A solution contains `Na_(2)CO_(3)` and `NaHCO_(3)` `20 cm^(3)` of this solution required `5.0 cm^(3)` of 0.1 M `H_(2)SO_(4)` solution for neutralisation using phenolphthalein as an indicator . Methyl orange is then added when a further `5.0 cm^(3)` of `0.2 M H_(2)SO_(4)` was required. Calculate the masses of `Na_(2)CO_(3)` and `NaHCO_(3)` in 1 litre of this solution.

Explanation

`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`