In a conductometric titration, small volume of titrant of higher concentration is added stepwise to

Question

In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition. The limiting ionic conductivity $\left(\Lambda_{0}ight)$ values $\left(ight.$ inmSm $\left.{ }^{2} \mathrm{~mol}^{-1}ight)$ for different ions in aqueous solutions are given below: \ For different combinations of titrates and titrants given in List-I, the graphs of 'conductance' versus 'volume of titrant' are given in List-II. \Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

Professor PC Thomas Classes · Answer

(c) $: \operatorname{In}(P)$\ $\mathrm{KCl}+\mathrm{AgNO}_{3} ightarrow \mathrm{AgCl} \downarrow+\mathrm{KNO}_{3}$\ Here, $\mathrm{Cl}^{-}$is replaced by $\mathrm{NO}_{3}^{-}$, so conductance initially decreases slowly, then rise rapidly due to formation of $\mathrm{KNO}_{3}$.\ Thus, $P ightarrow 3$.\ In $(Q)$\ $\mathrm{AgNO}_{3}+\mathrm{KCl} ightarrow \mathrm{AgCl} \downarrow+\mathrm{KNO}_{3}$\ Here $\mathrm{AgNO}_{3}$ is present in lower concentration, also conductivity of $\mathrm{NO}_{3}^{-}$is lesser than $\mathrm{Cl}^{-}$, so on addition of KCl , conductance increases slowly in the starting and after equivalence point it increases rapidly. Thus, $Q ightarrow 4$.\ In (R)\ $\mathrm{NaOH}+\mathrm{HCl} ightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NaCl}$ Conductivity of NaCl is much lower than NaOH , so initially the conductance decreases rapidly, but after equivalence point, it increases rapidly due to high conductivity of HCl .\ So, $R ightarrow 2$.\ In $(S)$\ NaOH is a strong electrolyte while $\mathrm{CH}_{3} \mathrm{COOH}$ is a weak electrolyte. Initial conductance of solution is very high.\ $\mathrm{NaOH}+\mathrm{CH}_{3} \mathrm{COOH} ightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$ On addition of $\mathrm{CH}_{3} \mathrm{COOH}$ conductance decreases, but after equivalence point, the solution will be a buffer solution. Thus, conductance will become constant. So, $S ightarrow 5$.\