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Arrhenius Acid

Oxalic acid ...

Question

Oxalic acid $(H_{2} C_{2} O_{4})$ is present in many plants and vegetables. If $24.0$ mL of $0.01$ M $K M n O_{4}$ solution is needed to titrate $1.0$ g of $H_{2} C_{2} O_{4}$ to the equivalence-point, the mass percentage (as nearest integer) of $H_{2} C_{2} O_{4}$ in the sample is :

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Solution

$2$ moles of $K M n O_{4}$ $\equiv$ $5$ moles of $H_{2} C_{2} O_{4}$ .
Molar mass of oxalic acid = $90$ g/mol.
Number of moles of $K M n O_{4} = \frac{24.0}{1000} \times 0.01 = 0.00024$ mol.
Number of moles of $H_{2} C_{2} O_{4} = \frac{5}{2} \times 0.00024 = 0.0006$ mol.
Mass of $H_{2} C_{2} O_{4} = 0.0006 \times 90 = 0.054$ g.
Mass percentage of $H_{2} C_{2} O_{4} = \frac{0.054}{1} \times 100 = 5.4 %$ .
The nearest integer is $5.$

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