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 wt.of `H_(2)C_(2)O_(4)=`ag in `1` litre
wt.of `NaHC_(2)O_(4)=` bg in `1` litre
For acid base reaction:
Now ( Meq.of `H_(2)C_(2)O_(4)+`Meq.of `NaHC_(2)O_(4))` in `10 mL=3xx0.1`
`:.` Meq.of `H_(2)C_(2)O_(4)+` Meq.of `NaHC_(2)O_(4)` in one litre
`=3xx0.1xx100=30`
`(a)/(45)xx1000+(b)/(112//1)xx1000=30`
`:. (1000a)/(45)+(1000b)/(112)=30`....(1)
`( "Eq.wt.of" H_(2)C_(2)O_(4)=(M)/(2)=(90)/(2)=45 "and Eq.wt.of" NaHC_(2)HO_(4)=(M)/(1)=(112)/(1)=112)` (as acid salt)
For redox change:
`C_(2)^(3+)rarr2C^(4+)+2e`
`5e+Mn^(7+)rarrMn^(2+)`
Meq.of `H_(2)C_(2)O_(4)+"Meq.of" NaHC_(2)O_(4)` in `10 mL=4xx0.1`
`:. "Meq.of" H_(2)C_(2)O_(4)+"Meq.of"NaHC_(2)O_(4)` in `1` litre `= 4xx0.1xx100=40`
`:. (a)/(45)xx1000+(b)/(112//2)xx1000=40`
`:. (1000a)/(45)+(2000b)/(112)=40`.....(2)
`( because "Eq.wt."H_(2)C_(2)O_(4)=M//2 "and eq.wt of" NaHC_(2)O_(4)=M//2)` ( as reductant)
Solving eqs. (1) and (2), we get
`a=0.90g, b=1.12g`
Note: Also given `a+b=2.02` and thus eq.(1) or (2) can be used to find `a` and `b` by using `a+b=2.02`