Calculate Molarity of Thiosulphate in Redox Titration with Permanganate

To find the value of $x$, we analyze the given redox process step-by-step using stoichiometry and equivalence principle.

Step 1: Reaction between $\text{MnO}_4^-$ and $\text{I}^-$ in basic medium

In a basic/faintly alkaline medium, permanganate ion ($\text{MnO}_4^-$) is reduced to manganese dioxide ($\text{MnO}_2$): $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \longrightarrow \text{MnO}_2 + 4\text{OH}^-$

The change in oxidation state of manganese is from $+7$ to $+4$, so the $n$-factor of $\text{MnO}_4^-$ in this reaction is $3$.

The given question specifies that iodide ion ($\text{I}^-$) is oxidized to liberate molecular iodine ($\text{I}_2$): $2\text{I}^- \longrightarrow \text{I}_2 + 2e^-$

Step 2: Determine the amount of $\text{I}_2$ liberated

Calculate the moles of reactants initially present: $\text{Moles of } \text{MnO}_4^- = 0.5\text{ L} \times 0.2\text{ M} = 0.1\text{ mol}$ $\text{Moles of } \text{KI} = 0.5\text{ L} \times 1.5\text{ M} = 0.75\text{ mol}$

The equivalents of electrons supplied by the complete reduction of $\text{MnO}_4^-$ are: $\text{Equivalents of } e^- = 0.1\text{ mol} \times 3 = 0.3\text{ eq}$

Since $\text{KI}$ is present in excess ($0.75\text{ mol} > 0.3\text{ eq}$), $\text{MnO}_4^-$ is the limiting reagent.

Since $2\text{ moles of } e^-$ produce $1\text{ mole of } \text{I}_2$, the moles of molecular iodine ($\text{I}_2$) formed are: $n_{\text{I}_2} = \frac{\text{Equivalents of } e^-}{2} = \frac{0.3}{2} = 0.15\text{ mol}$

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Step 3: Titration of liberated $\text{I}_2$ with thiosulphate ($\text{S}_2\text{O}_3^{2-}$)

The reaction between iodine and thiosulphate solution in the presence of starch indicator is given by: $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \longrightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$

From the stoichiometry of the reaction, $1\text{ mole of } \text{I}_2$ reacts with $2\text{ moles of } \text{S}_2\text{O}_3^{2-}$.

Thus, the number of moles of thiosulphate consumed is: $n_{\text{thiosulphate}} = 2 \times n_{\text{I}_2} = 2 \times 0.15\text{ mol} = 0.3\text{ mol}$

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Step 4: Calculate the molarity ($x$) of the thiosulphate solution

The volume of thiosulphate solution consumed is $300\text{ mL} = 0.3\text{ L}$.

Using the formula $n = \text{Molarity} \times \text{Volume (in L)}$: $0.3\text{ mol} = x\text{ M} \times 0.3\text{ L}$

$x = \frac{0.3}{0.3} = 1$

Final Answer: The value of $x$ is 1.