The molar solubility of magnesium carbonate (MgCO_3) is 1.8 × 10^{-4} mol/L. What is the K_{sp} for MgCO_3?

• A. 6.5 × 10^{-7}
• B. 3.2 × 10^{-8}
• C. 1.8 × 10^{-4}
• D. 9.0 × 10^{-8}

Answer: (B)

Verified step by step guidance

1. Write the dissociation equation for magnesium carbonate (MgCO_3) in water: \(\mathrm{MgCO_3 (s) \rightleftharpoons Mg^{2+} (aq) + CO_3^{2-} (aq)}\).
2. Express the molar solubility as the concentration of ions in solution. Since 1 mole of MgCO_3 produces 1 mole of \(\mathrm{Mg^{2+}}\) and 1 mole of \(\mathrm{CO_3^{2-}}\), the concentrations are \([\mathrm{Mg^{2+}}] = s\) and \([\mathrm{CO_3^{2-}}] = s\), where \(s\) is the molar solubility.
3. Write the expression for the solubility product constant \(K_{sp}\): \(K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{CO_3^{2-}}]\).
4. Substitute the concentrations in terms of \(s\) into the \(K_{sp}\) expression: \(K_{sp} = s \times s = s^2\).
5. Calculate \(K_{sp}\) by squaring the given molar solubility value: \(K_{sp} = (1.8 \times 10^{-4})^2\) (do not compute the final number here, just set up the expression).