The molar solubility of manganese (II) hydroxide, Mn(OH)2, is 3.48 × 10^-3 M. What is the Ksp of Mn(OH)2 at this temperature?

• A. 2.42 × 10^-6
• B. 5.67 × 10^-8
• C. 4.21 × 10^-7
• D. 1.21 × 10^-5

Answer: (C)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the dissolution of manganese (II) hydroxide in water: Mn(OH)_2(s) ⇌ Mn^{2+}(aq) + 2OH^{-}(aq).
2. Identify the expression for the solubility product constant (Ksp) for Mn(OH)_2. The Ksp expression is given by: Ksp = [Mn^{2+}][OH^{-}]^2.
3. Since the molar solubility of Mn(OH)_2 is given as 3.48 × 10^-3 M, this represents the concentration of Mn^{2+} ions in solution at equilibrium.
4. For every mole of Mn(OH)_2 that dissolves, 2 moles of OH^{-} ions are produced. Therefore, the concentration of OH^{-} ions is 2 × 3.48 × 10^-3 M.
5. Substitute the concentrations of Mn^{2+} and OH^{-} into the Ksp expression: Ksp = (3.48 × 10^-3) × (2 × 3.48 × 10^-3)^2. Calculate this expression to find the Ksp value.