Question

Which of the following acts as a self – indicator?
A. \[{K_2}C{r_2}{O_7}\]
B. \[KMn{O_4}\]
C. Oxalic acid
D. Iodine

Answer 1

Hint: The molecule which acts as a self – indicator from the above options is a compound of a group 7 element. It has the transition metal in a +7 oxidation state.

Complete step by step answer:
\[KMn{O_4}\] acts as a self – indicator. \[KMn{O_4}\] solutions are dark purple in colour. It is due to +7 oxidation state of Manganese (Mn). Mn is in its highest oxidation state (+7). So it can act only as an oxidising agent. \[KMn{O_4}\] is usually titrated against solutions like Ferrous Ammonium Sulphate (\[{(N{H_4})_2}Fe{(S{O_4})_2}{({H_2}O)_6}\]) and Oxalic acid (\[{C_2}{H_2}{O_4}\]) which are acidic in nature. During the titration of say \[{C_2}{H_2}{O_4}\] with \[KMn{O_4}\], \[Mn{O_4}^ - \] (\[M{n^{ + 7}}\]) is reduced to \[M{n^{ + 2}}\] while \[{C_2}{O_4}^{ - 2}\] (\[{C^{ + 3}}\]) is oxidised to form \[C{O_2}\]. \[M{n^{ + 2}}\]is light pink in colour. When all the \[{C_2}{O_4}^{ - 2}\] ions are oxidised and the end point is reached, more and more \[KMn{O_4}\] is still being added to the solution which does not get reduced. Hence, we see a change in colour of the solution from light pink or colourless to dark purple. This indicates the end point of titration. As no other indicator is required to mark the end point of titration, \[KMn{O_4}\] acts as a self-indicator in its redox titrations.

Hence, the correct answer is (B).

Additional information:
Oxalic acid is heated before titration with \[KMn{O_4}\] as a certain temperature is required to initiate the reaction.

Note: A student might not observe the change in colour of \[KMn{O_4}\] if it is added very fast as the end point would come before our first stop of titration. In that case, one can think that the above statement is false. \[KMn{O_4}\] has to be added very slowly during titration.