For the diprotic weak acid H2A, Ka1 = 2.9 × 10-6 and Ka2 = 7.9 × 10-9. What is the pH of a 0.0750 M solution of H2A?

HA⁻ and H⁺ ions are formed from H₂A in the first dissociation.
As Ka₁ ≫ Ka₂, we can assume that in the second dissociation the changes of [H₂A], [HA⁻] and [H⁺] are negligible.

Consider the first dissociation of H₂A :
____________ H₂A(aq) __ ⇌ __ HA⁻(aq) __ + __ H⁺(aq) ___ Ka₁ = 2.9 × 10⁻⁶
Initial: _____ 0.0750 M _______ 0 M __________ 0 M
Change: ______ -y M ________ +y M _________ +y M
At eqm: __ (0.0750 - y) M _____ y M __________ y M

As Ka is very small, the dissociation of H₂A would be to a very small extent.
It is assumed that 0.0750 ≫ y, i.e. [H₂A] at equilibrium = (0.0750 - y) M ≈ 0.0750 M

Ka₁ = [HA⁻] [H⁺] / [H₂A]
2.9 × 10⁻⁶ = y² / 0.0750
y = √[0.0750 × (2.9 × 10⁻⁶)]
y = 4.7 × 10⁻⁴

Consider the second dissociation of H₂A :
____________ HA⁻(aq) __ ⇌ __ A²⁻(aq) __ + __ H⁺(aq) ___ Ka₂ = 7.9 × 10⁻⁹
At eqm: __ 4.7 × 10⁻⁴ M _____ z M ______ 4.7 × 10⁻⁴ M

Ka₂ = [A²⁻] [H⁺] / [HA⁻]
7.9 × 10⁻⁹ = z (4.7 × 10⁻⁴) / (4.7 × 10⁻⁴)
z = 7.9 × 10⁻⁹

Answer :
pH = -log[H⁺] = -log(4.7 × 10⁻⁴) = 3.3
[H₂A] at equilibrium = 0.0750 M
[A²⁻] at equilibrium = z M = 7.9 × 10⁻⁹ M