A solution contains `Na_(2)CO_(3) " and " NaHCO_(3)`. 10 mL of the solution required 2.5 mL of 0.1 M`H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then 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

2.5 mL of `0.1 M H_(2)SO_(4)=2.5 mL " of" 0.2 N H_(2)SO_(4)`
`=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture
So,
`5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture
`-=5 mL " of" 0.2 N Na_(2)CO_(3)`
`-=(0.2xx53)/(1000)xx5=0.053 g`
Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture
Between first and second end points,
=2.5 mL of `0.2 M H_(2)SO_(4)` used
=2.5 mL of `0.4N H_(2)SO_(4)` used
=5 mL of `0.2 N H_(2)SO_(4)` used
`-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture
`(5-2.5)mL 0.2 N H_(2)SO_(4)`
`-=NaHCO_(3)` present in 10 mL of mixture
`-=2.5 mL 0.2 N NaHCO_(3)`
`-=(0.2xx84)/(1000)xx2.5=0.042 g`
Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture.