A solution contained Na_(2)CO_(3) and NaHCO_(3). 25 mL of this solution required 5 mL 0.1 N HClfor titration with phenophthalein as indicator . The titrationwas 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 .

A solution contained Na_(2)CO_(3) and NaHCO_(3). 25 mL of this solutio

1.	A solution contained Na_(2)CO_(3) and NaHCO_(3). 25 mL of this solution required 5 mL 0.1 N HClfor titration with phenophthalein as indicator . The titrationwas 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 .
Answer» Solution :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 toNaCl +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.eof " 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.eof " 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`)