Question

Consider the following diprotic acid. H2AH20 Ho+HA Kal-5.0x105 +A2- The initial concentration of H2A = 1.0 × 10-5 M. HA" + H20 H30" Kai = 4.5 x 10-5 Lal 1. Since the Ka and Ka2 values are very close, treat both reactions simultaneously. (a) Determine the H3+ at equilibrium 2 Now, as a comparison, ignore the second reaction (a) Determine the [H0. 3. Is the approximation of ignoring the second reaction valid?

Answer 1

Case a: Assuming simultaneous reaction, we maintain the complete reaction with an average Ka of 4.75x10 ^ -5.

H2A + 2H2O = 2H3O + + A-2

[H2A] = 1x10 ^ -5 -x

[H3O +] = 2x

[A-2] = x

so:

Ka = [H3O +] * [A-2] / [H2A] = 4.75x10 ^ -5

4.75x10 ^ -5 = 4x ^ 3 / (1x10 ^ -5 - x)

clearing x, we have x = 10 ^ -5

for [H3O +] = 2x = 2x10 ^ -5

Case b: Taking into account only the first reaction:

H2A + H2O = H3O + + HA-

[H2A] = 1x10 ^ -5 - x

[H3O +] = x

[HA-] = x

we have:

Ka = x ^ 2 / 1x10 ^ -5 - x

clearing x, x = 8.5x10 ^ -6 = [H3O +]

We can see that doing that consideration is not correct, since, the concentrations of H3O + in the equilibrium are not equal.