1.

51 mLof a solution , containing 1 g each of Na_(2)CO_(3) ,NaHCO_(3) and NaOHwas titrated with N HCl . What will be the titre readings if only methyl orange is used as indicator from the very beginning ?

Answer»

Solution :The reactions in this case are ,
`Na_(2)CO_(3) +HCl to NaHCO_(3) +NaCl `
`NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)`
(produced)
`NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)`
(originally present )
and `NaOH +HCl to NaCl +H_(2)O `
THUS , we have
m.e of `Na_(2)CO_(3) + m.e of NaHCO_(3) + m.e of NaHCO_(3) + m.e of NaOH `
(produced )(originally present )
` = " m.e of " v_(2)"mL (say ) of N HCl " `
`{:(1/106, xx 1000 +1/106 xx 1000 + 1/84 xx 1000 + 1/40 xx 1000 = 1 xx v_(2)),( :.,""v_(2)=55.8):}`.


Discussion

No Comment Found

Related InterviewSolutions