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.
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
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Determine the molar solubility of CaCO3 in a solution of 0.500 M Na2CO3. Ksp for CaCO3 is 4.96 × 10−9.
A
5.00 × 10−1
B
7.04 × 10−5
C
9.96 × 10−5
D
4.96 × 10−9
E
9.92 × 10−9
Verified step by step guidance1
Start by writing the balanced chemical equation for the dissolution of calcium carbonate (CaCO3) in water: CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq).
Identify the expression for the solubility product constant (Ksp) of CaCO3, which is given as 4.96 × 10⁻⁹. The Ksp expression is: Ksp = [Ca²⁺][CO₃²⁻].
Since the solution already contains 0.500 M Na2CO3, the concentration of CO₃²⁻ ions from Na2CO3 is 0.500 M. Assume the additional concentration of CO₃²⁻ from the dissolution of CaCO3 is negligible compared to 0.500 M.
Let the molar solubility of CaCO3 be 's'. Therefore, the concentration of Ca²⁺ ions in the solution will be 's'. Substitute these values into the Ksp expression: Ksp = [Ca²⁺][CO₃²⁻] = s × 0.500.
Solve for 's' using the Ksp value: s = Ksp / 0.500. This will give you the molar solubility of CaCO3 in the solution.
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