JEE Main 2022ChemistryIonic EquilibriumMediumNumerical

JEE Main 2022 — Ionic Equilibrium Question with Solution

JEE Main 2022 (27 Jul Shift 1)

Question

At 310 K, the solubility of CaF2 in water is 2.34×10-3 g/100 mL. The solubility product of CaF2 is ---- ×10-8mol/L3(nearest integer). (Given molar mass : CaF2=78 g mol-1)

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Step-by-step explanation

The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp.

Solubility of CaF2=S mole//L

S=2.34×10-30.1×78=2.3478×10-2=3×10-4 mol/L

KspCaF2=4 S3=43×10-43

=108×10-12

=0.0108×10-8mol/L3

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About this question

This is a previous-year question from JEE Main 2022, 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.