equilibrium chem

Question

A chemist finds that the solubility of silver carbonate is 1.3 x 10 -4 mol/L at 25 C. Calculate the Ksp for silver carbonate.

Answer 1

Silver carbonate = Ag₂CO₃

Ag₂CO₃(s) <==> 2Ag⁺(aq) + CO₃^2-(aq)

Ksp = [Ag⁺]²[CO₃^2-]

If the [CO₃^2-] = 1.3x10^-4 M, then the [Ag⁺] = 2 x 1.3x10^-4 = 2.6x10^-4 M

Substituting these values and solving for the Ksp, we have...

Ksp = (2.6x10^-4)²(1.3x10^-4)

Ksp = 7.5x10^-12