Question

A solution of $N{a}_2{S}_2{O}_3$ is saturated iodometrically against $0.334$ g of pure $KBr{O}_3$(converted to $B{r}^{-}$) requiring $40$ ml. $N{a}_2{S}_2{O}_3$ solution. What is the molarity of $Na{S}_2{O}_3$ solution? (Molar mass of $KBr{O}_3=167$g $mo{l}^{-1}$)

Answer 1

${BrO}_3^{-}+6I^{(-)}+6H^{(+)}\to {Br}^{(-)}+3I_2+3H_{2}O$

${3I}_2+6S_2{O}_3^{=}\to 6I^{(-)}+3S_4{O}_6^{2-}$

So $6$ moles of $S_2{O}_3^{2-}\equiv$ $1$ mole of ${BrO}_3^{-}$

$\therefore$ no. of moles of $S_2{O}_3^{2-}\equiv \frac{0.334}{167}\times 6=0.012$

$\therefore$ Concentration of ${Na}_2{S}_2{O}_3$ solution $=\frac{0.012\times 1000}{40}M=0.3M$