Calculate the pH and [S2− ] in a 0.17 M H2S solution. Assume Ka1 = 1.0 ✕ 10−7; Ka2 = 1.0 ✕ 10−19.?

As Ka₁ ≫ Ka₂, HS⁻ and H⁺ are almost completely formed in the first dissociation of H₂S.
It can be assumed that in the second dissociation the changes of [H₂S], [HS⁻] and [H⁺] are negligible.

Consider the first dissociation of H₂S :
____________ H₂S(aq) __ ⇌ __ HS⁻(aq) __ + __ H⁺(aq) ___ Kₐ₁ = 1.0 × 10⁻⁷
Initial: ______ 0.17 M ________ 0 M __________ 0 M
Change: ______ -y M ________ +y M _________ +y M
At eqm: ___ (0.17 - y) M ______ y M __________ y M

As Kₐ₁ is very small, the dissociation of H₂S would be to a very small extent.
It is assumed that 0.17 ≫ y, i.e. [H₂S] at equilibrium = (0.17 - y) M ≈ 0.17 M

Kₐ₁ = [HS⁻] [H⁺] / [H₂S]
1.0 × 10⁻⁷ = y² / 0.17
y = √[0.17 × (1.0 × 10⁻⁷)]
y = 1.3 × 10⁻⁴
pH = -log[H⁺] = -log(1.3 × 10⁻⁴) = 3.9

Consider the second dissociation of H₂S :
____________ HS⁻(aq) __ ⇌ __ S²⁻(aq) __ + __ H⁺(aq) ___ Kₐ₂ = 1.0 × 10⁻¹⁹
At eqm: __ 1.3 × 10⁻⁴ M _______ ? M _____ 1.3 × 10⁻⁴ M

Ka₂ = [S²⁻] [H⁺] / [HS⁻]
1.0 × 10⁻¹⁹ = [S²⁻] (1.3 × 10⁻⁴) / (1.3 × 10⁻⁴)
[S²⁻] = 1.0 × 10⁻¹⁹ M

Answer :
pH = 3.9
[S²⁻] = 1.0 × 10⁻¹⁹ M