Question

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 4.08× 10–4 and Ka2 = 2.89× 10–12, calculate the pH and molar concentrations of H2A, HA–, and A2– for each of the solutions below. (a) a 0.167 M solution of H2A (b) a 0.167 M solution of NaHA (c) a 0.167 M solution of Na2A

Answer 1

The reactions are

H2A ⇔ HA- + H+;................ Ka1 = 4.08*10^-4
HA ⇔ A(^2-) + H+;................Ka2 = 2.89*10^-12

Because Ka2 is very low, very little HA- will dissociate, so that reaction is ignored and the first reaction treated as monoprotic

If x moles of H2A dissociate, then the remaining concentration of H2A is
[H2A] = 0.167 - x
and the concentration of the products is
[HA-] = x
[H+] = x

4.08*10^-4 = [HA-]*[H+]/[H2A] = x²/(0.167 - x)

Ka1 is small, only a small amount of H2A dissociates so x << 0.167 and we can use the approximation 0.167 - x ≈ 0.67 then = √[4.08*10^-4 *.0167

[H+] = x =0.00261 M
pH = 2.58

[HA-] also = x
[HA-] = 0.00261 M
[H2A] = 0.167 - x = 0.167-0.00261 = 0.164

Now consider the second reaction

Ka2 = [H+]*[A(2^-)]/[HA-]

from above we know the [H+] = [HA-] so Ka2 = [A-]

[A(^2-)] = 2.89*10^-12 M

(a) a 0.167 M solution of H2A pH= 2.58

b) a 0.167 M solution of NaHA pH =0.78

c)  a 0.167 M solution of Na2A pH= 11.53