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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 A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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

01

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#
02

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 \(4.3\times10^{-29}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hg2I2
Mercury(I) iodide, also known as \(\text{Hg}_2\text{I}_2\), is a fascinating compound. It's an example of a sparingly soluble salt, which means it doesn't dissolve easily in water. Instead, most of it remains as a solid when mixed with water.
When \(\text{Hg}_2\text{I}_2\) is dissolved to some extent, it can separate into its ions, \(\text{Hg}_2^{2+}\) and \(\text{I}^-\). These ions are crucial when it comes to understanding the solubility product constant \(K_{\text{sp}}\).
  • \(\text{Hg}_2\text{I}_2\) is often used in equilibrium studies to explore how substances dissolve.
  • It's known for having distinct properties, such as changing color from yellow to red when heated, indicating a change in its crystalline structure.
Understanding \(\text{Hg}_2\text{I}_2\) is key to mastering the concepts of solubility, equilibrium, and chemical calculations.
Radioactivity
Radioactivity involves the decay of unstable atomic nuclei, which releases energy in the form of particles or electromagnetic waves. In the context of this exercise, the isotope \(^{131}\text{I}\) was used.
\(^{131}\text{I}\) is a radioactive isotope of iodine often used in scientific studies because its decay can be measured as counts per minute. These measurements help determine the concentration of ions in solutions.
  • The use of \(^{131}\text{I}\) allows for precise determination of concentration by tracking radioactive decay.
  • Radioactive isotopes are helpful in tracing reactions and calculating exact amounts of substances present.
This exercise uses the radioactivity of iodine to understand chemical reactions and equilibrium processes better.
Chemical Equilibrium
Chemical equilibrium refers to a state in which the rate of the forward reaction matches the rate of the reverse reaction. For \(\text{Hg}_2\text{I}_2\), equilibrium is reached when the solid hasn't dissolved further, but some ions are present in the solution.
In our case, \(\text{Hg}_2\text{I}_2(s) \rightleftharpoons \text{Hg}_2^{2+}(aq) + 2\text{I}^-(aq)\), this equilibrium is critical for solving the problem.
  • At equilibrium, the concentration of each ion doesn't change, even though the reactions are ongoing.
  • This balance is expressed using the solubility product constant \(K_{\text{sp}}\), which quantifies the concentration of ions in solution.
Understanding chemical equilibrium is crucial for predicting how much of a compound will dissolve in a solution.
Concentration Calculation
Concentration calculations involve determining how much of a substance is present in a given volume of solution. In this exercise, the concentration of ions from \(\text{Hg}_2\text{I}_2\) was calculated.
By using radioactivity data, we first determined the moles of \(^{131}\text{I}\) and then used that to find the concentration of \(\text{I}^-\) ions.
  • Concentration is often expressed in molarity (M), which is moles per liter.
  • Knowing the concentration of \(\text{I}^-\) helped determine the concentration of \(\text{Hg}_2^{2+}\) ions.
Finally, these concentrations were used to calculate the solubility product constant \(K_{\text{sp}}\), allowing us to understand the extent to which \(\text{Hg}_2\text{I}_2\) dissolves in water.

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Most popular questions from this chapter

Many transuranium elements, such as plutonium-232 , have very short half- lives. (For \(^{232} \mathrm{Pu}\) , the half-life is 36 minutes.) However, some, like protactinium- 231 (half-life \(=3.34 \times 10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \(^{232} \mathrm{Pu}\) nuclei and 1 mole of \(^{231} \mathrm{Pa}\) nuclei are each formed from their respective number of protons and neutrons. (Since the masses of \(^{232} \mathrm{Pu}\) and \(^{231} \mathrm{Pa}\) are atomic masses, they each include the mass of the electrons present. The mass of the nucleus will be the atomic mass minus the mass of the electrons.)

Assume a constant \(1^{14} \mathrm{C} /^{12} \mathrm{C}\) ratio of 13.6 counts per minute per gram of living matter. A sample of a petrified tree was found to give 1.2 counts per minute per gram. How old is the tree? (For \(^{14} \mathrm{C}, t_{1 / 2}=5730\) years.)

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, Sg, in honor of Glenn T. Seaborg, discoverer of the transuranium elements. a. \(^{263}\) Sg was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) g decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg}\) ?

Do radiotracers generally have long or short half-lives? Explain.

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{24} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am}\) ?

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