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 weight of `H_(2)C_(2)O_(4) = a g` in 1 L and weight of `NaHC_(2)O_(4) = b g` in 1L for acid base reaction now
`"Meq. of" H_(2)C_(2)O_(4) + "Meq. of" NaH_(2)C_(2)O_(4) in 10 mL = 3 xx 0.1`
`"Meq. of" H_(2)S_(2)O_(4) + "Meq. of" NaH_(2)C_(2)O_(4) in 1 L = 3 xx 0.1 xx 100 = 30`
`a/45 xx 1000 + b/(112//1) xx 1000 = 30 `........ (i)
For redox change :
`C_(2)^(6+) rarr 2C^(4+) + 2e^(-)`
`Mn^(7+) + 5e^(-) rarr Mn^(2+)`
`"Meq. of" H_(2)C_(2)O_(4) + "Meq. of" NaH_(2)C_(2)O_(4) in 10 mL = 4 xx 0.1`
`"Meq. of" H_(2)C_(2)O_(4) + "Meq. of" NaH_(2)C_(2)O_(4) in 1 L = 4 xx 0.1 xx 100 = 40`
`( : "Eq. wt. of" H_(2)C_(2)O_(4) = M/2 "and Eq. wt. of" NaHC_(2)O_(4) = M/2 "as reductant")` ........ (ii)
Solving, equations (i) and (ii), we get, a = 0.9 g and b = 1.12 g