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40 gm NaOH, 106 gm Na_(2)CO_(3) and 84 gm NaHCO_(3) is dissolved in water and the solution is made 1 lit, 20 ml of this stock solution is titrated with 1 N HCl, hence which of the followign statements are correct?

Answer» <html><body><p>The burette reading of HCl will be 40 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>, if phenolphthalein is used as indiator from the beginning <br/>The burette reading of HCl will be 60 ml, if phenolphthalein is used as indicator form the beginning. <br/>The burette readin of HCl will be 40ml, if methyl orange is used as indicator after the first end point <br/>The burette reading of HCl will be 80 ml, if methyl <a href="https://interviewquestions.tuteehub.com/tag/organe-2898437" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANE">ORGANE</a> is used as indicator from the very beginning. </p>Solution :`n_(NaOH)=n_(Na_(2)CO_(3))=n_(NaHCO_(3))=1` <br/> a) In presence of Hph, <br/> Eqts of `NaOH+(1)/(2)` <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> `N_(2)CO_(3)` = Eq HCl <br/> `(200xx1xx1)+(1)/(2)(20xx1xx2)=1xxV` <br/> `V_(HCl)=40ml` <br/> c) In presence of methyl orange at first end point <br/> `(1)/(2)` eq `NA_(2)CO_(3)+` eq `NAHCO_(3)=` eq HCl <br/> `((1)/(2)xx20xx1xx2)+(20xx1xx1)=V_(HCl)<a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a>` <br/> `V_(HCl)=40ml` <br/> d) If methyl orange used <br/> Eq `NAOH+` Eq `NAl_(2)CO_(3)+` Eq `NaHCO_(3)` <br/> = Eq HCl <br/> `(20xx1xx1)+(20xx1xx2)+(20xx1xx1)` <br/> `=V_(HCl)xx1impliesV_(HCl)=80`</body></html>


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