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When I was a teenager, I played with a chemistry set. I tried adding potassium permanganate, alum, and sodium hydrogen sulphite (not sodium hydrogen sulphate) to water.

Upon adding the third ingredient, the water would rapidly turn clear, and a brown precipitate would form. After filtering this out, I allowed the water to evaporate, which left a white crystalline deposit. All three ingredients were definitely needed in order for the reaction to happen, which is why I remember it, since none of the other reactions I knew about needed three reactants. (This was a long time ago, so there's a very slight possibility I've mis-remembered the ingredients, but I'm 99% sure they're right.)

At the time, neither my teachers nor my parents could help me work out what kind of reaction it might be, so out of sheer curiosity, I thought I'd ask here. I'm no chemist, so simple explanations would be appreciated.

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up vote 4 down vote accepted

The fact that the solution turned clear is a pretty good guarantee that this was a redox reaction: the permanganate oxidised some other species, and in turn was reduced, probably to manganese dioxide, the brown precipitate you noticed. Unfortunately, neither of the other reactants you mention are reducing agents!

Do you think it's possible that the third reagent was actually sodium hydrogen sulphite, NaHSO$_3$? If so, this would make a lot more sense: the (hydrogen)sulfite ion would have been oxidised to (hydrogen)sulfate. This reaction requires an acidic medium, and the reason the alum (potassium aluminium sulfate) was necessary was to lower the pH.

These individual steps have the following equations:

Al$^{3+}$ + 4 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 4 H$^+$

MnO$_4^-$ + 4 H$^+$ + 3 $e^-$ $\rightarrow$ MnO$_2\downarrow$ + 2 H$_2$O

HSO$_3^-$ + H$_2$O $\rightarrow$ HSO$_4^-$ + 2 H$^+$ + 2 $e^-$

(note that the last two are half-equations -- they can’t occur on their own, but I’ve written them separately for clarity) giving an overall reaction of

Al$^{3+}$ + 2 MnO$_4^-$ + 3 HSO$_3^-$ + 3 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 2 MnO$_2\downarrow$ + 3 HSO$_4^-$

The white crystalline deposit in this case would just be alum again, more or less. You have some sodium in your solution now, from the sodium hydrogensulfite, so you'd get some sodium aluminium sulfate as well, but the only anion in your solution will still be (hydrogen)sulfate, and my guess is that with aluminium this will crystallise out as the sulfate.

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Yes, I am actually pretty sure it was sodium hydrogen sulphite, not sulphate. My curiosity has been satisfied - many thanks! –  Nathaniel Aug 28 '12 at 19:54
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