Question

Help please

For the diprotic weak acid H2A, Ka1-2.8 x10 and Ka28.9 x109 What is the pH of a 0.0450 M solution of H2A? What are the equilibrium concentrations of H2A and A in this solution? Number Number Number 2-

Answer 1

H₂A <------> HA^- + H+;..................K_a1 = 2.8x10^-6
HA- <------> A^2- + H+;...................K_a2 = 8.9x10^-9

For the concentrations of H⁺ and HA^- you consider only the first reaction since the HA^- product will only slightly dissociate (K_a2 is very small).

If x moles of H₂A dissociate, then x moles of HA^- and H⁺ are produced, and the concentration of H₂A is reduced by x:

[H₂A] = 0.0450 - x
[H⁺] = x
[HA^-] = x

K_a1 = [H⁺]x[HA^-] / [H₂A] = x² / (0.0450 - x)

2.8x10^-6 x (0.0450 - x) = x²
x² + 2.8 x10^-6x – (0.0450 x 2.8x10^-6) = 0
x = 0.0017

[H⁺] = x
pH = -log(x) = 2.8
[H₂A] = 0.045 - x = 0.043 M

For the second reaction, K_a2 = [A^2-]x[H⁺] / [HA^-] = ([A^2-] x) / x;

   [A^2-] = K_a2
   Therefore    [A^2-]= 8.9*10^-9 M