A mixture of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` weighing `2.02 g` was dissolved in water and the solution made uptp one litre. `10 mL` of this solution required `3.0 mL` of `0.1 N NaOH` solution for complete neutralization. In another experiment `10 mL` of same solution in hot dilute `H_(2)SO_(4)` medium required `4 mL` of `0.1N KMnO_(4) KMnO_(4)` for compltete neutralization. Calculate the amount of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` in mixture.

Explanation

For `H_(2)CO_(2)O_(4)` n-factor`=2` for neutralisation
For `NaHC_(2)O_(4)`, n-factor`=1` for neutralisation
Let `H_(2)C_(2)O_(4)`, be a gram
Let `NaHC_(2)O_(4)` be b gram.
m" Eq of "`H_(2)O_(2)O_(4)` in `10 mL=(a)/((90)/(2))xx1000xx(1)/(100)=(10a)/(45)`
m" Eq of "`NaHC_(2)O_(4)` in `10mL=(b)/((112)/(1))xx1000xx(1)/(100)=(10b)/(112)`
m" Eq of "`NaOH=3xx0.1=0.3`
`(10a)/(45)+(10b)/(112)=0..3` ..(i)
For `H_(2)CO_(2)O_(4)`, n factor`=2` with `KMnO_(4)`
or `NaHC_(2)O_(4)`, n factor`=2` with `KMnO_(4)`
`m" Eq of "H_(2)C_(2)O_(4)` in `10mL=(a)/((90)/(2))xx1000xx(1)/(100)=(10a)/(45)`
m" Eq of "`NaHC_(2)O_(4)` in `10mL=(b)/((112)/(2))xx1000xx(1)/(100)=(10b)/(56)`
`m" Eq of "KMnO_(4)=4xx0.1=0.4`
`(10a)/(45)+(10b)/(56)=0.4`
Solving equations (i) and (ii) we get
`a=0.90g`
`b=2.2-0.90=1.12g`