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The only cations present in a slightly acidic solution are `Fe^(3+),Zn^(2+) and Cu^(2+)`. The reagent that when added in excess to this solution would identify the separate `Fe^(3+)` in one step isA. 2M HClB. 6M `NH_(3)`C. 6 NaOHD. `H_(2)S` gas |
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Answer» Correct Answer - B `Fe^(3+)` is a third group radical, whose reagent is `NH_(4)OH` in presence of `NH_(4)Cl`. So if 6M `NH_(3)` is added in the slightly acidic (HCl) solution of ions it will lead to the `Fe^(3+)+Zn^(2+)+Cu^(2+) overset("6 M"NH_(3))to underset("Brown ppt.")(Fe(OH)_(3))+underset("Soluble")([Zn(NH_(3))_(4)]^(2+))+underset("Soluble")([Cu(NH_(3))_(4)]^(2+))` |
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