Question

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).

Hint:

In iodometric titrations, always convert the oxidising agent to equivalent moles of iodine first, then relate iodine to thiosulphate.

Answer 1

Correct Answer: 3

Concept: This problem involves {redox reactions} followed by {iodometric titration}. Key ideas used:
In basic medium, \( \mathrm{KMnO_4} \) oxidises iodide ions to iodine. 
Liberated iodine is titrated with sodium thiosulphate. 
Stoichiometry of redox reactions determines the amount of iodine formed. 
Step 1: Write the redox reaction in basic medium In basic medium, permanganate ion is reduced to manganese dioxide: \[ 2\mathrm{MnO_4^-} + 6\mathrm{I^-} + 4\mathrm{H_2O} \longrightarrow 2\mathrm{MnO_2} + 3\mathrm{I_2} + 8\mathrm{OH^-} \] From the equation: \[ 2 \text{ mol } \mathrm{KMnO_4} \longrightarrow 3 \text{ mol } \mathrm{I_2} \] 
Step 2: Calculate moles of reactants {Moles of KI:} \[ n(\mathrm{KI}) = 0.5 \times 1.2 = 0.6 \text{ mol} \] {Moles of \( \mathrm{KMnO_4} \):} \[ n(\mathrm{KMnO_4}) = 0.5 \times 0.2 = 0.1 \text{ mol} \] 
Step 3: Identify the limiting reagent From the balanced equation: \[ 2 \mathrm{KMnO_4} : 6 \mathrm{I^-} \Rightarrow 1 \mathrm{KMnO_4} : 3 \mathrm{I^-} \] For \(0.1\) mol \( \mathrm{KMnO_4} \), required iodide: \[ 0.1 \times 3 = 0.3 \text{ mol} \] Available iodide \(=0.6\) mol \(>\) required \(0.3\) mol \[ \Rightarrow \mathrm{KMnO_4} \text{ is the limiting reagent} \] 
Step 4: Calculate moles of iodine liberated From stoichiometry: \[ 2 \mathrm{KMnO_4} \rightarrow 3 \mathrm{I_2} \] \[ 0.1 \mathrm{KMnO_4} \rightarrow \frac{3}{2}\times0.1 = 0.15 \text{ mol } \mathrm{I_2} \] 
Step 5: Titration of iodine with sodium thiosulphate Reaction: \[ \mathrm{I_2} + 2\mathrm{S_2O_3^{2-}} \rightarrow 2\mathrm{I^-} + \mathrm{S_4O_6^{2-}} \] \[ 1 \text{ mol } \mathrm{I_2} \rightarrow 2 \text{ mol } \mathrm{Na_2S_2O_3} \] \[ \text{Moles of } \mathrm{Na_2S_2O_3} = 2 \times 0.15 = 0.30 \text{ mol} \] 
Step 6: Calculate volume of \( \mathrm{Na_2S_2O_3} \) \[ M = \frac{n}{V} \Rightarrow V = \frac{n}{M} = \frac{0.30}{0.1} = 3.0 \text{ L} \] Final Answer: \[ \boxed{3\ \text{L}} \]