To determine the \(K_{\text {sp }}\) 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 \(\mathrm{I}\). 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 A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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Determine the moles of radioactive Iodine in the withdrawn sample

Using the given count rate of iodine and the radioactivity measured in the withdrawn sample, we can calculate the moles of radioactive Iodine present: \[\text { Moles of radioactive } I^{-} = \frac{\text{counts per minute of sample}}{\text{counts per minute per mole of I}}\] \[\text { Moles of radioactive } I^{-} = \frac{33 \,\text{cpm}}{5.0 \times 10^{11}\, \text{cpm/mol} }\]

02

Calculate the concentration of Iodine ions

Now that we have the moles of radioactive Iodine, we can find the overall concentration of Iodine ions in the withdrawn sample: \[\left[\mathrm{I}^{-}\right] = \frac{\text{moles of radioactive }I^{-}}{\text{volume of sample in L}}\] \[\left[\mathrm{I}^{-}\right] = \frac{33 \,\text{cpm}\div5.0 \times 10^{11}\, \text{cpm/mol}}{0.150\, \text{L}}\] Calculate the value for the concentration of Iodine ions.

03

Determine the concentration of Hg2+ ions at equilibrium

From the balanced equation, for every one Hg2+ ion, there are two I- ions. Therefore, the concentration of Hg2+ ions is half of the concentration of Iodine ions: \[\left[\mathrm{Hg}_{2}^{2+}\right] = \frac{1}{2}\left[\mathrm{I}^{-}\right]\] Calculate the concentration of Hg2+ ions by using the value for Iodine ion concentration from Step 2.

04

Calculate the solubility product constant, \(K_{sp}\)

Having found the concentrations of the ions at equilibrium, we can now use them to calculate the \(K_{sp}\) value. From the given equation: \[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\] Substitute the values for the concentrations of Hg2+ and I- ions from Steps 2 and 3, and calculate the \(K_{sp}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\).

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!