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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 MH_(2)SO_(4) forneutralisation 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.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.5 mL of `0.1 M H_(2)SO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)=2.5 mL " of" 0.2 N H_(2)SO_(4)` <br/> `=(1)/(2)Na_(2)CO_(3)` present in 10 mL of mixture <br/> So, <br/> `5 mL " of" 0.2 N H_(2)SO_(4)=Na_(2)CO_(3)` present in 10 mL of mixture <br/> `-=5 mL " of" 0.2 N Na_(2)CO_(3)` <br/> `-=(0.2xx53)/(1000)xx5=0.053 g` <br/> Amount of `Na_(2)CO_(3)=(0.053)/(10)xx1000=5.3 g//L` of mixture <br/> Between first and second end <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a>, <br/> =2.5 mL of `0.2 M H_(2)SO_(4)` used <br/> =2.5 mL of `0.4N H_(2)SO_(4)` used <br/> =5 mL of `0.2 N H_(2)SO_(4)` used <br/> `-=(1)/(2)Na_(2)CO_(3)+NaHCO_(3)` present in 10 mL of mixture <br/> `(5-2.5)mL 0.2 N H_(2)SO_(4)` <br/> `-=NaHCO_(3)` present in 10 mL of mixture <br/> `-=2.5 mL 0.2 N NaHCO_(3)` <br/> `-=(0.2xx84)/(1000)xx2.5=0.042 g` <br/> Amount of `NaHCO_(3)=(0.042)(10)xx1000=4.20 g//L` of mixture.</body></html>


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