solubility-product constant help please?

Question

a 1.00L solution saturated at 25 degrees Celcius with lead(II) iodide contains .54g of PbI2. Calculate the solubility-product constant for this salt at 25 degrees Celcius.

Answer 1

Ksp = [Pb++][I-]^2

Atomic weights: Pb=207 I=127 PbI2=461

0.54gPbI2 x 1molPbI2/461gPbI2 = 1.2 x 10^-3 mol PbI2

Then [Pb++] = 1.2 x 10^-3M and [I-] = 2.4 x 10^-3M

K2p = (1.2x10^-3)(2.4x10^-3)^2 = 6.9 x 10^-9