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The first ionization constant of `H_(2)S` is `9.1xx10^(-8)`. Calculate the concentration of `HS^(Θ)` ion in its `0.1 M` solution. How will this concentration be affected if the solution is `0.1 M` in `HCl` also? If the second dissociation constant if `H_(2)S` is `1.2xx10^(-13)`, calculate the concentration of `S^(2-)` under both conditions. |
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Answer» i. To calculate `[HS^(Θ)]` `{:(,H_(2)S,hArr,H^(o+),+,HS^(Θ)),("Initial",0.01 M,,,,),("After disso",0.01-x,,x,,x),(,~~0.1,,,,):}` `{:(Initial,0.01 M,,,),(After disso,0.01-x,,x,x),(,~~0.1,,,):}` `K_(a)=(x xx x)/0.1=9.1xx10^(-8) or x^(2)=9.1xx10^(-9)` or `x=9.54xx10^(-5)=[HS^(Θ)]=[H^(o+)]` ii. In `0.1 M HCl:` `{:(HCl,rarr,H^(o+),+,Cl^(Θ)("completely ionised")),(0.1,,0,,0),(0,,0.1,,0.1):}` Due to common ion effect `(H^(o+))`, the disociation of `H_(2)S` is supresed. `:. K_(a_(1))=([H^(o+)][HS^(o+)])/([H_(2)S])=((0.1)[HS^(o+)])/0.1` `:. K_(a_(1))=[HS^(Θ)]=9.1xx10^(-8) M`. iii. `HS^(Θ)rarr H^(o+)+S^(2-)` `K_(a_(2))=([H^(o+)][S^(2-)])/([HS^(Θ)])` `1.2xx10^(-3)=((9.54xx10^(-5))[S^(2-)])/((9.54xx10^(-5)))` `:. [S^(2-)]=1.2xx10^(-13) M`. iv. In presence of `0.1 M HCl`. `K_(a_(1))xxK_(a_(2))=([H^(o+)]^(2)[S^(2-)])/([H_(2)S])` `9.1xx10^(-8)xx1.2xx10^(-13)=((0.1)^(2)[S^(2-)])/((0.1))` `:. [S^(2-)]=(9.1xx10^(-8)xx1.2xx10^(-13))/0.1` `=1.092xx10^(-19) M` |
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