Why the colour of KMnO4 at the end point of titration may disappear after some time?

Question

In the redox titration of $\ce{FeSO4}$ with $\ce{KMnO4}$, the colour change of the solution at the end point is colourless to light pink. Most laboratory manuals and books advice us to check whether the light pink colour stays or not for 30 seconds. If the colour stays after 30 seconds, then end point is reached. If the colour disappears before 30 seconds, then the titration is not finished, so more $\ce{KMnO4}$ should be added.

As far as I understand, at the end point, all the reducing agent is depleted. So, the colour of $\ce{KMnO4}$ stays. And, if I have interpreted my book correctly, the colour of $\ce{KMnO4}$ is to stay for more than 30 seconds so that we can say that the end point is reached.

This means, that the colour of $\ce{KMnO4}$ might disappear after 30 seconds even though the end point is reached. But why? All reducing agents are depleted. Aren't they?

Now, I have found some reactions which might possibly cause the discharge of colour after the end point:

\begin{align} \ce{3 MnSO4 + 2 KMnO4 + 2 H2O &-> 5 MnO2 + K2SO4 + 2 H2SO4}\\ \ce{3 MnSO4 + 2 KMnO4 + 8 H2SO4 &-> 5 Mn(SO4)2 + K2SO4 + 8 H2O}\\ \ce{4 KMnO4 + 2 H2O &-> 4 MnO2 + 4 KOH + 3 O2} \end{align}

I am not sure which of them occur in reality. So, which of the reactions are the reasons for the 30 second rule? And why is it 30 second anyway? Why not 40 second or 1 minute?

Answer 1

The most likely reason for the colour of permanganate to vanish after a few seconds even though you thought you have the end point is: you have not reached the end point.

Acid/base reactions are typically extremely fast reactions; their rate is not even limited by the speed of diffusion since protons can be shuffled along via the Grotthuss mechanism much more rapidly. Thus, you can assume each acid/base reaction to instantly reach completion.

With practically all other reactions that is not the case. They are typically slower and can be much slower. The reacting species must meet each other in solution which can take time. At the beginning of the titration, there are still many iron(II) ions floating around ready to be oxidised by any incoming permanganate. Towards the end, iron(II) is getting more scarce. So a permanganate ion may linger around much longer before it meets an iron ion. You should give it a certain time in case it does find one. The value of $\pu{30s}$ is arbitrary. Ten seconds or two minutes would probably do equally well (except that ten seconds may or may not be slightly on the short side). The book/method just needs to present any ballpark value.

So the reaction occuring that causes the discolouring of the solution is indeed exactly the reaction you are observing: the oxidation of iron(II). It is merely slowed down so it can take a few seconds to proceed to completion when close to the equivalence point.

Answer 2

In redox titrations, we titrate a substance with a reducing nature against a substance with an oxidising nature.

• $\ce{KMnO_4}$ is an oxidant. It reacts and reduces the other substance.
• The colour of manganese compounds largely depends on the oxidation state. For instance, $\ce{K\overset{+VII}{Mn}O4}$ is intensely purple while $\ce{\overset{+II}{Mn}SO4}$ is pale pink.

The reactions mentioned in the question involve reaction of products of $\ce{KMnO4}$ with $\ce{KMnO4}$ and the third one is the decomposition(?) of $\ce{KMnO4}$ in water, which looks absurd.

While titrating with oxalic acid or $\ce{FeSO4}$, we typically dilute the samples, add some dil. $\ce{H2SO4}$ and heat the mixture.

The reaction which most likely takes place in acidic media involves the conversion of $\ce{\overset{+VII}{Mn}O4^-}$ to $\ce{\overset{+II}{Mn}^{2+}}$. The half-reaction for this is: $$\ce{MnO_4^- +8H^+ +5e^-\to Mn^{2+} +4H2O}$$

Oxalic acid is oxidised to carbon dioxide as follows: $$\ce{C2H2O4 \to 2CO2 + 2H^+ + 2e^-}$$ $\ce{FeSO4}$, i.e. $\ce{Fe^{2+}}$ is oxidised to $\ce{Fe^{3+}}$ as: $$\ce{Fe^{2+} \to Fe^{3+} + e^-}$$

Combining these ionic equations and adding the right ions, we get the complete balanced equations.

We can now see that -

1. Dilution - So that the "pinkness" of $\ce{MnSO4}$, though only slight, does not interfere with the endpoint.
2. Addition of acid - To provide the $\ce{H^+}$ and the $\ce{SO4^{2-}}$ as counterion.
3. Heating - To accelerate the reaction.

The fact that heating is advised shows that the colour change is not as "rapid and conclusive" as in the case of normal acid-base titrations - we need to wait a bit.

Answer 3

Permanganate can oxidize water if left untouched for several days, this is why it should not be stored in a water solution too long even if sealed.

However, in this case you should consider faster reactions and setup. Typically redox titration is performed in a lab, that has quite a bit of organic dust and organic solvents in the air. Both might react with permanganate. Permangate solutions used for redox titration are quite diluted, so the reaction might affect the color rather fast. At the same time, redox reaction typically have limited speed and during titration color might be preserved just because permanganate didn't have time to react with reducer that is in the solution.

So, we need to draw a line when we expect permanganate to be completely reduced by reducer present in the solution and not diffused from air. The exact timing is a subject of trial and error.