The pH a solution obtained by mixing 5 mL of 0.1 M NH 4 OH solution with 250 mL of 0.1 M NH 4 Cl solution is \ \ \ \ × 10 -2 . (Nearest integer) Given : pK b ( NH 4 OH )=4.74 begin aligned & log 2=0.30 & log 3=0.48 & log 5=0.70 end aligned

Chemistry · Ionic Equilibrium

JEE Main 2026 — 4 April, Morning Shift — Question 66

The pH a solution obtained by mixing 5 mL of $0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}$ solution with 250 mL of 0.1 M $\mathrm{NH}_{4} \mathrm{Cl}$ solution is $\_\_\_\_$ $\times 10^{-2}$ . (Nearest integer)

Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=4.74$

\begin{aligned} & \log 2=0.30 & \log 3=0.48 & \log 5=0.70 \end{aligned}

Answer: 756

Numerical answer — enter this value.

Step-by-step solution

On mixing, final volume $= 255\,\mathrm{mL}$

$\left[\mathrm{NH_4OH}\right] = \frac{5 \times 0.1}{255}$ , $\left[\mathrm{NH_4Cl}\right] = \frac{250 \times 0.1}{255}$

$\mathrm{pOH} = pK_b + \log \frac{[\mathrm{NH_4Cl}]}{[\mathrm{NH_4OH}]}$

$\mathrm{pOH} = 4.74 + \log \frac{250 \times 0.1}{5 \times 0.1}$

$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 6.44$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Buffer solution & pH of Special Buffers - Zwitter ions ++