PLEASE HELP WITH THIS CHEM QUESTION!! Need help finding:
initial Pb2+ moles
initial I- moles
equilibrium I- moles
precipitated Pb2+ moles
precipitated I- moles
[Pb2+] at equilibrium
Ksp of PbI2

Given: 5ml of 0.0120M Pb(NO3)2, 2ml of 0.0300M KI, 3ml of 0.20M KNO3, (10ml total volume), 0.349 absorbance, and 5.21x 10^-3 M I- concentration at equilibrium

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Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Iam pretty sure that the answer is d