Ionic Product of Water (Edexcel International A Level (IAL) Chemistry): Revision Note

Ionic Product of Water, Kw

• In all aqueous solutions, an equilibrium exists in water where a few water molecules dissociate into protons and hydroxide ions

• We can derive an equilibrium constant for the reaction:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

K_c = [H⁺][OH^-] / [H₂O]

The concentration of water is effectively constant, so K_c x [H₂O] = [H⁺][OH^-] = K_w = 1 x 10^-14 mol² dm^-6 at 298 K.

• This is a specific equilibrium constant called the ionic product for water

• The product of the two ion concentrations at 298 K is 1 x 10^-14 mol² dm^-6

• This makes it straightforward to see the relationship between the two concentrations and the nature of the solution:

[H⁺] & [OH^–] Table

The relationship between K_w and pK_w is given by the following equation:

pK_w = -logK_w

pH Calculation of a Strong Base

• Strong bases are completely ionised in solution

BOH (aq) → B⁺ (aq) + OH^- (aq)

• Therefore, the concentration of hydroxide ions [OH^-] is equal to the concentration of base [BOH]

  • Even strong alkalis have small amounts of H⁺ in solution which is due to the ionisation of water

• The concentration of OH^- in solution can be used to calculate the pH using the ionic product of water

• Once the [H⁺] has been determined, the pH of the strong alkali can be found using pH = -log[H⁺]

• Similarly, the ionic product of water can be used to find the concentration of OH^- ions in solution if [H⁺] is known, simply by dividing K_w by the [H⁺]