question_answer 1)

The concentration of sulphide ion in \[0.1\,M\,HCl\] solution saturated with hydrogen sulphide is \[1\times {{10}^{-19}}M.\] If 10 mL of this is added to 5 mL of 0.04 M solution of the following : \[FeS{{O}_{4}},\,MnC{{l}_{2}},\,ZnC{{l}_{2}}\,and\,CdC{{l}_{2}}.\] In which of these solutions precipitation will take place? Given \[{{K}_{sp}}FeS=6.3\times {{10}^{-18}};\,\,{{K}_{sp}}MnS=2.5\times {{10}^{-13}}\] \[{{K}_{sp}}ZnS=1.6\times {{10}^{-24}};\,\,\,{{K}_{sp}}CdS=8\times {{10}^{-27}}\]

Answer:

We know that, precipitation takes place, when the ionic product exceeds solubility product. Concentration of \[{{S}^{2-}}\] ion after mixing will be: \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] \[1\times {{10}^{-19}}\times 10={{M}_{2}}\times 15\] \[{{M}_{2}}=6.67\times {{10}^{-20}}\,M\] Concentration of metal ions on mixing will be: \[[F{{e}^{2+}}]=[M{{n}^{2+}}]=[Z{{n}^{2+}}]=[C{{d}^{2+}}]\] \[=\frac{5\times 0.04}{15}=1.33\times {{10}^{-2}}M\] Ionic product of these metal sulphides will be \[=1.33\times {{10}^{-2}}\times 6.67\times {{10}^{-20}}\] \[=8.87\times {{10}^{-22}}\,{{M}^{2}}\] Then, only Zns and CdS will precipitate because their ionic products are greater than corresponding solubility products.