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The gas liberated on heating a mixture of two salts with NaOH, gives a reddish brown precipitate with an alkaline solution of K_(2)Hgl_(4). The aqueous solution of the mixture on treatment with BaCl_(2) gives a white precipitate which is sparingly soluble in conc. HCI. On heating the mixture with K_(2)Cr_(2)O_(7) and conc. H_(2)SO_(4) red vapours A are produced. The aqueous solution of the mixture gives a deep blue colouration B with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of A and B. |
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Answer» Solution :`NH_(4^(*)),Fe^(2+),SO_(4^(2-)),CI^(-)` Formation of a reddish brown PPT with alkaline `K_(2)Hgl_(4)` solution indicates the presence of radical ammonium `NH_(4^(+))` and the gas liberated in ammonia. The REACTIONS are as: `NH_(4)^(+)OH^(-)toNH_(3)+H_(2)O` `K_(2)Hgl_(4)to2KI+Hgl_(2)` `Hgl_(2)+2NH_(3)toHg(NH_(2))l+NH_(4)l` `Hg(NH_(2))l+HgltoNH_(2)Hg_(2)ł_(3)` On treatment with Bacla, a white ppt is formed which indicates the presence of sulphate anion `(SO_(4)^(2-))` sparingly soluble in conc. HCI. With `K_(2)Cr_(2)O_(7)` and conc. `H_(2)SO_(4)` RED vapours are evolved which indicates the presence of chloride ION `(Cl^(-))`. The reaction is as: `K_(2)Cr_(2)O_(7)+4Cl^(-)+underset(conc.)(6H_(2)SO_(4))tounderset(A)(2CrO_(2)Cl_(2))+2KHSO_(4)+4HSO_(4)^(-)+3H_(2)O` On treatment with potassium ferricyanide indicates the presence of ferrous ions. `[Fe(CN)_(6)]^(3-)toFe^(3+)+[Fe(CN)_(6)]^(4)` `4Fe^(3+)+3K_(4)[Fe(CN)_(6)]toFe_(4)underset(deep blue)([Fe(CN)_(6)]_(3))+12K^(+)` |
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