Excess Ca(OH)₂ is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00-mL sample is titrated with HCl, requiring 11.23 mL of 0.0983 M HCl to reach the endpoint. Calculate Ksp for Ca(OH)₂. Compare your result with that in Appendix D. Suggest a reason for any differences you find between your value and the one in Appendix D.

Verified step by step guidance

1. Step 1: Write the balanced chemical equation for the dissolution of Ca(OH)_2 in water: Ca(OH)_2 (s) ⇌ Ca^{2+} (aq) + 2OH^{-} (aq).
2. Step 2: Determine the moles of HCl used in the titration. Use the formula: \( \text{moles of HCl} = M \times V \), where \( M \) is the molarity and \( V \) is the volume in liters.
3. Step 3: Since HCl reacts with OH^- in a 1:1 ratio, the moles of OH^- in the solution are equal to the moles of HCl used.
4. Step 4: Calculate the concentration of OH^- ions in the 50.00 mL sample. Use the formula: \( [OH^-] = \frac{\text{moles of OH^-}}{\text{volume of solution in liters}} \).
5. Step 5: Use the concentration of OH^- to find the concentration of Ca^{2+} ions. Since the stoichiometry of the dissolution is 1:2, \( [Ca^{2+}] = \frac{1}{2} [OH^-] \). Then, calculate the K_{sp} using the expression: \( K_{sp} = [Ca^{2+}][OH^-]^2 \).