Problem 2 - Chemical Principles in the Laboratory

Question

Magnesium hydroxide is only very slightly soluble in water. The reaction by which it goes into solution is: $$ \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) $$ a. Formulate the expression for the equilibrium constant, \(K_{\text {top }}\) for the above reaction. b. It is possible to dissolve significant amounts of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in solutions in which the concentration of either \(\mathrm{Mg}^{2+}\) or \(\mathrm{OH}^{-}\) is kept very, very small. Explain, using \(K_{\varphi,}\) why this is the case. c. Explain why \(\mathrm{Mg}(\mathrm{OH})_{2}\) might have very appreciable solubility in \(1 \mathrm{M}\) HCl. (Hint: Consider the effect of Reaction 3 on the \(\mathrm{Mg}(\mathrm{OH})_{2}\) solution reaction.)

Step-by-Step Solution

Answer

a. \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \); b. Reducing \(\text{Mg}^{2+}\) or \(\text{OH}^-\) shifts equilibrium to dissolve more Mg(OH)2; c. HCl reacts with \(\text{OH}^-\) increasing Mg(OH)2 solubility.

Step 1: Formulate the Equilibrium Constant Expression (Part a)

The equilibrium constant for a solubility reaction, known as the solubility product constant or \( K_{sp} \), is expressed based on the concentrations of the ions in solution. For the given reaction, \[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \], the equilibrium expression is: \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]. Note that solid reactants are not included in the equilibrium expression.

Step 2: Understand the Effect of Ion Concentration on Solubility (Part b)

Lowering the concentration of \(\text{Mg}^{2+}\) or \(\text{OH}^-\) ions will shift the equilibrium to the right according to Le Chatelier's principle, thus dissolving more \(\text{Mg(OH)}_2\) to maintain the constant \(K_{sp}\) value. This can be achieved by removing one or both ions from the solution, allowing more solid to dissolve to compensate.

Step 3: Solubility in Acidic Solutions (Part c)

In an acidic solution such as \(1 \text{ M HCl}\), the \(\text{OH}^-\) ions react with \(\text{H}^+\) to form water, reducing \([\text{OH}^-]\) concentrations. This consumption of \(\text{OH}^-\) ions shifts the equilibrium towards dissolution of more \(\text{Mg(OH)}_2\) to counterbalance the \(\text{OH}^-\) ion reduction, enhancing its solubility.

Key Concepts

Equilibrium Constant Expression

In a solubility equilibrium, the equilibrium constant expression is an essential concept. It helps in understanding how substances dissolve in a solution, especially those that only slightly dissolve, like magnesium hydroxide. For the reaction \( \mathrm{Mg(OH)}_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq) \), the equilibrium constant is known as the solubility product constant, represented as \( K_{sp} \). This expression involves only the concentrations of the ions in the solution, not the solid reactants themselves, since solids are considered to have constant activity and do not appear in the equilibrium constant expression. In this case, the expression is:

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

By considering the concentrations of \( \mathrm{Mg}^{2+} \) and \( \mathrm{OH}^- \), you can deduce how the reaction shifts to establish equilibrium.

Solubility Product Constant

The solubility product constant \( K_{sp} \) is a specific type of equilibrium constant used in the context of dissolving sparingly soluble salts. For magnesium hydroxide, which dissolves minimally in water, the \( K_{sp} \) provides crucial insights into its solubility limits. The lower the \( K_{sp} \), the less soluble the compound is.To increase solubility, concentrations of ions in the solution can be reduced, which leads to an adjustment in the establishment of equilibrium. This is based on maintaining the \( K_{sp} \) value within the system. An important strategy might include:

• Removing \( \mathrm{Mg}^{2+} \) or \( \mathrm{OH}^- \) ions from the solution to allow more solid to dissolve, thereby maintaining the constant \( K_{sp} \).

Adjusting ion concentrations is key when managing the solubility of barely soluble salts like magnesium hydroxide.

Le Chatelier's Principle

Le Chatelier's principle is fundamental for predicting the reaction's behavior upon changes in conditions, like concentration shifts in a chemical equilibrium. Applying this principle helps us understand why lowering concentrations of \( \mathrm{Mg}^{2+} \) or \( \mathrm{OH}^- \) in the solution increases the solubility of \( \mathrm{Mg(OH)}_2 \). As ions are removed from the solution, the equilibrium shifts to the right which implies:

• More \( \mathrm{Mg(OH)}_2 \) solid will dissolve.

This shift aims to counteract the change enforced by the reduction in ion concentration, highlighting the reason behind increased solubility.

Acid-Base Reactions

In the context of solubility, acid-base reactions can have a significant impact on dissolving seemingly insoluble compounds like magnesium hydroxide. When \( \mathrm{Mg(OH)}_2 \) is introduced into an acidic environment, such as \( 1 \text{ M HCl} \), the \( \mathrm{OH}^- \) ions react with \( \mathrm{H}^+ \) ions to form \( \mathrm{H_2O} \). This reaction reduces the \( [\mathrm{OH}^-] \) concentration, effectively shifting the equilibrium to dissolve more \( \mathrm{Mg(OH)}_2 \).Steps involved in this process might include:

• Addition of a strong acid leads to the neutralization of base ions.
• The decrease in \( \mathrm{OH}^- \) concentration shifts equilibrium towards more dissolution.

Thus, the application of an acid-base reaction indirectly enhances the solubility of magnesium hydroxide by manipulating the ion concentrations.