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 (Λ0) values (in mS m²mol⁻¹ ) for different ions in aqueous solutions are given below:

Ions	Ag+	K+	Na+	H+	\(\text{NO}_{3}^{-}\)	Cl-	\(\text{SO}^{2-}_{4}\)	OH-	CH3COO-
\(\Lambda_0\)	6.2	7.4	5.0	35.0	7.2	7.6	16.0	19.9	4.1

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: AgNO3	(1)
(Q)	Titrate: AgNO3 Titrant: KCl	(2)
(R)	Titrate: NaOH Titrant: HCl	(3)
(S)	Titrate: NaOH Titrant: CH3COOH	(4)
		(5)

• A. P-4, Q-3, R-2, S-5
• B. P-2, Q-4, R-3, S-1
• C. P-3, Q-4, R-2, S-5
• D. P-4, Q-3, R-2, S-1

Answer 1

The Correct Option is C

Step 1: Analyzing Each Titration Reaction 

• (P) KCl + AgNO₃
  • Ag⁺ reacts with Cl^- to form AgCl precipitate, removing conductive ions.
  • Conductance initially decreases as Cl^- is removed, then increases due to excess Ag⁺.
  • Corresponds to (3) - V-shape (Decrease, then increase).
• (Q) AgNO₃ + KCl
  • Ag⁺ is removed as AgCl precipitates.
  • Conductance steadily decreases due to the removal of mobile ions.
  • Corresponds to (4) - Steady decline.
• (R) NaOH + HCl
  • Neutralization reaction: \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
  • Conductance decreases initially as OH^- is replaced by less conductive Cl^-.
  • Then increases as excess H⁺ is added.
  • Corresponds to (2) - Decrease, minimum, then increase.
• (S) NaOH + CH₃COOH
  • Weak acid titration: \( \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \)
  • Conductance first increases as CH₃COO^- is formed, then decreases as the reaction completes.
  • Corresponds to (5) - Increase then decrease (Inverse V-shape).

Step 2: Conclusion

• (P) → (3)
• (Q) → (4)
• (R) → (2)
• (S) → (5)

Thus, the correct answer is (C) P-3, Q-4, R-2, S-5.

Answer 2

To solve the problem, we need to match each titration combination with the correct conductance versus volume of titrant graph, using the given limiting ionic conductivities.

1. Understanding Ionic Conductivities:
The limiting ionic conductivities (\(\Lambda_0\)) indicate how well ions conduct electricity in aqueous solution.
- Higher \(\Lambda_0\) means higher conductance.
- When titrant is added, the conductance changes depending on the mobility and concentration of ions formed or removed.

2. Analyzing Each Combination:

(P) Titrate: KCl, Titrant: AgNO₃
- Initial ions: K⁺ (7.4) and Cl⁻ (7.6) with moderate conductance.
- Adding Ag⁺ replaces K⁺ with Ag⁺ (6.2), and precipitates Cl⁻ as AgCl, reducing conductance.
- Conductance decreases initially then increases after equivalence.
- Matches graph (3).

(Q) Titrate: AgNO₃, Titrant: KCl
- Initial ions: Ag⁺ (6.2) and NO₃⁻ (7.2).
- Adding K⁺ replaces Ag⁺ with higher mobility K⁺ (7.4), increasing conductance.
- Matches graph (4).

(R) Titrate: NaOH, Titrant: HCl
- Na⁺ (5.0), OH⁻ (19.9) initially.
- Adding H⁺ (35.0) reacts with OH⁻ to form water, reducing conductance.
- After equivalence, excess H⁺ increases conductance.
- Matches graph (2).

(S) Titrate: NaOH, Titrant: CH₃COOH
- Similar to (R), but acetate ion (4.1) replaces hydroxide (19.9), lowering conductance steadily.
- Conductance decreases and then levels off.
- Matches graph (5).

3. Final Matching:
P - 3, Q - 4, R - 2, S - 5

Final Answer:
Option (C)