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A solution contians Na_(2)CO_(3) and NaHCO_(3), 20cm^(3) of this solution requires 5.0 cm^(3) of 0.1 M H_(2)SO_(4) Solution for neutralization using phenolphthalein as the indicator. Methylorange is then added when a further 5.0cm^(3) of 0.2 M H_(2)SO_(4) was required. Calculate the masses of Na_(2)CO_(3) and NaHCO_(3) is 1L of this solution. |
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Answer» Solution :`Na_(2)CO_(3) + ( 1)/( 2) H_(2)SO_(4) rarr NaHCO_(3) + ( 1)/( 2) Na_(2) SO_(4)`phenolphthalein would change the colour after this reaction. Meq. of `H_(2)SO_(4)` used for 5ml mixture using phenolpthalein as indicator. `= 2 xx 0.1 xx 5 = 1` ` :. ( 1)/( 2) ` Meq. of `Na_(2) CO_(3) = 1 ` Now METHYL orange is ADDED in this solution after 1 end point. Meq. of `H_(2)SO_(4)` used for solution after 1 end point using methyl orange as Indicator `= 5 xx 0.2 xx 2 = 2 ` `:. ( 1)/( 2) ` Meq. of `Na_(2) O_(3) +` Meq. of `NaHCO_(3) = 2` Meq. of `NaHCO_(3) = 2-1= 1 ` `( W)/( 84) xx 1000 =1:. W = 0.084 g ` `:.` Weight of `NaHCO_(3)` in one litre `= ( 0.084 xx 1000)/( 20) = 4.2 g ` `:.` Meq. of `Na_(2) CO_(3) = 2 :. ( W)/( 53) xx 1000 = 2 ` or `W = ( 106)/( 1000) = 0.106 ` `:. ` Weight of `NaHCO_(3)` in litre |
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