A mixture of `n_(1)` moles of `Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` is titrated separately with `H_(2)O_(2)` and `KOH`, to reach at equivalence point. Which of the following statement is/are correct?

Options

A. Moles of `H_(2)O_(2)` and `KOH` are `n_(1)+n_(2)` and `n_(2)`
B. Moles of `H_(2)O_(2)` and `KOH` is `n_(1)+(n_(2))/(2)` and `n_(1)`
C. n-factors of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)`, respectively, are 1 and 2.
D. n-factors of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and KOH, respectively, are 2 and 1.

Answer

A, C

Explanation

`Na_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` both reacts with `H_(2)O_(2)` as reducing agent only. (n factor for both `=2`)
With `H_(2)O_(2)`:
" Eq of "`H_(2)O_(2)=" Eq of "Na_(2)C_(2)O_(4)+" Eq of "NaHC_(2)O_(4)`
`2xx` moles of `H_(2)O_(2)=n_(1)xx2+n_(2)xx2`
Moles of `H_(2)O_(2)=(2(n_(1)+n_(2))/(2)=(n_(1)+n_(2))`
With KOH: Only `NaHC_(2)O_(4)` reacts with KOH as acid base titration n factor `=1(one H^(o+)` ion)
" Eq of "`KOH=" Eq of "NaHCO_(3)`
`1xx` Moles of `KOH=n_(2)xx1`
Moles of KOH`=n_(2)`
`therefore` Moles of `H_(2)O_(2)` and KOH are: `(n_(1)+n_(2))` and `n_(2)`.
n-factor of `NaHC_(2)O_(4)` with KOH and `H_(2)O_(2)=1 and 2`
n-factor of `Na_(2)C_(2)O_(4)` with `H_(2)O_(2)` and `KOH=2 and 2`.