Question

For the diprotic weak acid H2A, Ka1 = 2.5 × 10-6 and Ka2 = 5.5 × 10-9. What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

                           H2A (aq) ------------> H^+ (aq) + HA^- (aq)

   I                      0.08                           0                 0

   C                     -x                               +x               +x

   E                0.08-x                              +x               +x

                       Ka1 = [H^+][HA^-]/[H2A]

                    2.5*10^-6   = x*x/0.08-x

                   2.5*10^-6 *(0.08-x) = x^2

                     x   = 0.000446

                  [H^+]   = x    = 0.000446M

                  [HA^-]   = x    = 0.000446M

                 [H2A]   = 0.08-x   = 0.08-0.000446   = 0.079554M

                     HA^- (aq) ---------------> H^+ (aq) + A^2- (aq)

                      Ka2   = [H^+][A^2-]/[HA^-]

                      5.5*10^-9   = 0.000446*[A^2-]/0.000446

                      [A^2-]    = 5.5*10^-9 M

                    PH   = -log[H^+]

                             = -log0.000446

                              = 3.3506 >>>>answer