Question

A saturated solution of lead(ii) fluoride, pbf2, was prepared by dissolving solid pbf2 in water. the concentration of pb2+ ion in the solution was found to be 2.08×10−3 m. calculate ksp for pbf2

Answer 1

Equation :

PbF2 ------> Pb2+ + 2F-

Solubility product is given as:

Ksp = [Pb2+][F-]^2

let x = [Pb2+]

Ksp = x(2x)^2 = 4(x)^3 = 4(2.08*10^-3)^3

Ksp = 3.60*10^-8