Question

An important oxidation-reduction reaction used to determine very small amounts of $O_3$ and other oxidants in polluted air involves oxidation of iodide ion :
$2I^{-}+O_3+H_{2}O⟶2O{H}^{-}+I_2+O_2$
The solution is acidified and the $I_2$ is titrated with sodium thiosulphate solution using starch indicator. One litre of a mixture of $O_3$ and $O_2$ at STP was allowed to react with an excess of acidified solution of $KI$ and $I_2$ liberated required $40mL$ of $M/10$ sodium thiosulphate solution. What is the mole or volume percentage of ozone in the mixture?

Answer 1

Answer: 4.48

$2I^{-}+O_3+H_{2}O⟶2O{H}^{-}+I_2+O_2$
$I_2+2N{a}_2{S}_2{O}_3⟶N{a}_2{S}_4{O}_6+2NaI$
From the above reaction:
$2N{a}_2{S}_2{O}_3\equiv I_2\equiv O_3$
mol of $N{a}_2{S}_2{O}_3=\frac{40\times 0.1}{1000}=0.004$
mol of $I_2=$ mol of $O_3=0.002$
volume of $O_3$ at $STP=0.002\times 22.4$
$=0.0448L$ present in $1L$ mixture
Percentage of $O_3=4.48\%$ (by volume)