Question

Two equations can be written for the dissolution of \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) in acidic solution. $$\begin{aligned} \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons & \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta G^{\circ}=-95.5 \mathrm{kJ} \mathrm{mol}^{-1} \\ (c) Will the solubilities of \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) in a buffer solution at \(\mathrm{pH}=8.5\) depend on which of the two equations is used as the basis of the calculation? Explain. \frac{1}{2} \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons & \frac{1}{2} \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta G^{\circ}=-47.8 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned}$$ (a) Explain why these two equations have different \(\Delta G^{\circ}\) values. (b) Will \(K\) for these two equations be the same or different? Explain.

Short answer

The \(\Delta G^{\circ}\) values are different because this is an extensive property, and they depend on the amount of substance in the reaction. On the other hand, the \(K\) values for these two equations will be the same, because it is an intensive property and it does not depend on the amount of substance. The solubilities of \(Mg(OH)_2\) at \(pH = 8.5\) would be same in both cases since the effects of \(H^+\) on the reaction would affect both reactions the same way.

Step-by-step solution

Understanding why \(\Delta G^{\circ}\) values are different

The \(\Delta G^{\circ}\) of a reaction, also known as the standard Gibbs free energy change, is generally different when you have half or double of a reaction because it is an extensive property.An extensive property of a system depends on the system size or the amount of material in the system, as it is directly proportional to the amount of substance in a reaction. So, if you halve the reaction, you will get half the energy change.

Understanding if K values are the same or different

The equilibrium constant \(K\) is an intensive property, meaning it does not depend on the amount of substance undergoing a reaction. It remains constant regardless of whether the reaction is doubled or halved. Thus, \(K\) for these two reactions will be the same.

Analyzing the effect of pH

pH is a measure of the concentration of hydrogen ions (\(H^+\)) in a solution. A change in pH can impact the equilibrium position of a reaction involving \(H^+\). In both equations, \(Mg(OH)_2\) is reacting with \(H^+\), so the equilibrium position, and hence the solubility, would be affected. However, as both reactions involve the same substances, the change in \(H^+\) concentration would affect both reactions in the same way, and thus the calculated solubilities would be the same for both equations under the same \(pH = 8.5\).

Key concepts

Gibbs Free Energy

The concept of Gibbs Free Energy is pivotal in determining the spontaneity of a chemical reaction. It is represented as \( \Delta G \). If the value of \( \Delta G \) is negative, the process is spontaneous under standard conditions, meaning it can occur without external energy input.
\( \Delta G^{\circ} \) values differ between reactions due to its nature as an extensive property. An extensive property depends on the quantity of the material involved. For two reactions derived from the same original reaction but differing by a factor of two, as with the dissolution of Mg(OH)\(_{2}\), the \( \Delta G^{\circ} \) value will differ accordingly as you are essentially scaling the energy change with the reaction size. When the reaction is halved, \( \Delta G^{\circ} \) is halved as well, as seen in the provided equations with values of -95.5 kJ/mol and -47.8 kJ/mol.
Understanding Gibbs Free Energy helps in predicting reaction outcomes under given conditions. However, keep in mind that the actual direction of a reaction depends on the conditions such as temperature, pressure, and concentration of reactants and products.

Equilibrium Constant

The equilibrium constant, represented as \( K \), describes the ratio of concentrations of products to reactants at equilibrium in a chemical reaction. Unlike \( \Delta G^{\circ} \, \) the equilibrium constant \( K \) is an intensive property, meaning it is not affected by changes in the amount of substances in the reaction.
Intensive properties remain unchanged when the system is divided or multiplied, unlike extensive properties. Therefore, in the case of both versions of the dissolution reaction of \( \text{Mg(OH)}_2 \, \) though they have different \( \Delta G^{\circ} \, \) the equilibrium constants \( K \) for these reactions remain identical as the fundamental chemical species and their ratios at equilibrium do not change with reaction slicing or scaling.
The importance of the equilibrium constant lies in its utility for predicting how the equilibrium of a particular reaction will shift with varying conditions. It's a crucial parameter when considering reaction directionality.

pH and Solubility

The pH of a solution is a measure of its acidity or basicity, quantified as the negative logarithm of the concentration of hydrogen ions (\( H^+ \)). When dealing with the solubility of compounds like Mg(OH)\(_2\), changes in pH can significantly impact the degree of solubility.
As seen in the reactions from the exercise, Mg(OH)\(_2\) interacts with hydrogen ions, forming Mg\(^{2+}\) and water. Increasing the concentration of \( H^+ \) in the solution by lowering the pH will push the equilibrium toward dissolution, enhancing its solubility. Conversely, a higher pH reduces solubility as there are fewer \( H^+ \) ions to drive the reaction forward.
Under a buffer solution of pH 8.5, the change in the concentration of \( H^+ \) affects both reactions equally. Despite the different ways \(\Delta G^{°}\) is calculated in the full and half-reaction scenarios, the strong dependency on \( H^+ \) concentration means their solubility outcomes will mirror each other at the same pH level, here being 8.5.

Dissolution Reaction

A dissolution reaction describes the process where a solid substance dissolves in a solvent, forming a solution. It involves two main processes: the breaking apart of solid bonds and the integration into the solvent. The case of \( \text{Mg(OH)}_2 \text{(s)} \, \) illustrates such a reaction within an aqueous environment.
In the given reactions, the dissolution of magnesium hydroxide occurs in the presence of \( H^+ \) ions from an acidic solution, forming magnesium ions \( \text{Mg}^{2+} \) and water. This reaction demonstrates not only the principle of solubility equilibrium but also showcases how the surrounding pH can markedly affect the dissolution process.
Two reactions described in varying stoichiometry illuminate how the overall energy change (represented by \( \Delta G^{°} \)) differs due to scaling, although the intrinsic characteristics of the dissolution remain constant. Understanding this reaction in both chemical and mathematical terms provides insight into how solids interact with solutions and the conditions under which they dissolve.