Aluminum ion reacts with aqueous ammonia to produce a white gelatinous precipitate of Al(OH)3: \[\ce{Al^{3+}(aq) + 3NH3(aq) + 3H2O(aq) <=> Al(OH)3(s) + 3NH4+(aq)}\]
A strong base, such as \(\ce{NaOH}\), precipitates \(\ce{Al(OH)3}\), which is amphoteric and dissolves in an excess of hydroxide or in acids.
\[\ce{Al^{3+}(aq) + 3OH^-(aq) <=> Al(OH)3(s)}\]
\[\ce{Al(OH)3(s) + OH-(aq) <=> Al(OH)4-(aq)}\]
\[\ce{Al(OH)3(s) + 3H+(aq) <=> Al3+(aq) + 3H2O(l)}\]
The dye aluminon is adsorbed by the gelatinous \(\ce{Al(OH)3}\) precipitate to form a red "lake" and a colorless solution. Although this reaction is not suitable for separation of aluminum ion, it can be used as a confirmatory test for \(\ce{Al^{3+}}\) after precipitation of \(\ce{Al(OH)3}\) with aqueous ammonia.
\(\ce{Cl^-}\), \(\ce{SO_4^{2-}}\)
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Solution:
Part A:
Aluminum chloride react with sodium hydroxide to produce aluminium hydroxide and sodium chloride.
The balanced chemical equation:
AlCl3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq)
Ionic equation: Al3+(aq) + 3Cl-(aq) + 3Na+(aq) + 3OH-(aq) → Al(OH)3(s) + 3Na+(aq) + 3Cl-(aq)
Net ionic equation: Al3+(aq) + 3OH-(aq) → Al(OH)3(s)
Part B:
The balanced chemical equation:
AlCl3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq)
According to the chemical equation:
1) n(NaOH) = n(NaCl) = 2.0 mol
Moles of NaCl = 2.0 mol
2) n(NaOH)/3 = n(Al(OH)3)
Moles of Al(OH)3 = Moles of NaOH / 3 = (2.0 mol) / 3 = 0.67 mol
0.67 mol of aluminum hydroxide (Al(OH)3) and 2.0 mol of sodium chloride (NaCl) would be produced.
Aqueous aluminium chloride solution reacts with aqueous sodium hydroxide and give results according to the amount of available
sodium hydroxide to react with Aluminium chloride. This reaction can be used to identify aluminium ion from other several cations
in qualitative analysis. In this tutorial, we will learn how these reactions happen in detail.
Slowly adding sodium hydroxide to aluminium chloride solution
If, sodium hydroxide is slowly added to aqueous aluminium chloride solution, a white precipitate of aluminium hydroxide (Al(OH)3) is given. This reaction is not a redox reaction because oxidation numbers of atoms are not changed during the reaction.
AlCl3(aq) + NaOH(aq) → Al(OH)3(s) + NaCl(aq)
Balanced equation can be written as below.
AlCl3(aq) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq)
If more sodium hydroxide is added to the solution which contains aluminium hydroxide precipitate, precipitate will dissolve and give a sodium aluminate, a colourless solution.
Aluminium kozijnen
Al(OH)3(s) + NaOH(aq) → NaAl(OH)4
- You can observe the same reaction of solid sodium hydroxide pellets or powder is added to the aqueous aluminium chloride solution. But, keep in mind that adding large amount of sodium hydroxide will not give the observation of white precipitate forming because large amount of hydroxyl ion causes to form sodium aluminate, colourless aqueous solution.
- As well as, you can add solid state aluminium chloride to aqueous sodium hydroxide solution. Most times, it will not give the white precipitate because, there are excess hydroxyl ion to form sodium aluminate.
- All alkali metal cations except Lithium can be identified because those alkali metal cations do not make precipitates with NaOH. Lithium form a white precipitate, but not soluble in excess NaOH as Aluminium cation does.
- Because hydroxides of magnesium and calcium is not soluble in excess NaOH, magnesium and calcium also can be identified from aluminium +3 ion.
- Zinc +2 cation cannot be identified from aluminium +3 because zinc hydroxide is a white precipitate and soluble in excess NaOH.
Questions
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