Question

Discuss the use of K_sp to separate AgCl from AgBr and AgI, by using NH₃ (find the necessary K_sp and K_f values).

Answer 1

Silver chloride is very thermodynamically stable in ammonia solution. The K for the dissolution of silver chloride in ammonia solution is only 0.0029. suggesting that the product - silver ammonia complex ion - is much less thermodynamically stable than the reactant - silver chloride.

AgCl+2H3N↽−−⇀[Ag(NH3)2]++Cl−

K=Ksp(AgCl)⋅Kf([Ag(NH3)2]+)=1.8⋅10⁻¹⁰⋅(1.6⋅10−7)=0.0029

However, silver chloride will dissolve in concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule Ksp value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good sigma electron pair donor ligand, combines with the silver ion.

AgBr : precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colourless solution

AgI : precipitate is insoluble in ammonia solution of any concentration