Question

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

Short answer

The solubility of \(\mathrm{MgF}_{2}\) in moles per liter is \(1.3 \times 10^{-3} \mathrm{mol/L}\) and the solubility product constant (\(K_{\mathrm{sp}}\)) for \(\mathrm{MgF}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.8 \times 10^{-9}\).

Step-by-step solution

Find the molar mass of \(\mathrm{MgF}_{2}\)

To find the molar mass of \(\mathrm{MgF}_{2}\), we will sum the molar masses of Magnesium (Mg) and two Fluorine ions (F): Molar mass of Mg = 24.31 g/mol Molar mass of F = 19.00 g/mol Molar mass of \(\mathrm{MgF}_{2}\) = 24.31 + (2 × 19.00) = 62.31 g/mol

Calculate the solubility in moles per liter

We are given the solubility in grams per liter (\(8.0 \times 10^{-2} \mathrm{g} / \mathrm{L}\)), and we want to find the solubility in moles per liter. We can achieve this by dividing the solubility in grams per liter by the molar mass of \(\mathrm{MgF}_{2}\): Solubility (moles per liter) = \(\frac{8.0 \times 10^{-2} \mathrm{g} / \mathrm{L}}{62.31 \mathrm{g/mol}} = 1.3 \times 10^{-3} \mathrm{mol/L} \)

Set up the equilibrium expression for \(K_\text{sp}\)

Using the dissolution equation: \[\mathrm{MgF}_{2}(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{F}^{-}(aq)\] The equilibrium expression for the solubility product constant \(K_\text{sp}\) is given by: \[K_\text{sp} = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2\] We are given the solubility of \(\mathrm{MgF}_{2}\) as 1.3 × 10⁻³ mol/L. According to the stoichiometry, for every 1 mole of \(\mathrm{MgF}_{2}\) that dissolves, 1 mole of \(\mathrm{Mg}^{2+}\) and 2 moles of \(\mathrm{F}^{-}\) are formed. Thus, the concentration of \(\mathrm{Mg}^{2+}\) is equal to the solubility in moles per liter (1.3 × 10⁻³ mol/L), while the concentration of \(\mathrm{F}^{-}\) is double this value (2 × 1.3 × 10⁻³ mol/L).

Calculate \(K_\text{sp}\)

Now that we have the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{F}^{-}\), we can plug them into the equilibrium expression to calculate \(K_\text{sp}\): \[K_\text{sp} = [\mathrm{Mg}^{2+}][\mathrm{F}^{-}]^2 = (1.3 \times 10^{-3})(2 \times 1.3 \times 10^{-3})^2 = 6.8 \times 10^{-9}\] The solubility product constant (\(K_\text{sp}\)) for \(\mathrm{MgF}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.8 \times 10^{-9}\).

Key concepts

Equilibrium expressions

In chemistry, equilibrium expressions help us understand the balance between the reactants and products in a chemical reaction that has reached equilibrium. When a sparingly soluble salt like magnesium fluoride, \(\text{MgF}_2\), dissolves in water, it partially dissociates into its ions. The equilibrium expression for its solubility can be represented by an equation that considers both the solid salt remaining and the ions formed in solution.

For \(\text{MgF}_2\), the reaction can be written as: \[\text{MgF}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{F}^-(aq)\]This tells us that one molecule of \(\text{MgF}_2\) dissociates into one magnesium ion, \(\text{Mg}^{2+}\), and two fluoride ions, \(2\text{F}^-\). The equilibrium constant for this dissolution, called the solubility product constant \(K_{\text{sp}}\), is given by:\[K_{\text{sp}} = [\text{Mg}^{2+}][\text{F}^-]^2\]This expression indicates that \(K_{\text{sp}}\) depends on the concentrations of \(\text{Mg}^{2+}\) and \(\text{F}^-\) at equilibrium. To calculate \(K_{\text{sp}}\), you must know the solubility of the salt in mol/L.

Stoichiometry

Stoichiometry plays a crucial role in predicting the proportions of reactants and products involved in chemical reactions. It is particularly vital when calculating the concentrations of ions formed from dissolving substances in water. For the dissolution of \(\text{MgF}_2\) in water, understanding stoichiometry helps us accurately determine the concentrations of the ions \(\text{Mg}^{2+}\) and \(\text{F}^-\) formed from a known amount of dissolved \(\text{MgF}_2\).

• The dissolution equation reveals a 1:2 ratio between \(\text{Mg}^{2+}\) and \(\text{F}^-\).
• For every mole of \(\text{MgF}_2\) that dissolves, one mole of \(\text{Mg}^{2+}\) and two moles of \(\text{F}^-\) are produced.

Thus, if the solubility of \(\text{MgF}_2\) is given as \(1.3 \times 10^{-3}\) mol/L, the concentration of \(\text{Mg}^{2+}\) will also be \(1.3 \times 10^{-3}\) mol/L, and the concentration of \(\text{F}^-\) will be twice that, \(2.6 \times 10^{-3}\) mol/L. Understanding these proportions helps us calculate the product concentrations required to determine the \(K_{\text{sp}}\). Knowing stoichiometric relationships simplifies complex calculations, making them more manageable.

Molar mass calculation

Before performing calculations involving substances in chemical reactions, it is vital to determine their molar masses. Molar mass serves as a critical factor in converting between grams and moles, facilitating comparisons and calculations. For \(\text{MgF}_2\), calculating the molar mass allows us to convert its solubility from grams per liter to moles per liter.The steps to calculate the molar mass of \(\text{MgF}_2\) include:

• Taking the atomic mass of magnesium (Mg), which is 24.31 g/mol.
• Adding twice the atomic mass of fluorine (F), each with a molar mass of 19.00 g/mol.
• Summing these values: \(24.31 + 2 \times 19.00 = 62.31\) g/mol.

With this molar mass, we can convert \(0.08\) g/L of \(\text{MgF}_2\) solubility from grams to moles: \[\text{Solubility (mol/L)} = \frac{0.08 \text{ g/L}}{62.31 \text{ g/mol}} = 1.3 \times 10^{-3} \text{ mol/L}\]This conversion is essential for setting up further calculations related to the equilibrium expression and \(K_{\text{sp}}\) determination. Such calculations make use of fundamental concepts and molar mass to reach meaningful conclusions in chemistry.