A mixed solution of KOH and Na_(2)CO_(3) required 20 mL of N/20 HCl solutionwhen titrated with phenophthalein as indicator .Butthe same amountof solution when titrated with methylorange as indicator required 30 mL of the same acid . Calculate the amount of KOH and Na_(2)CO_(3) .

A mixed solution of KOH and Na_(2)CO_(3) required 20 mL of N/20 HCl so

1.	A mixed solution of KOH and Na_(2)CO_(3) required 20 mL of N/20 HCl solutionwhen titrated with phenophthalein as indicator .Butthe same amountof solution when titrated with methylorange as indicator required 30 mL of the same acid . Calculate the amount of KOH and Na_(2)CO_(3) .
Answer» Solution :NEUTRALISATION reactions are `{:(KOH+HCl to KCl +H_(2)O),(Na_(2)CO_(3)+HCl to NaHCO_(3)+NaCl ),(NaHCO_(3)+HCl to NaCl +H_(2)O +CO_(2)):}}`phenolphthalein is used methyl orange is used . As DISCUSSED in the previous example we have with phenophthalein , m.e of 20 mL of N/20 HCl = m.e of KOH + m.e of `Na_(2)CO_(3)` or m.e or KOH + m.e of `Na_(2)CO_(3)` or m.e of KoH + m.e of `Na_(2)CO_(3) = 20 xx 1/20 = 1 ""...(1)` Now , with methyl orange , m.e of 30 mL of N/20 hCl m.e of HOH + m.e of `Na_(2)CO_(3) + " m.e of " Na_(2)CO_(3) = 20 xx 1/20 = 1 ""..(1)` Now , with methyl orange , m.e of 30 mLof N/20 HCl = m.e of KOH + m.e of `Na_(2)CO_(3) + " m.e of "NaHCO_(3)` produced . Since m.e of `Na_(2)CO_(3) `= m.e of `Na_(2)CO_(3) ` + m.e of `Na_(2)CO_(3)` or m.eof KOH + `2 xx ` m.e of `Na_(2)CO_(3) = 1.5 ""...(2)` Subtracting Eqn. (1) From Eqn. (2) , we get , m.e of `Na_(2)CO_(3) = 1.5 - 1 = 0.5` ` :. " EQUIVALENT of " Na_(2)CO_(3) = (0.5)/1000` ` :." wt of " Na_(2)CO_(3) = 106`) From eqn.s (3) and (1) , m.e of KOH ` = 1- 0.5 = 0.5` Equivalent of KoH ` = (0.5)/1000` WEIGHT of KOH = `(0.5)/1000 xx 56 = 0.28 g . `( eq. wt of KOH = 56) .