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) HNO3 H+ + NO3

-

C 2.5 M 2.5 M

Kw = [H+][OH-] [OH-]= Kw / [H+]

= (1x10-14)/(2.5)= 4x10-15 M

(b) Ba(OH)2 Ba2+ + 2 OH-

C 0.16 M 0.32 M

Kw = [H+][OH-] [H+] = Kw / [OH-]

= (1x10-14)/(0.32)= 3.1x10-14 M

pH and pOHpH: 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) = 7pOH = -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 #1-4p390 #9-12

p392 #13-18