JEE Main 2023 — Redox Reactions Question with Solution

${\mathrm{C}}_6{\mathrm{H}}_5\underset{\left(\mathrm{WA}\right)}{\mathrm{C}}\mathrm{OOH}+\mathrm{Na}\underset{\left(\mathrm{SB}\right)}{\mathrm{O}}\mathrm{H}\to {\mathrm{C}}_6{\mathrm{H}}_5\underset{\left(\mathrm{Salt}\right)}{\mathrm{C}}\mathrm{OONa}+{\mathrm{H}}_2\mathrm{O}$

 Initially the conductance is low due to the feeble ionisation of acetic acid. On the addition of base, there is decrease in conductance($\mathrm{A}\to \mathrm{B}$) not only due to the replacement of${\mathrm{H}}^{+}$ by ${\mathrm{Na}}^{+}$ but also suppresses the dissociation of benzoic acid due to common ion benzoate. But very soon, the conductance increases($\mathrm{B}\to \mathrm{C}$) on adding $\mathrm{NaOH}$ as $\mathrm{NaOH}$ neutralises the un-dissociated ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$ to ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$ which is the strong electrolyte. This increase in conductance continues raise up to the equivalence point. The graph near the equivalence point is curved due the hydrolysis of salt ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$. Beyond the equivalence point, conductance increases($\mathrm{C}\to \mathrm{D}$) more rapidly with the addition of $\mathrm{NaOH}$ due to the highly conducting ${\mathrm{OH}}^{-}$ ions