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~~