EQUILBRIUM OF
Acids and Bases
Chapter 17
1
Water
AUTOIONIZATION
H2O can function as both an ACID
and a BASE.
Equilibrium constant for water
= Kw
Kw = [H3O+] [OH-] =
1.00 x 10-14 at 25 oC
2
Water
3
OH-
H3O+
Autoionization
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
so [H3O+] = [OH-] = 1.00 x 10-7 M
[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 25 oC
pH = -log (1.00 x 10-7)
= - (-7) = 7
4
[H3O+],
[OH-]
and pH
What is the pH of the
0.0010 M NaOH solution?
[H3O+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
General conclusion —
Basic solution pH > 7
Neutral
pH = 7
Acidic solution pH < 7
5
[H3O+], [OH-] and pH
If the pH of Coke is 3.12, it is ________.
log [H3O+] = - pH
Take antilog and get
[H3O+] = 10-pH
[H3O+] = 10-3.12 =
7.6 x 10-4 M
6
pX Scales
In general
pX = -log X
pOH = - log [OH-]
pH = - log [H+]
pKw = 14 = pH + pOH
7
Weak Base
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
8
Equilibria Involving
Weak Acids and Bases
9
Aspirin is a good
example of a
weak acid,
Ka = 3.2 x 10-4
10
Weak Acids and Bases
Acid
Conjugate Base
acetic, CH3CO2H CH3CO2-, acetate
ammonium, NH4+ NH3, ammonia
bicarbonate, HCO3CO32-, carbonate
A weak acid (or base) is one
that ionizes to a VERY small
extent (< 5%).
Weak Acids and Bases
11
acetic acid, CH3CO2H (HOAc)
HOAc + H2O D H3O+ + OAcAcid
Conj. base
+
[H3O ][OAc ]
-5
Ka 
 1.8 x 10
[HOAc]
(K is designated Ka for ACID)
[H3O+] and [OAc-] are SMALL, Ka << 1.
Equilibrium Constants
for Weak Acids
12
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
13
0.0001 M
0.003 M
0.06 M
2.0 M
a pH meter, Screen 17.9
Calculations with
Equilibrium
Constants
pH of an acetic
acid solution.
What are your
observations?
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the pH.
And the equilibrium concs. of EACH
Step 1. ICE table.
[HOAc]
I
C
E
[H3O+]
[OAc-]
14
Equilibria Involving A Weak Acid
[HOAc]
I
1.00
C
-x
E
1.00-x
[H3O+]
[OAc-]
0
0
+x
+x
x
x
Note that we neglect [H3O+] from H2O.
15
Equilibria Involving A Weak Acid
16
Step 2. Write Ka expression
+ ][OAc- ]
2
[H
O
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
This is a quadratic.
Use quadratic formula or
method of approximations
(see Appendix A).
HOWEVER
Equilibria Involving A Weak Acid
Assume x is very small because Ka is so small.
+
[H3O ][OAc ]
-5
Ka  1.8 x 10
=

[HOAc]
-5
Ka  1.8 x 10
=
2
x
1.00 - x
2
x
1.00
Now we can more easily solve this
approximate expression.
17
18
Equilibria Involving A Weak Acid
Step 3. Solve Ka approximate expression
2
x
-5
Ka  1.8 x 10
=
1.00
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = -log (4.2 x 10-3) = 2.37
Equilibria Involving A Weak Acid
19
For many weak acids
[H3O+] = [conj. base] = [Ka • Co]1/2
where C0 = initial conc. of acid
Useful Rule of Thumb:
If 100•Ka < Co,
then [H3O+] = [Ka•Co]1/2
Equilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of
formic acid, HCO2H.
HCO2H + H2O D HCO2- + H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.4
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.5
20
Weak Bases
Equilibrium Constants
for Weak Bases
22
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O D NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1; ICE table
[NH3]
I
C
E
[NH4+]
[OH-]
23
Weak Base
24
Step 1. ICE table
[NH3]
[NH4+]
[OH-]
0
0
I
0.010
C
-x
+x
+x
E
0.010 - x
x
x
Weak Base
25
Step 2. Solve the equilibrium expression
2
+ ][OH- ]
[NH
x
-5
4
Kb  1.8 x 10 =
=
[NH3 ]
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
[NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid
!
26
Acids
Conjugate
Bases
27
Relation
of Ka, Kb,
[H3O+]
and pH
28