Question

The only cations present in a slightly acidic solution are $F{e}^{3+},Z{n}^{2+}$ and $C{u}^{2+}$ . The reagent that when added in excess to this solution would identify and separate $F{e}^{3+}$ in one step is

• A. $2MHCl$
• B. $6MN{H}_3$
• C. $6MNaOH$
• D. $H_{2}S$ gas

Answer 1

Answer: (B)

$F{e}^{3+},Z{n}^{2+}$ and $C{u}^{2+}$ ions are present in slightly acidic solution. On adding $6MN{H}_3$ solution, i.e., $6MN{H}_{4}OH$ we get the following reactions
$F{e}^{3+}+3O{H}^{-}⟶\underset{\text{dark brown ppt.}}{Fe(OH{)}_3}$
$Z{n}^{2+}+4N{H}_3⟶\underset{\text{colourless solution}}{[Zn(N{H}_3{)}_4{]}^{2+}}$
$C{u}^{2+}+4N{H}_3\to \underset{\text{deep blue solution}}{[Cu(N{H}_3{)}_4{]}^{2+}}$
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 $6MN{H}_3$.