Concept:The solubility product (\( K_{sp} \)) relates to the molar solubility (\( s \)) based on the stoichiometry of the dissolving salt.
Formula:$$ \text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^- $$
$$ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 $$
Solution:- Given \( K_{sp} = 6.5 \times 10^{-6} \).
- Set up the equation: \( 4s^3 = 6.5 \times 10^{-6} \).
- Divide by 4: \( s^3 = 1.625 \times 10^{-6} \).
- Take the cube root: \( s = \sqrt[3]{1.625 \times 10^{-6}} = \sqrt[3]{1.625} \times 10^{-2} \).
- Since \( 1^3 = 1 \) and \( 1.2^3 = 1.728 \), \( \sqrt[3]{1.625} \) is approximately 1.17.
- Therefore, \( s \approx 1.17 \times 10^{-2} \text{ M} \).
Why other options are incorrect:Option A is incorrect algebra. Options C and D have positive exponents, representing impossibly high solubility for a sparingly soluble salt.
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.