10 " mL of " `NaHC_(2)O_(4)` is oxidised by 10 " mL of " 0.02 M `MnO_(4)^(ɵ)`. Hence, 10 " mL of " `NaHC_(2)O_(4)` is neutralised by

Options

A. 10 " mL of " 0.1 M NaOH
B. 10 " mL of " 0.02 M NaOH
C. 10 " mL of " 0.1 M `Ca(OH)_(2)`
D. 10 " mL of " 0.05 N `Ba(OH)_(2)`

Answer

A

Explanation

In `NaHC_(2)O_(4),C_(2)O_(4)^(2-)` oxidised to `CO_(2)` and `H^(o+)` is neutralised.
`Ew_(NaHC_(2)O_(4))(as C_(2)O_(4)^(2-)=(M)/(2)`
`(as H^(o+))=(M)/(1)`
`10xxN_(1)(NaHC_(2)O_(4))=10xx0.1N MnO_(4)^(ɵ)`
`N_(1)=0.1N`