Question:medium
500 mL of \(1.2\) M KI solution is mixed with 500 mL of \(0.2\) M \( \mathrm{KMnO_4} \) solution in basic medium.
The liberated iodine is titrated with standard \(0.1\) M \( \mathrm{Na_2S_2O_3} \) solution in the presence of starch indicator
till the blue colour disappears.
The volume (in L) of \( \mathrm{Na_2S_2O_3} \) consumed is ________ (nearest integer).
500 mL of \(1.2\) M KI solution is mixed with 500 mL of \(0.2\) M \( \mathrm{KMnO_4} \) solution in basic medium.
The liberated iodine is titrated with standard \(0.1\) M \( \mathrm{Na_2S_2O_3} \) solution in the presence of starch indicator
till the blue colour disappears.
The volume (in L) of \( \mathrm{Na_2S_2O_3} \) consumed is ________ (nearest integer).
Show Hint
In iodometric titrations, always convert the oxidising agent to equivalent moles of iodine first, then relate iodine to thiosulphate.
Updated On: Jun 6, 2026
Show Solution
Correct Answer: 3
Solution and Explanation
Step 1: Understanding the Concept:
This problem involves a sequential redox reaction.
First, Potassium permanganate (\(\text{KMnO}_4\)) oxidizes Iodide (\(I^-\)) to Iodine (\(I_2\)) in a basic medium.
Second, the liberated Iodine is titrated with Sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)).
The principle used is the Law of Equivalence, where the equivalents of oxidizing and reducing agents are equal.
Step 2: Key Formula or Approach:
1. Number of equivalents = Molarity \(\times\) Volume (L) \(\times\) \(n\)-factor.
2. Law of Equivalence: Equivalents of \(\text{KMnO}_4\) used = Equivalents of \(I_2\) liberated.
3. Equivalents of \(I_2\) liberated = Equivalents of \(\text{Na}_2\text{S}_2\text{O}_3\) consumed.
Step 3: Detailed Explanation:
1. Identify \(n\)-factors:
In basic or neutral medium, \(\text{MnO}_4^-\) is reduced to \(\text{MnO}_2\).
The oxidation state of Mn changes from \(+7\) to \(+4\), so the \(n\)-factor for \(\text{KMnO}_4 = 3\).
For the titration of Iodine with Thiosulfate:
\(I_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2I^- + \text{S}_4\text{O}_6^{2-}\).
The oxidation state of Sulfur in \(\text{Na}_2\text{S}_2\text{O}_3\) is \(+2\) and in \(\text{Na}_2\text{S}_4\text{O}_6\) it is \(+2.5\).
Change per sulfur atom = \(0.5\). For two sulfur atoms in the formula, \(n\)-factor = \(1\).
2. Calculate Moles and Equivalents:
Moles of \(\text{KMnO}_4 = \text{Molarity} \times \text{Volume} = 0.2 \times 0.5 = 0.1 \text{ moles}\).
Equivalents of \(\text{KMnO}_4 = 0.1 \times 3 = 0.3 \text{ Eq}\).
3. Equate with Thiosulfate:
By Law of Equivalence:
Equivalents of \(\text{KMnO}_4 = \text{Equivalents of } \text{Na}_2\text{S}_2\text{O}_3\).
\[ 0.3 = 0.1 \times 1 \times V \]
\[ V = \frac{0.3}{0.1} = 3 \text{ L} \]
Step 4: Final Answer:
The volume of \(\text{Na}_2\text{S}_2\text{O}_3\) consumed is 3 L.
This problem involves a sequential redox reaction.
First, Potassium permanganate (\(\text{KMnO}_4\)) oxidizes Iodide (\(I^-\)) to Iodine (\(I_2\)) in a basic medium.
Second, the liberated Iodine is titrated with Sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)).
The principle used is the Law of Equivalence, where the equivalents of oxidizing and reducing agents are equal.
Step 2: Key Formula or Approach:
1. Number of equivalents = Molarity \(\times\) Volume (L) \(\times\) \(n\)-factor.
2. Law of Equivalence: Equivalents of \(\text{KMnO}_4\) used = Equivalents of \(I_2\) liberated.
3. Equivalents of \(I_2\) liberated = Equivalents of \(\text{Na}_2\text{S}_2\text{O}_3\) consumed.
Step 3: Detailed Explanation:
1. Identify \(n\)-factors:
In basic or neutral medium, \(\text{MnO}_4^-\) is reduced to \(\text{MnO}_2\).
The oxidation state of Mn changes from \(+7\) to \(+4\), so the \(n\)-factor for \(\text{KMnO}_4 = 3\).
For the titration of Iodine with Thiosulfate:
\(I_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2I^- + \text{S}_4\text{O}_6^{2-}\).
The oxidation state of Sulfur in \(\text{Na}_2\text{S}_2\text{O}_3\) is \(+2\) and in \(\text{Na}_2\text{S}_4\text{O}_6\) it is \(+2.5\).
Change per sulfur atom = \(0.5\). For two sulfur atoms in the formula, \(n\)-factor = \(1\).
2. Calculate Moles and Equivalents:
Moles of \(\text{KMnO}_4 = \text{Molarity} \times \text{Volume} = 0.2 \times 0.5 = 0.1 \text{ moles}\).
Equivalents of \(\text{KMnO}_4 = 0.1 \times 3 = 0.3 \text{ Eq}\).
3. Equate with Thiosulfate:
By Law of Equivalence:
Equivalents of \(\text{KMnO}_4 = \text{Equivalents of } \text{Na}_2\text{S}_2\text{O}_3\).
\[ 0.3 = 0.1 \times 1 \times V \]
\[ V = \frac{0.3}{0.1} = 3 \text{ L} \]
Step 4: Final Answer:
The volume of \(\text{Na}_2\text{S}_2\text{O}_3\) consumed is 3 L.
Was this answer helpful?
0
Top Questions on Electrochemistry
- Alkyl halide and aryl halide react in the presence of sodium metal and dry ether. The name of the reaction is?
- MHT CET - 2024
- Chemistry
- Electrochemistry
- The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is:
- MHT CET - 2024
- Chemistry
- Electrochemistry
- What is the standard electrode potential for the reduction half-reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)?
- MHT CET - 2025
- Chemistry
- Electrochemistry
- What is the standard electrode potential of the half-reaction: \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \, (\text{solid})? \] Given that the standard electrode potential for the half-reaction: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \, (\text{solid}) \quad \text{is} \, +0.80 \, \text{V}. \] Also, the cell potential for the following reaction is: \[ \text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^+ \] is \( 0.46 \, \text{V} \).
- MHT CET - 2025
- Chemistry
- Electrochemistry
- Want to practice more? Try solving extra ecology questions todayView All Questions
