CHEMISTRY: Equilibrium Help (Acids and Bases-Diprotic Acid)?

Consider the first dissociation of H₂A:
__________________ H₂A(aq) ____ + ____ H₂O(l) ____ ⇌ ____ HA⁻(aq) ____ + ____ H₃O⁺(aq) ______ Kₐ₁ = 10⁻⁴˙²⁶
Initial (mol L⁻¹): _____ 1.314 ______________________________ 0 _______________ 0
Change (mol L⁻¹): _____ -x _______________________________ +x _______________ +x
Eqm (mol L⁻¹): ____ (1.314 - x) ____________________________ x ________________ x

As Kₐ₁ is very small, H₂A dissociates to a very small extent.
Hence, [H₂A] at eqm = (1.314 - x) mol L⁻¹ ≈ 1.314 mol L⁻¹

At eqm: Kₐ₁ = [HA⁻] [H₃O⁺] / [H₂A]
10⁻⁴˙²⁶ = x² / 1.314
x = √(10⁻⁴˙²⁶ × 1.314) = 0.00850
Hence, [HA⁻] = [H₃O⁺] = 0.00850 mol L⁻¹

Consider the second dissociation of H₂A:
__________________ HA⁻(aq) ____ + ____ H₂O(l) ____ ⇌ ____ A²⁻(aq) ____ + ____ H₃O⁺(aq) ______ Kₐ₂ = 10⁻⁸˙⁷⁸
Initial (mol L⁻¹): ____ 0.00850 ____________________________ 0 ______________ 0.00850
Change (mol L⁻¹): _____ -y _______________________________ +y ________________ +y
Eqm (mol L⁻¹): ___ (0.00850 - y) ___________________________ y ___________ (0.00850 + y)

As Kₐ₂ ≪Kₐ₁, the second dissociation is to a much smaller extent the first dissociation.
Hence, [HA⁻] at eqm = (0.00850 - y) mol L⁻¹ ≈ 0.00850 mol L⁻¹
and [H₃O⁺] at eqm = (0.00850 + y) mol L⁻¹ ≈ 0.00850 mol L⁻¹

At eqm: Kₐ₂ = [A²⁻] [H₃O⁺] / [HA⁻]
10⁻⁸˙⁷⁸ = y * 0.00850 / 0.00850
y = 1.66 × 10⁻⁹
Hence, [A²⁻] = 1.66 × 10⁻⁹ mol L⁻¹