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A solution contains Na_(2)CO_(3) and NaHCO_(3) . 10 mL of the solution requires 2.5 mL of 0.1 " M " H_(2)SO_(4) for neutralisation using phenlphthalein as an indicator . Methyl orange is added when a further 2.5 mL of 0.2 " M " H_(2)SO_(4) was required . calculate the amountof Na_(2)CO_(3) and NaHCO_(3)in one litre of the solution . |
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Answer» SOLUTION :The NEUTRALISATION reactions are ` 2Na_(2)CO_(3) +H_(2)SO_(4) to 2NaHCO_(3) +na_(2)SO_(4)` `2NaHCO_(3) +H_(2)SO_(4) to Na_(2)SO_(4) +H_(2)O +CO_(2)` The volume of `H_(2)SO_(4)` (2.5 mL ) , used while using phenophthalein correspondsto the volume required for conversion of `NaHCO_(3) " to " Na_(2)HC_(3)` .THUS at the end pointwith phenolphthalein , we have , m.eof `2.5 " mL of " 0.1 M (i.e , 0.2 N) H_(2)SO_(4) = m.e of Na_(2)CO_(3)` or m.eof `Na_(2)CO_(3) = 0.5` Equivalent of `na_(2)CO_(3) =(0.5)/(1000)` Wtof `Na_(2)CO_(3)//10 m L = (0.5)/1000 XX 106 = 0.053 g ` equivalent wt . of `Na_(2)CO_(3)`is 106 according to given reaction ) ` :. ` wtof `Na_(2)CO_(3) =(0.5)/(1000)` Wt . of `Na_(2)CO_(3)//10 mL = (0.5)/1000 xx 106 = 0.053 g ` (equivalent wt . of `Na_(2)CO_(3)` is 106 according to given reaction ) ` :. ` wt of `Na_(2)CO_(3)` per litre = 5.3 g . Againwith methyl ORANGE , we have , m.e of 2.5 mL of `0.2 ` M (i.e., 0.4 N ) `H_(2)SO_(4)` solution = m.e of `NaHCO_(3)` produced from `na_(2)CO_(3)of NaHCO_(3)` originally present . ` :. ` m.e of `NaHCO_(3)` originally present = `1- 0.5 = 0.5` ` :. ` equivalentof `NaHCO_(3) = (0.5)/1000` ` :. ` wt . of `NaHCO_(3) ` per mL = `(0.5)/1000 xx 84 = 0.042 g ` ( eq. wt of `NaHCO_(3) = 84` according to given reaction Wt. of`NaHCO_(3)` per mL = `(0.5)/1000 xx 84 = 0.042 g ` (eq.wt f `NaHCO_(3) =84 `according to given reaction ) Wt. of `NaHCO_(3) ` per litre = 4.2 g . |
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