1. The Ion Product Constant for Water (Kw)

2. 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)

3. 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

4. Ex 1) Find the [H+] and [OH-] in:
(a) 2.5 M nitric acid (b) M Barium hydroxide
(a) HNO  H NO3-
C 2.5 M M
Kw = [H+][OH-] [OH-] = Kw / [H+]
= (1x10-14)/(2.5)
= 4x10-15 M
(b) Ba(OH)  Ba OH-
C 0.16 M M
Kw = [H+][OH-] [H+] = Kw / [OH-]
= (1x10-14)/(0.32)
= 3.1x10-14 M

5. 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

6. 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

7. 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-] = 6.8x > 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]
= = 4.17
check: = 14 !

8. HOMEWORK
P382 #1-4
p390 #9-12
p392 #13-18