Question

Calculate the value of the equilibrium constant for the reaction below. Ksp for Ag2S is 6.0...

Calculate the value of the equilibrium constant for the reaction below.

Ksp for Ag2S is 6.0 × 10−51...... Kf for Ag(NH3)2+ is 1.7 × 107.

Ag2S(s) + 4 NH3(aq) ↔ 2 Ag(NH3)2+(aq) + S2−(aq)

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Homework Answers

Answer #1

The equilibrium reaction is:

Ag2S(s) + 4 NH3(aq) ↔ 2 Ag(NH3)22+(aq) + S2−(aq)

Keq = [Ag(NH3)2+]2 [S2−] / [Ag2S] [NH3]4

This equilibrium reaction can be had from other two reactions as:

Ag2S(s) ↔ 2 Ag2+(aq) + S2−(aq)  

Ksp = [S2−] [Ag2+]2 / [Ag2S] = 6.0 × 10−51 ========(1)

2Ag2++ 4NH3(aq) ↔ 2Ag(NH3)22+(aq)

Kf = [Ag(NH3)22+]2 /[NH3]4 [Ag22+]2 = 1.7 × 107.==========(2)

Dividing (1) by (II)

Ksp / Kf = [Ag(NH3)2+]2 [S2−] / [Ag2S] [NH3]4 = Keq = 6.0 × 10−51 / 1.7 × 107 = 3.52 X 10−44

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