Question

\(\mathrm{Ca}(\mathrm{OH})_{2}\) has a \(K_{\mathrm{sp}}\) 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 \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Mg}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) Explain why this reaction can be used in the commercial isolation of magnesium from seawater.

Short answer

The equilibrium constant is \(1.02 \times 10^{-7}\), allowing \(\mathrm{Ca(OH)}_2\) to precipitate \(\mathrm{Mg}^{2+}\) from seawater.

Step-by-step solution

Identify Relevant Equilibrium Expressions

For the given reaction, we start by writing the dissolution equilibrium expressions for each compound. For \(\mathrm{Ca(OH)}_2\), we have the equilibrium: \(\mathrm{Ca(OH)}_2(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{OH}^-(aq)\) with \(K_{sp} = 5.5 \times 10^{-5}\). For \(\mathrm{Mg(OH)}_2\), we have \(\mathrm{Mg(OH)}_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\mathrm{OH}^-(aq)\) with \(K_{sp} = 5.6 \times 10^{-12}\).

Write the Reaction and Express \(K\)

The target reaction is \(\mathrm{Ca(OH)}_2(s) + \mathrm{Mg}^{2+}(aq) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{Mg(OH)}_2(s)\). The equilibrium constant \(K\) for this reaction can be expressed using the relation involving the \(K_{sp}\) values of the individual dissolution processes: \(K = \frac{K_{sp}(\text{Mg(OH)}_2)}{K_{sp}(\text{Ca(OH)}_2)}\).

Calculate the Equilibrium Constant \(K\)

Substitute the given \(K_{sp}\) values into the equation: \(K = \frac{5.6 \times 10^{-12}}{5.5 \times 10^{-5}}\). Simplify the expression to find \(K = 1.02 \times 10^{-7}\).

Interpret the Value of \(K\)

The calculated \(K\) value \(1.02 \times 10^{-7}\) is very small, indicating that the formation of \(\mathrm{Mg(OH)}_2\) solid and \(\mathrm{Ca}^{2+}(aq)\) is heavily favored, driving the reaction towards the right side. This is why \(\mathrm{Ca(OH)}_2\) can be used to remove \(\mathrm{Mg}^{2+}\) ions from seawater by precipitating them as \(\mathrm{Mg(OH)}_2\).

Key concepts

Solubility Product (Ksp)

The solubility product constant, abbreviated as \(K_{sp}\), is a special type of equilibrium constant. It is used in chemistry to understand the dissolution of slightly soluble compounds. For a given solid substance that dissolves in water, its \(K_{sp}\) represents the level of solubility at equilibrium. For instance, \(\mathrm{Ca(OH)}_2\) has a \(K_{sp}\) of \(5.5 \times 10^{-5}\). This tells us that in a saturated solution, the product of the concentration of \(\mathrm{Ca}^{2+}\) ions and \(\mathrm{OH}^-\) ions raised to their stoichiometric coefficients, equals this constant.
In practical terms, a higher \(K_{sp}\) value indicates greater solubility of the compound in water. Conversely, a lower \(K_{sp}\) suggests the compound is less soluble and more likely to form a precipitate. \(\mathrm{Mg(OH)}_2\), having a \(K_{sp}\) of \(5.6 \times 10^{-12}\), is far less soluble compared to \(\mathrm{Ca(OH)}_2\). This difference in solubility is crucial in processes like purifying or isolating elements from solutions.
The \(K_{sp}\) value of a compound helps predict whether a precipitate will form under given concentrations of ions. Understanding the solubility product is essential in fields like analytical chemistry, geochemistry, and environmental science.

Precipitation Reaction

Precipitation reactions occur when two soluble compounds in aqueous solution react and form an insoluble solid, known as a precipitate. These reactions are driven by the low solubility of the product, which is typically represented by its \(K_{sp}\) value. In the case of the reaction between \(\mathrm{Ca(OH)}_2\) and \(\mathrm{Mg}^{2+}\) ions, the formation of \(\mathrm{Mg(OH)}_2\) demonstrates a classic precipitation reaction.
When \(\mathrm{Ca(OH)}_2\) is introduced into a solution containing \(\mathrm{Mg}^{2+}\), the insolubility of \(\mathrm{Mg(OH)}_2\) causes it to precipitate out of the solution. This is because the product of the concentrations of \(\mathrm{Mg}^{2+}\) ions and \(\mathrm{OH}^{-}\) ions exceeds the \(K_{sp}\) value for \(\mathrm{Mg(OH)}_2\), resulting in the solid forming. Such reactions are important in industries that deal with metal recovery and wastewater treatment, as they help to separate particular ions by removing them as solids.
Precipitation reactions are useful not only in isolating or recovering certain substances but also in understanding the extent of solubility and the conditions necessary for precipitate formation.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs because the rate of the forward reaction equals the rate of the reverse reaction. In our context with \(\mathrm{Ca(OH)}_2\) and \(\mathrm{Mg(OH)}_2\), the given equilibrium constant \(K\) tells us about the position of this equilibrium.
The small \(K\) value calculated from the \(K_{sp}\) values of \(\mathrm{Ca(OH)}_2\) and \(\mathrm{Mg(OH)}_2\) indicates that the system strongly favors the formation of \(\mathrm{Mg(OH)}_2(s)\) and \(\mathrm{Ca}^{2+}(\text{aq})\). In this equilibrium, only a tiny fraction of the \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg(OH)}_2\) will revert to the original reactants under standard conditions. This is why \(\mathrm{Ca(OH)}_2\) is an effective agent for removing \(\mathrm{Mg}^{2+}\) from solutions like seawater.
Chemical equilibrium concepts are foundational for understanding many reactions and processes in chemistry. It helps scientists predict how substances will behave under different conditions, crucial in fields ranging from pharmacology to industrial chemistry.