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NaHC(2)O(4) is 0.1 M when neutralised wi...

`NaHC_(2)O_(4)` is 0.1 M when neutralised with NaOH. Hence, it is ...... when oxidised with `MnO_(4-)//H^(+)` .

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The correct Answer is:
B
`NaHC_(2)O_(4) to n=2` (as a reducing agent)
So, `0.1xx2=0.2N NaHC_(2)O_(4)` as a RA.
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