A solution contains `Na_(2)CO_(3)` and `NaHCO_(3). 10 mL` of the solution required `2.5 mL "of" 0.1M H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further `2.5 mL "of" 0.2M H_(2)SO_(4)`was required. The amount of `Na_(2)CO_(3)` and `NaHCO_(3)` in `1 "litre"` of the solution is:

Explanation

`Na_(2)CO_(3)` `NaHCO_(3)`
Let a meq. b meq.
when HPh is used as indicator
`Na_(2)CO_(3) + H_(2) SO_(4) rarr NaHCO_(3) + NaHSO_(4)` then `(1)/(2)` meq. of `Na_(2)Cl_(3)`= meq. of `H_(2)SO_(4) `
`(a)/(2) = 2.5 xx 0.1 xx 2 rArr a = 1 `
MeOH is added after the first end point the solution Contains `NaHCO_(3)` original & `NaHCO_(3)` produced.
`NaHCO_(3) + H_(2)SO_(4) rarr H_(2) CO_(3) + NaHSO_(4)`
meq. of `NaHCO_(3)` original `+` meq. of `NaHCO_(3)` produced
`2.5 xx 0.2 xx 2 = b + 1//2 `meq. of `Na_(2) CO_(3) = b + a//2 `
`b + a//2 = 1 `
`b = 1- 0.5 = 0.5 `
wt of `Na_(2) CO_(3) //` lit `= a xx 10^(-3) xx ( 106)/( 2) xx ( 1)/( 10) xx 1000`
` = 1 xx ( 53)/( 10) = 5.3 gm`
wt of `NaHCO_(3) //` lit `= b xx 10^(-3) xx 84 xx ( 1)/( 10) xx 1000 = 4.2 gm`