A solution contains `Na_(2)CO_(3) and NaHCO_(3)` . 10 mL of the solution requires `2.5` mL of `0.1 " M " H_(2)SO_(4)` for neutralisation using phenlphthalein as an indicator . Methyl orange is added when a further 2.5 mL of `0.2 " M " H_(2)SO_(4)` was required . calculate the amount of `Na_(2)CO_(3) and NaHCO_(3) ` in one litre of the solution .

Explanation

The neutralisation reactions are
` 2Na_(2)CO_(3) +H_(2)SO_(4) to 2NaHCO_(3) +na_(2)SO_(4)`
`2NaHCO_(3) +H_(2)SO_(4) to Na_(2)SO_(4) +H_(2)O +CO_(2)`
The volume of `H_(2)SO_(4)` (2.5 mL ) , used while using phenophthalein corresponds to the volume required for conversion of `NaHCO_(3) " to " Na_(2)HC_(3)` .Thus at the end point with phenolphthalein , we have ,
m.e of `2.5 " mL of " 0.1 M (i.e , 0.2 N) H_(2)SO_(4) = m.e of Na_(2)CO_(3)`
or m.e of `Na_(2)CO_(3) = 0.5`
Equivalent of `na_(2)CO_(3) =(0.5)/(1000)`
Wt of `Na_(2)CO_(3)//10 m L = (0.5)/1000 xx 106 = 0.053 g `
equivalent wt . of `Na_(2)CO_(3)` is 106 according to given reaction )
` :. ` wt of `Na_(2)CO_(3) =(0.5)/(1000)`
Wt . of `Na_(2)CO_(3)//10 mL = (0.5)/1000 xx 106 = 0.053 g `
(equivalent wt . of `Na_(2)CO_(3)` is 106 according to given reaction )
` :. ` wt of `Na_(2)CO_(3)` per litre = 5.3 g .
Again with methyl orange , we have ,
m.e of 2.5 mL of `0.2 ` M (i.e., 0.4 N ) `H_(2)SO_(4)` solution
= m.e of `NaHCO_(3)` produced from `na_(2)CO_(3) of NaHCO_(3)` originally present .
` :. ` m.e of `NaHCO_(3)` originally present = `1- 0.5 = 0.5`
` :. ` equivalent of `NaHCO_(3) = (0.5)/1000`
` :. ` wt . of `NaHCO_(3) ` per mL = `(0.5)/1000 xx 84 = 0.042 g `
( eq. wt of `NaHCO_(3) = 84` according to given reaction
Wt. of`NaHCO_(3)` per mL = `(0.5)/1000 xx 84 = 0.042 g `
(eq.wt f `NaHCO_(3) =84 `according to given reaction )
Wt. of `NaHCO_(3) ` per litre = 4.2 g .