Question

Calculate the molar solubility of Ag ion in AgCl? (Ksp = 1.6x10-10)

AgCl (s) ↔ Ag+ (aq) + Cl- (aq)

Answer 1

Answer:

The molar solubility of Ag+ ion is 1.265 x 10^-5

Explanation:

AgCl (s) ↔ Ag+ (aq) + Cl- (aq)

Given above is the AgCl dissociation equation. It gives Ag+ and Cl-. If we assume the molar solubility to be S, solubility of Ag+ will be 1 x S and that of Cl- will be 1 x S.

The solubility product Ksp = Sol(Ag+) x Sol(Cl-) = S^2

Ksp = 1.6 x 10^-10 = S^2

This gives S= Square root of 1.6 x 10^-10 = 1.265 x 10^-5