Question

Using sodium hydroxide solution, how would you distinguish:
(a) Ammonium sulphate from sodium sulphate.
(b) Zinc nitrate solution from calcium nitrate solution.
(c) Iron (II) chloride from Iron (III) chloride.

Answer 1

(a) Ammonium sulphate will react with sodium hydroxide to form sodium sulphate, water and ammonia gas. The odour of ammonia can be easily detected and will turn moist red litmus paper blue. On the other hand, sodium sulphate will not give any visible reaction with sodium hydroxide.

$$\begin{gathered} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{NH}}_3+2{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Pungent}\mathrm{smell} \\ \mathrm{Turns}\mathrm{red}\mathrm{litmus}\mathrm{blue} \\ {\mathrm{Na}}_2{\mathrm{SO}}_4+\mathrm{NaOH}\to \mathrm{No}\mathrm{reaction} \end{gathered}$$

(b) Zinc nitrate will react with sodium hydroxide to yield zinc hydroxide, which is white in colour, gelatinous in texture and will remain insoluble in excess sodium hydroxide. On the other hand, calcium nitrate will react with sodium hydroxide to form white calcium hydroxide precipitate, which is not gelatinous in texture but will dissolve in excess sodium hydroxide.

$$\begin{gathered} \mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NaOH}\to \mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2{\mathrm{NaNO}}_3 \\ \mathrm{White}\mathrm{precipitate} \\ \mathrm{Ca}{\left(\mathrm{OH}\right)}_2+2\mathrm{NaOH}\mathit{\left(}excess\mathit{\right)}\to \mathrm{Insoluble} \\ \mathrm{Zn}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NaOH}\to \mathrm{Zn}{\left(\mathrm{OH}\right)}_2+2{\mathrm{NaNO}}_3 \\ \mathrm{White}\mathrm{precipitate} \\ \mathrm{Zn}{\left(\mathrm{OH}\right)}_2+2\mathrm{NaOH}\mathit{\left(}excess\mathit{\right)}\to {\mathrm{Na}}_2\left[\mathrm{Zn}{\left(\mathrm{OH}\right)}_4\right] \end{gathered}$$

(c) Iron (II) chloride will react with sodium hydroxide to give ferrous hydroxide (Fe(OH)₂), which is dirty-green in colour. On the other hand, iron (III) chloride will react with sodium hydroxide to give ferric hydroxide (Fe(OH)₃), which is reddish-brown in colour.

$$\begin{gathered} {\mathrm{FeCl}}_2+2\mathrm{NaOH}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2+2\mathrm{NaCl} \\ \mathrm{Dirty}\mathrm{green}\mathrm{precipitate} \\ {\mathrm{FeCl}}_3+3\mathrm{NaOH}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3+3\mathrm{NaCl} \\ \mathrm{Reddish}\mathrm{brown}\mathrm{precipitate} \end{gathered}$$