For the diprotic weak acid H2A Kat 33 x108 and Ka2 77 x10 Wh

For the diprotic weak acid H2A, Kat -3.3 x108 and Ka2 7.7 x10 What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A in this solution? Number Number Number

Solution

Solution:-

H2A   <---> H+ + HA-
0.08 0          0   (initial)
0.08-x           x         x   (at equilibrium)

Ka1 = [H+]*[HA-] / [H2A]
3.6*10^-6 = x*x / (0.08-x)
Since Ka is small, x will be small and it can ignored as compared to 0.055
3.3*10^-6 = x*x / 0.08
x= 6.42*10^-3 M

So,
[H+] = x= 6.42*10^-3 M
[HA-]=x= 6.42*10^-3 M
[H2A]=0.08-x= 0.08- 6.42*10^-3 = 0.073 M
HA-   ---> A2- + H+
Ka2 =[A2-][H+]/[HA-]
Ka2 =[A2-]
[A2-] = Ka2
    = 7.7*10^-9 M

pH = -log [H+]
    = -log (6.42*10^-3 )
    = 2.19