Chemistry Practical
The cation that will not be precipitated by $$H$$_2S$$ in the presence of dil $$HCl$$ is:
In $$IV$$ group analysis, $$NH_{4}OH$$ is added before passing $$H_{2}S$$ gas because:
Solubility product of sulphides of group (II) radicals are less than solubility product of sulphides of (IV) group. So to precipitate (II) group radicals as sulphides $$S^{2-}$$ are sufficient. In presence of Dil $$HCl$$ $$S^{2-}$$ are released due to the common ion effects. In (IV) group the ionisation of $$H_2S$$ is favoured by the use of $$NH_4OH$$
Hence option (C) is correct.
The cation that will not be precipitated by $$H$$_2S$$ in the presence of dil $$HCl$$ is:
A solution containing a group-IV cation gives a precipitate on passing $$H_2S$$. A solution of this precipitate in $$dil.HCl$$ produces a white precipitate with basic $$NaOH$$ solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is:
In Nessler's reagent for the detection of ammonia the active species is:
Reagent S is:
A reddish-pink substance on.heating gives off a vapor which condenses on the sides of the test tube and the substance turns blue. If on cooling water is added to the residue, it turns to its original color. The substance is :
Reagent used in the qualitative analysis of IVth group is
Group reagent for analytic group $$IV$$ is
Precipitate of group $$IV$$ cations takes place when $$H_2S$$ is
Mark the correct statement(s)