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Standard XII

Chemistry

Redox Reactions

An aqueous so...

Question

An aqueous solution containing $0.10 g K I O_{3} (formula weight = 214.0 g / m o l)$ was treated with an excess of $K I$ solution. The solution was acidified with $H C l .$ The liberated $I_{2}$ consumed $55 m L$ of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate molarity of the sodium thiosulphate solution.

0.05 M

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0.25 M

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0.01 M

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0.05 M

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Solution

The correct option is D $0.05 M$
$K I O_{3} + 5 K I \to 3 K_{2} O + 3 I_{2}$
Moles of $K I O_{3} = \frac{0.10}{214.0} = 0.00047$
$∴$ Moles of $I_{2}$ liberated $= 0.00047 \times 3 = 0.00141$
$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$
Moles of $N a_{2} S_{2} O_{3} = 2 \times 0.00141 = 0.00282$
Molarity $= \frac{0.00282 \times 1000}{55} = 0.0512 M$

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Redox Reactions

Standard XII Chemistry

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The correct option is D 0.05 MKIO3+5KI→3K2O+3I2 Moles of KIO3=0.10214.0=0.00047 ∴ Moles of I2 liberated =0.00047×3=0.00141 2Na2S2O3+I2→2NaI+Na2S4O6 Moles of Na2S2O3=2×0.00141=0.00282 Molarity =0.00282×100055=0.0512 M