Question

A diprotic acid solution H2A has a molarity of .95 M. the concentrations of the species present at equilibrium are as follows: [H+] = .25 M, [HA-]=.25M, [A^2-] =4.6x10^-4 M. what are the first and second ionization constants for this acid?

Answer 1

H2A ==> H^+ + HA^-
HA^- ==> H^+ + A^-2

k1 = (H^+)(HA^-)/(H2A)
(H^+) = 0.25 from the problem.
(HA^-) = 0.25 from the problem.
(H2A) = 0.95-.25 = ??
Substitute and solve for k1.

k2 = (H^+)(A^=)/(HA^-).
(H^+) = 0.25 from the problem.
(HA^-) = 0.25 from the problem.
(A^=) = 4.6E-4. Note (H^+) and (HA^-) cancel (although you CAN put those numbers in and crunch them on the calculator) but the bottom line is k2 = (A^=) = 4.6E-4.