JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Equilibrium

A basic buffer obeys $\mathrm{pOH}=\mathrm{p}{K}_b+\log \frac{[\text{salt}]}{[\text{base}]}$, and $\mathrm{pH}=9.25$ gives $\mathrm{pOH}=4.75=\mathrm{p}{K}_b$, which requires equal amounts of unreacted $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ and the $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ formed, so the base taken must be exactly twice the acid. Mixing $0.2\mathrm{M}\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ ($0.5$ L, $100$ mmol) with $0.1\mathrm{M}\mathrm{HCl}$ ($0.5$ L, $50$ mmol) leaves $50$ mmol of $\mathrm{N}{\mathrm{H}}_4\mathrm{OH}$ alongside $50$ mmol of $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$, so $[\text{salt}]=[\text{base}]$ and $\mathrm{pOH}=\mathrm{p}{K}_b$. The required buffer is the mixture of $0.2\mathrm{M}\mathrm{N}{\mathrm{H}}_4\mathrm{OH}(0.5\mathrm{L})$ with $0.1\mathrm{M}\mathrm{HCl}(0.5\mathrm{L})$.