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.
You need to login to perform this action.
You will be redirected in
3 sec