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.

Answer

`[HS^(-)] = 9.54 xx 10^(-5)`, in `0.1M HCl [HS^(-)] = 9.1 xx 10^(-8)M,[S^(2-)] = 1.2 xx 10^(-13) M`, in `0.1 M HCl [S^(2-)] = 1.09 xx 10^(-19) M`

Explanation

(i) To calculate the concentration of `HS^(-)` ion:
Case I (in the absence of HCl):
Let the concentration of `HS^(-)` be xM.
`{:(,H_(2)S,harr,H^(+),,HS^(-)),(C_(f),0.1,,0,,0),(C_(f),0.1-x,,x,,x):}`
Then , `K_(a_(1)) = ([H^(+)][HS^(-)])/([H_(2)S])`
`9.1 xx 10^(-8) = ((x)(x))/([H_(2)S])`
`(9.1 xx 10^(-8)) (0.1-x) = x^(2)`
Taking `0.1 - xM , 0.1M.`, We have `(9.1 xx 10^(-8))(0.1) = x^(2)`,
`9.1 xx 10^(-9) = x^(2)`
`x = sqrt(9.1 xx 10^(-9))`
`= 9.54 xx 10^(-5) M`
`rArr [HS^(-)] = 9.54 xx 10^(-5) M`
Case II (In the presence of `HCl`) :
In the prensence of `0.1 M` of `HC`l,let `[HS^(-)]` be y M.
`{:("Then",H_(2)S,harr,HS^(-),+,H^(+)),(C_(f),0.1,,0,,0),(C_(f),0.1-y,,y,,y):}`
Also, `{:(HClharr,H^(+),+,Cl^(-)),(,0.1,,0.1):}`
Now, `K_(a_(1)) = ([HS^(-)][H^(+)])/([H_(2)S])`
`K_(a_(2)) = ([y](0.1+y)]/((0.1-y))`
`{:(9.1xx10^(-8)=(yxx0.1)/(0.1),(':'0.1-y,0.1M)),(,(and 0.1+y, 0.1M)):}`
`9.1 xx 10^(-8) = y`
`rArr [HS^(-)] = 9.1 xx 10^(-8)`
(ii) To calculate the concentration if `[S^(2-)]`:
Case I (In the absence of `0.1 M HCl`):
`HS^(-) harr H^(+) + S^(2-)`
`[HS] = 9.54 xx 10^(-5) M` , (From first ionization, case I)
Let `[S^(2-)]` be X.
Also, `[H^(+)] = 9.54 xx 10^(-5) M` , (From first ionization, case I)
`K_(a_(2)) = ([H^(+)][S^(2-)])/([HS^(-)])`
`K_(a_(2)) = ((9.54 xx 10^(-5))(X))/(9.54 xx 10^(-5))`
`1.2 xx 10^(-13) = X = [S^(2-)]`
Case II (in the presence of `0.1 M HCl`) :
Again, let the concentration of `HS^(-)` be X' M.
`[HS^(-)] = 9.1 xx 10^(-8) M` , (From first ionization, case II)
`[H^(+)] = 0.1M` , (From HCl case II)
`[S^(2-)] = X'`
Then, `K_(a_(2)) = ([H^(+)][S^(2-)])/([HS^(-)])`
`1.2 xx 10^(-13) = ((0.1)(X'))/(9.1 xx 10^(-8))`
`10.92 xx 10^(-21) = 0.1 X'`
`(10.92 xx 10^(-21))/(0.1) = X'`
`X' = (1.092 xx 10^(-20))/(0.1)`
`= 1.092 xx 10^(-19) M`
`rArr K_(a_(1)) = 1.74 xx 10^(-5)`