JEE Advanced 2024 Paper 1 · Q14 · Conductometric Titration

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 ($\Lambda_0$) values (in mS m$^2$ mol$^{-1}$) for different ions in aqueous solutions are given below:
Ions: $Ag^+$ ($\Lambda_0 = 6.2$), $K^+$ (7.4), $Na^+$ (5.0), $H^+$ (35.0), $NO_3^-$ (7.2), $Cl^-$ (7.6), $SO_4^{2-}$ (16.0), $OH^-$ (19.9), $CH_3COO^-$ (4.1).
For different combinations of titrates and titrants given in $\mathbf{List	ext{-}I}$, the graphs of 'conductance' versus 'volume of titrant' are given in $\mathbf{List	ext{-}II}$.
Match each entry in $\mathbf{List	ext{-}I}$ with the appropriate entry in $\mathbf{List	ext{-}II}$ and choose the correct option.

$\mathbf{List	ext{-}I}$:
(P) Titrate: KCl, Titrant: $AgNO_3$
(Q) Titrate: $AgNO_3$, Titrant: KCl
(R) Titrate: NaOH, Titrant: HCl
(S) Titrate: NaOH, Titrant: $CH_3COOH$

$\mathbf{List	ext{-}II}$ (graphs):
(1) Steep decrease then plateau (conductance falls then becomes flat).
(2) Decrease then sharp increase (deep V-shape).
(3) Roughly flat (small dip) then increase (a shallow V).
(4) Slow rise then sharper rise (two-segment increasing curve).
(5) Decrease then plateau (steep decrease followed by flat) — alternative shape.

SolveKar AI · Accepted Answer

(P) $KCl$ titrated with $AgNO_3$: $K^+ + Cl^- + Ag^+ + NO_3^- 	o AgCl\downarrow + K^+ + NO_3^-$. $K^+$ ($\Lambda_0=7.4$) replaced by similar $K^+$ (stays); $Cl^-$ (7.6) replaced by $NO_3^-$ (7.2) — very small dip; after equivalence, excess $Ag^+ + NO_3^-$ adds → conductance rises. Shallow V → graph (3).
(Q) $AgNO_3$ titrated with $KCl$: removes $Ag^+$ as $AgCl\downarrow$; $Ag^+$ (6.2) replaced by $K^+$ (7.4) — slight rise pre-equivalence, then steeper rise after equivalence due to excess $K^+ + Cl^-$. Two-stage rising curve → (4).
(R) NaOH titrated with HCl: high-conductivity $OH^-$ (19.9) consumed by $H^+$ → $H_2O$, leaving $Na^+ + Cl^-$. Conductance decreases sharply; after equivalence, excess $H^+$ (35) → very steep rise. Sharp V → (2).
(S) NaOH titrated with weak acid $CH_3COOH$: $OH^-$ (19.9) replaced by $CH_3COO^-$ (4.1) → conductance decreases; after equivalence, excess $CH_3COOH$ is weakly ionised → conductance nearly flat (small change). Steep decrease then plateau → (5).
Mapping P-3, Q-4, R-2, S-5 = option (C).