Diprotic acids and pH?

Kₐ₁ = 10⁻⁵˙³² = 4.79 × 10⁻⁶
Kₐ₂ = 10⁻⁸˙²³ = 5.89 × 10⁻⁹

Since Kₐ₁ ≫ Kₐ₂, H₃O⁺ ions are almost completely formed in the first dissociation of H₂A.
Hence, we can neglect the second dissociation in the calculation of [H₃O⁺].

Consider the first dissociation of H₂A:
__________________ H₂A(aq) ____ + ____ H₂O(l) ____ ⇌ ____ HA⁻(aq) ____ + ____ H₃O⁺(aq) ______ Kₐ₁ = 4.79 × 10⁻⁶
Initial (mol L⁻¹): _____ 1.177 ______________________________ 0 _______________ 0
Change (mol L⁻¹): _____ -y _______________________________ +y _______________ +y
Eqm (mol L⁻¹): ____ (1.177 - y) ____________________________ y ________________ y

As Kₐ₁ is very small, H₂A dissociates to a very small extent in the first dissociation.
Hence, [H₂A] at eqm = (1.177 - y) mol L⁻¹ ≈ 1.177 mol L⁻¹

At eqm: Kₐ₁ = [HA⁻] [H₃O⁺] / [H₂A]
4.79 × 10⁻⁶ = y² / 1.177
x = √[(4.79 × 10⁻⁶) × 1.177) = 0.00237
Hence, pH = -log[H₃O⁺] = -log(0.00237) = 2.63