Problem 76 - Chemistry: Principles and Reactions

Question

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 A 150.0-mL sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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

Step-by-Step Solution

Answer

The $K_{sp}$ value of $\mathrm{Hg}_{2} \mathrm{I}_{2}$ is approximately $4.26 \times 10^{-29}$.

Step 1: Calculate the concentration of I- in the saturated solution

Given the radioactivity of the sample withdrawn is 33 counts per minute, we can first determine the moles of radioactive I-131 in the sample: $$ \text{Moles of I-131} = \frac{\text{radioactivity of the withdrawn sample}}{\text{count rate per mole of I}} $$ Plug in the given values: $$ \text{Moles of I-131} = \frac{33\, \text{counts/minute}}{5.0 \times 10^{11}\, \text{counts/(minute*mole)}} \approx 6.6 \times 10^{-11}\, \text{moles} $$ Then, calculate the concentration of I- in the saturated solution: $$ [\text{I-}] = \frac{\text{moles of I-131}}{\text{volume of solution}} = \frac{6.6 \times 10^{-11}\, \text{moles}}{150.0\, \text{mL}} \times \frac{1\, \text{L}}{1000\, \text{mL}} \approx 4.4 \times 10^{-10}\, \text{M} $$

Step 2: Determine the concentrations of other species in the equilibrium

According to the balanced chemical equation, we have: $$ \mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg_{2}^{2+}}(aq) + 2\mathrm{I^-}(aq) $$ At equilibrium, the concentration of \(\mathrm{Hg_2^{2+}}\) is half the concentration of I- since there is a 1:2 ratio between them: $$ [\mathrm{Hg_2^{2+}}] = \frac{1}{2} [\text{I-}] = \frac{1}{2} \times 4.4 \times 10^{-10} = 2.2 \times 10^{-10}\, \text{M} $$

Step 3: Calculate the \(K_{sp}\) value

Now that we have the concentrations of I- and \(\mathrm{Hg_2^{2+}}\), we can calculate the \(K_{sp}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) using the expression: $$ K_{sp} = [\mathrm{Hg_2^{2+}}] \times [\text{I-}]^2 $$ Plug in the calculated concentrations: $$ K_{sp} = (2.2 \times 10^{-10}) \times (4.4 \times 10^{-10})^2 = 4.26 \times 10^{-29} $$ Thus, the \(K_{sp}\) value of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) is approximately \(4.26 \times 10^{-29}\).

Key Concepts

Chemical Equilibrium

When we talk about chemical equilibrium, we are focusing on the state of a chemical reaction where the concentrations of reactants and products remain constant over time. This doesn't mean the reactions have stopped. Instead, the forward and reverse reactions occur at the same rate, creating a balanced system.
In the case of the exercise, the equilibrium involves solid \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) dissolving and re-precipitating in water. The solid reaches an equilibrium with its ions in solution, forming \(\mathrm{Hg_2^{2+}}\) and \(\mathrm{I^-}\). The solubility product constant (\(K_{sp}\)) is a measure of the ion product at equilibrium and is unique for each compound at a given temperature.
Studying chemical equilibrium helps us understand how substances dissolve, react, and precipitate under specific conditions, providing insights into their ionic interactions.

Radioactivity

Radioactivity refers to the spontaneous emissions of energy or particles from unstable nuclei. This occurs naturally in some isotopes, such as Iodine-131 (\(\text{I-131}\)), which is used in the exercise to measure the concentration of iodide ions. It emits radiation that can be measured, providing a way to trace and quantify substances in chemical reactions.
The ability of radioactive isotopes to help monitor concentrations is invaluable, especially when dealing with changes in small quantities. In this scenario, the radioactivity of the iodide helps us to deduce the concentration of \(\mathrm{I^-}\) ions in solution as part of determining the \(K_{sp}\).
Safety is crucial when working with radioactivity. Always handle these materials using proper safety protocols to avoid unnecessary exposure.

Concentration Calculations

Calculating concentrations is a fundamental skill in chemistry, necessary for determining how much of a substance is present in a given volume. In this problem, we used the counts from radioactivity to find the concentration of \(\mathrm{I^-}\) in the solution. To calculate concentration, we use the formula:

• Concentration = Moles of solute / Volume of solution.
  

This allows us to express concentration in molarity (M), which is moles per liter. In the exercise, converting between counts and moles was critical. We took the measure of radioactivity and calculated the count per unit volume.
These calculations are crucial for assessing the solubility and establishing the solubility product constant (\(K_{sp}\)), essential for predicting how much solute can dissolve in a solution at equilibrium.