NEET CHEMISTRY: "pH of a saturated solution of $Ca(OH)_2$ is 9. The solubility product ($K_{sp}$) of $Ca(OH)_2$ is:..." — Solved MCQ | TopperSquare

Back to Directory

NEET CHEMISTRYGeneralHard

Question

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

1

$0.5 \times 10^{-15}$

2

$0.25 \times 10^{-10}$

3

$0.125 \times 10^{-15}$

4

$0.5 \times 10^{-10}$

Step-by-Step Solution

$Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-$ . pH = 9, Hence pOH = 14 - 9 = 5, $[OH^-] = 10^{-5} M$ . Hence $[Ca^{2+}] = \frac{10^{-5}}{2}$ . Thus $K_{sp} = [Ca^{2+}][OH^-]^2 = (\frac{10^{-5}}{2})(10^{-5})^2 = 0.5 \times 10^{-15}$ .

Explanation Unlocked on Signup

Join free — takes 30 seconds

Reveal Answer Free

Practice Mode Available

Practice All CHEMISTRY Questions

Analytics · Leaderboards · Full NEET Mock Tests · 15,000+ MCQs

Get Started Free

This NEET CHEMISTRY practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. Browse all NEET practice questions →