Question

For the diprotic weak acid h2a, ka1 = 4.0 × 10-6 and ka2 = 7.2 × 10-9. what is the ph of a 0.0450 m solution of h2a? what are the equilibrium concentrations of h2a and a2– in this solution?

Answer 1

For the first reaction :
the ICE table:
                  H2A      ↔      HA-      +      H+
initial       0.045 m             0                  0
change    -X                       +X               +X
Equ        (0.045-X)              X                  X

when Ka1 = [HA-][H+]/[H2A]
4x10^-6   = X^2 / (0.045-X) by solving for X
∴X= 4.2x10^-4 
∴[ H2a] = 0.045- (4.2x10^-4) = 0.045 m
[HA-] = [H+] = 4.2x10^-4

when PH = -㏒[H+]
 by substitution :
PH = -㏒ (4.2x10^-4)
      = 3.377 

For the second reaction :
by using ICE table :
                     HA-      ↔    A^2-   +   H+         
when
Ka2 = [A^2-] [H+] / [HA-]
by substitution 
7.2 x 10^-9 = [A^2-]*(4.2x10^-4) / (4.2x10^-4)
∴[A^2-] = (7.2x10^-9)*(4.2x10^-4) / (4.2x10^-4)
∴ [A^2-] = 7.2x10^-9 m

Answer 2

that's right how you did it