Question

500 mL of 0.2 M \(MnO_4^-\) solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard \(M\) thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, the value of \(x\) is _____.

Answer 1

Correct Answer: 0.5

Step 1: Understanding the Question:
This is a two-step redox titration problem. First, permanganate reacts with iodide. Then, the produced iodine reacts with thiosulphate. We need to find the molarity of thiosulphate.
Step 2: Key Formula or Approach:
Apply the principle of equivalence: \(\text{Milli-equivalents (mEq) of } MnO_{4}^{-} = \text{mEq of } I^{-} \text{ oxidized}\).
Then, \(\text{mEq of produced } I_{2} = \text{mEq of thiosulphate}\).
Step 3: Detailed Explanation:
1. Permanganate Reaction (Basic Medium):
In basic medium, \(MnO_{4}^{-}\) is reduced to \(MnO_{2}\). The oxidation state changes from +7 to +4.
n-factor for \(MnO_{4}^{-}\) (\(n_{1}\)) \(= 3\).
mEq of \(MnO_{4}^{-}\) \(= M \times V(mL) \times n = 0.2 \times 500 \times 3 = 300\).

2. Iodide Oxidation:
Iodide (\(I^{-}\)) is oxidized to \(I_{2}\). n-factor for \(I^{-}\) (\(n_{2}\)) \(= 1\).
mEq of \(I^{-}\) reacted must be 300.
Check moles of \(I^{-}\): \(1.5 \text{ M} \times 500 \text{ mL} = 750 \text{ mmol}\). Since only 300 mEq are needed, KI is in excess.
mEq of \(I_{2}\) produced \(= 300\).

3. Thiosulphate Titration:
Reaction: \(I_{2} + 2S_{2}O_{3}^{2-} \rightarrow 2I^{-} + S_{4}O_{6}^{2-}\).
n-factor for thiosulphate (\(n_{3}\)) \(= 1\) (per sulfur it is 0.5, for the whole molecule it is 1).
Equivalents of \(I_{2}\) liberated \(= 300 \text{ mEq}\).
mEq of thiosulphate \(= x \text{ M} \times 300 \text{ mL} \times 1 = 300 x\).
Equating both:
\[ 300 x = 300 \Rightarrow x = 1 \]
Step 4: Final Answer:
The value of \(x\) is 1.