Question

To determine the \(K_{\mathrm{sp}}\) 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 A 150.0 -mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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 \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\).

Short answer

Answer: The solubility product, \(K_{\mathrm{sp}}\), for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) in this saturated solution is \(1.4 \times 10^{-31}\).

Step-by-step solution

Calculate the concentration of I- in the sample

To determine the concentration of I- ions, we need to use the given radioactivity measurements. The fraction of radioactive I- ions is: $$\frac{\text{counts/min of solution sample}}{\text{counts/min per mole of I}}$$ So, plugging in the given values, we get: $$\frac{33 \text{ counts/min}}{5.0 \times 10^{11} \text{ counts/min per mole of I}}$$ This gives us the number of moles of I- ions per mole of total iodide ions in the solution. Next, we need to find the total moles of \(\mathrm{I}^{-}\) ions in the 150 mL solution.

Determine the total moles of I- ions

To find the total moles of I- ions present in the solution, we need to multiply the fraction we found in Step 1 by the volume of the solution: $$\text{Total moles of I- ions} = \left(\frac{33 \text{ counts/min}}{5.0 \times 10^{11} \text{ counts/min per mole of I}}\right) \times 0.150 \, \text{L}$$ For this case: $$\text{Total moles of I- ions} = \frac{33}{5.0 \times 10^{11}} \times 0.150 \, \text{L} = 9.9 \times 10^{-12} \, \text{moles}$$ Now we have the total moles of iodide ions in the 150 mL solution.

Calculate the concentration of I- ions and the concentration of Hg2+(aq) ions

To find the concentration of I- ions, we divide the total moles of I- ions by the volume of the solution (in liters): $$[\mathrm{I}^-] = \frac{9.9 \times 10^{-12}\, \text{moles}}{0.150\, \text{L}} = 6.6 \times 10^{-11}\, \mathrm{M}$$ Since the formula of the precipitate is \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), we know that there are two moles of I- ions for each mole of Hg2+ ions. Therefore, the concentration of Hg2+ ions in the saturated solution is half the concentration of I- ions: $$[\mathrm{Hg}_2^{2+}] = \frac{[\mathrm{I}^-]}{2} = \frac{6.6 \times 10^{-11}\, \mathrm{M}}{2} = 3.3 \times 10^{-11}\, \mathrm{M}$$

Calculate the solubility product, \(K_{\mathrm{sp}}\), for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\)

Now that we have the concentrations of \(\mathrm{Hg}_2^{2+}\) and \(\mathrm{I}^-\) ions in the saturated solution, we can find the solubility product, \(K_{\mathrm{sp}}\): $$K_{\mathrm{sp}} = [\mathrm{Hg}_2^{2+}][\mathrm{I}^-]^2$$ Plugging in the concentrations we found earlier: $$K_{\mathrm{sp}} = (3.3 \times 10^{-11}\, \mathrm{M})(6.6 \times 10^{-11}\, \mathrm{M})^2 = 1.4 \times 10^{-31}$$ Therefore, the solubility product, \(K_{\mathrm{sp}}\), for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) is \(1.4 \times 10^{-31}\).