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In equalitative analysis, cations of graph `II` as well as group `IV` both are precipitated in the form of sulphides. Due to low value of `K_(sp)` of group `II` sulphides, group reagent is `H_(2)S` in the presence of dil. `HC1`, and due to high value of `K_(sp)` of group `IV` sulphides, group reagent is `H_(2)S` in the presence of `NH_(4)OH` and `NH_(4)C1`. In a solution containing `0.1M` each of `Sn^(2+), Cd^(2+)`, and `Ni^(2+)` ions, `H_(2)S`gas is passed. `K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 15 10^(-28), K_(sp) of NiS - 3 xx 10^(-21), K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14)` If `H_(2)S` is passed into the above mixture in the presence of `HC1`, which ion will be precipitated first?A. `SnS`B. `CdS`C. `NiS`D. `SnS` and `CdS` (both together) |
Answer» Correct Answer - C `[S^(2-)]_(min)` for `SnS = (K_(sp)(SnS))/([Sn^(2+)])`, `[S^(2-)]_(min)` for `CdS = (K_(sp)(CdS))/([Cd^(2+)])`, `[S^(2-)]_(min)` for `NiS = (K_(sp)(NiS))/([Ni^(2+)])`, and `K_(sp) (NiS) lt K_(sp) (SnS) lt K_(sp) (CdS)` `:. NiS` will precipitate first. |
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