A) \[P{{b}^{2+}}\]
B) \[F{{e}^{3+}}\]
C) \[Z{{n}^{2+}}\]
D) \[C{{u}^{2+}}\]
Correct Answer: A
Solution :
As \[P{{b}^{2+}}\] is present in both I and II group of qualitative inorganic, analysis hence it gives ppt with both \[HCl\] and \[{{H}_{2}}S\]. \[P{{b}^{2+}}+HCl\xrightarrow{{}}\underset{white\,ppt.}{\mathop{PbC{{l}_{2}}}}\,+{{H}^{+}}\] \[P{{b}^{2+}}+{{H}_{2}}S\xrightarrow{dil\,\,HCl}PbS+{{H}^{+}}\] \[HCl\] suppresses the ionisation of weakly dissociated \[{{H}_{2}}S\]. \[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\] and thus, only the ionic product of the sulphides of group II radicals exceeds their corresponding solubility product and hence, only these are precipitated out.
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