Question

Using ammonium hydroxide, how would you distinguish between the following pairs of ions in solution:
(a) Ferrous ion and Ferric ion.
(b) Zinc ion and Lead ion.

Answer 1

(a) Ferrous and ferric ions in salt solutions are distinguished with the help of reaction with ammonium hydroxide. Let us consider their sulphate salt solution in the reaction.

Reaction with ammonium hydroxide
Ferric ion	Ferrous ion
Ferric sulphate (yellow solution) reacts with ammonium hydroxide to form ferric hydroxide, releasing reddish brown precipitate. $$\begin{gathered} {\mathrm{Fe}}_2({\mathrm{SO}}_4{)}_3\left(\mathrm{aq}\right)+6{\mathrm{NH}}_4\mathrm{OH}(\mathrm{aq})\stackrel{}{\to }2\mathrm{Fe}(\mathrm{OH}{)}_3\left(\mathrm{s}\right)+3({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4(\mathrm{aq}) \\ \mathrm{Ferric}\mathrm{sulphate}\mathrm{Ferric}\mathrm{hydroxide} \\ (\mathrm{yellow})(\mathrm{Reddish}\mathrm{brown}\mathrm{ppt}.) \end{gathered}$$	​Ferrous sulphate (light green solution) reacts with ammonium hydroxide to form ferrous hydroxide, releasing dirty green precipitate. $$\begin{gathered} {\mathrm{FeSO}}_4\left(\mathrm{aq}\right)+2{\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)\stackrel{}{\to }\mathrm{Fe}(\mathrm{OH}{)}_2(\mathrm{s})+({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right) \\ \mathrm{Ferrous}\mathrm{sulphate}\mathrm{Ferrous}\mathrm{hydroxide} \\ (\mathrm{light}\mathrm{green})(\mathrm{Dirty}\mathrm{green}\mathrm{ppt}) \end{gathered}$$

(b) Zinc and lead ions in salt solutions are distinguished with the help of reaction with ammonium hydroxide. Let us consider their sulphate salt solution in the reaction.

Reaction with ammonium hydroxide
Zinc ion	Lead ion
Zinc sulphate (colourless solution) reacts with ammonium hydroxide to form zinc hydroxide, releasing gelatinous white precipitate. $$\begin{gathered} {\mathrm{ZnSO}}_4\left(\mathrm{aq}\right)+2{\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)\stackrel{}{\to }\mathrm{Zn}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)+({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right) \\ \mathrm{Zinc}\mathrm{sulphate}\mathrm{Zinc}\mathrm{hydroxide} \\ \left(\mathrm{colourless}\right)(\mathrm{White}\mathrm{gelatinous}\mathrm{ppt}) \end{gathered}$$ By adding excess of ammonium hydroxide, the precipitate formed dissolves and forms complex compound. $$\begin{gathered} \mathrm{Zn}(\mathrm{OH}{)}_2(\mathrm{s})+({\mathrm{NH}}_4{)}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)\stackrel{}{\to }\left[\mathrm{Zn}\right({\mathrm{NH}}_3{)}_4]{\mathrm{SO}}_4(\mathrm{aq})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{l}) \\ \mathrm{Zinc}\mathrm{hydroxide}(\mathrm{in}\mathrm{excess})\mathrm{Tetrammine} \\ (\mathrm{White}\mathrm{gelatinous}\mathrm{ppt})\mathrm{zinc}(\mathrm{II})\mathrm{sulphate} \\ (\mathrm{colourless}) \end{gathered}$$	Lead nitrate (colourless solution) reacts with ammonium hydroxide to form lead hydroxide, releasing white precipitate. $$\begin{gathered} \mathrm{Pb}({\mathrm{NO}}_3{)}_2(\mathrm{aq})+2{\mathrm{NH}}_4\mathrm{OH}(\mathrm{aq})\stackrel{}{\to }\mathrm{Pb}(\mathrm{OH}{)}_2\left(\mathrm{s}\right)+2{\mathrm{NH}}_4{\mathrm{NO}}_3\left(\mathrm{aq}\right) \\ \mathrm{Lead}\mathrm{nitrate}\mathrm{Lead}\mathrm{hydroxide} \\ \left(\mathrm{colourless}\right)(\mathrm{White}\mathrm{ppt}) \end{gathered}$$ The precipitate formed is insoluble in excess of ammonium hydroxide.