Question

Uranium can be separated from its daughter isotope thorium by dissolving a sample in acid and adding sodium iodide, which precipitates thorium(III) iodide: \(\mathrm{Th}^{3+}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{ThI}_{3}(\mathrm{~s})\) If \(0.567 \mathrm{~g}\) of \(\mathrm{Th}^{3+}\) were dissolved in solution, how many milliliters of \(0.500 \mathrm{M}\) NaI(aq) would have to be added to precipitate all the thorium?

Short answer

14.64 mL of 0.500 M NaI solution is required.

Step-by-step solution

Calculate Molar Mass of Th3+

Determine the molar mass of Th (Thorium):Thorium has an atomic mass of 232.04 g/mol, so the molar mass of \(\mathrm{Th}^{3+}\) is also 232.04 g/mol since the charge does not alter the mass significantly.

Calculate Moles of Th3+

Calculate the moles of \(\mathrm{Th}^{3+}\) in the solution using the given mass:\[ \text{Moles of } \mathrm{Th}^{3+} = \frac{0.567 \text{ g}}{232.04 \text{ g/mol}} \approx 0.00244 \text{ mol} \]

Determine Moles of Iodide Needed

Using the balanced chemical reaction:\[ \mathrm{Th}^{3+} + 3 \mathrm{I}^{-} \rightarrow \mathrm{ThI}_3 \]It shows that 3 moles of \(\mathrm{I}^-\) are needed to react with 1 mole of \(\mathrm{Th}^{3+}\). Hence, the moles of \(\mathrm{I}^-\) needed are:\[ 3 \times 0.00244 \text{ mol} = 0.00732 \text{ mol} \]

Calculate Volume of NaI Solution Required

The concentration of NaI solution is given as 0.500 M, meaning there are 0.500 moles of NaI in every liter of solution.Find the volume of solution required to obtain 0.00732 moles of NaI:\[ \text{Volume (L)} = \frac{0.00732 \text{ mol}}{0.500 \text{ mol/L}} = 0.01464 \text{ L} \]Convert liters to milliliters:\[ 0.01464 \text{ L} = 14.64 \text{ mL} \]

Key concepts

Stoichiometry

Stoichiometry is a cornerstone of chemistry, focusing on the quantitative relationships between reactants and products in a chemical reaction. In this context, stoichiometry utilizes the balanced chemical equation to ensure that the proportions of reactants relate appropriately to the products they form.
In our problem regarding the precipitation of thorium(III) iodide, stoichiometry helps us calculate exactly how much iodide is necessary for a complete reaction with a given amount of thorium. By understanding that one mole of thorium ions reacts with three moles of iodide ions to form thorium iodide, we can directly relate the amounts of these substances.
It emphasizes the importance of balanced chemical equations and mole ratios, which allow us to predict product quantities or determine the required amount of a reactant.

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds react in a solution to form an insoluble compound, known as the precipitate. This type of reaction is crucial for separating components in a solution into distinct phases.
In the given exercise, thorium(III) ions (\(\mathrm{Th}^{3+}\)) react with iodide ions (\(\mathrm{I}^-\)) to form thorium(III) iodide, which precipitates out of the solution as a solid. This reaction is advantageous for isolating pure thorium since the solid can be easily separated from the liquid.
The precipitation happens because thorium iodide is insoluble in water and forms a distinct phase. Precipitation reactions like these are not only crucial in chemical labs but are also vital in many industrial applications, including water purification and metal recovery processes.

Concentration Calculations

Understanding concentration calculations is central for determining the precise volume or mass of solutions involved in reactions. Concentration indicates how much solute is dissolved in a given volume of solvent and is commonly expressed in molarity (M).
Molarity is defined as moles of solute per liter of solution. Here, the iodide solution has a concentration of 0.500 M, meaning there are 0.500 moles of sodium iodide in every liter.
To find out how much of this solution is necessary for the reaction, we first determine the moles needed using stoichiometry. Then, using the molarity, we calculate the required volume to ensure complete precipitation of thorium. Such calculations are essential for efficient and effective experimental and industrial processes, ensuring that materials are used optimally without excess.