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 $$x$$ M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of $$x$$ is __________.

The moles of permanganate present are

$$n(MnO_4^-)=M\times V=0.2\times0.5=0.1\text{ mol}.$$

The moles of iodide present are

$$n(I^-)=1.5\times0.5=0.75\text{ mol}.$$

In basic medium, the relevant reaction is

$$2MnO_4^-+6I^-+4H_2O\rightarrow2MnO_2+3I_2+8OH^-.$$

From the stoichiometry,

$$2\text{ mol }MnO_4^- \rightarrow 3\text{ mol }I_2.$$

Hence, the moles of iodine liberated are

$$n(I_2)=\frac{3}{2}\times0.1=0.15\text{ mol}.$$

The liberated iodine is titrated with sodium thiosulphate according to

$$I_2+2S_2O_3^{2-}\rightarrow2I^-+S_4O_6^{2-}.$$

Thus,

$$1\text{ mol }I_2\rightarrow2\text{ mol }S_2O_3^{2-}.$$

Therefore, the moles of thiosulphate required are

$$n(S_2O_3^{2-})=2\times0.15=0.3\text{ mol}.$$

Since (300\text{ mL}=0.3\text{ L}),

$$x=\frac{n}{V}=\frac{0.3}{0.3}=1\text{ M}.$$

Hence, the value of (x) is 1.