A) \[2M\,HCl\]
B) \[6M\,N{{H}_{3}}\]
C) \[6M\,NaOH\]
D) \[{{H}_{2}}S\,gas\]
Correct Answer: B
Solution :
\[F{{e}^{3+}},\text{ }Z{{n}^{2+}}\]and\[C{{u}^{2+}}\]ions are present in slightly acidic solution. On adding 6M\[N{{H}_{3}}\]solution, i.e., 6M\[N{{H}_{4}}OH\]we get the following reactions \[F{{e}^{3+}}+O{{H}^{-}}\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} dark\text{ }brown \\ \,\,\,\,\,\,\,\,ppt. \end{smallmatrix}}{\mathop{Fe{{(OH)}_{3}}}}\,\] \[Z{{n}^{2+}}+4N{{H}_{3}}\xrightarrow[{}]{{}}\underset{colourless\text{ }solution}{\mathop{{{[Zn{{(N{{H}_{3}})}_{4}}]}^{2+}}}}\,\] \[C{{u}^{2+}}+4N{{H}_{3}}\xrightarrow[{}]{{}}\underset{deep\text{ }blue\text{ }solution}{\mathop{{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}}}\,\] In this way, dark brown ppt. of\[Fe{{(OH)}_{3}}\]can be separated from\[C{{u}^{2+}}\]and\[Z{{n}^{2+}}\]amine complex solutions in a single step by adding 6M\[N{{H}_{3}}\].
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