Determine the mass in grams of lead (II) iodide that will dissolve in 500.0 mL of a solution containing 1.03 grams of lead (II) nitrate. Ksp of lead (II) iodide is 1.4 x 10-8. 1.03 g x mol = .00622 mol ------ ------ ------------- 0.5 L 331.2 L PbI2 (s) ==> PbI2(aq) ==>Pb2+ + 2I- x +.00622 2x (.00622)(2x)^2=1.4 x 10^-8 x = 7.5 x 10^-4 M 7.5 x 10^-4 m/L x .5 L x 461.01 g/mole = .173 g