Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :
$2 I ^{-}+ O _{3}+ H _{2} O \longrightarrow 2 OH ^{-}+ I _{2}+ O _{2}$
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 $40\, mL$ of $M / 10$ sodium thiosulphate solution. What is the mole or volume percentage of ozone in the mixture?

Redox Reactions

Solution:

$2 I ^{-}+ O _{3}+ H _{2} O \longrightarrow 2 OH ^{-}+ I _{2}+ O _{2} $
$ I _{2}+2 Na _{2} S _{2} O _{3} \longrightarrow Na _{2} S _{4} O _{6}+2 NaI$
From the above reaction:
$2 Na _{2} S _{2} O _{3} \equiv I _{2} \equiv O _{3} $
mol of $ Na _{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 $STP =0.002 \times 22.4$
$=0.0448 \,L$ present in $ 1 \,L $ mixture
Percentage of $ O _{3}=4.48 \% $ (by volume)