[Gammagirl]

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

                                 (x + .00622)(2x)^2=1.4 x 10^-8
                                     reduces to
                                               4x^2=1.4 x 10^-8
                                                     x=5.92 x 10^-5 M
                                                         5.92 x 10^moles/L  x .5 L x 461.01 g/mole= .0136 grams

[Borek]

                                 (x + .00622)(2x)^2=1.4 x 10^-8
                                     reduces to
                                               4x^2=1.4 x 10^-8

It doesn't.

[Gammagirl]

reduces to: 4x^3=1.4 x 10^-8 ?

[Gammagirl]

4x ^3=1.4 x 10^-8
x = 1.52 x 10^-3 M
1.52 x 10^-3 m/L x .5 L x 461.01 g/mole = .35 g

[Corribus]

reduces to: 4x^3=1.4 x 10^-8 ?

Still nope. You are missing an x² term.

[Gammagirl]

 I am starting over.

Ksp=[Pb2+] [I-]^2
1.4 X 10^-8 = s x s^2
1.4 x 10^-8 = .00622 M (2x)^2
x = 7.5 x 10^-4 M
7.5 x 10-^4 moles/L x .5 L x 461.01 g/mole = .173 g