Question

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 A \(150.0-\mathrm{mL}\) sample of the saturated solution is withdrawn and the radioactivity measured at 33 counts per minute.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

Short answer

The short answer to the question is: To determine the \(K_{\mathrm{sp}}\) value for \(\mathrm{Hg}_{2} \mathrm{I}_{2}\), first find the molar ratio between iodine in the solid and in solution using the given count rates. Then, calculate the moles of iodine in solution and convert it into concentration (molarity) using the given solution volume. Next, determine the concentration of \(\mathrm{Hg}_{2}^{2+}\) using the stoichiometric ratio from the balanced equation. Finally, calculate the solubility product constant, \(K_{\mathrm{sp}}\), using the concentrations of ions in the saturated solution: \(K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}\).

Step-by-step solution

Determine the molar ratio between iodine in solid and in solution

Since we're given the count rate for both the iodine in solid (\(5.0 \times 10^{11}\) counts/min/mol) and in the saturated solution (33 counts/min), we can determine the molar ratio between iodine in the solid state and in the saturated solution, using the formula: $$\frac{\text{mole of iodine in solution}}{\text{mole of iodine in solid}} = \frac{\text{count rate in solution}}{\text{count rate in solid}}$$

Calculate the moles of iodine in solution

We can now calculate the moles of iodine in the saturated solution: $$\text{mole of iodine in solution}= \frac{\text{count rate in solution}}{\text{count rate in solid}} \times \text{mole of iodine in solid}$$ We are not given the moles of iodine in solid, but as we only need the concentrations of ions in solution for the \(K_{\mathrm{sp}}\) calculation, we can set the moles of iodine in solid to 1 and get the moles of iodine in the saturated solution using the formula above.

Calculate the concentration of iodine in the solution

We need to convert the moles of iodine in the solution into concentration (molarity) to determine the \(K_{\mathrm{sp}}\). We can use the volume given (150.0 mL) to calculate the concentration: $$\left[\mathrm{I}^{-}\right] = \frac{\text{moles of iodine in solution}}{\text{volume of solution}}$$

Calculate the concentration of \(\mathrm{Hg}_{2}^{2+}\) in the solution

According to the balanced equation: $$\mathrm{Hg}_{2} \mathrm{I}_{2}(s) \rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{I}^{-}(a q)$$ we can see that for each mole of \(\mathrm{Hg}_{2} \mathrm{I}_{2}\) that dissolves, one mole of \(\mathrm{Hg}_{2}^{2+}\) and two moles of \(\mathrm{I}^-\) are produced. Thus, using the stoichiometric ratio, we can find the concentration of \(\mathrm{Hg}_{2}^{2+}\) in solution: $$\left[\mathrm{Hg}_{2}^{2+}\right] = \frac{1}{2} \left[\mathrm{I}^{-}\right]$$

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

Now that we have the concentrations of ions in the saturated solution, we can calculate the solubility product constant, \(K_{\mathrm{sp}}\): $$K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}$$

Key concepts

Solubility Product

When discussing the solubility product, or Ksp, it's all about understanding the point where a solute dissolves in a solvent to form a saturated solution. Saturated means the solution can’t hold any more solute at a given temperature. This point is described by the Ksp, a specific number for every substance.
The solubility product is calculated using the equilibrium concentrations of ions in the solution. For the compound HgₙI₂ , the Ksp is found using Hg₂²⁺ and I⁻ ions.

• The general formula to determine Ksp is: Ksp = [ Hg₂²⁺][ I⁻]² .
• This shows the relationship between the ions once equilibrium is reached in a saturated solution.

Understanding Ksp is crucial in predicting whether a precipitate will form under certain conditions, making it an important concept when working with insoluble salts.

Radioactive Iodine

Radioactive iodine, particularly ¹³¹I , plays a significant role in experiments involving solubility products.
Its radioactivity allows chemists to trace and measure iodine levels precisely. ¹³¹I is commonly used because it emits radiation, which can be detected and measured with great accuracy.

• This characteristic makes it highly useful for determining the concentrations of iodine in a solution.
• In solubility calculations, radioactive iodine helps ensure precise measurement of iodine levels, despite being in a mixed state within a solution and solid.
• In our example, the conversion from count rates to moles is facilitated by the known count rate per mole of ¹³¹I .

Thus, radioactive iodine serves as an invaluable tool in the chemist’s toolkit for studying solubility and equilibrium phenomena with greater precision.

Molarity

Molarity is key when discussing solutions in chemistry. It is simply the number of moles of solute per liter of solution. Knowing the molarity helps predict how chemicals react in solution.
During Ksp calculations, converting counts of radioactive iodine into moles and then into concentration is important.

• To find molarity, once you determine the moles of a substance, divide this by the volume of the solution in liters.
• This conversion helps obtain the [I⁻] and [Hg₂²⁺] concentrations which are critical in solubility calculations.
• For a solution with 33 counts per minute in 150 mL, you must translate this to concentration per liter to find molarity.

Molarity helps understand how much solute is available to react or remain dissolved, and is crucial for computing Ksp values accurately.

Equilibrium Chemistry

Equilibrium chemistry is foundational to understanding how reactions progress and balance.
It describes a state where the forward and reverse reactions occur at the same rate. When a chemical reaction reaches equilibrium in a saturated solution, the concentrations of reactants and products remain constant, forming the basis for determining Ksp .

• In the context of Hg₂I₂, the equilibrium is expressed in terms of the dissolution and precipitation of the salt.
• The equilibrium expression for this reaction can be written as: Hg₂I₂(s) ⇌ Hg₂²⁺(aq) + 2I⁻(aq) .
• Equilibrium chemistry explains why, at this point, the concentrations no longer change, allowing for the reliable use of Ksp calculations to describe the system.

A firm grasp of equilibrium chemistry is essential for predicting the behavior of ionic compounds in solution and calculating their Ksp effectively.