Multiple Choice
Determine the molar solubility of Al(OH)₃ in a solution containing 0.05 M AlCl₃. Ksp of Al(OH)₃ = 1.3 x 10⁻³³.
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18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
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}
Verified step by step guidance1
Write the dissociation equation for magnesium carbonate (MgCO_3) in water: \(\mathrm{MgCO_3 (s) \rightleftharpoons Mg^{2+} (aq) + CO_3^{2-} (aq)}\).
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.
Write the expression for the solubility product constant \(K_{sp}\): \(K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{CO_3^{2-}}]\).
Substitute the concentrations in terms of \(s\) into the \(K_{sp}\) expression: \(K_{sp} = s \times s = s^2\).
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).
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