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 A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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

Step by step solution

01

Determine the molar ratio between iodine in solid and in solution

Since we're given the count rate for both the iodine in solid (\(5.0 \times 10^{11}\) counts/min/mol) and in the saturated solution (33 counts/min), we can determine the molar ratio between iodine in the solid state and in the saturated solution, using the formula: $$\frac{\text{mole of iodine in solution}}{\text{mole of iodine in solid}} = \frac{\text{count rate in solution}}{\text{count rate in solid}}$$

02

Calculate the moles of iodine in solution

We can now calculate the moles of iodine in the saturated solution: $$\text{mole of iodine in solution}= \frac{\text{count rate in solution}}{\text{count rate in solid}} \times \text{mole of iodine in solid}$$ We are not given the moles of iodine in solid, but as we only need the concentrations of ions in solution for the \(K_{\mathrm{sp}}\) calculation, we can set the moles of iodine in solid to 1 and get the moles of iodine in the saturated solution using the formula above.

03

Calculate the concentration of iodine in the solution

We need to convert the moles of iodine in the solution into concentration (molarity) to determine the \(K_{\mathrm{sp}}\). We can use the volume given (150.0 mL) to calculate the concentration: $$\left[\mathrm{I}^{-}\right] = \frac{\text{moles of iodine in solution}}{\text{volume of solution}}$$

04

Calculate the concentration of \(\mathrm{Hg}_{2}^{2+}\) in the solution

According to the balanced equation: $$\mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q)$$ we can see that for each mole of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) that dissolves, one mole of \(\mathrm{Hg}_{2}^{2+}\) and two moles of \(\mathrm{I}^-\) are produced. Thus, using the stoichiometric ratio, we can find the concentration of \(\mathrm{Hg}_{2}^{2+}\) in solution: $$\left[\mathrm{Hg}_{2}^{2+}\right] = \frac{1}{2} \left[\mathrm{I}^{-}\right]$$

05

Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\)

Now that we have the concentrations of ions in the saturated solution, we can calculate the solubility product constant, \(K_{\mathrm{sp}}\): $$K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$$

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