Will a precipitate of lead (II) occur if there drops of 0.20M KI are added to 100.0 mL of 0.010M Pb (NO3)2?

Question

Assume that 1 drop= 0.05mL.

Answer 1

Pb(NO₃)₂(aq) + 2KI(aq) ==> PbI₂(s) + 2KNO₃(aq)

PbI₂(s) <==> Pb₂₊(aq) + 2I^-(aq)

Q = [Pb²⁺][I^-]²

Assuming the question reads that THREE drops of 0.20 M KI are added to 100.0 ml, then we have the following:

3 drops x 0.05 ml/drop = 0.15 ml added to 100.0 ml for a total volume of 100.15 ml

[Pb²⁺] = (100 ml)(0.010 M) = (100.15 ml)(x M) and x = 0.009985 M Pb²⁺

[I^-] = (0.15 ml)(0.20 M) = (100.15 ml)(x M) and x = 0.0002996 M I^-

Plug these concentrations into the equation for Q and compare the value of Q to the Ksp (look it up if not given). If Q > Ksp, then a precipitate will form, otherwise it will not.