Question

During estimation of oxalic acid Vs $KMnO_4$, self-indicator is.
A. $KMnO_4$
B. Oxalic acid
C. $K2SO4$
D. $MnSO4$

Answer 1

Hint: The titration of potassium permanganate against oxalic acid is one of the examples of redox reaction. Potassium permanganate is an oxidising agent which works in acidic medium more strongly than alkaline medium. Potassium permanganate is purple in colour.

Complete step-by-step answer:In redox titrations, both oxidation and reduction take place simultaneously.
During titration, one of the chemical compounds will get oxidised at the same time the other reactant will get reduced.
In the presence of sulphuric acid KMnO4 is a powerful oxidising agent and it can be used for the estimation of oxalic acid(oxalates).
Oxalic acid reacts with the acidic $KMnO_4$
In acidic medium the oxidising stability of KMnO4 is represented by the equation,
\[MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O + {K_2}S{O_4}\]
 Solution containing MnO4- ions are purple in colour and the solution containing $Mn$2+ colourless and therefore permanganate solution is decolourised when added to a solution of a reducing agent.
The moment there is an excess of potassium permanganate present the solution becomes purple.
The overall reaction in acidic medium can be written as,
$2KMnO4 + 5(COOH)2 + 3H2SO4 \to K2SO4 + 2MnSO4 + 8H2O + 10CO2 \uparrow $

Therefore, the self-indicator during the estimation of oxalic acid Vs $KMnO_4$ is
Option (A), $KMnO_4$

Note: Oxalic acid is oxidised to carbon dioxide by $KMnO_4$ which itself gets reduced to $MnSO4$. This titration cannot be carried out by nitric acid or hydrochloric acid because itself is an oxidising agent. Therefore, hydrochloric acid reacts with potassium permanganate which results in the formation of chlorine which is also an oxidising agent. Potassium permanganate is dark in colour, so always read the upper meniscus.