29@Manmay..the answer shud be D..coz wen Co2+ and Ni2+ will be precipitated the Sulphides of the preceeding grp will also get precipitated due to their less solubility product...
An aqueous solution of a given salt was made alkaline with solid NH4cl and NH4OH solution. H2S gas was then passed through the test tube. A black precipitate appeared. Which of the following conclusions is most appropriate about the precipitate. It could be
A) CuS or PbS B) CoS or N i S
C) FeS or N i S or CoS D) FeS or N i S or CoS or CuS or PbS
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7 Answers
1ITS (b)
Co 2+ and Ni 2+ belongs to group 4 basic radicals which r precipitated by the reagent as u mentioned in the question.
NH4Cl supresses the ionization of NH4OH by common ion effect to avoid precipitation of the hydroxides
NH4OH enhances the ionization of H2S by fixing up OH- ion and the conc. of S2- ion is increased
291BUT THE REAGENT IS DIFFERENT FOR PRECEEDING GROUPS TO BE PRECIPITATED.....
4ANS IS D
But even I have the same doubt as Manmay's
Govind plzz elaborate a bit
29See the basic reason behind making the cationic grps in Qualitative Analysis is that the solubility product of different compounds are different...so they are arranged on the basis of their solubility product..
the halide compounds in grp 1 have low solubility product compared to the halide compounds of other grp so they are precipitated first..the process is followed step by step..first grp 1 reagents are added then grp 2..if we get precipitate at any step we stop there and check fr confirmatory test...so solubility product of sulphides of grp 2 and grp 3 is less than the solubility product of sulphides of grp 3..so wen S2- is added in excess as in this case so sulphides of grp 4 and preceeding grp will get precipitated....
Wen H2S is added in the presence of HCl the conc of S2- is less due to commom ion effect
but wen H2S is added in the presence of NH4OH and NH4Cl the conc of S2- is more coz the equilibrium is forward shifted
