What is the solubility of Mg(OH)₂ (K_sp = 8.9 × 10^-12) in 1.0 L of a solution buffered (with large capacity) at pH 10.0?

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At pH 10, the concentration of OH- ions is 10^-4 M (since pOH = 14 - pH = 4). For Mg(OH)2, the solubility product (Ksp) is given as 8.9 × 10^-12. The dissolution can be represented as: Mg(OH)2(s) ⇌ Mg^2+(aq) + 2OH^-(aq). Let the solubility of Mg(OH)2 be s mol/L, then [Mg^2+] = s and [OH^-] = 2s. However, since the solution is buffered at pH 10, the [OH^-] is fixed at 10^-4 M, not dependent on s. Thus, Ksp = [Mg^2+][OH^-]^2 = s(10^-4)^2. Solving for s gives s = Ksp / (10^-4)^2 = 8.9 × 10^-12 / (10^-8) = 8.9 × 10^-4 M.

What Is the Solubility of Mg(OH)2 (Ksp = 8

What Is the Solubility of Mg(OH)₂ (K_sp = 8