This is in the context of the various changes in oxidation state the permangananate ion can undergo with the variation of the pH of the medium. Besides the cases of strongly basic and strongly acidic media, the case for neutral media (with reduction to $\ce{MnO2}$) in my notes has a footnote; that this medium is in fact slightly alkaline, due to it being a solution of $\ce{KMnO4}$.

I tried to reason this out. It looks like $$\ce{K+ + MnO4- + H2O -> K+ + OH- + HMnO4}$$ is how it works out.

A quick google search establishes the pKa of HMnO4 to be -2.25. So it makes a pretty good acid in comparison to water.

Thing is, in that case, the product side should have an $\ce{H2O}$ instead of the constituent ions, with the potassium permanganate dissociated. That means the solution's neutral. Which apparently it isn't, going with my notes.

Where did I mess up here?


1 Answer 1


The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does not work with the latter.

$\ce{2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O}$ (in molten KOH solvent; reactants are fused together)

$\ce{2 K2MnO4 + 2 H2O → 2 KMnO4 + 2 KOH + H2}$ (electrolysis in alkaline media; a direct chemical reaction would convert some material back to $\ce{MnO2}$)

Samples are of course purified but must be assumed to contain some leftover alkali from this preparation. Some alkali may also be left in solution on purpose because potassium permanganate is most stable in alkaline solution and shelf life would be improved by providing a buffer against absorption of carbon dioxide from the air.

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    $\begingroup$ Excess alkali is probably a good thing, as acidic permanganate would degrade to oxygen and MnO2, making the KMnO4 a little dirty-looking in solution. $\endgroup$ Commented May 9, 2021 at 13:39
  • $\begingroup$ I assume alkalization is due the decomposition. $\ce{4 KMnO4(aq) + 2 H2O -> 4 KOH(aq) + 4 MnO2(s) + 3 O2(g)}$ $\endgroup$
    – Poutnik
    Commented Sep 28, 2023 at 3:43

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