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Heat of Reaction

50 mL sample ...

Question

$50 m L$ sample of ozonised oxygen at $N T P$ was passed through a solution of potassium iodide. The liberated iodine required $15 m L$ of $0.08 N N a_{2} S_{2} O_{3}$ solution for complete titration. Calculate the volume of ozone at $N T P$ in the given sample.

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Solution

Reactions involved may be given as:
$2 K I + H_{2} O + O_{3} \to 2 K O H + I_{2} + O_{2} ↑$
$I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$
$1 m o l e O_{3} = 2 m o l e N a_{2} S_{2} O_{3} . . . . (i)$
No. of moles of hypo $= \frac{M a s s}{M o l e c u l a r m a s s (158)}$
$= \frac{E \times N \times V}{1000 \times 158}$
where, $E_{N a_{2} S_{2} O_{3}} = 158, N = 0.08, V = 15$
$∴$ No. of moles of hypo $= \frac{158 \times 0.08 \times 15}{1000 \times 158} = 1.2 \times 10^{- 3}$
No. of moles of $O_{3} = \frac{1}{2}$ mole of hypo [from eq. (i)]
$= \frac{1}{2} \times 1.2 \times 10^{- 3}$
$= 6 \times 10^{- 4}$ mole
Volume of $O_{3}$ at $N T P =$ No. of moles $\times 22400$
$= 6 \times 10^{- 4} \times 22400$
$= 13.44 m L$ at NTP.

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