Question

Amount of oxalic acid present in a solution can be determined by its titration with $KMn{O_4}\;$ solution in the presence of ${H_2}S{O_{4.}}$ The titration gives an unsatisfactory result when carried out in the presence of $HCL$because $HCl$:
A ) reduces permanganate to $M{n^{2 + }}$
B) oxidises oxalic acid to carbon dioxide and water
C) gets oxidised by oxalic acid to chlorine
D) omits ${H^ + }$ ions in addition to those from oxalic acid

Answer 1

The titration of potassium permanganate $\left( {KMn{O_4}} \right)$against oxalic acid $\left( {{C_2}{H_2}{O_4}} \right)$ is an example of redox titration. Redox titration is based on an oxidation-reduction reaction between the titrant and the analyte.

Complete Step by step answer: $\left( {KMn{O_4}} \right)$ is a strong oxidising agent and in the presence of sulfuric acid it acts as a powerful oxidising agent. In an acidic medium the oxidising ability of $\left( {KMn{O_4}} \right)$ is represented by the following equation.
In acidic solution,
$8{H^ + } + Mn{O_4}^{ - } + 5e\;\; \to \;M{n^{2 + }} + 4{H_{2}}O$
Solutions containing $Mn{O_4}^-\;ions$ are purple in colour and the solution containing $M{n^{2 + }}\;ions$ are colourless . The moment there is an excess of potassium permanganate present the solution becomes purple. Thus $KMn{O_4}$ serves as a self indicator in acidic solution.
Titration of oxalic acid with$KMn{O_4}$ when $HCl{\text{ }}$is used –
$HCl{\text{ }}$being a strong electrolyte dissociates in water to give ${H^ + }$ and $C{l^ - }$ ions.
$KMn{O_4}$ being a strong oxidising agent. It oxidises into Chloride ion$\left( {C{l^ - }} \right){\text{ }}to{\text{ }}C{l_2}$. Hence some amount of$\;KMnO4$ is used up in oxidising $C{l^ - }{\text{ }}to{\text{ }}C{l_{2.}}$ Side by side$\;KMn{O_4}$ is oxidising oxalate ion to$C{O_2}$ . $HCl$ is much stronger than Oxalic acid, it will react with $KMn{O_4}$ to produce the corresponding chlorides of Potassium and Manganese along with liberation of Chlorine gas. This results in a deficiency of the total amount of $KMn{O_4}$ available for the reaction with Oxalic acid. The half equation for this is: $8{H^ + } + Mn{O_4}^{ - } + 5e\;\; \to \;M{n^{2 + }} + 4{H_{2}}O$
In an acidic solution, $(Mn{O_4}^{ - \;})$ is reduced to the colourless $\left( {M{n^{2 + }}} \right){\text{ }}ion$ .
Due to $HCl{\text{ }}$the permanganate ions would oxidize the$C{l^{ - {\text{ }}}}ions$ to form $C{l_2}\left( g \right)$and so would not be
available to react with the oxalic acid. The equation for the reaction is as given below: $KMn{O_4}{\text{ }} + {\text{ }}16HCl{\text{ }} \to {\text{ }}2KCl{\text{ }} + {\text{ }}2MnC{l_{2{\text{ }}}} + {\text{ }}8{H_2}O{\text{ }} + {\text{ }}5C{l_2}$
Hence the option (A) is correct .

Note: For $HCl$, the chloride ion will be oxidised to chlorine gas by manganate (VII) ion. $HN{O_3}$ is also an oxidising agent and hence would compete with permanganate. The atoms which see an increment in the oxidation number oxidize and serv as the reducing agent too simultaneously.