If a solution of lead iodide has [I-] = 1.1x10-4 M, what is the lead(II)ion concentration at equilibrium?

Answer 1

Ksp = (Pb^2+)(I^-)^2
Substitute Ksp and I (I^- = 1.1E-4M) and solve for (Pb^2+)

Answer 2

5.86*10^-9