What is the molar solubility of AgCl (Ksp = 1.80 × 10⁻¹⁰) in 0.390 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⁺(aq) .. Kf = 1.7x10⁷

------------------------------------------------------------------------------------------

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

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

3.06x10^-3 = (x)(x) / (0.390 - 2x)²

3.06x10^-3 = x² / 4x² -1.56x + 0.1521

x² = 0.0122x² - 4.77x10^-3x + 4.65x10^-4

0.988x² + 4.77x10^-3x - 4.65x10^-4 = 0

x = 0.0194 M = solubility of AgCl in 0.390 M NH₃

(be sure to check all of the math)