Assuming that no equilibria other than dissolution is involved, calculate the concentration of iodide ions, [I-(aq)], in the following saturated solution of lead iodide, PbI2(s).
Solubility product PbI2 Ksp = 9.8 x 10-9
A) [I-] = 0.0043M
B) [I-] = 9.8 x 10-9M
C) [I-] = 0.0013M
D) [I-] = 0.0026M
Answer D