Into 2 liter of water 0,6 g of lead iodide is dissolved and formed saturated solution. Calculate KSP of this salt, if molecular mass of PbI2 is 461 g/mole.
PbI2 = Pb2+ + 2I-
Solubility of PbI2 = (.6/2)=0.3g/L
Molar Mass of PbI2 = 461g/mol
［PbI2］= (.3/461)mol/L = 6.5 ×10-4
Solubility product Ksp = x(2x)2=4x3
Ksp=4(6.5×10-4)3 =1.0985 ×10-9
Hence the solubility product of PbI2 is 1.0985×10-9 mol2/lit2