Green Potassium Manganate and Sodium Hydroxide Equation

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.




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


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


 3