In a lab, we added NH3(aq) to CuSO4(aq), and the solution turned dark blue with a precipitate.
The problem is that the products ( Cu(NH3)4 and SO4) should be soluble, no? The question is then, what is the percipitate?
The ppt is Cu(OH)2. The dark blue solution is Cu(NH3)42+ but if the (OH)- is high enough the hydroxide can (and does) form. USUALLY, adding an excess of NH3(aq) will dissolve the hydroxide and leave just the complex copper ion.
Sorry, I'm a bit confused. Can you explain how the ppt is formed? I don't really understand where the oh- comes from. is the NH3(aq) actually NH4+ and OH-?
As I understand it, doesn't CuSO4 + NH3 turn to Cu(NH3)4SO4 and then turns to Cu(NH3)4 and SO4?
An aqueous solution of NH3 is a weak base, forming ammonium ion and hydroxide ion.
NH3 + HOH <===> NH4^+ + OH^-
Copper sulfate is an ionic solid and dissociates in water as follows:
CuSO4(s) + HOH(l) ==> Cu^+2(aq) + SO4^-2(aq)
Then some of the copper(II) in solution reacts with the OH^- to form Cu(OH)2. Since that is sparingly soluble in water, it forms a precipitate.
Also the copper(II) reacts with ammonia to form the complex ion.
Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
At the end, we have the ppt of Cu(OH)2, the dark blue copper ammine complex ion [Cu(NH3)4^+2], the sulfate ion [SO4^-2], unionized aqueous NH3, plus very small amounts of copper(II) ion, ammonium ion, hydroxide ion, and hydrogen ion (from the water). I hope I didn't misplace something. Check my thinking.
Usually, the copper hydroxide will dissolve and turn completely into the ammine complex if an excess (sometimes it takes a LARGE excess) of NH3 is added.
Thanks alot. Understood it perfectly.
[Cu(NH3)4]SO4 + (NH4)2S
Hi, I don´t understand it very well but I was taught that if NH3 is concentrated, the solution is a complex which is soluble. But if NH3 is dilluted (dillute solution), a sediment (coagulation) will create. It is because a dilluted NH3 behaves as a hydroxide:))
To obtain the precipitate, you can add (NH4)2S to the mixture. The reaction would proceed as follows:
[Cu(NH3)4]SO4 + (NH4)2S → CuS(s) + 4NH4+ + SO4^2-
The addition of (NH4)2S will cause a double displacement reaction, where the Cu(NH3)4 ion from the original solution combines with the S^2- ion from (NH4)2S to form a precipitate of CuS. This precipitate is black in color. The ammonium ions (NH4+) and sulfate ions (SO4^2-) remain in the solution.