The Ionic Product of Water, Kw (Oxford AQA International A Level (IAL) Chemistry): Revision Note

Ionic Product of Water Calculations

• In all aqueous solutions, some water molecules dissociate into protons and hydroxide ions

• An equilibrium is established:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

• K_w is the ionic product of water

  • It is the equilibrium constant for the dissociation of water at 298 K

  • Its value is 1.00 x 10^-14 mol² dm^-6

• For the ionisation of water, the equilibrium expression to find K_w is:

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

K_w = $\frac{\left[H^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[H_{2}O\right]}$

• As the extent of ionisation is very low, only small amounts of H⁺ and  OH^- ions are formed

• The concentration of H₂O can therefore be regarded as constant and removed from the K_w expression

• The equilibrium expression therefore becomes:

K_w = [H⁺] [OH^-]

• As the [H⁺] = [OH^-] in pure water, the equilibrium expression can be further simplified to:

K_w = [H⁺]²

The effect of temperature on Kw

• The dissociation of water to form hydrogen and hydroxide ions is an endothermic process so absorbs heat energy

H₂O (l) ⇌ H⁺ (aq) + OH^- (aq)

• If temperature is increased, the forward reaction will be favoured to counteract the change and lower the temperature

• Th equilibrium will shift to the right and more H⁺ and OH^- ions will be formed causing the value of K_w to increase

• So, as temperature increases the value of K_w increase

• If the value of K_w increases, the pH decreases as shown below:

  T (°C)	Kw (mol2 dm-6)	pH
  0	0.114 x 10-14	7.47
  10	0.293 x 10-14	7.27
  20	0.681 x 10-14	7.08
  25	1.008 x 10-14	7.00

Calculating the pH 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⁺]

$\left[H^{+}\right]\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{\left[{\mathrm{OH}}^{-}\right]}$

$\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\left[{\mathrm{OH}}^{-}\right]}$

$\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}$

• 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⁺]