A solution contians `Na_(2)CO_(3)` and `NaHCO_(3), 20cm^(3)` of this solution requires 5.0` cm^(3)` of 0.1 M `H_(2)SO_(4)` Solution for neutralization using phenolphthalein as the indicator. Methylorange is then added when a further `5.0cm^(3)` of 0.2 M `H_(2)SO_(4)` was required. Calculate the masses of `Na_(2)CO_(3)` and `NaHCO_(3)` is 1L of this solution.

Explanation

`Na_(2)CO_(3) + ( 1)/( 2) H_(2)SO_(4) rarr NaHCO_(3) + ( 1)/( 2) Na_(2) SO_(4)` phenolphthalein would change the colour after this reaction. Meq. of `H_(2)SO_(4)` used for 5ml mixture using phenolpthalein as indicator.
`= 2 xx 0.1 xx 5 = 1`
` :. ( 1)/( 2) ` Meq. of `Na_(2) CO_(3) = 1 `
Now methyl orange is added in this solution after 1 end point. Meq. of `H_(2)SO_(4)` used for solution after 1 end point using methyl orange as
Indicator `= 5 xx 0.2 xx 2 = 2 `
`:. ( 1)/( 2) ` Meq. of `Na_(2) O_(3) +` Meq. of `NaHCO_(3) = 2`
Meq. of `NaHCO_(3) = 2-1= 1 `
`( W)/( 84) xx 1000 =1 :. W = 0.084 g `
`:.` Weight of `NaHCO_(3)` in one litre `= ( 0.084 xx 1000)/( 20) = 4.2 g `
`:.` Meq. of `Na_(2) CO_(3) = 2 :. ( W)/( 53) xx 1000 = 2 `
or `W = ( 106)/( 1000) = 0.106 `
`:. ` Weight of `NaHCO_(3)` in litre