The Ion Product Constant for Water ( Kw )

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# The Ion Product Constant for Water ( Kw ) - PowerPoint PPT Presentation

The Ion Product Constant for Water ( Kw ). Pure water dissociates according to the following reaction: H 2 O (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

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### The Ion Product Constant for Water (Kw)

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)

@ 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 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 !

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