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

Let mass of `H_(2)C_(2)O_(4)` present in the mixture be =a g in 1 litre
and mass of `NaHC_(2)O_(4)` present in the mixture be = b g in 1 litre
For acid-base reaction
`H_(2)C_(2)O_(4)+2NaOH to Na_(2)C_(2)O_(4)+2H_(2)O`
Eq. mass of `H_(2)C_(2)O_(4)=("Mol.mass")/(2)=(90)/(2)=45`
`NaHC_(2)O_(4)+NaOH to Na_(2)O_(4)+H_(2)O`
Eq. mass of `NaHC_(2)O_(4)=("Mol. mass")/(1)=112`
Now,
Equivalents of `H_(2)C_(2)O_(4) +` Equivalents of `NaHC_(2)O_(4)=(3xx01)/(1000)`
in 10 mL solution `" "` in 10 mL solution
`(axx10)/(45xx1000)+(bxx10)/(112xx1000)=(3xx0.1)/(1000)`
or `112a+45b=(3xx0.1xx45xx112)/(10)=151.2`
For redox reaction
Eq. mass of `H_(2)C_(2)O_(4)=(90)/(2)=45`
Eq. mass of `NaHC_(2)O_(4)=(112)/(2)=56`
(Change in oxidation number of carbon per molecule=2, `C_(2)^(3+) to 2C^(4+))`
Now,
Equivalents of `H_(2)C_(2)O_(4)+` Equivalents of `NaHC_(2)O_(4)=(4xx0.1)/(1000)`
in 10 mL solution `" " ` in 10 mL solution
`(axx10)/(45xx1000)+(bxx10)/(56xx1000)=(4xx0.1)/(1000)`
or 56a+45b=100.8
Solving equations (i) and (ii),
a=0.9g and b=1.12g