Question

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A 0.050 M H₂S solution contains 0.35 M NiCl₂ and 0.35 M Hg(NO₃)₂. What pH is required to precipitate the maximum amount of HgS but none of the NiS?

Answer 1

Here tge reaction of NiS with water will be

NiS (s) + H₂O (l) <==> Ni²⁺ (aq) + HS^- (aq) + OH^-(aq)
Ksp (NiS) = 3 x 10^-16 = [Ni²⁺][HS^-][OH^-]
Here [Ni²⁺] = 0.0.35 M [ because NiCl2 salt splits into Ni2+ and 2Cl-]
but,

H₂S (s) + H₂O (l) <==> HS^- (aq) + H₃O⁺ (aq)
Ka1 = {[HS-][H₃O⁺]} / [H₂S]
here [H₂S] = 0.05M, Ka₁ = 9 x 10^-8
9 x 10^-8 = [HS-][H₃O⁺] / 0.05M
[HS-][H₃O⁺] = 4.5 x 10^-9
[HS-] = 4.5 x 10^-9 / [H₃O⁺]
but Kw = [OH-][H₃O⁺]
[OH-]= Kw / [H₃O⁺]
Ksp (NiS) = 3 x 10-6
3 x 10^-16 = 0.35 M x (4.5 x 10^-9 / [H₃O⁺]) x (1 x 10^-14/[H₃O⁺])
[H₃O⁺] = 2.29 x 10^-4
pH = -log [H₃O⁺] = -log[2.29 x 10^-4]
pH = 3.64