For the diprotic weak acid H2A, Ka1 = 2.2 × 10-5 M and Ka2 = 7.8 × 10-7 M. What is the pH of a 0.0650 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Question

Thank you for the help!!

Answer 1

With Ka1 and Ka2 as close as they are, there is some interaction between the two but I will assume you can take the easy way out, assume there is no interaction, and solve the problem this way. The REAL answer won't be that far away (perhaps 3-5%).

..........H2A ==> H^+ + HA^-
I......0.0650.....0......0
C.........-x......x......x
E....0.0-650-x....x......x

Ka1 = (H^+)(HA^-)/(H2A)
Substitute the E line and solve for x = H^+), then convert to pH. You can then evaluate 0.0650-x = (H2A).

For A^2- I would do this
.........HA^- ==> H^+ + A^2-

Ka2 = (H^+)(A^2-)/(HA^-)
(H^+) = (HA^-); therefore,
(A^2-) = Ka2.
Remember you are making the assumption here that Ka1 and Ka2 act independently.