What is the molar solubility of AgCl (Ksp = 1.80 × 10⁻¹⁰) in 0.560 M NH₃? (Kf of Ag(NH₃)₂⁺ is 1.7 × 10⁷)

Answer 1

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

Ag⁺(aq) + 2NH₃(aq) ==> Ag(NH₃)₂⁺ ... Kf = 1.7x10⁷

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AgCl(s) + 2NH₃ <==> Cl^-(aq) + Ag(NH₃)₂⁺ ... Keq = Ksp x Kf = 1.80x10^-10 * 1.7x10⁷ = 3.06x10^-3

Keq = 3.06x10^-3 = [Cl^-][Ag(NH₃)₂⁺] / [NH₃]² - x

3.06x10^-3 = (x)(x) / (0.560)² = x² / 0.314 - x

x² = 9.6x10^-4 - 3.06x10^-3x

x = 0.029 M