PbI2 is: PbI2(s) ⇌ Pb2+(aq) + 2I–(aq) At a particular temperature, the concentration of lead ions in a saturated solution of lead iodide was found to be 2.02×10−3 M What is the solubility of lead iodide expressed as mol/L' molar solubility'?

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##PbI_2(s) rightleftharpoons Pb^(2+) + 2I^-##

You have been quoted the concentration of plumbic ions, ##Pb^(2+)##. It is easy to see why this concentration is equal to the required solubility. As a followup to this question, what do you think the examiner means by "saturated solution"?