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Is it possible for potassium permanganate to have color change (in a titration) from colorless to pink? If so, then in what circumstances (particularly for a titration)? I have been always been taught that color change for potassium permanganate is purple/pink to colorless. I have scoured the whole internet searching for an example and I couldn't find one.

The problem is a question in a recent A level paper which asked about the color change of $\ce{KMnO4}$ with an $\ce{Fe^2+}$ solution. The answer was colorless to pink. The examiner report further mentioned that many students wrote purple/pink to colorless, however it was marked incorrect.

Can someone explain the ambiguity here?

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  • $\begingroup$ "Potassium manganate", without further disambiguation, usually refers to Mn(VI) $\ce{K2MnO4}$. You need to use either "potassium permanganate" (much more common) or "potassium manganate(VII)". $\endgroup$ – orthocresol Mar 25 '17 at 12:58
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An aqueous solution of potassium permanganate, $\ce{KMnO4}$ is of purple colour, regardless if the solution is neutral, acidic, or basic. Potassium permanganate is a relative powerful oxidizer. (As a look-up in a table of standard electrode potentials may tell.) If one considers acidic conditions $\ce{KMnO4}$ may oxidize material, like

$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$

while being reduced from oxidation state +7 to oxidation state +2 in the form of $\ce{Mn^{2+}}$. Aqueous solutions of, say $\ce{MnSO4}$, however are almost colourless. On the other hand, $\ce{Fe^{2+}}$ alone is not a strong oxidizer, under normal conditions incapable to generate permanganate. That is why this analysis is widely used (application note, example 1, example 2 in lab classes and beyond).

The end point of the redox titration is reached if all $\ce{Fe^{2+}}$ is oxidized, so that under addition of additional $\ce{KMnO4}$, your analyte solution remains purple.

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  • $\begingroup$ But in the provided example of $\rm KMnO_4$ and $\rm Fe^{2+}$, it is not possible for the colour change to be from colourless to pink right (assuming no additional $\rm KMnO_4$ is added)? $\endgroup$ – mathnoob123 Mar 25 '17 at 12:35
  • $\begingroup$ @Faiq You have to add an extra bit of $\ce{KMnO4}$, otherwise you can't tell when the titration is done. $\endgroup$ – orthocresol Mar 25 '17 at 12:58
  • $\begingroup$ By observing for the colourless phase. The moment the colour disappears is my endpoint $\endgroup$ – mathnoob123 Mar 25 '17 at 13:00
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    $\begingroup$ @FaiqRaees During the titration, $\ce{KMnO4}$ is consumed, hence the purple colour of this added reagent solution, added to you analyte of $\ce{Fe^{2+}}$, vanishes. Once all the $\ce{Fe^{2+}}$ is consumed, the reaction with $\ce{KMnO4}$ is discontinued; hence, further addition of $\ce{KMnO4}$ will now yield the persistent purple colouration. $\endgroup$ – Buttonwood Mar 25 '17 at 13:03

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