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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 |
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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` |
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