On a hot day, a \(200.0-\mathrm{mL}\) sample of a saturated solution of \(\mathrm{Pb} \mathrm{I}_{2}\) was allowed to evaporate until dry. If \(240 \mathrm{mg}\) of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\text {sp }}\) value for \(\mathrm{Pb} \mathrm{I}_{2}\) on this hot day.

The balanced chemical equation for the dissociation of PbI2 in solution is: \(PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^- (aq)\)

First, we need to calculate the moles of PbI2. The problem states that 240 mg of solid PbI2 was collected after evaporation. - Convert the mass of PbI2 to grams by dividing 240 mg by 1000: \(240\,mg\, PbI_2 \times \frac{1\,g}{1000\,mg} = 0.24\,g\, PbI_2\) Next, we need the molar mass of PbI2. Using the periodic table, we find the molar mass of each constituent: - Lead (Pb): 207.2 g/mol - Iodine (I): 126.9 g/mol Summing these up for PbI2, we have: \(Molar\,mass\,of\,PbI_2 = 207.2\,g/mol + 2 \times 126.9\,g/mol = 460.0\,g/mol\) Now we can determine the moles of PbI2 by dividing the grams of PbI2 by its molar mass: \(\frac{0.24\,g\, PbI_2}{460.0\, g/mol} = 5.2 \times 10^{-4}\, mol\, PbI_2\)

Now that we have the moles of PbI2, we can find the concentrations of the Pb2+ and I- ions in the saturated solution. Since the initial volume of the solution was 200.0 mL, we can convert it to liters: \(200.0\,mL = 0.200\,L\) Using the balanced chemical equation provided, we see that 1 mole of PbI2 produces 1 mole of Pb2+ ions and 2 moles of I- ions in solution. Therefore, the concentrations of Pb2+ and I- ions are: - \(Pb^{2+}\): \(\frac{5.2 \times 10^{-4}\, mol\, PbI_2}{0.200\,L} = 2.6 \times 10^{-3}\, M\, Pb^{2+}\) - \(I^-\): Since 1 mole of PbI2 produces 2 moles of I- ions, multiply the PbI2 moles by 2: \(2 \times 5.2 \times 10^{-4}\, mol = 10.4 \times 10^{-4}\, mol\, I^-\) Now, calculate the concentration of I- ions: \(\frac{10.4 \times 10^{-4}\, mol}{0.200\,L} = 5.2 \times 10^{-3}\, M\, I^-\)

Now that we have the concentrations of Pb2+ and I- ions in the saturated solution, we can calculate the Ksp value for PbI2 using the formula: \(K_{sp} = [Pb^{2+}] [I^-]^2\) Plug in the values we calculated in the previous step: \(K_{sp} = (2.6 \times 10^{-3}\, M)(5.2 \times 10^{-3}\, M)^2\) Calculate the Ksp value: \(K_{sp} = 7.0 \times 10^{-9}\) The solubility product constant (Ksp) for PbI2 on this hot day is approximately \(7.0 \times 10^{-9}\).