InterviewSolution
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InterviewSolution
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A solution contains `Na_(2)CO_(3) " and " NaHCO_(3)`. 10 mL of the solution required 2.5 mL of 0.1 M`H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of `0.2 M H_(2)SO_(4)` was required. Calculate the amount of `Na_(2)CO_(3) " and" NaHCO_(3)` in one litre of the solution. |
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Answer» 2.5 mL of `0.1 M H_(2)SO_(4)=2.5 mL " of" 0.2 N H_(2)SO_(4)` `=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture So, `5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture `-=5 mL " of" 0.2 N Na_(2)CO_(3)` `-=(0.2xx53)/(1000)xx5=0.053 g` Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture Between first and second end points, =2.5 mL of `0.2 M H_(2)SO_(4)` used =2.5 mL of `0.4N H_(2)SO_(4)` used =5 mL of `0.2 N H_(2)SO_(4)` used `-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture `(5-2.5)mL 0.2 N H_(2)SO_(4)` `-=NaHCO_(3)` present in 10 mL of mixture `-=2.5 mL 0.2 N NaHCO_(3)` `-=(0.2xx84)/(1000)xx2.5=0.042 g` Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture. |
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