotic Acids

Notice that the ionization constants vary in the
following fashion:
K1  K2  K3

This is a general relationship.
Polyprotic Acids
Calculate the concentration of all species in
0.100 M arsenic acid, H3AsO4, solution.
1 Write the first ionization ionization step and
represent the concentrations.


3
4
2
4


H AsO  H  H AsO
0.100  x M
xM
xM
Polyprotic Acids
2 Substitute into the expression for K1.

H H AsO 

 2.5  10

K1
2

4
H 3AsO 4 

x  x 
4
K1 
 2.5  10
0.10  x 
4
4
5
x  2.5  10 x  2.5  10  0
2
simplifying assumption does not apply
Polyprotic Acids
3 Use the quadratic equation to solve for x,
and obtain two values
x
4
 2.5  10 
2.5 10 
4 2
21

 41 2.5  10 5
x  5.1 10 3 M and x  4.9 10 3 M
H   H AsO   xM  4.9 10

2

4
H 3AsO 4   0.100  x M  0.095M
3
M

Polyprotic Acids
4 Now we write the equation for the second
step ionization and represent the
concentrations.
H2AsO4
+
2
 H + HAsO4
from 1st step (4.9  10 M)
-3
algebraically (4.9  10  y) M yM
-3
yM
Polyprotic Acids
5 Substitute into the second step ionization
expression.
K2

=


2
H 3O HAsO 4

H 2 AsO 4


4.9  10  y  y 

=
 4.9  10  y 
-3
K2
  5.6  10
-3
apply assumption
8
Polyprotic Acids
K2

=
H 3 O  H A sO 24 
K2

=
4 .9  1 0 -3  y



H 2 A sO 4
4 .9  1 0 -3

  5 .6  1 0
8
 y 
 y
ap p ly assu m p tio n
4 .9  1 0   y 

K =
 5.6  1 0
 4 .9  1 0 
y  5.6  1 0 M   H 
  H A sO 
n o te  H     H 
-3
8
2
-3
8

2nd


1 st
2nd
2
4
Polyprotic Acids
6 Now we repeat the procedure for the third
ionization step.

HAsO24

5.6  10-8
H

 4.9 10  5.6 10 M
 z M
zM
1st and 2nd ionizations 5.6  10-8 M
algebraically


-3
AsO34
-8
zM
Polyprotic Acids
7 Substitute into the third ionization
expression.
K3

=


3
H 3O AsO 4
2
HAsO 4


  3.0  10
4.9  10  5.6  10  z  z 

=
5.6  10  z 
3
K3
13
apply assumption
8
8
Polyprotic Acids
H O  A s O 

K =
 3 .0  1 0
 H A sO 
4 .9  1 0  5 .6  1 0  z   z 

K =
 5 .6  1 0  z 

3
4
3
13
2
4
3
3
8
8
3
a p p ly a s s u m p tio n

  z   3 .0  1 0
 5 .6  1 0 
 z   3 .4  1 0 M   H 
4 .9  1 0  3
13
8
18

3 rd

 A s O 34 

Polyprotic Acids

Use Kw to calculate the [OH-] in the 0.100 M
H3AsO4 solution.


14
 H OH   10.  10
10
.  10
10
.  10
OH    H   4.9  10
14

14

OH   2.0  10

12
M
3
Polyprotic Acids

A comparison of the various species in 0.100
M H3AsO4 solution follows.
Species
H3AsO4
+
H
H2AsO4
2HAsO4
3AsO4
OH
Concentration
0.095 M
0.0049 M
0.0049 M
-8
5.6 x 10 M
-18
3.4 x 10 M
-12
2.0 x 10 M
Solvolysis


Solvolysis is the
reaction of a
substance with the
solvent in which it is
dissolved.
Hydrolysis refers to
the reaction of a
substance with
water or its ions.
Hydrolysis

…the reaction of
a substance with
water or its ions
A- + H2O HA + OHBH+ + H2O B + H3O+
Ah ha,
BASIC!
Hydrolysis


Hydrolysis refers to the reaction of a
substance with water or its ions.
Combination of the anion of a weak
acid with H3O+ ions from water to form
nonionized weak acid molecules.
A- + H2O HA + OHH2O H+ +OH-
Hydrolysis
NaCN
Na+ + H2O NR
CN- + H2O HCN + OHBasic
HCN + NaOH NaCN + H2O
Hydrolysis
NH4Cl
NH4+
Cl- + H2O NR
+ H2O NH3 + H3O+
Acidic
HCl + NH3 NH4Cl
Hydrolysis
NH4CN
NH4+ + H2O NH3 + H3O+
CN- + H2O HCN + OH-
Acidic or Basic
HCN + NH4OH NH4CN + H2O
Salts of Weak Bases and Weak
Acids

Acidic Solution
Ka > Kb
NH4F  NH4+ + FNH4 + H2O  NH3 +
+
F-
H3
+
O
OH
+ H2O  HF +
Ka = 7.2 x 10-4 and Kb = 1.8 x 10-5
Salts of Weak Bases and Weak
Acids

Neutral Solution
Ka = Kb
NH4CH3COO  NH4+ + CH3COO-
H3O+
CH3COO- + H2O  CH3COOH + OHNH4+ + H2O  NH3 +
Ka = 1.8 x 10-5 and Kb = 1.8 x 10-5
Salts of Weak Bases and Weak
Acids

Basic Solution
Ka < Kb
NH4CN  NH4+ + CNNH4+ + H2O  NH3 +
H3O+
+ H2O  HCN + OH
Ka = 4.0 x 10-10 and Kb = 1.8 x 10-5
CN-
Salts of Weak Bases and Weak
Acids



Acidic Solution
Ka > Kb
Basic solution
Kb > Ka
Neutral solution
Ka = Kb
Hydrolysis



The conjugate base of a strong acid is a
very weak base.
The conjugate base of a weak acid is a
stronger base.
Hydrochloric acid, a typical strong acid,
is essentially completely ionized in
dilute aqueous solutions.

HCl  H 2 O   H 3O  Cl
~100%

Hydrolysis


The conjugate base of HCl, the Cl- ion,
is a very weak base.

Cl  H 3O  No rxn. in dilute aqueous solutions

True of all strong acids and their anions.
Hydrolysis



HF, a weak acid, is only slightly ionized in dilute
aqueous solutions.
Its conjugate base, the F- ion, is a much stronger
base than the Cl- ion.
F- ions combine with H3O+ ions to form nonionized
+
HF.

HF + H 2 O 

H 3O  F
only slightly
F- + H 3O + 

 HF + H 2 O
nearly completely
Salts of Strong Soluble Bases
and Strong Acids


Salts made from strong acids and strong
soluble bases form neutral aqueous solutions.
An example is potassium nitrate, KNO3, made
from nitric acid and potassium hydroxide.

NO3 ( s)

NO3
  K 

+
H 2O  H 2O 

OH + H 3O
K
+
~100% in H 2 O

no rxn. to upset H 3O
+
+
OH   neutral
-
Salts of Strong Soluble Bases
and Weak Acids
Salts made from strong soluble bases and weak acids
hydrolyze to form basic solutions.


Anions of weak acids (strong conjugate bases) react with
water to form hydroxide ions
An example is sodium hypochlorite, NaClO, made
from sodium hydroxide and hypochlorous acid.


Na  ClO
+

H2O + H2O 
 OH + H3O
Na
+
ClO ( s)
~100% in H2O
-
Salts of Strong Soluble Bases
and Weak Acids

Na ClO ( s)   Na  ClO
 OH - + H O +
H O + H O 

+
~100% in H 2 O
-
2
2

-
3
ClO  H 3O  HClO  H 2O
-

Combine these equations into one single
equation that represents the reaction:


ClO  H 2 O 
 HClO  OH
-
Acid-Base Properties of Salts
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
Reaction of NH4+ with H2O
NH4+ + H2O  NH3 + H3O+
acid
base base acid
NH4+ ion is a moderate acid because its
conjugate base is weak.
Therefore, NH4+ is acidic solution
Salts of Weak Bases
and Strong Acids


Salts made from weak bases and strong acids
form acidic aqueous solutions.
An example is ammonium bromide, NH4Br,
made from ammonia and hydrobromic acid.

4

4
NH Br s   NH  Br


H 2O  H 2O 

OH  H3O
 NH  H O
NH  OH- 
4
-
H 2O ~100%
3
-
2

generates excess H3O solution is acidic
Hydrolysis Constant

Relationship of
conjugate acidbase pairs
Kw = KaKb
Ka
Kb
Example

The Ka for acetic acid, CH3COOH, is 1.8
x 10-5, calculate the Kb for the acetate
ion, CH3COO-.
14
Kw 1.00 x10
10
Kb 


5
.
6
x
10
5
Ka
1.8 x10
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O  NH3 +
H3O+
acid base
base
acid
Ka = 5.6 x 10-10
Step 1.
Set up concentration table
[NH4+]
[NH3]
[H3O+]
initial
0.10
0
0
change
-x
+x
+x
equilib
0.10-x
x
x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O  NH3 + H3O+
acid base base acid
Ka = 5.6 x 10-10
Ka  5.6 x10
10

NH 3 H 3O

NH 

4


2
x
0.10  x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O  NH3 + H3O+
acid base base acid
Ka = 5.6 x 10-10





NH
H
O
10
3
3
Ka  5.6 x10 

NH 

4
Assume 0.10-x = 0.10
x2
0.10  x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O is neutral
NH4++ H2O  NH3 + H3O+
acid base base acid
Ka = 5.6 x 10-10





NH
H
O
10
3
3
Ka  5.6 x10 

NH 

4
x2
0.10  x
Assume 0.10-x = 0.10
x = [NH3] = [H3O+] = 7.5 x 10-6 M
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NH4Cl.
Cl- + H2O neutral

NH4+ +
H2O
acid
base
Ka = 5.6 x 10-10
Calculate the pH
Step 3.
NH3
+
base
[H3O+] = 7.5 x 10-6 M
pH = -log [H3O+] = 5.13
H3O+
acid
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O neutral
CH3COObase
Step 1.
initial
change
equilib
H2O
 CH3COOH
acid
acid
Kb = 5.6 x 10-10
Set up concentration table
[CH3COO- ]
[CH3COOH]
+
+ OHbase
[OH-]
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O neutral
CH3COObase
Step 1.
initial
change
equilib
H2O
 CH3COOH
acid
acid
Kb = 5.6 x 10-10
Set up concentration table
[CH3COO- ]
[CH3COOH]
0.10
0
-x
+x
0.10-x
x
+
+ OHbase
[OH-]
0
+x
x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of
NaCH3COO.
Na+ + H2O neutral
+
base
acid
acid
base
Kb = 5.6 x 10-10
Solve the equilibrium expression
Step 2.
H2O

CH3COO-
CH3COOH + OH-
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of
NaCH3COO.
Na+ + H2O neutral
CH3COObase
Step 2.
Kb  5.6 x10
10
H2O
 CH3COOH + OHacid
acid
base
Kb = 5.6 x 10-10
Solve the equilibrium expression
+

CH 3COOH OH

CH COO 

3


2
x
0.10  x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O neutral
CH3COObase
Step 2.
H2O
 CH3COOH + OHacid
acid
base
Kb = 5.6 x 10-10
Solve the equilibrium expression
+





CH
COOH
OH
10
3
Kb  5.6 x10 

CH COO 

3
Assume 0.10-x = 0.10
x2
0.10  x
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O neutral
CH3COObase
Step 2.
H2O
 CH3COOH + OHacid
acid
base
Kb = 5.6 x 10-10
Solve the equilibrium expression
+
Kb  5.6 x10 10
CH COO OH   x

CH COO  0.10  x


3

3
Assume 0.10-x = 0.10
x = [CH3COO] = [OH-] = 7.5 x 10-6 M
2
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of NaCH3COO.
Na+ + H2O neutral
H2O
 CH3COOH
acid
acid
Kb = 5.6 x 10-10
Step 3.
Calculate the pH
[OH-] = = 7.5 x 10-6 M
pOH = -log [OH-] = 5.13
pH + pOH = 14.00
pH = 8.87
CH3COObase
+
+ OHbase