Inorganic answer 2

The reaction of copper with nitric acid is complicated and depends on the concentration of the acid and its temperature. With 50% acid the initial reaction, to give copper(II) nitrate A and nitrogen dioxide B is:

Cu + 4HNO3 à Cu(NO3)2 + 2NO2 + 2H2O.

If sodium hydroxide is added to a solution of copper(II) nitrate, a pale blue precipitate is given, usually represented as copper(II) hydroxide.

Cu(NO3)2 + 2NaOH à Cu(OH)2 + 2NaNO3

The reaction is actually with the hydrated copper ions themselves, which are deprotonated:

[Cu(H2O)6]2+ + 2OH- à Cu(OH)2 + 6H2O.

However, if as here the sodium hydroxide is added to the solution of the copper(II) salt, the precipitate is not the hydroxide, but the basic nitrate Cu(OH)2.Cu(NO3)2 (or sulphate, or chloride, depending on the copper(II) salt employed). To get the true hydroxide, the copper(II) salt has to be added to sodium hydroxide solution. The business of hydroxide precipitates is convoluted to say the least - take a look at my article on the subject. For the purposes of this answer, C is copper(II) hydroxide.

On heating (even in aqueous suspension) copper(II) hydroxide loses water to give G, copper(II) oxide, as a black precipitate:

C à G: Cu(OH)2 à CuO + H2O.


Copper(II) nitrate solution reacts with aqueous ammonia in a ligand replacement reaction to give the tetrammine complex, which is deep blue and is D:

A à D: [Cu(H2O)6]2+ + 4NH3 à [Cu(NH3)4(H2O)2]2+ + 4H2O


If a large excess of chloride ions is added to D, ligand replacement occurs to give the tetrahedral CuCl42- ion, E, which is bright green:

D à E: [Cu(NH3)4(H2O)2]2+ + 4H + + 4Cl- à [CuCl4]2- + 2H2O + 4NH4 +


Dilution of this solution E results in replacement of the chloride ligands by water, and solution F contains hexaquacopper(II), pale blue:

E à F: [CuCl4]2- + 6H2O à [Cu(H2O)6]2+ + 4Cl - .

 

Nitrogen dioxide reacts with water to give a mixture of nitrous and nitric acids; the nitrogen goes from oxidation state (+4) in NO2 to (+3) in nitrite and (+5) in nitrate. Nitrogen is simultaneously oxidised and reduced so the reaction is disproportionation:

B à H + I:           2NO2 + H2O à HNO2 + HNO3

 

© JRG Beavon 1999

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