Understanding the acid-base behaviour of aqueous transition metal ions

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In aqueous solutions of transition metal ion salts, the metal ion is invariably part of a complex known as an aqua ion. Six water molecules act as ligands forming an octahedral complex e.g. [Ni(H2O)6]2+(aq), [Fe(H2O)6]2+(aq), [Co(H2O)6]2+(aq).

Ions with an oxidation state equal to or greater than +3 have a high charge density. This causes polarisation of the O-H bond in one or more water ligands. H+ is ‘lost’ to a free water molecule in the solution and the H2O ligand becomes an OH– ligand.

[Fe(H2O)6]3+(aq) + H2O(l) ⇾ [Fe(H2O)5OH]2+(aq) + H3O+(aq)

This reaction is known as hydrolysis and the resulting aqueous solution is acidic because of the H3O+(aq).

Acid-base reactions

  1. with hydroxide ions

If we add hydroxide ions to a transition metal aqua ion complex, the OH– ion removes H+ from water ligands in a step-by-step manner until the complex is neutral. The insoluble metal hydroxide / uncharged complex precipitates out of solution.

[Fe(H2O)6]3+(aq) + OH–(aq) ⇾ [Fe(H2O)5OH]2+(aq) + H2O(l)

[Fe(H2O)5OH]2+(aq) + OH–(aq) ⇾ [Fe(H2O)4(OH)2]+(aq) + H2O(l)

[Fe(H2O)4(OH)2]+(aq) + OH–(aq) ⇾ [Fe(H2O)3(OH)3](s) + H2O(l)

[Fe(H2O)3(H)3](s) is effectively Fe(OH)3(s) which is an insoluble gelatinous orange-brown precipitate.

  1. with aqueous ammonia

Ammonia solution added to an aqua ion will have a similar effect to the addition of a hydroxide ion, with a proton being removed from water ligands in a step-by-step manner until the complex is neutral. Once again, the uncharged complex will precipitate out of solution.

[Fe(H2O)6]2+(aq) + NH3(aq) ⇾ [Fe(H2O)5OH]+(aq) + NH4+(aq)

[Fe(H2O)5OH]+(aq) + + NH3(aq) ⇾ [Fe(H2O)4(OH)2](s) + NH4+(aq)

[Fe(H2O)4(OH)2](s) is effectively Fe(OH)2(s) which is an insoluble gelatinous green precipitate.

  1. with carbonate ions

Metal carbonates such as sodium carbonate are weak bases and they will behave as such if added to the Fe3+ hexa aqua ion because, as we saw above, this aqua ion is acidic in solution.

This reaction also happens step-by-step with carbonate ions forming first hydrogen carbonate ions and then carbon dioxide.

[Fe(H2O)6]3+(aq) + 3CO32-(aq) ⇾ [Fe(H2O)3(OH)3](s) + 3HCO3–(aq)

[Fe(H2O)6]3+(aq) + 3HCO3–(aq) ⇾ [Fe(H2O)3(OH)3](s) + 3CO2(g) + 3H2O(l)

And if we combine these two equations, the HCO3– ions cancel out to give us …

2[Fe(H2O)6]3+(aq) + 3CO32-(aq) ⇾ 2[Fe(H2O)3(OH)3](s) + 3CO2(g) + 3H2O(l)

[Fe(H2O)6]3+(aq) is a yellow brown solution in which we would observe the brown precipitate of [Fe(H2O)3(OH)3] forming as the carbonate ions are added as well as effervescence (bubbles) due to the carbon dioxide.

The Fe2+ hexa aqua ion does NOT behave in the same way when aqueous sodium carbonate is added to it. The carbonate ion is not a strong enough base to remove H+ from water ligands in this complex ion (remember that the H+ is more strongly held in the ligand as Fe2+ is not as polarising as Fe3+).

[Fe(H2O)6]2+(aq) + CO32-(aq) ⇾ FeCO3(s) + 6 H2O(l)

We see the insoluble blue-green precipitate of FeCO3(s) form and there is NO effervescence.

Practice questions

  1. When aqueous sodium hydroxide is added dropwise to aqueous iron(II) nitrate, a pale green precipitate forms which changes colour to brown when left to stand in air.

(a) Identify the pale green precipitate and write an ionic equation for the reaction

(b) Explain why the pale green precipitate changes colour and give the formula of the brown precipitate

  1. A spatula of NH4Fe(SO4)2•12H2O is added to a solution of sodium hydroxide and the test tube is then warmed. Damp red litmus paper placed at the mouth of the test tube turns blue and a brown precipitate forms. Explain these observations.
  1. A solution of [Fe(H2O)6]2+ is split into two test tubes. Aqueous sodium carbonate is added to the first test tube. The second test tube is left to stand in air for several hours and then aqueous sodium carbonate is added. Describe the observations you would expect to see for each test tube and identify the products of any reactions.
  2. Explain why a solution containing [Al(H2O)6]3+ ions has a pH < 7.
Extension
  1. Vanadium chloride, VCl4, dissolves in water with a violent reaction. The resulting solution is strongly acidic and contains oxovanadium(IV) ions, VO2+. Suggest an explanation for these observations given that vanadium complexes have a coordination number of 6.
  2. Show how the manganate(VII) ion, MnO4–, could be derived from the interaction between the manganese(VII) ion, Mn7+, and water molecules. Why is the MnO4– ion NOT regarded as a complex ion?

Answers

  1. (a) the pale green precipitate is Fe(OH)2 or [Fe(H2O)4(OH)2]

Fe2+(aq) + 2OH–(aq) ⇾ Fe(OH)2(s) or [Fe(H2O)6]2+(aq) + 2OH–(aq) ⇾ [Fe(H2O)4(OH)2](s) + 2H2O(l)

(b) Fe2+ is oxidised to Fe3+ by the oxygen in the air. The red brown precipitate is Fe(OH)3 or [Fe(H2O)3(OH)3].

2. The salt dissolves to form NH4+ ions, Fe3+ / [Fe(H2O)6]3+ ions and SO42- ions. The Fe3+ forms a brown precipitate of Fe(OH)3 / [Fe(H2O)3(OH)3] and the litmus paper turning blue suggest ammonia gas is released.

3. In the first test tube we would see a blue-green precipitate of FeCO3 form. In the second test tube we would see a brown precipitate of Fe(OH)3 / [Fe(H2O)3(OH)3] form and bubbles of carbon dioxide.

4. The Al3+ ion has a high charge density so it will polarise and weaken the O-H bond in the water ligands. H+ ions are released forming [Al(H2O)5(OH)]2+ ions.

5. When vanadium chloride dissolves in water the V4+ ions form an octahedral complex with water ligands. However the very high charge density of the V4+ ion means that it is has an exceptionally strong polarising effect on the water ligands and two H+ are rapidly lost from the same water molecule.

[V(H2O)6]4+(aq) + H2O(l) ⇾ [V(H2O)5OH]3+(aq) + H3O+(aq)

[V(H2O)5OH]3+(aq) + H2O(l) ⇾ [V(H2O)4O]2+(aq) + H3O+(aq) + H2O(l)

[V(H2O)4O]2+ is more commonly written as VO2+ (it is not known why two H+ are lost from the same water ligand rather than from adjacent ligands as is the case when Fe3+ complexes dissolve in water).

  1. If we assume that Mn7+ coordinates with four water molecules in solution and then the very strong polarising effect of the ion leads to the loss of all 8 H+, the resulting ion would be MnO4–. In a complex ion the central metal cation and coordinated ligands can have an independent existence in solution (as is the case with ligand exchange reactions) but in MnO4– neither the Mn7+ nor the O2- ions are free in solution.