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Hydrogen Peroxide

A mixture of ...

Question

A mixture of $N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ required equal volumes of $0.2 M$ $K M n O_{4}$ and $0.2 M$ $N a O H$ separately for complete titration. The mole ratio of $N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ in the mixture is:

A

\frac{2}{11}

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B

\frac{15}{2}

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C

\frac{5}{2}

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D

\frac{7}{2}

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Solution

The correct option is B $\frac{15}{2}$
Assume that a mixture of $N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ required 1 L of 0.2 M $K M n O_{4}$ and 1 L of 0.2 M NaOH separately for complete titration.

1 L of 0.2 M $N a O H$ corresponds to $1 L \times 0.2 m o l / L = 0.2$ moles of $N a O H$ .
1 mole of $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ will react with 3 moles of $N a O H$ .
So 0.2 moles of $N a O H$ will react with $\frac{0.2}{3} = 0.0667$ moles of $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ .
1 L of 0.2 M $K M n O_{4}$ corresponds to $1 L \times 0.2 m o l / L = 0.2$ moles of $K M n O_{4}$
5 moles of $N a_{2} C_{2} O_{4}$ will react with 2 moles of $K M n O_{4}$ .
0.2 moles of $K M n O_{4}$ will react with $\frac{5}{2} \times 0.2 = 0.5$ moles of $N a_{2} C_{2} O_{4}$
The mole ratio $N a_{2} C_{2} O_{4}$ $:$ $K H C_{2} O_{4} {˙ H}_{2} C_{2} O_{4}$ $=$ $0.5 : 0.0667 = \frac{15}{2}$

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