JEE Main 2019 — Ionic Equilibrium Question with Solution
From: JEE Main 2019 (Online) 12th April Morning Slot
Question
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24
:
Choose an option
Show full solutionCorrect option: A
Correct answer
A3 × 10–22
Step-by-step explanation
[Al3+][OH-]3
2.4 × 10–24 = s(0.2)3
s =
s = 3 × 10–22
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This is a previous-year question from JEE Main 2019, covering the Ionic Equilibrium chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.