15.7 - Reactions of Transition Metal Ions with Aqueous Alkalis
- 1Precipitation reactions
- 2Amphoteric behaviour of chromium hydroxide
Precipitates of transition element hydroxides
In aqueous solutions, the transition elements exist as the hydrated ions [M(H2O)6]n+. These can also be written simply as Mn+(aq) where M is the metal.
When aqueous solutions containing transition element ions are mixed with sodium hydroxide (NaOH) or aqueous ammonia (NH3), coloured metal hydroxide precipitates form. In these reactions, the hydroxide or ammonia ligands substitute the water ligands in the metal aqua ions.
The equation for the precipitation of M2+ and M3+ aqua ions with NaOH(aq) is:
- M2+: [M(H2O)6]2+(aq) + 2OH-(aq) ➔ M(H2O)4(OH)2(s) + 2H2O(l)
- M3+: [M(H2O)6]3+(aq) + 3OH-(aq) ➔ M(H2O)3(OH)3(s) + 3H2O(l)
The equation for the precipitation of M2+ and M3+ aqua ions with NH3(aq) is:
- M2+: [M(H2O)6]2+(aq) + 2NH3(aq) ➔ M(H2O)4(OH)2(s) + 2NH4+(aq)
- M3+: [M(H2O)6]3+(aq) + 3NH3(aq) ➔ M(H2O)3(OH)3(s) + 3NH4+(aq)
For some metal ions like Cu2+, Co2+ and Cr3+, excess ammonia or excess sodium hydroxide causes the precipitate to react further to form soluble charged complex ions containing NH3 or OH- ligands.
You need to know the formulae and colours of the precipitates formed when Cu2+, Fe2+, Fe3+, Co2+ and Cr3+ metal aqua ions undergo precipitation reactions with aqueous NaOH and NH3.
Copper(II)
- With NaOH or NH3, a pale blue precipitate of Cu(H2O)4(OH)2 forms from the pale blue aqua ion.
- The precipitate remains insoluble with excess NaOH.
- With excess NH3, a deep blue solution of [Cu(NH3)4(H2O)2]2+ forms: Cu(H2O)4(OH)2(s) + 4NH3(aq) ➔ [Cu(NH3)4(H2O)2]2+(aq) + 2OH-(aq) + 2H2O(l)
Iron(II)
- With NaOH or NH3, a dark green precipitate of Fe(H2O)4(OH)2 forms from the pale green aqua ion.
- The precipitate remains insoluble with excess NaOH and excess NH3.
Iron(III)
- With NaOH or NH3, a brown precipitate of Fe(H2O)3(OH)3 forms from the yellow aqua ion.
- The precipitate remains insoluble with excess NaOH and excess NH3.
Cobalt(II)
- With NaOH or NH3, a blue precipitate of Co(H2O)4(OH)2 forms from the pale pink aqua ion.
- The precipitate remains insoluble with excess NaOH.
- With excess NH3, a yellow-brown solution of [Co(NH3)6]2+ forms: Co(H2O)4(OH)2(s) + 6NH3(aq) ➔ [Co(NH3)6]2+(aq) + 2OH-(aq) + 4H2O(l)
Chromium(III)
-
With NaOH or NH3, a dark green precipitate of Cr(H2O)3(OH)3 forms from the green aqua ion.
-
With excess NaOH, a dark green solution of [Cr(OH)6]3- forms: Cr(H2O)3(OH)3(s) + 3OH-(aq) ➔ [Cr(OH)6]3-(aq) + 3H2O(l)
-
With excess NH3, a purple solution of [Cr(NH3)6]3+ forms: Cr(H2O)3(OH)3(s) + 6NH3(aq) ➔ [Cr(NH3)6]3+(aq) + 3OH-(aq) + 3H2O(l)
Amphoteric metal hydroxides
Some metal hydroxide precipitates formed in these reactions display amphoteric behaviour:
- They can react with acids by accepting H+ ions (acting as a Brønsted-Lowry base).
- They also react with excess OH- ions by donating H+ ions (acting as a Brønsted-Lowry acid).
Chromium hydroxide, Cr(H2O)3(OH)3, is amphoteric because it reacts with both acids and bases:
- Reaction with acids: Cr(H2O)3(OH)3(s) + 3H+(aq) ➔ [C(H2O)6]3+(aq)
- Reaction with bases: Cr(H2O)3(OH)3(s) + 3OH-(aq) ➔ [Cr(OH)6]3-(aq) + 3H2O(l)
This amphoteric behaviour explains why Cr(OH)3 dissolves in excess sodium hydroxide.
Summary of precipitation reactions
The observations of the precipitation reaction of metal aqua ions with OH- and NH3 are summarised in the table below:
| Metal aqua ion | With OH-(aq) or NH3(aq) | With excess OH-(aq) | With excess NH3(aq) |
|---|---|---|---|
| Cu2+ | Blue precipitate of Cu(H2O)4(OH)2 | No change | Deep blue solution of [Cu(NH3)4(H2O)2]2+ |
| Fe2+ | Green* precipitate of Fe(H2O)4(OH)2 | No change | No change |
| Fe3+ | Brown precipitate of Fe(H2O)3(OH)3 | No change | No change |
| Co2+ | Blue* precipitate of Co(H2O)4(OH)2 | No change | Yellow-brown solution of [Co(NH3)6]2+ |
| Cr3+ | Dark green precipitate of Cr(H2O)3(OH)3 | Dark green solution of [Cr(OH)6]3- | Purple solution of [Cr(NH3)6]3+ |
*goes brown standing in air.