1.

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.

Answer»

SOLUTION :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 mLwill be `(0.0202 - x) g `.The weightof `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)` . Theeq.wtof `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.eof `H_(2)C_(2)O_(4) +" m.eof " 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 ` .


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