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Syllabus
Syllabus
(j) reactions, including ionic equations, and the accompanying colour changes of aqueous Cu2+, Fe2+, Fe3+, Mn2+ and Cr3+ with aqueous Sodium Hydroxide and aqueous Ammonia, including:
(i) precipitation reactions
(ii) complex formation with excess aqueous Sodium Hydroxide and aqueous Ammonia
{For precipitation, non-complexed formulae or complexed formulae, are acceptable e.g. Cu2+(aq) or [Cu(H2O)6] 2+; Cu(OH)2(s) or Cu(OH)2(H2O)4.}
{With excess NaOH, only Cr(OH)3 reacts further forming [Cr(OH)6] 3–.}
{With excess NH3, only Cr(OH)3 and Cu(OH)2 react forming [Cr(NH3)6] 3+ and [Cu(NH3)4(H2O)2] 2+ respectively}
What does this mean?
What does this mean?
We already saw in the Ligand substitution section that when ligands are exchanged it is possible to form complex ions with no overall charge.
These precipitate.
But if more ligand substitution occurs and the complex regains a charge it will re-dissolve
One way to do this is to react complex ions with bases, unfortunately you are required to recall not just the colours but the formulas of any complexes formed.
Copper (II)
Copper (II)
Aqueous Copper (II) ions form a Hexaqua complex ( [Cu(H2O)6]2+(aq)) with a 2+ overall charge because the water ligands are neutral:
[Cu(H2O)6]2+(aq) + 2OH-(aq)--> [Cu(H2O)4(OH)2](s) + 2H2O(l)
[Cu(H2O)6]2+(aq) + 2NH3(aq) --> [Cu(H2O)4(OH)2](s) + 2NH4+(aq)
Examiners are quite happy for you to write the formula as it is above, or simple as Cu(OH)2.
Notice that the same precipitate is formed in both cases - it will be the same blue colour as the hexaqua copper solution but often changes colour throughout on standing.
This tells us that the OH- ions are not being exchanged for water ligands in the top reaction - they are acting as bases (H+ acceptors) just like the Ammonia in the bottom equation.
OH- + H+ --> H2O & NH3 + H+ --> NH4+
In excess ammonia Le Chatlier's principle suggests that the high concentration of ammonia will be reduced by substituting some of the ligands in the precipitate. Why only four? Why doesn't this happen in excess OH-? Who cares, you just have to learn this and that the colour of the solution that is formed will be dark blue.
[Cu(H2O)4(OH)2](aq) + 4NH3(aq) --> [Cu(H2O)2(NH3)4]2+(aq) + 2 H2O(l) + 2OH-(aq)
Light Blue ppt Dark blue solution
Notice that the [Cu(H2O)2(NH3)4]2+ ion has a charge of 2+ because all the ligands are neutral and that this explains why it dissolves (though not why all the ligands aren't replaced.
Iron (II)
Iron (II)
Aqueous Iron (II) ions form a grey-green Hexaqua complex ( [Fe(H2O)6]2+(aq)) with a 2+ overall charge because the water ligands are neutral:
[Fe(H2O)6]2+(aq) + 2OH-(aq)--> [Fe(H2O)4(OH)2](s) + 2H2O(l)
[Fe(H2O)6]2+(aq) + 2NH3(aq) --> [Fe(H2O)4(OH)2](s) + 2NH4+(aq)
Notice that this is exactly the same as for Cu2+, although the precipitate will also be green
Examiners are still happy for you to write the formula as it is above, or simple as Fe(OH)2.
In excess ammonia you would expect it to redissolve but this does not happen for Fe2+.
However, the precipitate will darken and go brown at the surface as it is oxidised by the air. You won't be asked for a balanced equation.
Fe(OH)2](s) --> Fe(OH)3](s)
Grey-green ppt Orange-Brown ppt
Iron (III)
Iron (III)
Aqueous Iron (III) ions form a Orange-brown Hexaqua complex ( [Fe(H2O)6]3+(aq)) with a 3+ overall charge because the water ligands are neutral:
[Fe(H2O)6]3+(aq) + 3OH-(aq)--> [Fe(H2O)3(OH)3](s) + 3H2O(l)
[Fe(H2O)6]3+(aq) + 3NH3(aq) --> [Fe(H2O)3(OH)3](s) + 3NH4+(aq)
Notice that this is much the same as for Fe2+, although it takes three moles of base to precipitate because of the 3+ charge and the precipitate will be orange
Examiners are happy for you to write the formula as it is above, or simple as Fe(OH)3.
Again there is no re-dissolving for Fe3+, and no darkening because the precipitate is already in the +3 Oxidation state.
Manganese (II)
Manganese (II)
Aqueous Manganese (II) ions form a pale pink Hexaqua complex ( [Mn(H2O)6]2+(aq)) with a 2+ overall charge because the water ligands are neutral:
[Mn(H2O)6]2+(aq) + 2OH-(aq)--> [Mn(H2O)4(OH)2](s) + 2H2O(l)
[Mn(H2O)6]2+(aq) + 2NH3(aq) --> [Mn(H2O)4(OH)2](s) + 2NH4+(aq)
Notice that this is exactly the same as for Cu2+ and Fe2+, although this time the precipitate will be brown and darkens to black on standing as manganese oxide is made
Examiners are still happy for you to write the formula as it is above, or simple as Mn(OH)2.
It doesn't re-dissolve in excess Ammonia or Sodium Hydroxide.
Chromium (III)
Aqueous Iron (III) ions form a violet Hexaqua complex ( [Cr(H2O)6]3+(aq)) with a 3+ overall charge because the water ligands are neutral:
[Cr(H2O)6]3+(aq) + 3OH-(aq)--> [Cr(H2O)3(OH)3](s) + 3H2O(l)
[Cr(H2O)6]3+(aq) + 3NH3(aq) --> [Cr(H2O)3(OH)3](s) + 3NH4+(aq)
Notice that this is much the same as for Cr3+, although i the precipitate will be grey green
Examiners are happy for you to write the formula as it is above, or simple as Cr(OH)3.
This time the precipitate can be re-dissolved with either excess Sodium Hydroxide or Ammonia - but not in the same way!
[Cr(H2O)3(OH)3](s) + 3OH-(aq) --> [Cr(OH)6]3-(aq) + 3H2O(l)
[Cr(H2O)3(OH)3](s) + 6NH3(aq) --> [Cr(NH3)6]3+(aq) + 3H2O(l) + 3OH-(aq)
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