[Bjc51192]

The solubility product for lead II iodide (PbI2, MW =461.2g/mol) is ksp = 8.5*10^-9 at T=25 C. How many grams of PbI2 will dissolve in 100ml of each of the following solutions?

a) 100.0 ml of pure water.

Heres my work I want someone to check if my answer is correct since I dont have the solution.
    Pb2+    2I-                               PbI2------------------> Pb2+ 2I-
I  0.0       0.0
                                                  [Pb2+] [2I-] = 8.5*10^-9
C x          2x

E  x          2x                               (x)(2x)^2=8.5*10^-9


                                                 4x^3=8.5*10^-9

                                                  x=1.28*10^-3M

0.1l * 1.28*10^-3m/l * 461.2g/1mol = 0.06g


What do you guys think?

[Borek]

Looks OK, although I would report 0.059g (think about significant figures).

[BluePill]

Yep. All good expect for significant figures.