Determine the molar solubility of CaCO₃ in a solution of 0.500 M Na₂CO₃. K_sp 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

Answer: (E)

Verified step by step guidance

1. Start by writing the balanced chemical equation for the dissolution of calcium carbonate (CaCO3) in water: CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq).
2. 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⁻].
3. 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.
4. 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.
5. Solve for 's' using the Ksp value: s = Ksp / 0.500. This will give you the molar solubility of CaCO3 in the solution.