Calculate [S2-] of a 0.0533 M solution of hydrogen sulfide (H2S) Where:?

Consider the first dissociation of H₂S :
____________ H₂S(aq) __ ⇌ __ HS⁻(aq) __ + __ H⁺(aq) ___ Kₐ₁ = 1.0 × 10⁻⁷
Initial: _____ 0.0553 M _______ 0 M __________ 0 M
Change: ______ -x M ________ +x M _________ +x M
At eqm: __ (0.0750 - x) M _____ x M __________ x M

As Ka₁ is very small, the dissociation of H₂S would be to a very small extent.
It is assumed that 0.0553 ≫ x, i.e. [H₂S] at equilibrium = (0.0553 - x) M ≈ 0.0553 M

Kₐ₁ = [HS⁻] [H⁺] / [H₂S]
1.0 × 10⁻⁷ = x² / 0.0553
x = √[0.0553 × (1.0 × 10⁻⁷)]
x = 7.4 × 10⁻⁵

As Kₐ₁ ≫ Kₐ₂, [HS⁻] and [H⁺] are not affected by the second dissociation of H₂S.
Hence, [HS⁻] = [H⁺] = 7.4 × 10⁻⁵ M


Consider the second dissociation of H₂A :
____________ HS⁻(aq) __ ⇌ __ S²⁻(aq) __ + __ H⁺(aq) ___ Kₐ₂ = 1.0 × 10⁻¹⁹
At eqm: _ 7.4 × 10⁻⁵ M _____ y M ______ 7.4 × 10⁻⁵ M

Kₐ₂ = [S²⁻] [H⁺] / [HS⁻]
1.0 × 10⁻¹⁹ = y (7.4 × 10⁻⁵) / (7.4 × 10⁻⁵)
y = 1.0 × 10⁻¹⁹

[S²⁻] = 1.0 × 10⁻¹⁹ M