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
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In precipitation titrations where a highly mobile ion (like \( \text{Cl}^- \)) is replaced by another ion of similar mobility (like \( \text{NO}_3^- \)), the conductivity curve remains nearly horizontal up to the end point, followed by a sharp rise due to the excess added titrant.
Step 1: Understanding the Concept:
This problem involves a conductometric titration. The conductance (and molar conductivity) of a solution is determined by the total number of ions present and their individual ionic mobilities.
In this experiment, silver nitrate ($AgNO_3$) is added to potassium chloride ($KCl$).
The fundamental reaction is the precipitation of silver chloride:
\[ KCl(aq) + AgNO_3(aq) \rightarrow AgCl(s) \downarrow + KNO_3(aq) \]
The change in molar conductivity depends on whether the ions being removed from the solution are being replaced by ions of higher, lower, or equal mobility. Step 2: Key Formula or Approach:
The molar conductivity ($\lambda_m$) is proportional to the sum of the limiting ionic conductivities ($\lambda^\circ$) of the free ions in the solution.
Key ionic mobilities at $25^\circ C$ (in $10^{-4} S m^2 mol^{-1}$):
- $\lambda^\circ(K^+) = 73.5$
- $\lambda^\circ(Cl^-) = 76.3$
- $\lambda^\circ(NO_3^-) = 71.4$
- $\lambda^\circ(Ag^+) = 61.9$ Step 3: Detailed Explanation: 1. Before the Equivalence Point:
As $AgNO_3$ is added to the $KCl$ solution, $Ag^+$ ions react with $Cl^-$ ions to form solid $AgCl$. This removes $Cl^-$ from the solution.
For every $AgNO_3$ formula unit added, one $Cl^-$ ion is effectively replaced by one $NO_3^-$ ion.
The $K^+$ ions remain as "spectator ions" and their concentration does not change significantly (ignoring dilution).
Since the mobility of $NO_3^-$ (71.4) is very close to the mobility of $Cl^-$ (76.3), the total molar conductivity remains almost constant. It shows a very slight decrease because $NO_3^-$ is slightly less mobile than $Cl^-$.
The curve is essentially a horizontal line until the endpoint. 2. At the Equivalence Point:
All the $Cl^-$ ions have been precipitated. The solution contains primarily $K^+$ and $NO_3^-$ ions. 3. After the Equivalence Point:
Further addition of $AgNO_3$ does not lead to any reaction as there is no $Cl^-$ left.
Therefore, each addition simply adds more free $Ag^+$ and $NO_3^-$ ions to the solution.
The total concentration of ions increases rapidly, leading to a sharp and steep increase in the molar conductivity of the solution.
The resulting graph is a horizontal line followed by a sharp rise. This matches the behavior shown in Figure D. Step 4: Final Answer:
Based on the replacement of $Cl^-$ with $NO_3^-$ and the subsequent addition of excess $AgNO_3$, the plot in option (D) is correct.
