1.

What volume of 0.2 M KOH will be requried to neutralise 100 " mL of " 0.1 M `H_(3)PO_(4)` using methyl red indicator (change of colour pink `rarr` yellow) and then bromothymol blue indicator is added.A. 50 mLB. 100 mLC. 150 mLD. 200 mL

Answer» Correct Answer - B
Methyl red indicates the first step ionisation of `H_(2)PO_(4)` bromothymol blue indiates the second step ionisation of `H_(3)PO_(4)`, i.e.,
`H_(2)PO_(4)^(ɵ)toH^(o+)+HPO_(4)^(2-)` `(n=1)(N=Mxx1)`
First case: When methyl red is added (change of `H_(2)PO_(4)toH_(2)PO_(4)^(ɵ)+H^(o+))(n=1)(N=Mxx1)`
`KOH-=H_(3)PO_(4)`
`N_(1)V_(1)-=N_(2)V_(2)`
`0.2xx1xxV_(1)=0.1xx1xx100`
`V_(1)=50mL`
Second case:
When bromothymol blue is added
(change of `H_(2)PO_(4)^(ɵ)toH^(o+)+HPO_(4)^(2-))`
`(n=1)(N=Mxx1)`
`KOH-=H_(2)PO_(4)^(ɵ)`
`N_(1)V_(1)-=N_(2)V_(2)`
`0.2xx1xxV_(1)=0.1xx1xx100`
`V_(1)=50mL`
Total volume `=50+50=100mL`


Discussion

No Comment Found

Related InterviewSolutions