Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)]`& upon addition of `0.1`mole `HCl` to it will be :A. `10`B. `100`C. `1000`D. `10,000`

Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(

1.	Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)]`& upon addition of `0.1`mole `HCl` to it will be :A. `10`B. `100`C. `1000`D. `10,000`
Answer» Initial :`[HA^(2-)]=10(-8)M=K_(a_(2))` final:`K_(a_(1))xxK_(a_(2))=([H^(+)][HA^(2-)])/([H_(3)A]):.10^(-8)xx10^(-5)=([H^(+)]^(2)[HA^(2-)])/([H_(3)A])rArr[HA^(2-)]=10^(-12)M` So ratio`=(10^(-8))/(10^(-12))=10^(4)`

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Ratio of `[HA^(+)]` in `1L` of `0.1M H_(3)A` solution `[K_(a_(1))=10^(-5),K_(a_(2))=10^(-8)K_(a_(3))=10^(-11)]`& upon addition of `0.1`mole `HCl` to it will be :A. `10`B. `100`C. `1000`D. `10,000`

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