Question

26.

Kę for the complex ion Ag(NH3)2 + is 1.7 x 10”. Ksp for AgCl is 1.6 x 10-10. Calculate the molar solubility of AgCl in 0.70 M NH3. Solubility =

Answer 1

Solubility equilibrium of AgCl

AgCl(s) <-------> Ag⁺(aq) + Cl^-(aq)

Ksp = [Ag⁺] [ Cl^-] = 1.6 ×10^-10

Formation equilibrium of Ag(NH3)2⁺ is

Ag⁺(aq) + 2NH3(aq) <-------> Ag(NH3)2⁺(aq)

K_f = [Ag(NH3)2⁺]/( [Ag⁺][ NH3]²) = 1.7 ×10⁷

adding two equilibrium

AgCl(s) + 2NH3(aq) <-------> Ag(NH3)2⁺(aq) + Cl^-(aq)

K = [Ag(NH3)2⁺][Cl^-]/[NH3]²

K = Ksp × K^{​​​​​​}_f

= 1.6 ×10^-10 × 1.7 × 10⁷

= 2.72 × 10^-3

at equilibrium

[NH3] = 0.70 - 2x

[Ag(NH3)2⁺] = x

[Cl^-] = x

so,

x²/(0.70 - 2x)² = 2.72×10^-3

x/(0.70 - 2x) = 0.05215

x = 0.03651 - 0.1043x

1.1043x = 0.03651

x = 0.03306

Therefore,

molar solubility of AgCl in 0.70M NH3 = 0.03306M