Question

Calculate the KSP for lead (II) chloride if the solubility is 10.8 g/L.

Please show your work so I can better understand. Thanks! :)

Answer 1

S = 10.8 g/L

Solubility in mol/L = ( Solubility in g/L) / ( molar mass of PbCl2)

          = ( 10.8g/L) / ( 278 g/mol)

          = 0.03883 mol/L

Now we have equation   PbCl2 (s)   <----> Pb2+ (aq) + 2Cl- (aq)

at equilibrium per every PbCl2 dissociated we get Pb2+ and 2Cl-

solubility = 0.03883 mol/L means per 1L 0.03883 moles of PbCl2 dissociated

hence [Pb2+] = 0.03883 mol/L , [Cl-] = 2 x 0.03883 = 0.07767 mol/L

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

= ( 0.03883 M) ( 0.07767)^2

= 2.34 x 10^ - 4