Addition of ammonia solution to CuSO4 to give a pale blue precipitate followed by a clear, deep blue solution
I've asked a million question about precipates and I'm still really confused. What is going on in this example:
When ammonia is first added to a solution of copper(II) nitrate, a pale blue precipitate of copper(II) hydroxide formes. As more ammonia is added, however, this precipitate dissolves. Describe what is happening.
Pure ammonia is a gas; as a solution, ammonia exists as ammonium hydroxide, NH4OH. All ammonium compounds are soluble in water. You first form the Cu(OH)2 because of the hydroxide ions in solution. However, as more ammonium hydroxide is added, the equilibrium gets pushed so that Cu2+ is bound to six ammonia (NH3) ligands.
Actually, Cu+2 binds to 4, not 6, NH3 molecules to form [Cu(NH3)4]+2 (+2 is the charge on the whole complex).
My source says Cu(NH3)4]2+ is favored as a stable ion, but it can form six ligands.
ill answer this as soon as i get the BBC codes working.....dang
OK we are dealing with two equibriums here
When the copper nitrate is dissolved in the water it forms a hydrate
the two equilibirums are
Cu(H2O)4 2+ + 4NH3 Cu(NH3)4 2+ + 4H2O which has a kf of 5 x 1012
and the equilibrium with OH
Cu(H2O)4 2+ + 2OH- Cu(OH)2(H2O)2 2+(s) + 2H2O which has a kf of 5 x 1029
as we can see in the equilibrium competition if there are initially and with small or equal amounts of either NH3 or OH- that the precipitation of the copper hydroxide complex is favored.
However the dynamic changes when you begin to increase the NH3 concentration
using the laws of muliple equilibriums we combine the equilibriums and get
Cu(NH3)4 2+ + 2OH- + 2H2O Cu(OH)2(H2O)2 2+(s) + 4NH3
without going through all the math we can see here that buy significantly increasing the NH3 concentration we have to shift the equilibrium to the left..... adding NH3 excessivley will produce a greater amount of NH3 than OH-
conversely if we start adding OH- as a strong base we will see the precipitation again.
Even with four ammonia ligands, it still forms a pseudo-octahedron geometry. Two water ligands still bond to Cu2+ but the Cu-O bond is longer because of Jahn-Teller distortion of the d9 metal.