Kw, pH and pOH
Chemistry 12
◊
Chapter 14
Self-ionization of Water:
All aqueous solutions contain ions!
 Even pure water contains a few ions that are
produced by the dissociation of water molecules

Self-ionization of Water continued….
At 25°C only two water molecules in a billion
dissociate
 Chemists have determined the concentration of
H3O+ in pure water to be 1.0 x 10-7 M and OHions in pure water is 1.0 x 10-7 M as well.
 Because this dissociation is an equilibrium, we
can write an equilibrium expression….called….

Ion Product Constant of Water (Kw):
Kw = [H3O +][OH-]
= (1.0 x 10-7 M)(1.0 x 10-7 M)
= 1.0 x 10-14
(changes with temperature)
Note: water is not included because it is
liquid!
***Important because if we know the
concentration of either H3O + or OH-,
we can solve for the other using Kw!!!
[H3O+ ] and [OH- ] in aqueous solutions:

Acid solution: [H3O +] is greater than 1.0 x 10-7 M
and [OH-] less than 1.0 x 10-7 M

Basic solution: [OH -] is greater than 1.0 x 10-7 M
and [H3O +] less than 1.0 x 10-7 M

Neutral solutions concentrations of both equal 1.0
x 10-7 M
Acid – Base Concentrations:
concentration (moles/L)
10-1
pH = 3
pH = 11
OH-
H3O+
pH = 7
10-7
H3O+
OH-
OH-
H3O+
10-14
[H3O+] > [OH-]
[H3O+] = [OH-]
acidic
solution
neutral
solution
[H3O+] < [OH-]
basic
solution
pH Scale:
Used to describe acids and bases quantitatively
 This scale measures the strength of an acid or base
 The concentrations H3O+ ions range from 1o M to
10-15 M

Thus, this scale uses logarithms to convert to more “user
friendly” numbers
 Since, the pH scale is logarithmic 1 unit pH change means
tenfold change in the H+ ion concentration.


Uses Sørensen's pH definition, concentration of H+ is usually
confined to 1 to 10-14M range. Thus pH scale contains values
falling between 0 and 14. In some rare cases you may see pH
lower than 0 or higher than 14, when the concentration of H+
take some extreme values.
pH of Common Substances:
pH and pOH:

pH = -log10[H+]
or pH = -log10[H3O+]
(same thing!!)

[H+] aka.. [H3O +] = 10-pH

pOH = -log10[OH-]

[OH-] = 10-pOH
pH
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
[H1+]
[OH1-]
1 x 10-14
1 x 10-13
1 x 10-12
1 x 10-11
1 x 10-10
1 x 10-9
1 x 10-8
1 x 10-7
1 x 10-6
1 x 10-5
1 x 10-4
1 x 10-3
1 x 10-2
1 x 10-1
1 x 100
1 x 10-0
1 x 10-1
1 x 10-2
1 x 10-3
1 x 10-4
1 x 10-5
1 x 10-6
1 x 10-7
1 x 10-8
1 x 10-9
1 x 10-10
1 x 10-11
1 x 10-12
1 x 10-13
1 x 10-14
pOH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Example 1:

[H3O+] in a cola drink is about 5.0 x 10-3 M.
Calculate the pH of the drink. State whether
the drink is acidic or basic.

pH = -log [H3O+]
= -log (5.0 x 10-3)
= 2.30

Therefore, the cola drink is acidic.
Example 2:

Calculate the pH of a 0.0025M solution of
Ca(OH)2.
Ca(OH)2 <-> Ca2+ + 2OH-1
0.0025M
0.005M
 Kw = [H3O+] [OH-] = 1.0 x 10-14
(x) (0.005M) = 1.0 x 10-14
x =[H3O+] = 2.0 x 10-12


pH = -log [H3O+]
= -log (2.0 x 10-12)
= 11.7
Try it :
Worksheet
 Page 572 # 20-25
