Acid-Base Equilibria
Review:
In pure water & all aqueous solutions:
H2O + H2O = H3O+ + OHKw =
[H3O+][OH-] = 1 x 10-14
[H3O+] = [OH-] =
[H3O+] > [OH-] =
1x10
14
or =
1 x 10-7
Acidic solution
[H3O+] < [OH-] = Basic solution
[H3O+] = [OH-] = Neutral solution
Solutions of Strong Acids/ Bases:
0.10 M HNO3 actually consists of 0.10 M H3O+
or [H+] = 0.10 M
0.10 M NaOH…. Adds 0.10 OH- to the water solution
Ionization & dissociation are 100% for strong acids & bases
 H3O+ + NO3-
HNO3 + H2O
GONE
GONE

NaOH + H2O
In 0.10M Ba(OH-)
In 0.20 M H2SO4
2
Na+ (aq) + OH-(aq)
[OH-] = 0.20M
[H+] = 0.40 M
Solutions of Weak acids/bases
0.10 M HC2H3O2 : a weak acid - when it reacts with
water it does not ionize 100%.
What is the [H3O+] for this solution? Ka = 1.8 x 10-5
Equation for the ionization of HC2H3O2 in water:
HC2H3O2 + H2O = H3O+ + C2H3O2Write the MAE, equilibrium expression of this reaction:
Ka =
[H3O+] [C2H3O2-]
[HC2H3O2]
=
1.8 x 10-5
Weak acids (continue)
Ka =
[H3O+] [C2H3O2-]
[HC2H3O2]
= 1.8 x 10-5
[H3O+] = [C2H3O2-] = x
[HC2H3O2 ] = 0.10 - x
x2
.10  x
ignore
= 1.8 x 10-5
“rule of thumb”… if %ionization < 5%, the value
for x is so small it can be ignored. (otherwise you
would need to use quadratic equation to solve).
% ionization < 5% when: [HA] / Ka > 100
0.10  1.8 x 10-5 = 5,556 !
Weak acids (continue)
So….
x 2
.10  x
= 1.8 x 10-5
Also = [C2H3O2- ]
x = 1.3 x 10-3 = [H3O+]
% ionization =
1.3 x 10-3
.10
x 100 = 1.3%
< 5% as predicted
Weak acids ionize very little….
there is very little H3O+ present.
Weak bases….
B + H2O
NH3
= BH+
+ OH-
NH4+ + OH-
+ H2O =
Given: 0.35M NH3 solution; Kb = 1.8 x 10-5
What is the concentration of OH- ?
2
[NH4+][OH]
[NH3]
x=
=
x
.35  x
2.5 x 10-3 = [OH-]
=
1.8 x 10-5
ignore…<5%
Weak bases (continue)
If the [OH-] = 2.5 x 10 -3 , what is the [H3O+]?
Given: [H3O+][OH-] = 1 x 10-14
[H3O+][2.5 x 10 -3] = 1 x 10 -14
[H3O+] = 4.0 x 10-12
Are we saying that this solution has both H3O+ & OH - in it? Yes
All acid or base solutions have them….Why? They have H2O !
Why is this solution Basic?
[OH-] > [H3O+]
pH
A simple scale for ranking the H3O+ concentrations
of dilute acid/base solutions. (Sorenson).
pH = - Log [H3O+ ]
The logarithm of a number is that number expressed as
an exponent of the base 10.
For example, the logarithm of 1 is 0, 1 x 100.
If [H3O+] = 0.001 = 1x10-3
This is the pH
pH = -log 0.001 = 3
If this is 1
pH Scale
1M HA
1M BOH
Acidity increases
Basicity increases
neutral
pH
0 1 2 3 4 5 6
7
8
9
10
11
12
13
14
[H3O+] 100 10-1 10-2 10-310-410-5 10-6 1x10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14
[OH-]
10-12
pOH
12
Formulas to
remember
10-8
8
1x10-7 10-6
7
6
[H3O+][OH-]= 1x10-14
pH = - log[H3O]
pH + pOH = 14
10-3
3
100
More About the pH Scale
Given a 3.25 x 10-4 M HNO3 , solve for:
a. [H3O+]
HNO3 is a strong acid. Thus,
3.25 x 10-4M HNO3 = prediction
[H3O+] = 3.25 x 10- 4
b. pH = -log [3.25 x 10-4] = -(-3.488) = 3.49
-14
1
x
10
c. [OH-] =
3.25 x 10-4
= 3.08 x 10-11
d. pOH = -log [OH-] = -log [3.08 x 10-11] = 10.5
Or
pOH = 14.00 –3.49 = 10.51
Sample Problem
What is the pH of a 0.10 M HC2H3O2 solution?
Ka =
[H3O+] [C2H3O2-]
[HC2H3O2 ]
[H3O+] = [C2H3O2-]
[HC2H3O2 ] = 0.10
x2
.10  x
= 1.8 x 10-5
=
x
- x
= 1.8 x 10-5
x = 1.3 x 10-3 = [H3O+]
ignore
pH = - Log [1.3 x 10-3] = 2.9