Problem 30 - Chemistry and Chemical Reactivity

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 \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Mg}^{2+}(\mathrm{aq}) \rightleftarrows \mathrm{Ca}^{2+}(\mathrm{aq})+\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) 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 approximately \( 9.82 \times 10^6 \); the reaction favors magnesium precipitation, aiding its isolation from sea water.

Step 1: Understand the Given Solubility Products

We are given the solubility products, also known as the equilibrium constants for dissolution, for two reactions: \( Ca(OH)_2 \) and \( Mg(OH)_2 \). These equilibrium constants are \( K_{sp}^{Ca(OH)_2} = 5.5 \times 10^{-5} \) and \( K_{sp}^{Mg(OH)_2} = 5.6 \times 10^{-12} \).

Step 2: Write the Equilibrium Reactions

For the compound \( Ca(OH)_2 \), the dissolution reaction is: \( Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq) \) with \( K_{sp}^{Ca(OH)_2} = [Ca^{2+}][OH^{-}]^2 \). For \( Mg(OH)_2 \): \( Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq) \) with \( K_{sp}^{Mg(OH)_2} = [Mg^{2+}][OH^{-}]^2 \).

Step 3: Write the Given Reaction

The target reaction is: \( Ca(OH)_2(s) + Mg^{2+}(aq) \rightleftarrows Ca^{2+}(aq) + Mg(OH)_2(s) \). This reaction can be derived from manipulating the dissolution reactions of \( Ca(OH)_2 \) and \( Mg(OH)_2 \).

Step 4: Express the Equilibrium Constant for the Reaction

The equilibrium constant \( K \) for the given reaction can be calculated using the equation: \[ K = \frac{[Ca^{2+}][Mg(OH)_2(s)]}{[Mg^{2+}][Ca(OH)_2(s)]} = \frac{ [Ca^{2+}] }{ [Mg^{2+}] } \] This expression simplifies because the concentrations of the solids \( Mg(OH)_2(s) \) and \( Ca(OH)_2(s) \), which are pure substances, do not affect the chemical equilibrium.

Step 5: Calculate the Equilibrium Constant

Using the values of \( K_{sp} \) for both compounds: \[ K = \frac{K_{sp}^{Ca(OH)_2}}{K_{sp}^{Mg(OH)_2}} = \frac{5.5 \times 10^{-5}}{5.6 \times 10^{-12}} \]Calculate this expression to find \( K \): \[ K = \frac{5.5}{5.6} \times 10^{7} \approx 9.82 \times 10^{6} \].

Step 6: Explain the Commercial Isolation of Magnesium

The calculated \( K \) is very large (\( 9.82 \times 10^{6} \)), which means the reaction overwhelmingly favors the formation of \( Mg(OH)_2 \) solid and \( Ca^{2+} \) ions. Therefore, this equilibrium is strongly shifted to the right, making \( Mg^{2+} \) ions from sea water precipitate as \( Mg(OH)_2 \), allowing easy separation and extraction of magnesium.

Key Concepts

Solubility Product

The concept of solubility product, often abbreviated as \( K_{sp} \), is a crucial part of understanding chemical equilibrium in dissolution reactions. In simpler terms, \( K_{sp} \) helps us predict how much of a compound can dissolve in water before it starts to precipitate. For a slightly soluble compound like \(Ca(OH)_2\), the solubility product equation is \( K_{sp} = [Ca^{2+}][OH^{-}]^2 \). This equation tells us that as we increase the concentration of ions in solution beyond the solubility product, the excess will form a precipitate.

• In the given exercise, \(Ca(OH)_2\) has a \(K_{sp}\) of \(5.5 \times 10^{-5}\) which is relatively higher, indicating it dissolves more compared to \(Mg(OH)_2\).
• Meanwhile, \(Mg(OH)_2\), with its \(K_{sp}\) of \(5.6 \times 10^{-12}\), dissolves much less, allowing its ions to precipitate easily.

Understanding \(K_{sp}\) helps in predicting which compounds will precipitate under certain conditions based on ion concentrations.

Chemical Equilibrium

Chemical equilibrium refers to the state where the rates of the forward and backward reactions are equal, and the concentrations of the reactants and products remain constant over time. This concept is key when discussing solubility products because dissolution reactions must achieve equilibrium to accurately determine the solubility of a compound.

• In the context of \(Ca(OH)_2\) and \(Mg(OH)_2\), each has its distinct equilibrium constant reflecting its unique point of no net change in concentration.
• The equilibrium constant, \(K\), of the overall reaction \(Ca(OH)_2(s) + Mg^{2+}(aq) \rightleftharpoons Ca^{2+}(aq) + Mg(OH)_2(s) \), illustrates which side of the reaction is favored under equilibrium conditions.

Learning how to manipulate and calculate these constants allows one to predict the favorability of certain reactions and processes in various chemical systems.

Precipitation Reaction

Precipitation reactions involve the formation of a solid, or precipitate, within a solution following a reaction between two soluble salts. In our scenario with magnesium and calcium hydroxides, precipitation plays a central role.

• When \(Mg^{2+}\) ions from sea water react with \(Ca(OH)_2\), \(Mg(OH)_2\) forms as a precipitate.
• Given that \(Mg(OH)_2\) has a lower \(K_{sp}\), it precipitates out more readily, driving the reaction forward.

This behavior is pivotal for industrial processes where specific ions need to be removed or purified from mixtures, as it allows the separation of desired substances, like magnesium.

Isolation of Magnesium

The isolation of magnesium from sea water is an intriguing application of solubility products and precipitation reactions. Using the large equilibrium constant from the reaction \(Ca(OH)_2(s) + Mg^{2+}(aq) \rightleftharpoons Ca^{2+}(aq) + Mg(OH)_2(s)\), it is possible to infer why the reaction is ideal for magnesium extraction.

• The large value of \(K = 9.82 \times 10^6\) suggests that the formation of \(Mg(OH)_2\) is heavily favored, ensuring that magnesium ions are efficiently captured and precipitated.
• Once precipitated as \(Mg(OH)_2\), magnesium can be easily filtered and collected from the solution.

This method provides an efficient means to extract magnesium, a valuable metal, from large bodies of water where it is naturally abundant.

Dissolution Reactions

In dissolution reactions, solid chemicals dissolve into their constituent ions within a solvent until equilibrium is reached. This is a frequent process in various natural and industrial settings.

• An example is the dissolution of \(Ca(OH)_2\) into \(Ca^{2+}\) and \(OH^{-}\) ions, which is key to understanding the leverages in our exercise.
• By understanding and analyzing the dissolution processes of different substances like \(Ca(OH)_2\) and \(Mg(OH)_2\), we assess their equilibrium points and how they influence other reactions, such as precipitation.

Mastery of dissolution reaction principles allows chemists and engineers to predict outcomes, create reaction pathways, and optimize reactions for maximum efficiency.