The Ion Product Constant for
Water (Kw)
• Pure water dissociates according to the following reaction:
H2O(l)

H+(aq) + OH-(aq)
• There is an equal amount of H+ and OH- ions in solution
(neutral, pH = 7)
• at 25°C [H+] = [OH-] = 1x10-7 mol/L
• equilibrium constant for the dissociation of water: Kw
Kw = [H+][OH-]
= (1x10-7)(1x10-7)
= 1x10-14
* small k, reactants are favoured
(does not go to completion)
Since strong acids and bases dissociate completely in
water, [H+] = [acid]
@ 25°C
acids: [H+] > [OH-]
[H+] > 1x10-7
[OH-] < 1x10-7
bases: [OH-] > [H+]
[H+] < 1x10-7
[OH-] > 1x10-7
We can use Kw to calculate [H+] and [OH-] in solutions
Ex 1) Find the [H+] and [OH-] in:
(a) 2.5 M nitric acid
(b) 0.16 M Barium hydroxide
(a)
C
HNO3
2.5 M

H+
2.5 M
Kw = [H+][OH-]
(b)
C
Ba(OH)2
0.16 M
Kw = [H+][OH-]

+
NO3-
[OH-]= Kw / [H+]
= (1x10-14)/(2.5)
= 4x10-15 M
Ba2+
+ 2 OH0.32 M
[H+] = Kw / [OH-]
= (1x10-14)/(0.32)
= 3.1x10-14 M
pH and pOH
pH: The Power of the Hydronium Ion
• A measure of the amount of H+ ions in a solution
• Convenient way to represent acidity since [H3O+] is
usually a very small number
• 2 factors determine pH
• ionization
• concentration
because they both contribute to the number of H+
or OH- molecules in a solution.
• The practical scale goes from 0  14
pH = -log [H3O+]
pOH = -log [OH-]
In neutral water,
pH = -log [H3O+] = -(log(1 x 10-7) = 7
pOH = -log [OH-] = -(log(1 x 10-7) = 7
Note: pH + pOH = 14, always, regardless of solution!
Another way to calculate [H3O+] & [OH-] in solution:
[H3O+] = 10-pH
[OH-] = 10-pOH
Ex)
A liquid shampoo has a [OH-] of 6.8x10-5 mol/L
(a) Is the shampoo acid, basic or neutral?
(b) What is [H3O+]?
(c) What is the pH and pOH of the shampoo?
(a) [OH-] = 6.8x10-5
> 1.0x10-7, basic
(b) [H3O+] = Kw / [OH-] = (1.0x10-14)/(6.8x10-5)
= 1.5x10-10 mol/L
(c) pH = -log [H3O+]
pOH = -log [OH-]
= -log [1.5x10-10]
= -log [6.8x10-5]
= 9.83
= 4.17
check: 9.83 + 4.17 = 14 !
HOMEWORK
P382 #1-4
p390 #9-12
p392 #13-18