Question

A diprotic acid, H2A, has acid dissociation constants of Ka1 = 3.94× 10–4 and Ka2 = 2.17× 10–12. Calculate the pH and molar concentrations of H2A, HA–, and A2– at equilibrium for each of the solutions below.

(a) a 0.147 M solution of H2A

(b) a 0.147 M solution of NaHA

Answer 1

a) 0.147 M H₂A solution.

Stepwise Ionization with corresponding ionization constants are given as,

H₂A (aq) HA^- (aq) + H⁺(aq) ............ Ka1 = [HA^-][H⁺]/[H₂A] =3.94× 10^–4.

HA^- (aq) A^2- (aq) + H⁺(aq) .............Ka2 = [A^2-][H⁺]/[HA^-] = 2.17× 10^–12.

i) Let's deal with first ionization and it's consequences,

ICE table:

H₂A (aq) HA^- (aq) + H⁺(aq)

Initial conc. 0.147 M 0 0

Change -X +X +X

Eqm conc. (0.147-X) X X

Using these equilibrium concentration values of

Ka1 = (X)(X)/(0.147-X) =3.94× 10^–4.

X²/(0.147-X) =3.94× 10^–4.

But being H₂A a weak acid obviously X << 0.147 hence "small concentration assumptions says 0.147-X 0.147 M.

X²/0.147 =3.94× 10^–4.

X² = 0.147 x 3.94× 10^–4.

X² = 0.58 × 10^–4.

X= 7.6 × 10^–3.

By observing ICE table we write,

[HA^-] = [H⁺] = 7.6 × 10^–3 M.

[H₂A] = 0.147-X = 0.147 - 7.6 × 10^–3 = 0.139 M

ii) Second ionization & conquences

Now initial concentration for ionizable species at 2nd stage [HA^-] = 7.6 × 10^–3 M.

ICE table:

HA^- (aq)    A^2- (aq) + H⁺(aq)

Initial conc.   7.6 × 10^–3 M. 0 0

Change -Y +Y +Y

Eqm conc. (7.6 × 10^–3 -Y) Y Y

Using these equilibrium concentrations in expresion for Ka2 we get,

Ka2 = (Y)(Y) / (7.6 × 10^–3 -Y) = 2.17× 10^–12.

HA- is far weak acid hence 7.6 × 10^–3  - Y 7.6 × 10^–3.

Y²/(7.6 × 10^–3 M) = 2.17× 10^–12.

Y² = 7.6 × 10^–3 x 2.17× 10^–12.

Y² = 1.64 × 10^–14.

Y= 1.28 × 10^–7.

By ICE table,

[A^2-] = [H⁺] = 1.28 x 10^-7 M

[HA^-] = 7.6 x 10^-3 - Y 7.6 x 10^-3 M.

Now for pH = ?

H2A is dibasic acid but overall acidity is only due to first ionization as second ionization is contributed very little to [H⁺] & hence

pH = -log[H⁺] = -log(7.6x10^-3)

pH = 2.12

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