Question

Determine the pH, [H₃A], [H₂A^-], [HA^-2], and [A^-3] for a 0.30M solution of H₃A. The respective pK_a1, pK_a2, and pK_a3 values are 2.81, 4.87 and 8.14.

Answer 1

The acid Ka are calculated:

Ka1 = 10 ^ -pKa1 = 10 ^ -2.81 = 1.55x10 ^ -3

Ka2 = 1.35x10 ^ -5

Ka3 = 7.24x10 ^ -9

The first dissociation has to be:

H3A = H + + H2A-

Ka1 = [H +] * [H2A-] / [H3A] = X ^ 2 / 0.3 - X

It is assumed that - X is negligible and clears:

X = [H +] = [H2A] = √Ka1 * 0.3 = √1.55x10 ^ -3 * 0.3 = 0.022 M

pH = - log [H +] = - log 0.022 = 1.66

For the second dissociation:

H2A- = H + + HA-2

Ka2 = [H +] * [HA-2] / [H2A-] = 0.022 * [HA-2] / 0.022 = 1.35x10 ^ -5

It clears [HA-2] = 1.35x10 ^ -5 M

For the third dissociation:

HA-2 = H + + A-3

Ka3 = [H +] * [A-3] / [HA-2] = 0.022 * [A-3] / 1.35x10 ^ -5 = 7.24x10 ^ -9

It clears [A-3] = 4.44x10 ^ -12 M

The answers are:

pH = 1.66

[H3A] = 0.278 M

[H2A-] = 0.022 M

[HA-2] = 1.35x10 ^ -5 M

[A-3] = 4.44x10 ^ -12 M

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.