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Standard XIII

Chemistry

Redox Reactions

An impure sam...

Question

An impure sample of $A s_{2} O_{3}$ weighing $20 g m$ was dissolved in water containing $10 g m o f N a H C O_{3}$ and the final solution was diluted to $400 m L$ . $20 m L$ of this solution was oxidised by $25 m L$ of a solution of $I_{2}$ . Another solution of hypo containing $1.24 g m$ of $N a_{2} S_{2} O_{3} .5 H_{2} O$ in $40 m L$ was used to exactly reduce $25 m L$ of same $I_{2}$ solution (impurities inert) calculate % of $A s_{2} O_{3}$ in the sample (at. wt. of As = 75)

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Solution

$A s_{2} O_{3}$ is oxidised to $A s_{2} O_{5}$

$M_{h y p o} = \frac{1.24 \times 1000}{248 \times 40} = 0.125$
$m e q . o f h y p o = m e q . o f I_{2} i n 25 m L = 0.125 \times 1 \times 40 = 5$
$m e q . o f I_{2} = m e q . o f A s_{2} O_{3} = 5$
$m e q . o f A s_{2} O_{3} i n 20 m L s o l u t i o n = 5 S o, m e q . o f A s_{2} O_{3} i n 400 m L = \frac{5 \times 400}{20} = 100$
$m o l e o f A s_{2} O_{3} = \frac{100 \times 10^{- 3}}{2} = 50 \times 10^{- 3} = 0.05$
$W e i g h t o f A s_{2} O_{3} = 0.05 \times 198 = 9.9 g$
$W e i g h t % = \frac{9.9}{20} \times 100 = 49.5 %$

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