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 \(^{131}\) 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 A 150.0 -mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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 \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) $$\mathrm{Hg}_{22}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \quad K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$$

Step by step solution

01

Determine the concentration of iodide ions in the sample.

We can calculate the concentration of \(\mathrm{I}^{-}\) in the 150 mL sample based on the count rate of the radioactivity in the sample. The ratio of the count rate in the solution (33 counts per minute) to the count rate in the solid \(\mathrm{Hg}_2\mathrm{I}_{2}\) sample (5.0e11 counts per minute per mole of I) can help us find the concentration of iodide ions in the solution. \[ \frac{33\,\text{counts/min}}{5.0 \times 10^{11}\,\text{counts/min/mol}} = \frac{[\mathrm{I}^-]}{1\,\text{mol}} \] Now, we can calculate the concentration of \(\mathrm{I}^-\): \[ [\mathrm{I}^{-}] = 33 \div 5.0 \times 10^{11} = 6.6 \times 10^{-11}\,\text{mol} \] Since the volume of the sample is 150 mL, we can convert the amount (in moles) to concentration (in mol/L): \[ [\mathrm{I}^{-}] = \frac{6.6 \times 10^{-11}\,\text{mol}}{0.150\,\text{L}} = 4.4 \times 10^{-10}\,\text{M} \]

02

Determine the concentration of mercuric ions in the sample.

From the balanced chemical equation, we can see that for each mole of \(\mathrm{Hg}_{2}^{2+}\) ions, there are 2 moles of \(\mathrm{I}^{-}\) ions formed in the solution. Therefore, we can calculate the concentration of \(\mathrm{Hg}_{2}^{2+}\) using the following relationship: \[ [\mathrm{Hg}_{2}^{2+}] = \frac{1}{2}[\mathrm{I}^{-}] \] Now, we can calculate the concentration of \(\mathrm{Hg}_{2}^{2+}\): \[ [\mathrm{Hg}_{2}^{2+}] = \frac{1}{2}(4.4 \times 10^{-10}\,\text{M}) = 2.2 \times 10^{-10}\,\text{M} \]

03

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

Now that we have the concentrations of \(\mathrm{Hg}_{2}^{2+}\) and \(\mathrm{I}^-\), we can calculate the \(K_{sp}\) using the following expression: \[ K_{sp} = [\mathrm{Hg}_{2}^{2+}][\mathrm{I}^-]^2 \] Substitute the values of the concentrations into the expression: \[ K_{sp} = (2.2 \times 10^{-10})(4.4 \times 10^{-10})^2 \] Now, we can calculate the \(K_{sp}\) value: \[ K_{sp} = 4.27 \times 10^{-29} \] So, the solubility product constant, \(K_{sp}\), for \(\mathrm{Hg}_{2}\mathrm{I}_{2}\) is approximately \(4.27 \times 10^{-29}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!