Chemistry Help!

Question

For the diprotic weak acid H2A, Ka1=3.4x10^-6 and Ka2=8.1x10^-9. What is the pH of 0.0700 M solution of H2A?

pH=

What are the equilibrium concentration of H2A and A2- in this solution?

[H2A]=

[A2-]

Answer 1

Looking at the dissociation of the diprotic acid:

H₂A <-==> H⁺ + HA^- ... Ka1 = 3.4x10^-6

HA^- <==> H⁺ + A^2- ... Ka2 = 8.1x10^-9

Ka1 = 3.4x10^-6 = [H⁺][HA^-] / [H₂A]

3.4x10^-6 = (x)(x) / 0.0700 - x (assume x is small and ignore it in denominator)

3.4x10^-6 = x² / 0.0700

x² = 2.38x10^-7

x = 4.88x10^-4 = [H⁺] = [HA^-] (note: this value is only ~ 0.7% of 0.0700 M so above assumption was valid)

Ka2 = 8.1x10^-9 = [H⁺][A^2-] / [HA^-]

8.1x10^-9 = (x)(x) / 4.88x10^-4 - x (assume x is small and ignore it in denominator)

8.1x10^-9 = x² / 4.88x10^-4

x² = 3.95x10^-12

x = 2.00x10^-6 = [H⁺] = [A^2-] (note: this value is only ~ 0.4% of 4.88x10^-4 M so above assumption ok)

pH = -log [H⁺]

[H+] = 4.88x10^-4 + 2.00x10^-6 = 4.9x10^-4

pH = -log 4.9x10^-4

pH = 3.31

[H₂A] = 0.0700 - 4.88x10^-4 = 0.0695 M

[A^2-] = 2.00x10^-6 M