Question

In qualitative analysis, cations of group II as well as group IV precipitated in the form of sulphides. Due to low value of `Ksp` of group `II` sulphides, group reagent is `H_(2)S` in presence of dil. `HCl` and due to high value of `Ksp` of group IV sulphides, group reagent is `H_(2)S` in presence of `NH_(4)OH` and `NH_(4)Cl`.
In `0.1M H_(2)S` solution, `Sn^(2+), Cd^(2+)` and `Ni^(2+)` ions are present in equimolar concentration `(0.1M)`. Given : `Ka_(1)(H_(2)S) = 10^(-7), Ka_(2)(H_(2)S) = 10^(-14), K_(sp)(SnS) = 8 xx 10^(-29) , K_(sp) (CdS) = 10^(-28), K_(sp)(NiS) = 3 xx 10^(-21)`
At what pH precipitate of NiS will form
A. `12.76`
B. `7`
C. `1.24`
D. `4`

Answer 1

Correct Answer - C
`k_(sp) (NiS) = 3 xx 10^(-21) :. [S^(-2)] = (3 xx 10^(-21))/(0.1)`
`{:(H_(2)ShArrH^(+)+HS^(-1),k_(a_(1))=10^(-7)),(HS^(-1)hArrH^(+)+S^(-2),k_(a_(2))=10^(-14)):}`
`k_(a_(2)) = ([H^(+)][S^(2)])/([HS^(-1)])`
`k_(a_(1)).k_(a_(2)) = 10^(-21)([H^(+)]^(2)[3xx10^(-20)])/([0.1])`
`[H^(+)] =sqrt((1)/(300)):.pH = 1.2388`