Question

The pH of a saturated solution of \( \text{Ca(OH)}_2 \) is 9. The solubility product (\( K_{sp} \)) of \( \text{Ca(OH)}_2 \) is:

• A. \( 5.0 \times 10^{-16} \)
• B. \( 4.0 \times 10^{-16} \)
• C. \( 6.5 \times 10^{-16} \)
• D. \( 8.0 \times 10^{-16} \)

Hint:

To solve for \( K_{sp} \), first determine ion concentrations using pH or pOH and apply stoichiometry.

Answer 1

The Correct Option is A

The dissociation of \( \text{Ca(OH)}_2 \) in water is: \[ \text{Ca(OH)}_2 \xrightarrow{} \text{Ca}^{2+} + 2 \text{OH}^-. \] Given the pH is 9, the \( \text{pOH} \) is: \[ \text{pOH} = 14 - \text{pH} = 14 - 9 = 5. \] The concentration of \( \text{OH}^- \) is: \[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-5}. \] From the stoichiometry of the reaction: \[ [\text{Ca}^{2+}] = \frac{[\text{OH}^-]}{2} = \frac{10^{-5}}{2} = 5 \times 10^{-6}. \] The solubility product (\( K_{sp} \)) is: \[ K_{sp} = [\text{Ca}^{2+}] [\text{OH}^-]^2 = (5 \times 10^{-6}) (10^{-5})^2 = 5 \times 10^{-6} \times 10^{-10} = 5.0 \times 10^{-16}. \] ---
Final Answer: \[ \boxed{5.0 \times 10^{-16}} \]