ph of 0.001M H2X OF 50%dissosiation

Answer 1

If we consider only the first dissociation,
(assuming that the concentration of H+ ions we get from the second dissociation is negligible)

H2X(aq)<===> HX^-(aq) + H^+(aq)

Equ. concentrations of
H2X(aq)=(10^-3)-[(10^-3)/2] moldm-3

H^+(aq)=[HX^-(aq)= (10^-3)/2 moldm-3

As 50℅ of the initial concentration of H2X is being dissociated ,[H^+] in the solution=(10^-3)/2

pH= -log10[H^+(aq)] = -log10{(10^-3)/2}