Calculate the molar solubility of leadII chloride in a 010 M

Calculate the molar solubility of lead(II) chloride in a 0.10 M calcium chloride solution. (Be sure you are starting with the correct formulas for these compounds!)

Solution

solubility product of CaCl2 =ksp=1210(highly soluble in aqueous as it has high Ksp)

ksp PbCl2=1.6*10^-6 (slightly soluble in aqueous solvent)

CaCl2 ---->Ca2+ +2Cl- (complete dissociation)

PbCl2 (s)<--->Pb2+(aq) +2Cl-(s)

ksp=[Pb2+][Cl-]^2

[Cl-]=0.10M

Let the molar solubility of lead(II) chloride be S

ksp=[Pb2+][Cl-]^2

or ,1.6*10^-6=S*(0.10M)^2

S=(1.6*10^-6)/(0.1M)^2=1.6*10^-4M

Thus,the molar solubility of lead(II) chloride =S=[Pb2+]=1.6*10^-4M