Question

Magnesium fluoride dissolves in water to the extent of \(8.0 \times 10^{-2} \mathrm{~g} / \mathrm{L}\) at \(25 \mathrm{C}\). Calculate the solubility of \(\mathrm{MgF}_{2}(s)\) in moles per liter, and calculate \(K_{\mathrm{sp}}\) for \(\mathrm{MgF}_{2}\) at 25 .

Short answer

The solubility of MgF₂ in moles per liter is \(1.28 \times 10^{-3} \mathrm{~M}\) and the solubility product constant (Kₛₚ) for MgF₂ at 25 °C is \(8.39 \times 10^{-9}\).

Step-by-step solution

Determine the molar mass of MgF₂

To convert the solubility from g/L to moles per liter, we need to know the molar mass of MgF₂. The molar masses of magnesium (Mg) and fluorine (F) are 24.31 g/mol and 19.00 g/mol, respectively. MgF₂ contains one magnesium atom and two fluorine atoms, so we can calculate the molar mass as follows: Molar mass of MgF₂ = (1 × 24.31 g/mol) + (2 × 19.00 g/mol) = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol

Convert solubility from g/L to moles per liter

We are given the solubility of MgF₂ as 8.0 × 10⁻² g/L. Using the molar mass calculated in step 1, we can convert the solubility to moles per liter: Solubility (in moles per liter) = (8.0 × 10⁻² g/L) / (62.31 g/mol) = 1.28 × 10⁻³ moles per liter

Write the balanced dissolution equation

Now, we will write the balanced equation for the dissolution of MgF₂ in water: MgF₂(s) ⇌ Mg²⁺(aq) + 2F⁻(aq)

Set up an expression for Kₛₚ

According to the balanced equation, the mole ratio of Mg²⁺ to F⁻ is 1:2. So, for every mole of MgF₂ that dissolves, one mole of Mg²⁺ and two moles of F⁻ are produced in the solution. We can write the solubility in terms of the molar concentrations of the ions: [Mg²⁺] = 1.28 × 10⁻³ M [F⁻] = 2 × 1.28 × 10⁻³ M = 2.56 × 10⁻³ M The expression for the Kₛₚ is the product of the molar concentrations of the ions, raised to the power of their coefficients in the balanced equation: Kₛₚ = [Mg²⁺] [F⁻]²

Calculate the Kₛₚ for MgF₂

Now, we can substitute the molar concentrations determined in Step 4 into the Kₛₚ expression: Kₛₚ = (1.28 × 10⁻³) × (2.56 × 10⁻³)² = (1.28 × 10⁻³) × (6.5536 × 10⁻⁶) = 8.3886 × 10⁻⁹ So, the solubility product constant (Kₛₚ) for MgF₂ at 25 °C is 8.39 × 10⁻⁹.

Key concepts

Solubility Calculation

Understanding solubility is fundamental in chemistry, especially when it comes to predicting whether a substance will dissolve in a given solvent and to what extent. Solubility calculation involves determining how much of a solute can be dissolved in a solvent to form a saturated solution at a defined temperature and pressure. In the context of the exercise provided, the solubility of magnesium fluoride (MgF₂) in water was initially presented in grams per liter, which is a common unit for expressing solubility.

To work with solubility in chemical calculations, however, we often need to express it in moles per liter (M), which is a unit of concentration. This requires us to convert the mass of the solute to moles, a step that necessitates the knowledge of the molar mass of MgF₂. Once we have the solute's solubility in moles per liter, we can engage in various calculations, including determining the solubility product constant (K_sp), which is a valuable expression of solubility in the realm of equilibrium chemistry.

Molar Mass

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's integral to converting between the mass of a substance and the number of moles, which is critical for stoichiometry and for understanding the composition of molecules and compounds. In our case, we calculated the molar mass of MgF₂ by summing the molar masses of one magnesium atom and two fluorine atoms.

Determining Molar Mass:

For an element, the molar mass directly corresponds to the atomic weight found on the periodic table. For a compound like MgF₂, you multiply the atomic weight of each element by the number of atoms of that element in the compound and then add the totals together. This process is essential for solubility calculations as it allows us to measure the amount of the substance in terms of its moles, which is essential for relating the mass of a solute to its corresponding number of particles.

Equilibrium Constant

The equilibrium constant is a crucial aspect of chemical equilibrium, expressing the ratio of product concentrations to reactant concentrations, with each raised to the power of their stoichiometric coefficients. The equilibrium constant provides insight into the extent of a reaction at equilibrium—and in the case of soluble ionic compounds, we specifically use the solubility product constant (K_sp).

K_sp is unique because it applies to the special case of a solid dissolving into its constituent ions in a solution. When we calculated K_sp for MgF₂, we focused on the ions produced when MgF₂ dissolves. The small value of K_sp (in this case, 8.39 × 10⁻⁹) indicates a low solubility, which is typical for sparingly soluble salts like MgF₂. Knowledge of K_sp values enables chemists to predict whether a precipitation reaction will occur and to calculate concentrations of ions in a solution at equilibrium.