Chapter 19: Problem 93

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 A $150.0-\mathrm{mL}$ sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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 $\mathrm{Hg}_{2} \mathrm{I}_{2}$ $$ \mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \qquad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2} $$

Short Answer

Expert verified

The Ksp value for Hg2I2 is $4.3\times10^{-29}$.

Step by step solution

We are given the radioactivity count rate of the solid Hg2I2 sample (5.0 x 10^11 counts per minute per mole of I), and after reaching equilibrium, the radioactivity was measured at 33 counts per minute in the 150 mL saturated solution. To find the moles of 131I in the 150 mL saturated solution, we use the ratio between the initial radioactivity count rate and the radioactivity count rate in the saturated solution: \[ \begin{aligned} \text{Moles of 131I in saturated solution} &= \frac{33 \ \text{counts/min}}{5.0\times10^{11}\ \text{counts/min/mol}}\\ &= 6.6\times10^{-11} \text{moles} \end{aligned} \] #Step 2: Determine the concentration of I- ions in the saturated solution# Now that we have the moles of 131I in the 150 mL saturated solution, we can calculate the concentration of I- ions in the saturated solution:

The volume of saturated solution is 150 mL, which is equal to 0.150 L. So, the concentration of I- ions in the solution can be calculated as: \[ \begin{aligned} \text{Concentration of I- ions} &= \frac{\text{Moles of 131I}}{\text{Volume of saturated solution}}\\ &= \frac{6.6\times10^{-11}\ \text{moles}}{0.150\ \text{L}} \\ &= 4.4\times10^{-10}\ \text{M} \end{aligned} \] #Step 3: Determine the concentration of Hg22+ ions in the solution#

According to the balanced chemical equation, 1 mole of Hg2I2 produces 1 mole of Hg22+ and 2 moles of I- ions. We know the concentration of I- ions. Hence, we can find the concentration of Hg22+ ions: \[ \text{Concentration of Hg}_{2}^{2+}\ \text{ions} = \frac{1}{2} \times \text{Concentration of I- ions} = \frac{1}{2} \times 4.4\times10^{-10}\ \text{M} = 2.2\times10^{-10}\ \text{M} \] #Step 4: Calculate the Ksp value for Hg2I2#

Now that we have the concentrations of Hg22+ and I- ions in the saturated solution, we can calculate the Ksp value: \[ \begin{aligned} K_{sp} &= \left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\\ &= \left(2.2\times10^{-10}\right)\left(4.4\times10^{-10}\right)^{2}\\ &= 4.3\times10^{-29} \end{aligned} \] The Ksp value for Hg2I2 is $4.3\times10^{-29}$.