chem
posted by dude .
The Ksp of PbI2 is 8.7x10^9. Find the solubility of lead(II) iodide on mol/L.
a. 1.3x10^3
b. 9.3x10^9
c. 6.6x10^5
d. 1.6x10^3

PbI2 ==>Pb^+2 + 2I^
Ksp = (Pb^+2)(I^)^2
Let S = solubility, then (Pb^+2) = S and I^) = 2S.
Substitute into the Ksp expression and solve for S. Post your work if you get stuck.