Question

How would you distinguish between ${\mathrm{Zn}}^{2+}$ and ${\mathrm{Pb}}^{2+}$ using ammonium hydroxide solution?

Answer 1

Distinguishing between ${\mathrm{Zn}}^{2+}$and ${\mathrm{Pb}}^{2+}$:-

• When ${\mathrm{Zn}}^{2+}$reacts with Ammonium hydroxide$\left({\mathrm{NH}}_4\mathrm{OH}\right)$, it produces a white gelatinous ppt of Zinc hydroxide$\mathrm{Zn}{\left(\mathrm{OH}\right)}_2$.
• The balanced reaction is:

$\underset{\mathrm{Zinc}\mathrm{ions}}{{\mathrm{Zn}}^{2+}\left(\mathrm{s}\right)}+\underset{\mathrm{Ammonium}\mathrm{hydroxide}}{{\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Zinc}\mathrm{hydroxide}}{\mathrm{Zn}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)(\downarrow )}+\underset{\mathrm{Ammonium}\mathrm{ions}}{2{{\mathrm{NH}}_4}^{+}\left(\mathrm{aq}\right)}$

• This white ppt formed of Zinc hydroxide dissolves in excess addition of Ammonium hydroxide, that is it is soluble in excess Ammonium hydroxide
• When ${\mathrm{Pb}}^{2+}$reacts with Ammonium hydroxide$\left({\mathrm{NH}}_4\mathrm{OH}\right)$, it produces a chalky white ppt of Lead hydroxide$\mathrm{Pb}{\left(\mathrm{OH}\right)}_2$.
• The balanced reaction is:

$\underset{\mathrm{Lead}\mathrm{ions}}{{\mathrm{Pb}}^{2+}\left(\mathrm{s}\right)}+\underset{\mathrm{Ammonium}\mathrm{hydroxide}}{{\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\mathrm{hydroxide}}{\mathrm{Pb}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)(\downarrow )}+\underset{\mathrm{Ammonium}\mathrm{ions}}{2{{\mathrm{NH}}_4}^{+}\left(\mathrm{aq}\right)}$

• This chalky white ppt of Lead hydroxide does not dissolve in excess Ammonium hydroxide, that is it's insoluble in excess Ammonium hydroxide.

Therefore, ${\mathrm{Zn}}^{2+}$when reacts with Ammonium hydroxide, it produces a white gelatinous ppt of Zinc hydroxide which is soluble in excess Ammonium hydroxide, whereas, ${\mathrm{Pb}}^{2+}$ when reacts with Ammonium hydroxide, produces a chalky white ppt of Lead hydroxide which is insoluble in excess Ammonium hydroxide.