JEE Main 2019ChemistryIonic EquilibriumMediumMCQ

JEE Main 2019 — Ionic Equilibrium Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

What is the molar solubility of AIOH3 in 0.2 M NaOH solution? Given that, solubility product of AlOH3=2.4×10-24 :

Choose an option

Show full solutionCorrect option: D
Correct answer
D3×10-22

Step-by-step explanation

Concept of common -ion will be applicable for OH- group.

AlOH3⇌Al3+ S        +3OH-3s+0.2

Ksp for AlOH3=2.4×10-24=s3s+0.23

Looking at Ksp, it is clear that s will be much lesser than 0.2.

Or, s≪0.2

s×0.23=2.4×10-24⇒s=2.4×10-24/0.008
or, s=3×10-22

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About this question

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.