Question

For the diprotic weak acid H2A, Ka1 = 4.0

Answer 1

Diprotic acid refers to acid that dissociates to give two H+ ion.

First, write down chemical equations. There will be two steps since this is "di"protic acid.

1. H2A >> H+ + HA-

2. HA- >> H+ + A2-

Then, write down expressions for Ka.

(1st equation)

1.

(2nd equation)

2.

As you can observe, there are four unknowns, i.e., [H2A], [H+], [HA-], and [A2-]. We already have two equations so we need two more to solve this. We can generate two more equations by making assumptions.

We can base one assumption on the fact that the value of K_a1 for this acid is a lot times larger than the value of K_a2.

This means that only a small fraction of the HA^- ions formed in the first step go on to dissociate in the second step. If this is true, most of the H⁺ ions in this solution come from the dissociation of H₂A, and most of the HA^- ions formed in this reaction remain in solution. As a result, we can assume that the H⁺ and HA^- ion concentrations are more or less equal.

(3rd equation)

First assumption:

[H+]=[HA-]

We need one more equation, and therefore one more assumption. Since H2A is a weak acid, we can assume that most of the H₂A that dissolves in water will still be present when the solution reaches equilibrium. In other words, we can assume that the equilibrium concentration of H₂A is approximately equal to the initial concentration of 0.0700 M.

(4th equation)

Second assumption:

[H2A] =CH2A=0.07 M

We now have four equations for the four unknowns.

Using the 1st,3rd and 4th equations...

Using the 2nd equation and the computed value above..The answers to the question "equilibrium concentrations of H2A and A2-" are encircled in red below.

We need to validate if our assumptions are correct. Our first assumption was that the dissociation of H2A is small compared to the initial concentration.. Comparing the concentration of [H+] computed and initial concentration given...

Result above shows that indeed a very small amount, 0.76%, dissociated to H+.

Answering the question "pH of acid solution"..