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.
$$\ce{ 3K2MnO4 + 2NaOH + 2H2O -&gt; MnO2 +2NaMnO4 + 6KOH }$$

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