When ${\mathrm{Na}}_2{\mathrm{MnO}}_4$ is kept in strongly acidic solution the products obtained are

When ${\mathrm{MnO}}_4^{2-}$ placed in  strong acidic medium, then it undergoes disproportionation reaction.

$\stackrel{+6}{\mathrm{Mn}}{\mathrm{O}}_4^{2-}\stackrel{{\mathrm{H}}^{+}}{\to }\stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}+\stackrel{+4}{\mathrm{Mn}}{\mathrm{O}}_2$

Therefore, the above reaction can be written as:

${\mathrm{Na}}_2{\mathrm{MnO}}_4\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{NaMnO}}_4+{\mathrm{MnO}}_2+{\mathrm{H}}_2\mathrm{O}$