10 g of a mixture of `NaHC_2O_4` and `Na_2C_2O_4` on heating gave `6.12 g` of `Na_2CO_3`. Another 10 g of the mixture was dissolved in 1.0 L of solution. 25" mL of " this solution was titrated with 0.1 N NaOH. Find the volume of NaOH required.

Explanation

`NaHC_2O_4overset(triangle)toNa_2C_2O_4`
`Na_2C_2O_4overset(triangle)toNa_2CO_3`
(No need to balance the equation)
`10g (NaHC_2O_4+Na_2C_2O_4)overset(triangle)to6.12gNa_2CO_3`
Let x g of `NaHC_2O_4` and `(1-x)g` of `Na_2C_2O_4` be taken
`(1)/(2)((x)/(112))+((10-x)/(134))=(6.12)/(106)`
Solving, we get `x=5.63g(NaHC_2O_4)`
`therefore10-5.63=4.57g of Na_2C_2O_4`
Second experiment: `NaOH` reacts only with `NaHC_2O_4` Due to the presence of one acidic H atom only .
`Eq(NaHC_2O_4)-=(Mw)/(1)`
`m" Eq of "NaOH=m" Eq of "NaHC_2O_4`
`Vxx0.1N-=((5.63)/((112)/(1)))xx1000xx(25)/(1000)`
`V=12.5 mL`