What is the value of the equilibrium constant, Kc, for the dissolution of PbCl2?

Question

Consider the reversible dissolution of lead(II) chloride.

PbCl2(s)−⇀↽−Pb2+(aq)+2Cl−(aq)

Suppose you add 0.2252g of PbCl2(s) to 50.0mL water. When the solution reaches equilibrium, you find that the concentration of Pb2+(aq) is 0.0159M and the concentration of Cl−(aq) is 0.0318M

What is the value of the equilibrium constant, K_c, for the dissolution of  PbCl2?

Answer 1

PbCl₂(s) <==> Pb²⁺(aq) + 2Cl^-(aq)

K = [Pb²⁺][Cl-]²

K = [0.0159][0.0318]²

K = 1.6x10^-5