JEE Main 2019ChemistryIonic EquilibriumMediumMCQ

JEE Main 2019 — Ionic Equilibrium Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

The pH of a 0.02 M NH4Cl solution will be [Given: KbNH4OH=10-5 and log⁡2=0.301]

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Show full solutionCorrect option: D
Correct answer
D5.35

Step-by-step explanation

NH4Cl is a salt of a strong acid and a weak base.

pH=12pKw-pKb-logC

=1214-5-log⁡2×10-2

=129-log⁡2-log⁡10-2

=129-log⁡2+2

=10.72=5.35

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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.