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Distinguish between the following pairs of compounds using the tests given within brackets:
A lead salt and a zinc salt (using excess ammonium hydroxide)

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Last updated date: 23rd Apr 2024
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Answer
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Hint: Consider the solubility of metal hydroxides in presence of excess ammonium hydroxide.

Complete step by step answer:
Both lead salt and zinc salt react with ammonium hydroxide to form white precipitate of \[{\rm{M}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\]. Here, M represents lead or zinc metal. But in presence of excess ammonium hydroxide, the precipitate of zinc hydroxide dissolves. But the precipitate of lead hydroxide does not dissolve in presence of excess of ammonium hydroxide.
Thus, zinc and lead salts can be distinguished from the solubility of their hydroxides in presence of excess of ammonium hydroxide.

Write the balanced chemical equations for the reaction of zinc and lead ions with ammonium hydroxide to form metal di hydroxides.

\[\begin{array}{l}
{\rm{P}}{{\rm{b}}^{2 + }}{\rm{ + 2 N}}{{\rm{H}}_4}{\rm{OH }} \to {\rm{ Pb}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \downarrow {\rm{ + 2 NH}}_4^ + {\rm{ }}\\
{\rm{Z}}{{\rm{n}}^{2 + }}{\rm{ + 2 N}}{{\rm{H}}_4}{\rm{OH }} \to {\rm{ Zn}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \downarrow {\rm{ + 2 NH}}_4^ + {\rm{ }}
\end{array}\]

Chalky white \[{\rm{Pb}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\] precipitate is insoluble in presence of excess ammonium hydroxide whereas gelatinous white \[{\rm{Zn}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}\] precipitate is soluble in presence of excess ammonium hydroxide.

Note:
Do not distinguish simply based on the colour of precipitate. The hydroxides of lead and zinc are white precipitates, with small difference in shades. To distinguish, consider different solubilities of zinc and lead hydroxide in presence of excess ammonium hydroxide.
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