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.

Explanation

To calculate `[HS^(-)]`
`{:(,H_(2)S,hArr,H^(+),+,HS^(-),),("Initial conc.",0.1M,,,,,),("After disso.",0.1-x~=0.1,,x,,x,):}`
`K_(s) = (x xx x)/(0.1)=9.1xx10^(-8) or x^(2)=9.1 xx 10^(-8) or x^(2)=9.1xx10^(-9) or x= 9.54xx10^(-5)`.
In presence of 0.1 M HCl, suppose `H_(2)S` dissociated is y. Then at equilibrium,
`[H_(2)S]=0.1-y~=0.1, [H^(+)]=0.1+y ~= 0.1 , [HS^(-)]=y M`
`K_(a)=(0.1xxy)/(0.1)=9.1xx10^(-8) ` (Given) or `y=9.1xx10^(-8)M`
To calculate `[S^(2-)]`
`H_(2)S overset(K_(a_(1)))hArr H^(+) + HS^(-) , HS^(-) overset(K_(a_(2)))hArr H^(+) + S^(2-)`
For the overall reaction, `H_(2)S hArr 2H^(+) + S^(2-)`
`K_(a)=K_(a_(1))xxK_(a_(2))=(9.1xx10^(-8))xx(1.2xx10^(-13))=1.092xx10^(-20)` ltbr. But `K_(a)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])`
In the absence of 0.1 M HCl, `[H^(+)]=2[S^(2-)]`
Hence, if `[S^(2-)]=x, [H^(+)]= 2x :. ((2x)^(2)(x))/(0.1)=1.092xx10^(-20) or 4x^(3) = 1.092xx10^(-21) or x^(3)=273xx10^(-24)`
` 3 log x = log 273 - 24 = 2.4362 - 24`
`log x = 0.8127 - 8 = bar(8).8127 or x = "Antilog" bar(8) . 8127 = 6.497 xx 10^(-8)= 6.5xx10^(-8)M`.
In presence of 0.1 M HCl, suppose `[S^(2-)]= y `, then `[H_(2)S] = 0.1 M [ H^(+)]=0.1+ 2y ~= 0.1 M `
`K_(a) = ((0.1)^(2)xxy)/(0.1) = 1.09 xx 10^(-20) or y = 1.09 xx 10^(-19) M`.