Calculate the solubility (in grams per 1.00×10² mL of solution) of magnesium hydroxide, Mg(OH)₂, in a solution buffered at pH = 12, given that the Ksp of Mg(OH)₂ is 1.8 × 10⁻¹¹.

• A. 0.0015 g
• B. 0.00015 g
• C. 0.15 g
• D. 0.015 g

Answer: (B)

Verified step by step guidance

1. Identify the chemical equilibrium involved: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). The solubility product constant (Ksp) expression for this equilibrium is Ksp = [Mg²⁺][OH⁻]².
2. Given the Ksp of Mg(OH)₂ is 1.8 × 10⁻¹¹, and the solution is buffered at pH = 12, calculate the concentration of OH⁻ ions. Since pH + pOH = 14, pOH = 14 - 12 = 2, and [OH⁻] = 10⁻² M.
3. Substitute the known [OH⁻] concentration into the Ksp expression: 1.8 × 10⁻¹¹ = [Mg²⁺](10⁻²)².
4. Solve for [Mg²⁺] by rearranging the equation: [Mg²⁺] = 1.8 × 10⁻¹¹ / (10⁻²)².
5. Convert the molar solubility of Mg²⁺ to grams per 100 mL of solution. Use the molar mass of Mg(OH)₂ to convert moles to grams, and adjust for the volume of solution (100 mL).