JEE Advance - Chemistry (2024 - Paper 1 Online - No. 14)
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 $\left(\Lambda_0\right)$ values (in $\mathrm{mS} \mathrm{m}{ }^2 \mathrm{~mol}^{-1}$ ) for different ions in aqueous solutions are given below:
$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text { Ions } & \mathrm{Ag}^{+} & \mathrm{K}^{+} & \mathrm{Na}^{+} & \mathrm{H}^{+} & \mathrm{NO}_3^{-} & \mathrm{Cl}^{-} & \mathrm{SO}_4^{2-} & \mathrm{OH}^{-} & \mathrm{CH}_3 \mathrm{COO}^{-} \\ \hline \Lambda_0 & 6.2 & 7.4 & 5.0 & 35.0 & 7.2 & 7.6 & 16.0 & 19.9 & 4.1 \\ \hline \end{array} $$
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.
| LIST-I | LIST-II |
|---|---|
| (P) Titrate: KCl Titrant: AgNO$_3$ |
|
| (Q) Titrate: AgNO$_3$ Titrant: KCl |
|
| (R) Titrate: NaOH Titrant: HCl |
|
| (S) Titrate: NaOH Titrant: CH$_3$COOH |
|
|
Explanation
Option (P) :
On adding $\mathrm{AgNO}_3$ solution to $\mathrm{KCl}$ solution precipitation of $\mathrm{AgCl}$ will occur due to which $\mathrm{Cl}^{-}$already present will be replaced by $\mathrm{NO}_3^{-}$ions. So conductance of solution will decrease till equivalence point. After complete precipitation of $\mathrm{AgCl}$, further added $\mathrm{AgNO}_3$ will increase the number of ions in resulting solution so conductance will increase.
Option (Q) :
On adding $\mathrm{KCl}$ solution to $\mathrm{AgNO}_3$ solution precipitation of $\mathrm{AgCl}$ will occur due to which already present $\mathrm{Ag}^{+}$ions will be replaced by $\mathrm{K}^{+}$ions in solution. So conductance of solution will increase. After complete precipitation of $\mathrm{AgCl}$ further added $\mathrm{KCl}$ will increase the number of ions in resulting solution so conductance will increase further.
Option (R) :
On adding $\mathrm{HCl}$ solution to $\mathrm{NaOH}$ solution, $\mathrm{OH}^{-}$will be replaced by $\mathrm{Cl}^{-}$ions so conductance of solution decreases. After complete neutralisation further added $\mathrm{HCl}$ will increase number of ions in the solution. So conductance will increase futher.
Option (S) :
On adding $\mathrm{CH}_3 \mathrm{COOH}$ solution to $\mathrm{NaOH}$ solution $\mathrm{OH}^{-}$will be replaced by $\mathrm{CH}_3 \mathrm{COO}^{-}$ions, so conductance of solution decreases. After complete neutralisation further added $\mathrm{CH}_3 \mathrm{COOH}$ will remain undissociated because it is a weak acid and there is also common ion effect on acetate ions. So number of ions in solution will remain almost constant therefore conductance of solution will remain constant.
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