Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.76×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.22×10−19, what is the equilibrium constant Kfinal for the following reaction?

Kf  = 8.4 x 10²⁶

Explanation:

H₂S(aq) ⇌ HS⁻(aq) + H⁺(aq)

K1 = [HS⁻] [H⁺] / [H₂S]

K1 = 9.76×10−8

9.76×10−8 =  [HS⁻] [H⁺] / [H₂S]

HS⁻(aq) ⇌ S²⁻(aq) + H⁺(aq)

K2 = 1.22×10−19

K2 = [S²⁻] [H⁺] / [HS⁻]

S²⁻ (aq)  + 2H⁺(aq)  ⇌ H₂S(aq)

Kf = [H₂S] /[S²⁻] [H⁺]²

To obtain the value of Kf, the relationship between Kf, k1 and k2 have to be determined;

K1K2 = ( [HS⁻] [H⁺] / [H₂S]) * [S²⁻] [H⁺] / [HS⁻]

K1K2 = [H⁺]² [S²⁻] / [H₂S]

K1K2 = 1 / Kf

Kf = 1  / (K1K2)

Kf = 1 / (9.76×10⁻⁸ * 1.22×10⁻¹⁹)

Kf = 1 / (11.91 x 10-27)

Kf = 0.084 x 10²⁸

Kf  = 8.4 x 10²⁶

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