The Ion Product Constant for Water (Kw)

Pure water dissociates according to the following reaction:

There is an equal amount of _____ and ______ ions in solution (neutral, pH = 7)

at 25°C [H+] = [OH-] = ______________________

equilibrium constant for the dissociation of water: _________
Kw
=
=
=
* ____________ k, ______________ are favoured
(_________ _________ _____ _____ ____________________)

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
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
Ex)
[OH-] = 10-pOH
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?