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) 
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:
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.
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 |
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) | | ||
Updated On: Mar 7, 2025
- P-4, Q-3, R-2, S-5
- P-2, Q-4, R-3, S-1
- P-3, Q-4, R-2, S-5
- P-4, Q-3, R-2, S-1
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The Correct Option is C
Solution and Explanation
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.
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