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Effect on Indicators by Acids and Bases

A 0.5 g sam...

Question

A $0.5$ g sample containing $M n O_{2}$ on treatment with $H C l$ liberates $C l_{2}$ . The $C l_{2}$ is passed into a solution of $K I$ and $30.0$ mL of $0.1 M N a_{2} S_{2} O_{3}$ are required to titrate the liberated iodine. Calculate the percentage of $M n O_{2}$ in the sample (as nearest integer).
[Given that the atomic weight of $M n = 55$ g/mol]

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Solution

The redox reactions are as shown below:
$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$

$C l_{2} + 2 K I \to 2 K C l + I_{2}$

$2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}$

Hence, $2$ moles of $N a_{2} S_{2} O_{3}$ is equal to $1$ mole of $C l_{2}$ or one mole of $I_{2}$ . They are equal to $1$ mole of $M n O_{2}$ .

The number of moles of $N a_{2} S_{2} O_{3}$ used is $\frac{30 \times 0.1}{1000} = 30 \times 10^{- 4}$ mol.

This corresponds to $\frac{30 \times 10^{- 4}}{2} \times 87$ g $= 0.1305$ of pure $M n O_{2}$

The mass of impure $M n O_{2}$ is $0.5$ g.

The percentage of $M n O_{2}$ in the sample $= \frac{0.1305}{0.5} \times 100 = 26.1 %$

The nearest integer is $26$ .

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