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\( \mathrm{KMnO}_{4} \) can be prepared from \( \mathrm{K}_{2} \mathrm{MnO}_{4} \) as per the reaction :
\[
3 \mathrm{MnO}_{4}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{MnO}_{4}^{2-}+\mathrm{MnO}_{2}+4 \mathrm{OH}^{-}
\]
\( \mathrm{P} \)
The reaction can go to completion by removing \( \mathrm{OH}^{-} \)ions by adding:
(NEET 2013)
W.
(a) \( \mathrm{KOH} \)
(b) \( \mathrm{CO}_{2} \)
(c) \( \mathrm{SO}_{2} \)
(d) \( \mathrm{HCl} \)
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