Chemistry
posted by Homework help .
For the diprotic weak acid H2A, Ka1 = 2.2 × 105 M and Ka2 = 7.8 × 107 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?
Thank you for the help!!

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 35%).
..........H2A ==> H^+ + HA^
I......0.0650.....0......0
C.........x......x......x
E....0.0650x....x......x
Ka1 = (H^+)(HA^)/(H2A)
Substitute the E line and solve for x = H^+), then convert to pH. You can then evaluate 0.0650x = (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.