Question

For the diprotic weak acid H2A, Ka1 = 2.8 × 10-6 and Ka2 = 8.7 × 10-9. What is the pH of a 0.0500 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

Given the concentration of H2A = 0.0500 M

The chemcial equation for the first dissociation is

------------- H2A ------- > HA^-(aq) + H⁺(aq) ; Ka1 = 2.8*10^-6

init.con: 0.0500M ------ 0 M -------- 0 M

change: - 0.0500x M, +0.0500xM, +0.0500xM

eqm.con:0.0500(1 - x)M, 0.0500xM, 0.0500xM

Ka1 = 2.8*10^-6 = [HA^-(aq)]*[H⁺(aq)] / [H2A] = (0.0500xM * 0.0500xM) / 0.0500(1 - x)M

since x <<1, (1 - x) is nearly equals to 1.

=> x = underroot( 5.6*10^-5) = 7.48*10^-3 M

Hence [H2A] = 0.0500(1 - x) M = 0.0500(1 - 7.48*10^-3)M = 0.0496 M (answer)

[HA^-(aq)] = [H⁺(aq)] =  0.0500x M = 0.0500*7.48*10^-3 M = 3.74*10^-4 M

The chemcial equation for the second dissociation is

------------- HA^-(aq) ------- > A^2-(aq) + H⁺(aq) ; Ka2 = 8.7*10^-9

init.con: 3.74*10^-4 M ----- 0 M ------------- 3.74*10^-4 M

change: - 3.74*10^-4 yM, +3.74*10^-4 yM, +3.74*10^-4 yM

eqm.con:3.74*10^-4(1 - y)M, 3.74*10^-4 yM, 3.74*10^-4 (1+y)M

Ka1 = 2.8*10^-6 = [A^2-(aq)]*[H⁺(aq)] / [HA^-(aq)] = [3.74*10^-4 yM*3.74*10^-4 (y+1)M] / 3.74*10^-4(1 - x)M

since y <<1, (1 - y) and (1+y) are nearly equals to 1.

=> y = 7.49x10^-11 M

Hence [A^2-(aq)] = 3.74*10^-4 yM = 3.74*10^-4 * 7.49x10^-11M = 2.80*10^-14 M(answer)

[H+(aq)] = 3.74*10^-4 (1+y)M =  3.74*10^-4 M

=> pH = - log[H+(aq)] = 3.43 (answer)