500 mL of 1.2 M KI solution is ,nixed with 500 mL of 0.2 M $$KMnO_{4}$$ solution in basic medium. The liberated iodine was titrated with standard 0.1 M $$Na_{2}S_{2}O_{3}$$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of $$Na_{2}S_{2}O_{3}$$ consumed is_________.(Nearest integer)

Given 500 mL of 1.2 M KI mixed with 500 mL of 0.2 M $$KMnO_4$$ in basic medium, the liberated iodine is titrated with 0.1 M $$Na_2S_2O_3$$.

We begin by calculating the moles of the reactants. Moles of KI = $$0.5 \times 1.2 = 0.6$$ mol and moles of $$KMnO_4$$ = $$0.5 \times 0.2 = 0.1$$ mol.

In basic medium, $$MnO_4^-$$ oxidizes $$I^-$$ to $$I_2$$ according to the reaction: $$2MnO_4^- + 6I^- + 4H_2O \to 2MnO_2 + 3I_2 + 8OH^-$$. Here Mn goes from +7 to +4 (gaining 3e each, total 6e for 2 Mn atoms) and I goes from -1 to 0 (losing 1e each, total 6e for 6 I atoms), so electrons are balanced.

From stoichiometry, 2 mol $$MnO_4^-$$ requires 6 mol $$I^-$$. Since we have 0.1 mol $$KMnO_4$$, it requires $$0.1 \times 3 = 0.3$$ mol $$I^-$$, whereas 0.6 mol $$I^-$$ is available, making $$KMnO_4$$ the limiting reagent.

Because 2 mol $$MnO_4^-$$ produces 3 mol $$I_2$$, 0.1 mol $$MnO_4^-$$ yields $$0.1 \times \frac{3}{2} = 0.15$$ mol $$I_2$$.

The liberated iodine is then titrated with thiosulfate according to $$I_2 + 2Na_2S_2O_3 \to 2NaI + Na_2S_4O_6$$, so the moles of $$Na_2S_2O_3$$ needed are $$2 \times 0.15 = 0.30$$ mol, which at 0.1 M corresponds to a volume of $$\frac{0.30}{0.1} = 3$$ L.

The answer is 3.