`0.1 M H_2SO_4=0.2 N H_2SO_4`
`0.2 M H_2SO_4=0.4 N H_2SO_4`
Volume of 0.2 N `H_2SO_4` used for half neutralisation of `Na_2CO_3=2.5mL`
Volume of 0.2 N `H_2SO_4` used for complete neutralisation of `Na_2CO_3=2xx2.5=5mL`
Total volume of `H_2SO_4` used
`=2.5 of 0.2 N H_2SO_4+2.5mL` of `0.4 N H_2SO_4`
`=25 mL` of `0.2 N H_2SO_4+5 ” mL of ” 0.2 N H_2SO_4`
`=075″ mL of ” 0.2 N H_2SO_4`
Volume of `0.2 N` `H_2SO_4` used for neutralisation of `NaHCO_3`
`=7.5-5.0=2.5mL`
`N_1V_1(Na_2CO_3)=N_2V_2(H_2SO_4)`
`N_1xx10=0.2xx5`
`N_1=(0.2xx5)/(10)=0.1N`
Amount of `Na_2CO_3` in 1 L `=0.1xx53=5.3g`
`N_1V_1(NaHCO_3)=N_2V_2(H_2SO_4)`
`N_1xx10=0.2xx2.5`
`N_1=(0.2xx2.5)/(10)=0.05N`
Amound of `NaHCO_3` in `1L=0.05xx84=4.2g`

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.

Weight of `NaCO_(3)` in 1 litre =5.3 g
Weight of `NaHCO_(3)` in 1 litre =4.2 g