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.

Step by step solution

01

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.

02

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.

03

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 \(pH = 8.5\).

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