Question

Calculate the pH at which an acid indicator with K_a = 1 × 10^–5 changes colour when indicator concentration is 1 × 10^–3 M. Also report the pH at which coloured ions are 60% present.

Answer 1

For indicator dissociation equilibrium

The mid - point of the colour change of an indicator Hl_n is the point at which [ln^– ] = [Hl_n], K_ln = [H⁺] = 1 × 10^–5  

∴ pH = 5 

Thus, at pH = 5 the indicator will change its colour.