Question

Calculate the solubility (in mol/L) of lead(II) chloride, PbCl2, at 25 degrees celsius. Ksp = 1.59 x 10^-5

Answer 1

Answer:

The given salt is PbCl2, it is an AB2 type electrolyte, hence the relation between solubility (s) and solubility product (Ksp) is given by Ksp = 4S³ [As it's AB2 type electrolyte it produces, AB2 A²⁺ +2B^- from which the relation is obtained ] substituting in the formula we get Ksp = 4S³ ===> S³ = Ksp /4   ====> S = 3 Ksp/4 mol/L

Given in the data, Ksp = 1.59 x10^-5, on substituting in the formula we get

                              S = 3[(1.59 x 10 ^-5) / 4] mol/L

                             S = 0.0158408mol/L

                  Hence solubility of PbCl2 at 25⁰C is, S = 0.0158408mol/L