Ionic Product of Water (Edexcel A Level 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], so K_c x [H₂O] = [H⁺][OH^-]

• The concentration of water is constant, so the left-hand side becomes K_w:

K_w = [H⁺][OH^-] = 1 x 10^-14 mol² dm^-6 at 298 K

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

• At 298 K the product of the two ion concentrations is 1 x 10^-14 mol² dm^-6 (K_w changes with temperature)

• 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

pK_a

• The range of values of K_a is very large and for weak acids, the values themselves are very small numbers

Table of K_a values

• For this reason it is easier to work with another term called pK_a 

• The pK_a  is the negative log of the K_a value, so the concept is analogous to converting [H⁺] into pH values

pK_a = -logK_a

• Looking at the pK_a  values for the same acids:

Table of pK_a values

• The range of pK_a values for most weak acids lies between 3 and 7

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

K_w = [H⁺] [OH^-]

• Step 1: Rearrange the equation

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

• Step 2: Substitute the K_w value

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

• Step 3: Substitute [H⁺] into the pH equation

pH = -log₁₀[H⁺]

pH = -log₁₀$\left(\frac{1\times {10}^{-14}}{\left[{\mathrm{OH}}^{-}\right]}\right)$

• 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⁺]