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10 ml of "0.02 M KMnO"(4) is required to...

10 ml of `"0.02 M KMnO"_(4)` is required to oxidize 20 ml of oxalic acid 25 ml of the same oxalic acid sol was used to neutralize 20 ml of `NaOH` of unknown strength. Calculate the amount of `NaOH` present per litre of solution.
`2KMnO_(4)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]`
`{:(COOH),(|),(COOH):}+[O]rarr 2CO_(2)+H_(2)O`
`{:(COOH),(|),(COOH):}+2NaOHrarr +{:(COOH),(|),(COOH):}+2H_(2)O`

Text Solution

Verified by Experts

`M xx n xxV" (oxalic acid)"=Mxx n xxV(KMnO_(4))`
`Mxx2xx20=0.02xx5xx10`
`M=(0.02xx5xx10)/(2xx20)=(1)/(40)=0.025`
`"Further, "Mxx n xx V"(oxalic acid)"=Mxx n xx V(NaOH)`
`0.025xx2xx25=Mxx1xx20`
`M=(0.025xx2xx25)/(20)=0.0625`
`M=(0.025xx2xx25)/(20)=0.0625`
`=0.0625 xx40=2.5" g/litre"`
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