2.5.3

# The Ionic Product of Water (A2 Only)

Test yourself

## The Ionic Product of Water

Water is very slightly dissociated into its ions.

### Dissociation of water

• Water very slightly dissociates into its ions according to the following equation:
• 2H2O(l) ⇋ H3O+(aq) + OH-(aq)

### Equilibrium constant

• The equilibrium constant for the dissociation of water is:
• Kc = $\frac{[H_3O^+][OH^-]}{[H_2O]}$

### Ionic product of water

• The equilibrium constant can be simplified because the concentration of water is pretty much constant.
• Water only dissociates slightly, so the concentration doesn’t really change.
• We write a new equation: Kw = [H3O+(aq)][OH-(aq)]
• It may also be written as: Kw = [H+(aq)][OH-(aq)]

### Temperature dependence of Kw

• Kw is just an equilibrium constant.
• Because of this, it behaves the same as any other equilibrium constant, Kc.
• Importantly, this means it has a temperature dependence. The hotter the water is, the more dissociated it is and the larger its Kw value.

## Use of Kw

Kw can be used to calculate the concentration of hydroxide ions given the pH.

### The value of Kw

• At room temperature, it just so happens that Kw is around 1 ×10-14.
• As the concentration of hydroxide ions must equal the concentration of hydrogen ions for pure water, we know the concentration of both must be 1 × 10-7.
• If the [H+] is 1 ×10-7, then the pH is 7.
• This is why we say a neutral solution has a pH of 7.
• A neutral solution is one with the same concentration of hydroxide ions and hydrogen ions.

### Use of Kw

• The Kw is a constant at a specific temperature.
• This means that at a specific temperature, we can relate the [H+] to the concentration of [OH].
• If we know the pH of a solution, we can calculate the concentration of [OH].

### Example

• Say we dissolve one mole of HCl into a litre of water:
• The concentration of H+ ions is 1moldm-3.
• We can rearrange the ion product equation from:
• Kw = 1 × 10-14 = [H3O+(aq)][OH-(aq)]
• To:
• [OH-(aq)] = 1 × 10-14 ÷ 1moldm-3
So we can calculate that [OH-(aq)] = 1 × 10-14.
• The [OH-(aq)] is much lower in an acidic solution than in a neutral solution.