Question

Aqueous ammonia is used as precipitating reagent for ${ Al }^{ 3+ }$ ions as ${ Al }{ (OH) }_{ 3 }$ rather than aqueous NaOH because:
(a) ${ NH }_{ 4 }^{ + }$ is a weak base
(b) NaOH is a strong base
(c) NaOH forms $\left[ { Al(OH) }_{ 4 } \right] ^{ - }$ ions.
(d) NaOH forms $\left[ { Al(OH) }_{ 2 } \right] ^{ + }$ ions.

Answer 1

Hint: In order to solve this question we need to use our knowledge of the salt analysis. The aluminium metal cation lies in group 3 for the basic radicals of salt analysis. The preliminary test for the detection of the aluminium metal cations consists of precipitating the cations in the form of a white precipitate of aluminium hydroxide.

Complete step by step answer:
Before solving this question let us first understand some of the properties of aluminium metal. The most common oxidation state of aluminium is +3. Its melting point is $ 648^{ o }{ C }$ while its boiling point is $1800^{ o }{ C }$. It is a silvery, soft metal. Though it is highly active but it reacts with the atmospheric oxygen in order to form a coating of aluminium oxide which protects the metal.
Now, for the detection of metal ions, we follow the scheme of the salt analysis. The aluminium metal cation lies in the group 3 for the salt analysis scheme along with ferric and ferrous ions.
When the salt solution, a solution of aqueous ammonia is added, we get a gelatinous white precipitate of aluminium hydroxide. Aluminium hydroxide is amphoteric in nature. The reaction is given below:
$ { Al }^{ 3+ }(aq)+3{ NH }_{ 3 }(aq)+3{ H }_{ 2 }O(l)\rightleftharpoons { Al(OH) }_{ 3 }(s)\downarrow +3{ NH }_{ 4 }^{ + }(aq)$
If we add a strong alkali such as a solution of sodium hydroxide then again a white precipitate of aluminium hydroxide is formed but it dissolves if we add excess of the sodium hydroxide solution or an acid. The reactions are shown below:

$ { Al }^{ 3+ }(aq)+3{ OH }^{ - }(aq)\rightleftharpoons { Al(OH) }_{ 3 }(s)\downarrow $
$ { Al(OH) }_{ 3 }(s)+{ OH }^{ - }(aq)\rightleftharpoons [Al{ (OH) }_{ 4 }{ ] }^{ - }(aq)$
$ { Al(OH) }_{ 3 }(s)+3{ H }^{ + }(aq)\rightleftharpoons { Al }^{ 3+ }(aq)+3{ H }_{ 2 }O(l)$
Therefore, aqueous ammonia is used as a precipitating agent for ${ Al }^{ 3+ }$ ions as ${ Al }{ (OH) }_{ 3 }$ rather than aqueous NaOH.
So, the correct answer is “Option C”.

Note: For confirming the presence of the aluminium ions we can use the dye aluminon. The gelatinous precipitate aluminium hydroxide absorbs the dye to give a red lake floating over a colourless solution