JEE Main 2023ChemistryIonic EquilibriumMediumNumerical

JEE Main 2023 — Ionic Equilibrium Question with Solution

JEE Main 2023 (29 Jan Shift 1)

Question

Millimoles of calcium hydroxyide required to produce 100mL of the aqueous solution of pH 12 is x×10-1. The value of x is _____ (Nearest integer).
Assume complete dissociation.

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Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

We know,
pH = -log[H+] , pH = -log[OH-], pH + pOH = 14
∵  pH=12

∴   H+=10-12M

∴   OH-=10-2M
We also know,
Ca(OH)2 ⇌ 2OH- + Ca2+

∴ Ca(OH)2=5×10-3M

5×10-3=milli moles of CaOH2 100mL

milli moles of CaOH2=5×10-1

Ans. =5

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About this question

This is a previous-year question from JEE Main 2023, 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.