Problem 57 - Chemistry: Atoms First

Question

How many grams of Milk of Magnesia, Mg(OH) \(_{2}(s)\) (58.3 g/mol), would be soluble in 200 mL of water. \(K_{\mathrm{sp}}=\) \(7.1 \times 10^{-12} .\) Include the ionic reaction and the expression for \(K_{\mathrm{sp}}\) in your answer. \(\left(K_{\mathrm{w}}=1 \times 10^{-14}=\right.\) \(\left.\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\right)\)

Step-by-Step Solution

Answer

Approximately 1.60 mg of Mg(OH)\textsubscript{2} will be soluble in 200 mL of water.

Step 1: Write the Balanced Dissolution Reaction

First, write the balanced equation for the dissolution of Mg(OH)\textsubscript{2}: \[Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)\]

Step 2: Write the Expression for the Solubility Product Constant

Now, express the solubility product constant, \(K_{\mathrm{sp}}\), for the reaction: \[K_{\mathrm{sp}} = [Mg^{2+}][OH^{-}]^2\]

Step 3: Calculate the Concentration of Hydroxide Ions

Given the \(K_{\mathrm{sp}}\) value, let the solubility of Mg(OH)_2 be \(s\) mol/L, which means \([Mg^{2+}] = s\) and \([OH^{-}] = 2s\). Plug these into the expression for \(K_{\mathrm{sp}}\):\[7.1 \times 10^{-12} = s(2s)^2\]Solve for \(s\): \[7.1 \times 10^{-12} = 4s^3\]\[s = \sqrt[3]{\frac{7.1 \times 10^{-12}}{4}}\]

Step 4: Solve for the Solubility of Mg(OH)2

Calculate the cube root to find the value of \(s\): \[s = \sqrt[3]{\frac{7.1 \times 10^{-12}}{4}} \approx 1.37 \times 10^{-4} \text{ mol/L}\]

Step 5: Convert Moles to Grams

Multiply the moles of Mg(OH)2 by its molar mass to find the mass in grams that will dissolve in 200 mL of water: \[\text{Mass of Mg(OH)2} = s \times 200\, \text{mL} \times \frac{1\, \text{L}}{1000\, \text{mL}} \times 58.3 \, \text{g/mol}\]\[\text{Mass of Mg(OH)2} = 1.37 \times 10^{-4} \times 0.200 \times 58.3 \approx 1.60 \times 10^{-3} \text{ g}\]

Key Concepts

Dissolution of Compounds

The dissolution of compounds, particularly sparingly soluble ionic substances like Milk of Magnesia (Mg(OH)₂), is an intriguing process that involves breaking the solid into its constituent ions. The solubility product constant (K_sp) plays a central role in quantifying the extent of dissolution.

When Mg(OH)₂ dissolves in water, the equilibrium between the solid substance and its ions can be expressed by a balanced chemical equation. Assuming the dissolution occurs to a small extent, the concentrations of the ions can be used to determine K_sp, which is a unique value for each compound at a given temperature. For a generic salt AB that dissolves as A⁺(aq) and B^-(aq), the equilibrium expression would be K_sp = [A⁺][B^-].

In our exercise, the dissolution of Mg(OH)₂ follows this principle, and the solubility product expression considers both the stoichiometry of the ionic species and their respective concentrations in the solution. This information is crucial for determining the amount of compound that can dissolve in a given volume of water, a useful calculation when dealing with medications or chemicals with limited solubility.

Ionic Equilibrium

Ionic equilibrium is fundamental understanding in chemistry, referring to a state where the rate at which compounds dissociate into ions in a solution is equal to the rate at which these ions recombine to form the solid compound. The equilibrium is dynamic, meaning that the reactions continuously occur in both directions.

For the equilibrium of Mg(OH)₂ dissociation, it is represented as Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH^-(aq). The solubility product constant (K_sp) is essential for this equilibrium, as it gives us a constant value relating to the concentrations of the ions in a saturated solution. When the ionic product of the ions' concentrations exceeds K_sp, the excess ions will precipitate out of the solution until equilibrium is re-established.

This equilibrium concept helps predict whether a precipitate will form when two solutions are mixed and aids in the calculation of the solubility of a compound in a solvent, as demonstrated in the textbook exercise solution for Milk of Magnesia.

Stoichiometry of Reactions

Stoichiometry of reactions involves the quantitative relationships between reactants and products in a chemical reaction, based on the balanced chemical equation. It is a key concept for understanding reaction yields, reagent quantities, and concentrations.

In the discussed example, the stoichiometry indicates that one molecule of Mg(OH)₂ dissociates to form one Mg²⁺ ion and two OH^- ions. The stoichiometry of the balanced equation directly influences the expression of the solubility product constant (K_sp), as shown by the formula K_sp = [Mg²⁺][OH^-]^2. This square on the hydroxide concentration reflects the 1:2 ratio of Mg²⁺ to OH^- ions.

When performing calculations, stoichiometry allows for the transformation of moles to grams, as done in the final step of the provided exercise, where the moles of Mg(OH)₂ solubilized in water are converted to grams using its molar mass. This kind of calculation is vital in many areas of science and technology, including the preparation of solutions and manufacturing of products.