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A mixture of H(2)C(2)O(4) and NaHC(2)O(4...

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.

Text Solution

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Let us consider `10 mL` of the stock solution contain x millimol oxalic acid `H_(2)C_(2)O_(4)` and y millimol of `NaHC_(2)O_(4)`.
When titrated against `NaOH`, basicity of oxalic acid is 2 while that `NaHC_(2)O_(4)` is 1.
`rArr 2x + 2y = 3 xx 0.1 =0.3` ...(i)
When titrated against acidic `KMnO_(4)`, n-factors of both oxalic acid and `NaHC_(2)H_(4)` would be 2.
`rArr 2x+2y = 4 xx 0.1 = 0.4` ...(ii)
solving equations (i) and (ii) gives
`y = 0.1, x = 0.1`
`rArr` In `1.0L` solution, mole of `H_(2)C_(2)O_(4) = (0.1)/(1000) xx 100 = 0.01`
Mole of `NaHC_(2)O_(4) = (0.1)/(1000) xx 100 = 0.01`
`rArr` Mass of `H_(2)C_(2)O_(4) = 90 xx 0.01 =0.9 g`
Mass of `NaHC_(2)O_(4) = 112 xx 0.01 = 1.12 g`
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