Question

In a 0.5 M solution of a diprotic acid H₂A (K_a1 -3.6 x 10^-5 and K_a2 - 8 x 10^-12 at 25^*C) What is the equilirbium concnetration of A^2-?

Answer 1

Let us write down the stepwise dissociation of the diprotic acid as

H₂A (aq) + H₂O (I) <=======> H₃O⁺ (aq) + HA^- (aq) ; K_a1 = [H₃O⁺][HA^-]/[H₂A]

HA^- (aq) + H₂O (l) <=======> H₃O⁺ (aq) + A^2- (aq); K_a2 = [H₃O⁺][A^2-]/[HA^-]

In both the cases, we assume that the concentration of H₂O (l) is virtually large enough to affect the acid dissociation constants. We look at the values of K_a1 and K_a2 to find that K_a1 is much larger than K_a2 (more than a million times larger). Hence, we can use two approximations to deduce the concentrations of the species. Since K_a1›› K_a2, we can assume that only a very small amount of HA^- undergoes dissociation in the second step and that in the first dissociation, [H₃O⁺] is almost equal to that of [HA^-]. Also, we shall assume that the equilibrium concentration of H₂A is equal to its initial concentration, i.e, 0.5 M.

Hence, we have,

K_a1 = [H₃O⁺][HA^-]/[H₂A] = (x)(x)/(0.5)

or, 3.6 x 10^-5 = x²/0.5

or, x² = 1.8 x 10^-5

or, x = 0.00424 M.

Now, let the concentration of A^2- be y M.

We know, K_a2 = [H₃O⁺][A^2-]/[HA^-].

Since we have already assumed that only a small fraction of HA^- undergoes dissociation, we can safely say that [H₃O⁺] concentration remains the same from first step. Hence, we have,

8 x 10^-12 = (0.00424)(y)/(0.00424)

or, y = 8 x 10^-12 M.

The molar concentration of A^2- is found to be 8 x 10^-12 M with the assumptions that we have made. (ans)