When I prepare sodium perborate by mixing sodium hydroxide with metaborate ($\ce{NaBO}$) and hydrogen peroxide and cooling it I applied a test to it by acidifying it with $\ce{HCl}$ and adding $\ce{KMnO4}$. A precipitate appeared and it turned black. What is the reason?

I tested another sample of sodium perborate by acidifying it with $\ce{NH4Cl}$ and after that KI and this turned white. Again, why?


1 Answer 1


In your first test, the $\ce{KMnO4}$ probably reacted with the $\ce{H2O2}$ excess remaining after your preparation. The ionic equation for the reaction is

$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O +3O2 + 2OH-}$$

The $\ce{MnO2}$ formed is an insoluble black precipitate. Although there is an $\ce{OH-}$, the reaction apparently occurs in neutral and faintly acidic medium as well.

The result of your second test doesn't seem to be very clear. Sodium perborate apparently reacts very quickly with any water so if you added any water during the second test, it must have reacted to form oxygen gas bubbles (which appear white).

  • $\begingroup$ The second may be pure boric acid, that is weakly soluble $\endgroup$
    – permeakra
    Mar 11, 2013 at 11:03
  • $\begingroup$ @permeakra Thanks! I was not quite sure what that one was. Do you want to make that into another answer or shall I edit it into mine? $\endgroup$
    – kaliaden
    Mar 11, 2013 at 11:39
  • $\begingroup$ Edit: The reaction also occurs in neutral and faintly acidic medium not only basic medium. $\endgroup$
    – kaliaden
    Mar 11, 2013 at 13:40

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