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?