Which of the following mixture gives a buffer solution with pH =9.25 ? Given : pK b ( NH 4 OH )=4.75

0.2 M NH 4 OH (0.5 L )+0.1 M HCl (0.5 L )

0.2 M NH 4 OH (0.4 L )+0.1 M HCl (1 L )

0.5 M NH 4 OH (0.2 L )+0.2 M HCl (0.5 L )

0.4 M NH 4 OH (1 L )+0.1 M HCl (1 L )

Chemistry · Ionic Equilibrium

JEE Main 2026 — 22 January, Evening Shift — Question 56

Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ? Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=4.75$

Option A:
$0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.4 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$
Option B:
$0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.5 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$
Correct
Option C:
$0.5 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.2 \mathrm{~L})+0.2 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$
Option D:
$0.4 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(1 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$

Answer: B

Step-by-step solution

$\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\text { Salt }}{\text { Base }}$ $4.75=4.75+\log \frac{\text { Salt }}{\text { Base }}$ [0pt] Milimoles of [Salt] = milimoles of [Base] Option (D) :

\mathrm{NH}_{4} \mathrm{OH}+\mathrm{HCl} \rightarrow \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{O}

$0.2 \mathrm{M}, 0.5 \mathrm{~L} 0.1 \mathrm{M}, 0.5 \mathrm{~L}$ 100 mmole 50 mmole 50 mmole - 50 mmole Milimoles of $\mathrm{NH}_{4} \mathrm{OH}=$ milimoles of $\mathrm{NH}_{4} \mathrm{Cl}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Buffer solution & pH of Special Buffers - Zwitter ions ++