A solution contained `Na_(2)CO_(3) and NaHCO_(3)`. 25 mL of this solution required 5 mL 0.1 N HCl for titration with phenophthalein as indicator . The titration was repeated with the same volume of the solution but woth methyl orange .12.5 mL of 0.1 N HCl was required this time . Calculate the amount of `Na_(2)CO_(3) and naHCO_(3)` in the solution .

Explanation

Neutralisation reaction with phenolphthalein is
`na_(2)CO_(3) +HCl to NaHCO_(3) +NaCl`
while with methyl orange , the reactions are ,
`Na_(2)CO_(3) + HCl to NaHCO_(3) +NaCl `
`naHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)`
(produced)
and `naHCO_(3) +HCl to NaCl +H_(2)O +CO_(2)`
(originally present)
Thus ,
we have with phenolphthalein is
`na_(2)CO_(3) +HCl to NaCl +H_(2)O +CO_(2)`
( originally present)
Thus,
we have with phenolphthalein ,
m.e of `Na_(2)CO_(3) ` = " m.e of " 5 mL of `0.1 ` N HCl
` = 0.1 xx 5 = 0.5`
` :. ` eq . of `na_(2)CO_(3) = (0.5)/1000 = 0.005`
` :. ` wt of `Na_(2)CO_(3)= (0.0005 xx 106) g `
` = 0.053 g`
[ Eq. wt . of `na_(2)CO_(3)` in the given reactio is 106]
And with methyl orange ,
m.e of `Na_(2)CO_(3) +" m.e of " NaHCO_(3) + " m.e of " NaHCO_(3)`
(produced ) (originally present )
= m.e of `12.5` mL of `0.1 ` N HCl
or `0.5 + 0.5 + " m.e of " NaHCO_(3) = 0.1 xx 12.5 = 1.25`
or `" m.e of "NaHCO_(3) = (0.25)/(1000) xx 84 = 0.021 g `
( eq. wt of `NaHCO_(3) =84`)