JEE Main 2021ChemistryIonic EquilibriumMediumNumerical

JEE Main 2021 — Ionic Equilibrium Question with Solution

JEE Main 2021 (31 Aug Shift 2)

Question

The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x×10-4. The value of x is (Nearest integer)
log 2.5=0.3979

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Show full solutionCorrect answer: 6021
Correct answer
6021

Step-by-step explanation

Milli equivalents of HClNaVa=50×1=50
Milli equivalents of NaOHNbVb=30×1=30
Since NaVa>NbVb
H+=NaVa-NbVbVa+Vb=50-3080=2080=0.25=2.5×10-1
pH=-logH+=-log2.5×10-1=1-0.3979=0.6021
pH×104=0.6021×104=6021

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About this question

This is a previous-year question from JEE Main 2021, 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.