What is the molar solubility of AgCl (Ksp = 1.80 × 10⁻¹⁰) in 0.530 M NH₃? (Kf of Ag(NH₃)₂⁺ is 1 × 10⁷)

Question

What is the molar solubility of AgCl (Ksp = 1.80 × 10⁻¹⁰) in 0.530 M NH₃? (Kf of Ag(NH₃)₂⁺ is 1 × 10⁷)

Answer 1

Complex ion formation:

AgCl(s) <==> Ag⁺(aq) + Cl^-(aq) ... Ksp = 1.80x10^-10

Ag⁺(aq) + 2NH₃(aq) ==> Ag(NH₃)₂⁺(aq) ... Kf = 1x10⁷

____________________________

AgCl(s) + 2NH₃(aq) ==> Ag(NH₃)₂⁺_(aq) + Cl^-(aq) ... K = Ksp x Kf = 1.80x10^-3

K = 1.80x10^-3 = [Ag(NH₃)₂⁺][Cl^-] / [NH₃]²

1.80x10^-3 = (x)(x) / (0.530)² = x2 / 0.281

x² = 5.06x10^-4

x = 2.25x10^-2 M = molar solubility of AgCl in 0.530 M NH₃