the ionic product of water kwkw. ionic product of water, k w just because a solution contains [h + ]...

The Ionic Product of Water

Kw

Ionic Product of water, Kw

• Just because a solution contains [H+] it doesn’t necessarily mean it’s acidic.

• All aqueous solutions will contain H+ and OH- ions

• For an acid solution [H+] > [OH-]

• For an alkaline solution [H+] < [OH-]

• For an neutral solution [H+] = [OH-]

Ionic Product of water, Kw

H2O(l) + H2O (l) H3O+

(aq) + OH-(aq)

H2O(l) H+

(aq) + OH-(aq)

Kc = [H+][OH-] [H2O]

Ionic Product of water, Kw

H2O(l) H+

(aq) + OH-(aq)

Kc = [H+][OH-] [H2O]

Since [H2O] is effectively constant and in large excess

Kw = [H+][OH-] mol2dm-6

Ionic Product of water, Kw

At 298K the value of Kw is 1 x 10-14 mol2dm-6

Kw = [H+][OH-] mol2dm-6

In pure water [H+] = [OH-]

So Kw = [H+]2 Hence [H+] = √Kw

At 298K [H+] = √(1 x 10-14) = 1 x 10-7

So at 289K the pH of pure water is 7

Effect of temperature on Kw

Dissociation of water is endothermic

Increasing temp increases [H+]

Increasing temp increases Kw

At higher temps pH of water will decrease

H2O(l) H+

(aq) + OH-(aq)

At 321K the value of Kw is 4 x 10-14 mol2dm-6

calculate the pH of water at this tempKw = [H+][OH-] mol2dm-6

In pure water [H+] = [OH-]

So Kw = [H+]2 Hence [H+] = √Kw

At 321K [H+] = √(4 x 10-14) = 2 x 10-7

So at 321K the pH of pure water is 6.70

pH = -log10[H+]

The pH of Strong Bases

Kw gives us a method of calculating the pH of strong bases

Kw = [H+][OH-] mol2dm-6

So [H+] = Kw [OH-]

Calculate the pH of 0.1M KOH at 298K

Kw = [H+][OH-] = 1 x 10-14 mol2dm-6

Since KOH is a strong base [OH-] = 0.1

[H+] = Kw

[OH-]

pH = -log10[H+]

[H+] = 1 x 10-14

0.1= 1 x 10-13

pH = 13

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

Kw = [H+][OH-]

[H+] = 3.162 x 10-7

pH = -log10[H+]

Kw = [H+]2 for pure water

[H+] = [OH-] for pure water

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

Kw = (3.162 x 10-7)2

Kw = [H+]2 for pure water

Kw = 1x 10-13

Kw = [H+][OH-]

[H+] = Kw

[OH-]= 1 x 10 -13

0.01= 1 x 10 -11

At 338K the pH of pure water is 6.5. Calculate Kw at this temp. Hence calculate the pH of 0.01M NaOH at this temp

pH = -log10 [H+]

pH = 11

[H+] = 1 x 10 -11