Question

The solubllity of AgCI in water at 298 K is 1.06 x 10^-5 mole per litre. Calculate is solubility product at this temperature.

Answer 1

The solubility equilibrium in the saturated solution is 

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq) 

The solubility of AgCl is 1.06 x 10^-5 mole per litre.

 [Ag⁺ (aq)] = 1.06 x 10^-5 mol L^-1

[Cl^- (aq)] = 1.06 x 10-⁵ mol L^-1

K_sp = [Ag⁺(aq)] [Cl^- (aq)] 

= (1.06 x 10^-5 mol L^-1) x (1.06 x 10^-5 mol L^-1) 

= 1.12 x 10^-2 moI^-2 L^-2