`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution.
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Molecular mass of `Na_(2)CO_(3). 10 H_(2)O=286` Equivalent mass of `Na_(2)CO_(3).10H_(2)O=(286)/(2)=143` 100 mL solution of sodium carbone contains =1 g 1000 mL solution of sodium carbone contains=10 g Normality of the solution`=(10)/(143)` Applying the formula, Normality of acid solution` xx` its volume =Normality of sodium carbonate solution` xx` its volume, Normality of the acid solution `=(10xx42.9)/(143xx30)=0.1` Let V mL be the volume of `H_(2)SO_(4)` teken. `8xx5+4.8xx5+34xxV=0.1xx2000` V=4 mL Amount of `SO_(4)^(2-)=("Normality" xx "Eq. mass" xx"Volume")/(1000)` `=(34xx48xx4)/(1000)=6.528 g`
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