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25 mL of a mixture of NaOH+Na_(2)CO_(3), when titrated with (N)/(10)HCl using phenolphthalein indicator required 25 mL HCl to deccolourise phenolphthalein. At this stage methyl orange was added and addition of acid was continued. The second end point was reached after further addition of 5 mL of the acid. Calculate the amount of Na_(2)CO_(3) and NaOH 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> :Between first and second end points, <br/> `NaHCO_(3)+HCl to NaCl+H_(2)CO_(3)` <br/> `5 mL (N)/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)HCl-=(1)/(2)Na_(2)CO_(3)` present in <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mL of a mixture <br/> or `10 mL (N)/(10)HCl-=Na_(2)CO_(3)` present in 25 mL of a mixture <br/> `-=10 mL (N)/(10)Na_(2)CO_(3)=0.053 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> Na_(2)CO_(3)` <br/> Amount of `Na_(2)CO_(3)` in one litre of mixture `=(0.053)/(25)xx1000=2.12 g` <br/> `(25-5)mL(N)/(10)HCl-=NaOH` present in 25 mL of mixture <br/> `-=25 mL (N)/(10)NaOH` <br/> `-=0.08 gNaOH` <br/> Amount of NaOH in one litre of mixture`=(0.08)/(25)xx1000=3.2 g`</body></html>


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