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 the wt. Of `H_(2)C_(2)O_(4)` in 10 mL of the solution be x g . The weight of `NaHC_(2)O_(4)` in 10 mL will be `(0.0202 - x) g ` .The weight of `NaHC_(2)O_(4)` iin 10 mL will be `(0.0202 -x)` g
In the first experiment , `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` are neutralised by NaOH changing into `Na_(2)C_(2)O_(4)` . The eq.wt of `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` will therefore be 90/2 and 112 respectively .
Thus ,
m.e of `H_(2)C_(2)O_(4) + " m.e of " NaH_(2)O_(4) =" m.e of " NaOH `
`x/(90//2) xx 1000 + ((0.0202-x))/112 xx 100 = 0.1 xx 3 " "..(1)`
In the second experiment , both `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` are oxidised t `CO_(2)` by `KMnO_(4)` . the equivalent weight of `H_(2)C_(2)O_(4) and NaHC_(2)O_(4)` will , therefore be 90/2 and 112/2 respectively `({:(C_(2)O_(4)^(2-) to, 2CO_(2)),(+6,+8):})`
Thus ,
m.e of `H_(2)C_(2)O_(4) +" m.e of " NaHC_(2)O_(4) = " m.e of " KMnO_(4)`
`x/(90//2) xx 1000 + ((0.202-x))/(112//2) xx 1000 = 0.1 xx 4 " " ...(2)`
Subtracting (1) from (2) we get ,
`(0.202-x)/112 = (0.1)/1000`
` :. " " x = 0.009 ` g/10 mL of solution
The 1000 mL of solution contains
`H_(2)C_(2)O_(4) = 0.9 g `
and `NaHC_(2)O_(4) = 2.02 - 0.9 = 1.12 g ` .