Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3.57× 10–4 and Ka2 = 3.14× 10–12, calculate the pH and molar concentrations of H2A, HA–, and A2– for each of the solutions below.

(a) a 0.182 M solution of H2A
(b) a 0.182 M solution of NaHA
(c) a 0.182 M solution of Na2A

To find the pH and molar concentrations of H2A, HA–, and A2– for each solution, we need to calculate the ionization of H2A at the given concentrations and use the ionization constants Ka1 and Ka2. We'll go through each solution step by step:

(a) 0.182 M solution of H2A:
Since we have a diprotic acid, we need to consider both ionization steps.

1. Calculate the concentration of H2A ionized into HA–:
Using the ionization constant Ka1, we can set up the equilibrium expression:
Ka1 = [HA–][H+]/[H2A]
Since we have just dissociated H2A into HA– and H+, [HA–] = [H+]. So we can rewrite the equation as:
Ka1 = [H+]^2 / [H2A]
[H2A] = 0.182 M (given)
Substituting the values, we get:
3.57 × 10–4 = [H+]^2 / 0.182

Rearranging the equation and solving for [H+]:
[H+]^2 = (3.57 × 10–4) * 0.182
[H+]^2 = 6.504 × 10–5
[H+] ≈ 2.55 × 10–3 M

2. Calculate the concentrations of HA– and A2–:
[HA–] = [H+] = 2.55 × 10–3 M
[A2–] = 0 (since no ionization has occurred yet)

3. Calculate the pH:
pH = -log[H+]
pH ≈ -log(2.55 × 10–3)
pH ≈ 2.594

(b) 0.182 M solution of NaHA:
Since NaHA is a salt, it completely dissociates in water.

1. Calculate the concentrations of HA– and A2–:
[HA–] = 0.182 M (given)
[A2–] = 0 (since it hasn't ionized yet)

2. Calculate the pH:
Since HA– is already in its dissociated form and does not contain any H+, the pH will be determined by the autoionization of water.
pH ≈ 7 (neutral)

(c) 0.182 M solution of Na2A:
Similar to NaHA, Na2A is a salt and will dissociate completely.

1. Calculate the concentrations of HA– and A2–:
[HA–] = 0 (Since it hasn't ionized yet)
[A2–] = 0.182 M (given)

2. Calculate the pH:
Again, since there is no H+ contributed by the salt, the pH will be determined by the autoionization of water.
pH ≈ 7 (neutral)

So, to summarize:
For a 0.182 M solution of H2A:
[H2A] ≈ 0.182 M
[HA–] ≈ [H+] ≈ 2.55 × 10–3 M
[A2–] ≈ 0
pH ≈ 2.594

For a 0.182 M solution of NaHA:
[H2A] ≈ 0 (since it's already dissociated)
[HA–] ≈ 0.182 M
[A2–] ≈ 0
pH ≈ 7 (neutral)

For a 0.182 M solution of Na2A:
[H2A] ≈ 0 (since it's already dissociated)
[HA–] ≈ 0
[A2–] ≈ 0.182 M
pH ≈ 7 (neutral)