Answer
The mass of iodide ion must exceed 0.003g so that there is precipitation.
Work Step by Step
$PbI_{2}: Ksp=7.9\times10^{-9}$
It is given that we added "1 drop (0.2mL) of 0.10 M $Pb(NO_{3})_{2}$" into the solution. Indeed, we can't use 0.10 M as the concentration of $Pb^{2+}$, as it was its concentration in the drop. Instead, we must calculate its concentration in the new 10.2mL solution using the formula $M_{1}V_{1}=M_{2}V_{2}$.
$M_{1}V_{1}=M_{2}V_{2}$
$0.10\times0.2=M_{2}\times10.2$
$M_{2}=\frac{0.10\times0.2}{10.2}=1.96\times10^{-3}$ M
$[Pb^{2+}]=1.96\times10^{-3}$ M
$PbI_{2} (s)$-> $Pb^{2+} (aq) + 2I^{-} (aq)$
Therefore, $PbI_{2}$'s solubility-product expression is: $Ksp=[Pb^{2+}][I^{-}]^2=7.9\times10^{-9}$
To achieve precipitation, the product $[Pb^{2+}][I^{-}]^2$ must exceed the solubility constant.
$[Pb^{2+}][I^{-}]^2>7.9\times10^{-9}$
$[I^{-}]^2>\frac{7.9\times10^{-9}}{1.96\times10^{-3}}$
$[I^{-}]^2>4.03\times10^{-6}$
$[I^{-}]>\sqrt {4.03\times10^{-6}}=0.002$ M
number of $I^{-}$ moles $> C\times V=0.002 M \times0.012 L=2.4\times10^{-5} mol$
mass of $I^{-}$ $> 2.4\times10^{-5} mol \times 126.9 g/mol=0.003g$
