The exothermic dissolution of Mg(OH)2 (solid) in water is represented by the equation above. The Ksp of Mg(OH)2 is 1.8 x 10-11. Why will decreasing the pH increase the solubility of Mg(OH)2 in an aqueous solution?
The Correct Answer and Explanation is :
The solubility of magnesium hydroxide (Mg(OH)₂) in water is governed by the equilibrium:
[
\text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq)
]
The solubility product constant (Ksp) for this reaction is (1.8 \times 10^{-11}), which represents the equilibrium between solid magnesium hydroxide and its ions in solution.
The reason why decreasing the pH will increase the solubility of Mg(OH)₂ is because of the common ion effect, specifically involving the concentration of hydroxide ions ((OH^-)).
- When the pH decreases (which means the concentration of hydrogen ions ([H^+]) increases), the equilibrium is shifted to the right due to Le Chatelier’s Principle.
- In an acidic environment, the increase in ([H^+]) reacts with ([OH^-]) to form water, effectively reducing the concentration of (OH^-) ions in the solution.
- To restore equilibrium, the dissolution of Mg(OH)₂ increases, producing more (OH^-) and (Mg^{2+}), thus increasing the solubility of Mg(OH)₂.
In summary, when the pH is decreased (i.e., more acidic), the concentration of (OH^-) ions decreases, shifting the equilibrium to the right, which increases the solubility of magnesium hydroxide.