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40 gm NaOH, 106 gm Na(2)CO(3) and 84 gm ...

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?

A

The burette reading of HCl will be 40 ml, if phenolphthalein is used as indiator from the beginning

B

The burette reading of HCl will be 60 ml, if phenolphthalein is used as indicator form the beginning.

C

The burette readin of HCl will be 40ml, if methyl orange is used as indicator after the first end point

D

The burette reading of HCl will be 80 ml, if methyl organe is used as indicator from the very beginning.

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
`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`
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