A green solution of potassium manganate (VI) turns purple and a brown solid is precipitated, when $C O_{2}$ gas is bubbled through the solution.

The above change is observed when $C O_{2}$ gas is passed into an aqueous solution of the given ion. The function of $C O_{2}$ is:

It makes the solution acidic due to the formation of

H_{2} C O_{3}

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It makes the solution basic due to the formation of

O_{3}^{2 -}

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It is simply the medium of the reaction

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None of the above is correct

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Solution

The correct option is A It makes the solution acidic due to the formation of $H_{2} C O_{3}$
It makes the solution acidic due to the formation of $H_{2} C O_{3}$ .

$C O_{2} + H_{2} O \to H_{2} C O_{3} c a r b o n i c a c i d$

$3 K_{2} M n O_{4} + 2 C O_{2} \to 2 K M n O_{4} p u r p l e + M n O_{2} (b r o w n s o l i d) + 2 K_{2} C O_{3}$

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C O_{2} + H_{2} O ⇌ H_{2} C O_{3} ⇌ H^{+} + H C O_{3}^{-}

M n O_{2}

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