Amount of oxalic acid present in a solution can be determined by its titration with ${K M n O}_{4}$ solution in the presence of $H_{2} {S O}_{4}$ . The titration gives unsatisfactory result when carried out in the presence of HCL, because HCL:

gets oxidised by oxalic acid to chlorine

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furnishes

H^{+}

ions in addition to those from oxalic acid

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reduces permanganate to

{M n}^{2 +}

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oxidises oxalic acid to carbon dioxide and water

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Solution

The correct option is A gets oxidised by oxalic acid to chlorine
1) $H C l$ is also oxidised along with oxalic acid by $K M n O_{4}$ .

2) $2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 5 C l_{2} + 8 H_{2} O$

3) $2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} C_{2} O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$

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Similar questions

K M n O_{4}

H_{2} S O_{4}

H C l

H C l

K M n O_{4}

H_{2} S O_{4}

H C l

H C l

Why in the redox titration of $K M n O_{4}$ vs oxalic acid, we heat oxalic acid solution before starting the titration ?

K M n O_{4}

H_{2} S O_{4}

H C l

H C l

K M n O_{4}

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