Question

The solubility of \(\mathrm{Li}_{3} \mathrm{Na}_{3}\left(\mathrm{AlF}_{6}\right)_{2}\) is \(0.0744 \mathrm{~g}\) per \(100 \mathrm{ml}\) at \(298 \mathrm{~K}\). Calculate the solubility product of the salt. (Atomic masses: \(\mathrm{Li}=7, \mathrm{Na}=23, \mathrm{Al}=27, \mathrm{~F}=19)\) (a) \(2.56 \times 10^{-22}\) (b) \(2 \times 10^{-3}\) (c) \(7.46 \times 10^{-19}\) (d) \(3.46 \times 10^{-12}\)

Short answer

\(K_{sp} = 2.56 \times 10^{-22}\)

Step-by-step solution

Determine the Molar Mass of the Salt

Calculate the molar mass of \(\mathrm{Li}_{3}\mathrm{Na}_{3}\left(\mathrm{AlF}_{6}\right)_{2}\) by adding the atomic masses of its constituent elements: 3(Li) + 3(Na) + 2(Al) + 12(F).

Calculate the Molarity of the Salt

Convert the solubility from grams per 100 mL to moles per liter (Molarity, M) by dividing the mass of the salt that can be dissolved by the molar mass of the salt and multiplying by 10 to account for the volume.

Write the Dissociation Equation

Write the balanced dissociation equation for the salt \(\mathrm{Li}_{3}\mathrm{Na}_{3}\left(\mathrm{AlF}_{6}\right)_{2} \rightarrow 3\mathrm{Li}^+ + 3\mathrm{Na}^+ + 2\mathrm{Al}^{3+} + 12\mathrm{F}^-\).

Express the Solubility Product (Ksp)

Use the molar concentrations of the ions obtained from the solubility and the stoichiometry of the dissociation equation to calculate the solubility product \(K_{sp}\).

Key concepts

Dissociation Equation

Understanding the dissociation equation is fundamental to grasping how a salt breaks apart into its constituent ions when dissolved in water. This process is essential for calculating the solubility product of the substance. For instance, take the salt \(\mathrm{Li}_{3}\mathrm{Na}_{3}(\mathrm{AlF}_{6})_{2}\) used in our exercise, which dissociates according to the reaction: \[(\mathrm{Li}_{3}\mathrm{Na}_{3}(\mathrm{AlF}_{6})_{2} \rightarrow 3\mathrm{Li}^+ + 3\mathrm{Na}^+ + 2\mathrm{Al}^{3+} + 12\mathrm{F}^-)\].

Each molecule of the salt yields multiple ions of each type. Building the knowledge to balance these equations prepares students to anticipate the concentration of each ion in solution. For solubility products, the coefficients in the balanced equation directly influence the calculation since each ion contributes to the overall product, with its concentration raised to the power corresponding to its coefficient. This process lays the groundwork for understanding the relationship between dissolved substances and their respective ions in a solution.

Molar Mass Calculation

Calculating the molar mass of a compound is crucial before converting its mass to moles, which is a key step in finding the molarity and solubility product. The molar mass is the sum of the atomic masses of all the atoms in the formula of the compound. Using the atomic masses provided (Li = 7, Na = 23, Al = 27, F = 19) for the compound \(\mathrm{Li}_{3}\mathrm{Na}_{3}(\mathrm{AlF}_{6})_{2}\), we would add up the atomic masses like so: \[(3 \times 7) + (3 \times 23) + (2 \times 27) + (12 \times 19) = 21 + 69 + 54 + 228 = 372 \text{g/mol}.\]

This calculated molar mass enables us to further understand the conversion of mass solubility to molarity, a necessary step in determining the solubility product (Ksp). Molar mass calculations not only aid in understanding chemical compositions but also provide the means to quantitatively compare different substances.

Molarity Conversion

Molarity conversion is a key concept when it comes to solving chemistry problems involving solutions. It's the bridge between the mass of a substance and its role in the solution. Molarity, denoted as M, is defined as the number of moles of solute per liter of solution. To convert the solubility from grams per 100 mL to molarity, as done in our exercise, you follow these steps: take the solubility in grams, divide by the molar mass to find the moles, and then adjust for the volume of the solution. So for our example: \[(0.0744 \text{ g}) / (372 \text{ g/mol}) \times 10 = 2 \times 10^{-3} \text{ M}.\]

Recognizing this process allows students to transform a practical measurement (mass of the salt per 100 mL of water) into a concentration which chemists use to make predictions about the behavior of solutions. This value is especially important in the context of calculating the Ksp, where the concentration of the ions must be known.