To determine the \(K_{s p}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), a solid sample is used, in which some of the iodine is present as radioactive I-131. The count rate of the sample is \(5.0 \times 10^{11}\) counts per minute per mole of \(\mathrm{I}\). An excess amount of \(\mathrm{Hg}_{2} \mathrm{I}_{2}(s)\) is placed in some water, and the solid is allowed to come to equilibrium with its respective ions. A 150.0-mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute. From this information, calculate the \(K_{s p}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\).

Given the radioactivity of the sample withdrawn is 33 counts per minute, we can first determine the moles of radioactive I-131 in the sample: $$ \text{Moles of I-131} = \frac{\text{radioactivity of the withdrawn sample}}{\text{count rate per mole of I}} $$ Plug in the given values: $$ \text{Moles of I-131} = \frac{33\, \text{counts/minute}}{5.0 \times 10^{11}\, \text{counts/(minute*mole)}} \approx 6.6 \times 10^{-11}\, \text{moles} $$ Then, calculate the concentration of I- in the saturated solution: $$ [\text{I-}] = \frac{\text{moles of I-131}}{\text{volume of solution}} = \frac{6.6 \times 10^{-11}\, \text{moles}}{150.0\, \text{mL}} \times \frac{1\, \text{L}}{1000\, \text{mL}} \approx 4.4 \times 10^{-10}\, \text{M} $$

According to the balanced chemical equation, we have: $$ \mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg_{2}^{2+}}(aq) + 2\mathrm{I^-}(aq) $$ At equilibrium, the concentration of \(\mathrm{Hg_2^{2+}}\) is half the concentration of I- since there is a 1:2 ratio between them: $$ [\mathrm{Hg_2^{2+}}] = \frac{1}{2} [\text{I-}] = \frac{1}{2} \times 4.4 \times 10^{-10} = 2.2 \times 10^{-10}\, \text{M} $$

Now that we have the concentrations of I- and \(\mathrm{Hg_2^{2+}}\), we can calculate the \(K_{sp}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) using the expression: $$ K_{sp} = [\mathrm{Hg_2^{2+}}] \times [\text{I-}]^2 $$ Plug in the calculated concentrations: $$ K_{sp} = (2.2 \times 10^{-10}) \times (4.4 \times 10^{-10})^2 = 4.26 \times 10^{-29} $$ Thus, the \(K_{sp}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) is approximately \(4.26 \times 10^{-29}\).