Describe how to write the Ksp.

b2+ + 2 NO3- + 2 Na+ + 2 I- >> PbI2 (s) + 2 Na+ + 2 NO3-
net ionic
Pb2+ + 2 I- >> PbI2
Ksp = [Pb2+] [I-]^2
Ksp = 7.1 x 10^-9
molar solubility
let x = mol/L of Pb2+ that dissolve . This will give x mol/L Pb2+ and 2x mol/L I-
7.1 x 10^-9 = (x) (2x)^2 = 4x^3
molar solubility = 0.0012 M
~~thanks to google research~~