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.

Explanation

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`