JEE Main 2026 Apr 4 Shift 1, Chemistry Q24: Buffer Solutions

This is a basic buffer ($\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ + $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$). Use the Henderson-Hasselbalch form for a base:

$\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[\mathrm{salt}]}{[\mathrm{base}]}.$

Since volumes are common, the ratio of concentrations equals the ratio of millimoles:

mmol NH${}_4$OH $=5\times 0.1=0.5$; mmol NH${}_4$Cl $=250\times 0.1=25$.

$\frac{[\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}]}{[\mathrm{N}{\mathrm{H}}_4\mathrm{OH}]}=\frac{25}{0.5}=50.$

$\log 50=\log 100-\log 2=2-0.30=\mathrm{1.70.}$

$\mathrm{pOH}=4.74+1.70=\mathrm{6.44.}$

$\mathrm{pH}=14-6.44=7.56=756\times 10^{-2}$.