InterviewSolution
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InterviewSolution
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Assertion (A): A solution contains `0.1M` each of `pB^(2+), Zn^(2+),Ni^(2+)`, ions. If `H_(2)S` is passed into this solution at `25^(@)C`. `Pb^(2+), Ni^(2+), Zn^(2+)` will get precpitated simultanously. Reason (R): `Pb^(2+)` and `Zn^(2+)` will get precipitated if the solution contains `0.1M HCI`. `[K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]`A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - A Here `0.1M H_(2)S` is passed into the solution directly. Selectively precipitation of `Pb^(2+), Zn^(2+)` and `Ni^(2+)` is not possible, as the ionic product to their respective sulphides is more than their `K_(sp)` values and `S^(-2)` ion is not added slowely or drowise. If `H_(2)S` is passed into the solution contianing only `H_(2)O`, then `[S^(2-)] ~~K_(2) of H_(2)S = 10^(-14)M`. But in `0.1M HCI` solution. `K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/(H_(2)S)` `10^(-7)xx 10^(-14) = ((0.1)[S^(2-)])/(0.1) :. [S^(2-)] = 10^(-20)M`. `:. Q_(sp)` of `PbS` and `ZnS(0.1 xx 10^(-20)) gt K_(sp) of PbS` and `Zns` (Hence both precipitate) `Q_(sp) of NiS (10^(-21)) lt K_(sp)` of `NiS` (Hence do not precipitate). |
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