Mg(OH)2(s) ⇄ Mg2+(aq) + 2 OH−(aq) The exothermic dissolution of Mg(OH)2(s) in water is represented b — Equilibrium Chemistry Question

STEPS:

1. Analyze the solubility equilibrium equation: The dissolution of solid magnesium hydroxide is represented by the equilibrium process:
$$\text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\ \text{OH}^-(aq)$$
The solubility product constant expression for this process is:
$$K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = 1.8 \times 10^{-11}$$
2. Apply Le Chatelier's Principle to solubility: For a solid to become more soluble, the equilibrium must shift in the forward direction (to the right), which represents more solid dissolving into its constituent ions. According to Le Chatelier's Principle, a system at equilibrium will shift to counteract a disturbance. To force this system to the right, we must either remove one of the product ions ($[\text{Mg}^{2+}]$ or $[\text{OH}^-]$) or decrease temperature (since the reaction is exothermic, though temperature is not a variable in the options).
3. Evaluate the chemical effect of decreasing the pH:
* Decreasing the pH means increasing the concentration of hydronium ions ($\text{H}^+$ or $\text{H}_3\text{O}^+$) in the solution.
* Hydronium ions react completely and neutralization occurs when they encounter hydroxide ions ($\text{OH}^-$) in solution:
$$\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$$
* This neutralization reaction removes $\text{OH}^-$ ions from the solubility equilibrium.
4. Determine the equilibrium shift: Because the concentration of the product ion $\text{OH}^-$ is reduced, the system counteracts this change by shifting to the right to replenish the lost hydroxide. This shift causes more solid $\text{Mg(OH)}_2(s)$ to dissolve. Therefore, decreasing the pH increases the solubility of magnesium hydroxide, making Option A the correct choice.

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WHY_OTHERS_WRONG:

• Option B is incorrect: Increasing the pH means adding hydroxide ions ($\text{OH}^-$) to the solution. This increases the concentration of a product ion, which triggers the common-ion effect. The system shifts to the left to consume the excess product, which causes dissolved ions to precipitate back into solid $\text{Mg(OH)}_2(s)$, decreasing its solubility.
• Option C is incorrect: Ammonia ($\text{NH}_3$) is a weak base. When added to an aqueous solution, it reacts with water to produce ammonium and hydroxide ions ($\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$). The introduction of additional $\text{OH}^-$ ions increases the pH of the solution and drives the solubility equilibrium to the left via the common-ion effect, which decreases solubility.
• Option D is incorrect: Adding $\text{Mg(NO}_3)_2$ introduces a highly soluble salt that dissociates completely into magnesium ions ($\text{Mg}^{2+}$) and nitrate ions ($\text{NO}_3^-$). This drastically increases the concentration of the product ion $\text{Mg}^{2+}$ in the solution. Under the common-ion effect, the excess product shifts the equilibrium to the left, forcing $\text{Mg(OH)}_2$ out of solution and decreasing its overall solubility.