JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Redox Reactions and Electrochemistry

In the mixture the strong acid $\mathrm{HCl}$ is fully ionised while the weak acid $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ is barely ionised, so added $\mathrm{NaOH}$ first neutralises $\mathrm{HCl}$: the highly mobile ${\mathrm{H}}^{+}$ is replaced by $\mathrm{N}{\mathrm{a}}^{+}$ and the conductance falls until $2.0\mathrm{mL}$ has been added. From $2.0$ to $5.0\mathrm{mL}$ the acetic acid is converted to its salt, giving a gentle rise, and beyond $5.0\mathrm{mL}$ the excess $\mathrm{O}{\mathrm{H}}^{-}$ makes it climb steeply. For $\mathrm{HCl}$, the moles used are $0.1\times 2.0/1000=2\times 10^{-4}\mathrm{mol}$ in $40\mathrm{mL}$, so its concentration is $2\times 10^{-4}/0.040=0.005\mathrm{M}$. The concentration of $\mathrm{HCl}$ in the original mixture is $0.005\mathrm{M}$.