Question
In a conductance experiment, aqueous $\text{AgNO}_{3}$ solution is added to aqueous $\text{KCl}$ solution gradually and simultaneously the molar conductivity ($\lambda_{m}$) is measured. The correct plot of $\lambda_{m}$ versus volume of $\text{AgNO}_{3}$ solution is
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
To solve this question, we need to understand the conductometric titration involving the addition of aqueous \(\text{AgNO}_{3}\) to an aqueous \(\text{KCl}\) solution. The reaction involved is:
\(\text{AgNO}_{3(aq)} + \text{KCl}_{(aq)} \rightarrow \text{AgCl}_{(s)} + \text{KNO}_{3(aq)}\)
As silver nitrate (\(\text{AgNO}_{3}\)) is added to potassium chloride (\(\text{KCl}\)), silver chloride (\(\text{AgCl}\)) precipitates, and potassium nitrate (\(\text{KNO}_{3}\)) remains in solution.
- Initially, \(\text{KCl}\) is present in the solution contributing to the initial conductivity due to the mobility of \(\text{K}^{+}\) and \(\text{Cl}^{-}\) ions.
- As \(\text{AgNO}_{3}\) is added, \(\text{Ag}^{+}\) ions react with \(\text{Cl}^{-}\) ions to form the insoluble \(\text{AgCl}\), effectively removing these ions from contributing to conductivity.
- Since \(\text{AgCl}\) is precipitating, the number of free ions decreases, resulting in a decrease in molar conductivity (\(\lambda_{m}\)).
- Upon complete precipitation of \(\text{Cl}^{-}\) ions, the further addition of \(\text{AgNO}_{3}\) leads to an increase in molar conductivity again because now, only \(\text{Ag}^{+}\) and \(\text{NO}_{3}^{-}\) ions are present in the solution, which are good conductors.
Therefore, the correct plot should initially show a decrease in molar conductivity as \(\text{AgNO}_{3}\) is added, followed by a minimum point, and then an increase as the excess \(\text{AgNO}_{3}\) is added.
This corresponds to the correct answer plot, showing the expected behavior during the conductometric titration.
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