Problem 30 - Chemistry and Chemical Reactivity

Question

\(\mathrm{Ca}(\mathrm{OH})_{2}\) has a \(K_{* p}\) of \(5.5 \times 10^{-5},\) whereas \(K_{\mathrm{sp}}\) for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(5.6 \times 10^{-12} .\) Calculate the equilibrium constant for the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Mg}^{2+}(\mathrm{aq}) \rightleftarrows \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) Explain why this reaction can be used in the commercial isolation of magnesium from sea water.

Step-by-Step Solution

Answer

The equilibrium constant is \( K_{eq} \approx 9.82 \times 10^6 \). The reaction favors \( \mathrm{Mg(OH)}_2 \) formation, aiding magnesium isolation.

Step 1: Write the Dissolution Equilibria

The dissolution equilibria for \( \mathrm{Ca(OH)}_2 \) and \( \mathrm{Mg(OH)}_2 \) can be written as: 1. \( \mathrm{Ca(OH)}_2 (s) \rightleftharpoons \mathrm{Ca}^{2+} (aq) + 2\mathrm{OH}^- (aq) \quad K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2 \)2. \( \mathrm{Mg(OH)}_2 (s) \rightleftharpoons \mathrm{Mg}^{2+} (aq) + 2\mathrm{OH}^- (aq) \quad K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^-]^2 \)

Step 2: Set Up the Reaction

The given reaction is:\[ \mathrm{Ca(OH)}_2(s) + \mathrm{Mg}^{2+}(aq) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{Mg(OH)}_2(s) \]

Step 3: Express the Reaction in Terms of Dissolution Equilibria

Notice that the overall reaction can be expressed in terms of the dissolution reactions you wrote earlier. It essentially involves the dissolution of \( \mathrm{Ca(OH)}_2 \) and the precipitation of \( \mathrm{Mg(OH)}_2 \).

Step 4: Calculate the Equilibrium Constant

For the transformation, we use the formula:\[ K_{eq} = \frac{K_{sp}(\mathrm{Ca(OH)}_2)}{K_{sp}(\mathrm{Mg(OH)}_2)} \]Substitute the given values:\[ K_{eq} = \frac{5.5 \times 10^{-5}}{5.6 \times 10^{-12}} \approx 9.82 \times 10^{6} \]

Step 5: Explain the Reaction Use

Since the equilibrium constant \( K_{eq} = 9.82 \times 10^6 \) is very large, the reaction heavily favors the formation of \( \mathrm{Mg(OH)}_2(s) \). Thus, it "pulls" \( \mathrm{Mg}^{2+} \) ions from solution, allowing magnesium to be isolated efficiently from sea water.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, abbreviated as Ksp, is a helpful tool in chemistry for predicting the solubility of compounds. Specifically, it’s used when dealing with ionic compounds that dissociate in water. Ksp is the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients from the balanced chemical equation.
For example, when calcium hydroxide, \( \text{Ca(OH)}_{2} \), dissolves in water, it dissociates into calcium ions and hydroxide ions. The Ksp expression for this would be:

• \( K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2 \)

Using the known values of Ksp, scientists can calculate how much of a compound will dissolve before excess forms a precipitate, which is crucial in predicting the outcomes of chemical reactions.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions are mixed, leading to the formation of an insoluble solid, or precipitate. This process is governed by the ionic product of the reacting ions exceeding the Ksp of the precipitate.
Consider the reaction involving magnesium hydroxide, \( \text{Mg(OH)}_{2} \). When the concentration of \( \text{Mg}^{2+} \) ions and \( \text{OH}^- \) ions exceeds its Ksp, \( \text{Mg(OH)}_{2} \) precipitates out of the solution. Such reactions are vital in chemical production and analysis because they allow for purification and selective isolation of specific ions.
Precipitation can also be exploited practically, like in the removal or recovery of unwanted ions from wastewater, ensuring industrial processes are more environmentally friendly.

Magnesium Isolation

Isolating magnesium from seawater is possible due to its precipitation reaction. In the solution, magnesium ions react with hydroxide ions to form \( \text{Mg(OH)}_{2} \), a solid precipitate. This reaction occurs because the equilibrium constant for the given reaction significantly favors the formation of this precipitate.
The efficient isolation relies on the principle that certain conditions will "pull" magnesium ions out of the aqueous solution, focusing on the vast difference in Ksp values between \( \text{Ca(OH)}_{2} \) and \( \text{Mg(OH)}_{2} \). Such processes make it economically feasible to extract magnesium, a light and useful metal, from the immense quantities of seawater available.

Equilibrium Constant Calculation

Equilibrium constants (\( K_{eq} \)) can provide insights into the direction and extent of chemical reactions. Calculating these constants is straightforward when you know the conditions under which the reactions occur.
For the specific reaction given in the original problem, the equilibrium constant is determined using the Ksp values of the involved compounds. By dividing the Ksp of \( \text{Ca(OH)}_{2} \) by the Ksp of \( \text{Mg(OH)}_{2} \), we obtain:

• \( K_{eq} = \frac{5.5 \times 10^{-5}}{5.6 \times 10^{-12}} = 9.82 \times 10^{6} \)

A high \( K_{eq} \) value indicates that the reaction heavily favors the products' formation, implying most reactants are converted. This process forms the basis for understanding reactions of different compounds in solution, predicting product yields, and determining reaction feasibility under given circumstances.