When a sample of green potassium manganate (VI), $\ce{K2MnO4}$, is added to aqueous sodium hydroxide, a brown black solid A and a purple solution of B are obtained.

I know that the brown black solid A is $\ce{MnO2}$, but I can't seem to write out the entire equation.

Can someone help me with this?

Thanks! Really appreciated.

  • $\begingroup$ What are your thoughts on the identity of B? $\endgroup$ Feb 15, 2015 at 2:26
  • $\begingroup$ definitely a Mno4 ion, possibly NaMno4 $\endgroup$ Feb 15, 2015 at 2:41
  • $\begingroup$ Great, that's correct. To make balancing simpler, I suggest removing the spectator ions, or if you find that difficult, imagine for now that all alkali metal ions are the same type, for example using $\ce{KOH}$ instead of $\ce{NaOH}$. So put $\ce{K2MnO4}$ in the reagents, and $\ce{KMnO4}$ and $\ce{MnO2}$ in the products. On which side does the $\ce{KOH}$ go? Wherever you put it, notice that only one side will have hydrogen, and the equation cannot be balanced. What do you think is a reasonable compound to put on the other side that also contains hydrogen? $\endgroup$ Feb 15, 2015 at 3:09
  • $\begingroup$ i managed to use half equations in basic sln to solve this question. Removed all the spectators $\endgroup$ Feb 15, 2015 at 3:45
  • 2
    $\begingroup$ Showing your work by answering your own question is highly encouraged! $\endgroup$ Feb 15, 2015 at 11:45

1 Answer 1


$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -> MnO2 +2NaMnO4 + 6KOH }$$

As you said the brown solid A is $\ce{MnO2}$ while the purple solution is due to the permanganate ion .


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