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?

40 gm NaOH, 106 gm Na_(2)CO_(3) and 84 gm NaHCO_(3) is dissolved in wa

1.	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» The burette reading of HCl will be 40 ML, if phenolphthalein is used as indiator from the beginning The burette reading of HCl will be 60 ml, if phenolphthalein is used as indicator form the beginning. The burette readin of HCl will be 40ml, if methyl orange is used as indicator after the first end point The burette reading of HCl will be 80 ml, if methyl ORGANE is used as indicator from the very beginning. Solution :`n_(NaOH)=n_(Na_(2)CO_(3))=n_(NaHCO_(3))=1` a) In presence of Hph, Eqts of `NaOH+(1)/(2)` EQ `N_(2)CO_(3)` = Eq HCl `(200xx1xx1)+(1)/(2)(20xx1xx2)=1xxV` `V_(HCl)=40ml` c) In presence of methyl orange at first end point `(1)/(2)` eq `NA_(2)CO_(3)+` eq `NAHCO_(3)=` eq HCl `((1)/(2)xx20xx1xx2)+(20xx1xx1)=V_(HCl)XX1` `V_(HCl)=40ml` d) If methyl orange used Eq `NAOH+` Eq `NAl_(2)CO_(3)+` Eq `NaHCO_(3)` = Eq HCl `(20xx1xx1)+(20xx1xx2)+(20xx1xx1)` `=V_(HCl)xx1impliesV_(HCl)=80`