How do I find the concentration of A2− for a weak diprodic acid?

Question

A solution contains the weak diprotic acid H2A in water. Given that the concentration of HA− generated from the first ionization of the parent acid, H2A, is 0.00137 M, what is the concentration of A2−? 

Use Ka2=9.10×10−11

Answer 1

The sequential dissociation/ionization of the weak acid is represented as ...

H₂A ==> H⁺ + HA^- first ionization

HA^- ==> H⁺ + A^2- second ionization Ka₂ = 9.10x10^-11

All we are concerned with is

HA^- ==> H⁺ + A^2-

0.00137.....0...........0.......Initial

-x.............+x.........+x.......Change

0.00137-x...x........x.........Equilibrium

Ka₂ = 9.10x10^-11 = [H⁺][A^2-] / [HA^-]

9.10x10^-11 = (x)(x) / 0.00137 - x and if we assume x is small compared to 0.00137 we can ignore it...

9.10x10^-11 = x² / 0.00137

x² = 1.25x10^-13

x = 3.54x10^-7 M which is only ~ 0.03% of 0.00137 M so our above assumption was valid

[A^2-] = 3.54x10^-7 M