Ionisation of Weak Electrolytes

When acetic acid is dissolved in water, it dissociates partly into H⁺ and H₃O⁺ and CH₃COO‾ ions as:

CH₃COOH + H₂O  CH₃COO‾ + H₃O⁺

In dilute solution , concentration of water is constant. The product of K and constant 

The product of K and   is denoted by K_a, the ionization constant or dissociation constant of the acid.

If C represents the initial concentration of the acid in moles L^-1

and α , the degree of dissociation , then equilibrium concentration of the ions ( H₃O⁺ and CH₃COO‾ ) is equal to Cα and that of the undissociated acetic acid = C ( 1- α ) i.e. we have

In case of weak electrolyte, The value of α is very small and can be neglected in comparison to 1 i.e. 1-α =1.Hence we get

α = √ Ka / C

If V is the volume of the solution in litres containing 1 mole of the electrolyte , C = 1/ V.Hence, we have,

α = √ Ka × V

For a weak base like NH₄OH we have

α = √ K_b / C

α = √ K_b × V

For a weak electrolyte , the degree of ionisation is inversely proportional to the square root of molar concentration  or directly proportional to the square root of volume containing one mole of solute.This is called Ostwald’s dilution law.