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pH the Power of H

Fe2+ as well ...

Question

$F e^{2 +}$ as well as $H N O_{2}$ can be used to estimate $M n O_{4}^{-}$ in acidic medium. If 100 mL of 1 M $F e^{2 +}$ can be oxidised by $V$ mL of 1 M $M n O_{4}^{-} / H^{+}$ solution, calculate volume of 1 M $H N O_{2}$ that can be oxidises by $V$ mL of 1 M $M n O_{4}^{-} / H^{+}$ solution.

A

40 m L

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B

25 m L

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C

50 m L

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D

none of these

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Solution

The correct option is B $50 m L$
The balanced chemical equations are as shown.
$10 F e S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O + K_{2} S O_{4} + 2 M n S O_{4} 5 H N O_{2} + 2 K M n O_{4} + 3 H_{2} S O_{4} \to 2 M n S O_{4} + 5 H N O_{3} + K_{2} S O_{4} + 3 H_{2} O$
From above equations, we can say that 10 mole $F e S O_{4} =$ 5 mole $H N O_{2}$
or 2 mole $F e S O_{4} =$ 1 mole $H N O_{2}$
Since same volume and same concentration of $K M n O_{4}$ is used for titration of $F e^{2 +}$ ions and $H N O_{2}$ , the volume of $H N O_{2}$ required will be $\frac{100 m L \times 1 M}{\frac{2 m o l e F e^{2 +}}{1 m o l e H N O_{2}} \times 1 M} = 50 m L$

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{M n O_{4}}^{-}

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0.1

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M n O_{4} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

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{F e}^{2 +}

C_{2} O_{4}^{2 -}

.

100

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