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The first ionization constant of H(2)S i...

The first ionization constant of `H_(2)S` is `9.1 xx 10^(-8)`. Calculate the concentration of `HS^(-)` ion in its `0.1M` solution. How will this concentration be affected if the solution is 0.1M in HCl also ? If the second dissociation constant of `H_(2)S` is `1.2 xx 10^(-13)`, calculate the concentration of `S^(2-)` under both conditions.

Text Solution

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(i) `H_(2)S + H_(2)O to H_(3)O^(+) + HS^(-)`
`K_(a)=([H_(3)O^(+)][HS^(-)])/[H_(2)S]`
`implies[H_(3)O^(+)]=[HS^(-)]=sqrt(K_(a1).C)=sqrt(9.1xx10^(-8)xx0.1)=9.54xx10^(-5)M`
(ii) In the presence of 0.1 M HCl,` [H_(3)O^(+)]` = 0.1 M
`K_(a1)=([H_(3)O^(+)][HS^(-)])/([H_(2)S])` `implies 9.1xx10^(-8)= ([0.1][HS^(-)])/[0.1]implies [HS^(-)]=9.1xx10^(-8)M`
Hence, concentration of `[HS^(-)]` is decreased in the presence of due to common-ion effect.
(iii) For second dissociation constant,`HS^(-) + H_(2)O to H_(3)O^(+) + S^(2-)` (In absence of HCl)
`[HS^(-)]=9.54xx10^(-5)M`, `K_(a2)=([H_(2)O][S^(2-)])/[HS^(-)]`
`[H_(3)O^(+)]=[S^(2-)]`=`sqrt(K_(a2).C)`=`sqrt(1.2xx10^(-12)xx9.54xx10^(-5))`=`3.38xx10^(-9) M`
(iv) In the presence of 0.1 M HCl, `K_(a2)`=`([H_(3)O^(+)][S^(2-)])/[HS^(-)]`=`1.2xx10^(-13)=([0.1][S^(-2)])/([9.1xx10^(-8)])`
`implies[S^(2-)]=1.092xx10^(-19)M`
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