In basic medium $$CrO_4^{2-}$$ oxidises $$S_2O_3^{2-}$$ to form $$SO_4^{2-}$$ and itself changes into $$Cr(OH)_4^-$$. The volume of 0.154 M $$CrO_4^{2-}$$ required to react with 40 mL of 0.25 M $$S_2O_3^{2-}$$ is ______ mL. (Rounded-off to the nearest integer)

In basic medium, $$CrO_4^{2-}$$ oxidises $$S_2O_3^{2-}$$ to $$SO_4^{2-}$$, and $$CrO_4^{2-}$$ itself is reduced to $$Cr(OH)_4^-$$.

First, we determine the change in oxidation states. In $$CrO_4^{2-}$$, chromium is in the +6 oxidation state, and in $$Cr(OH)_4^-$$, chromium is in the +3 oxidation state. So each Cr atom gains 3 electrons.

In $$S_2O_3^{2-}$$, the average oxidation state of sulfur is +2, and in $$SO_4^{2-}$$, sulfur is in the +6 oxidation state. Each sulfur atom loses 4 electrons, and since there are 2 sulfur atoms per $$S_2O_3^{2-}$$, each $$S_2O_3^{2-}$$ ion loses a total of 8 electrons.

Using the equivalence relation (milliequivalents of oxidant = milliequivalents of reductant): $$n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2$$, where $$n_1 = 3$$ (n-factor of $$CrO_4^{2-}$$) and $$n_2 = 8$$ (n-factor of $$S_2O_3^{2-}$$).

Substituting: $$3 \times 0.154 \times V_1 = 8 \times 0.25 \times 40$$

$$0.462 \times V_1 = 80$$

$$V_1 = \frac{80}{0.462} = 173.16 \text{ mL} \approx 173 \text{ mL}$$

The volume of $$0.154$$ M $$CrO_4^{2-}$$ required is $$\mathbf{173}$$ mL.