What is the molar solubility of AgCl (Ksp = 1.80 × 10⁻¹⁰) in 0.750 M NH₃? (Kf of Ag(NH₃)₂⁺ is 1.7 × 10⁷)

Question

Please answer with the quadratic equation if needed for the problem

Answer 1

This is a problem dealing with complex ion formation:

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

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

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

K = 3.06x10^-3 = [Ag(NH₃)₂⁺][Cl^-] / [NH₃]² = (x)(x)/ (0.750 - x)² (assume x is small relative to 0.750), then ...

x² = 3.06x10^-3 (0.563) = 1.72x10^-3

x = 0.0414 (compared to 0.750 this is ~5.5% so we need to go back and use the quadratic)

3.06x10^-3 = (x)(x) / (0.750 - x)²

3.06x10^-3 = (x)(x) / x²-1.5x + 0.563

0.997x² + 4.59x10^-3 - 1.72x10^-3 = 0

x = 0.0393 M = molar solubility

(be sure to check the math)