# What reactions occur when mixing copper sulfate and sodium hydroxide?

I tried reacting copper sulfate with sodium hydroxide to get copper hydroxide, which should precipitate, according to the following equation: $$\ce{2NaOH + CuSO4 -> Cu(OH)2(s) + Na2SO4}$$ I looked on Google Images, and the color of copper hydroxide is light-blue, but something interesting happened when I mixed these two solutions: the precipitate formed - and was originally light-blue, as expected - however, it turned into this black-green sludge within a few seconds. I'm thinking a complex ion might have formed between the $\ce{Na2SO4}$, and the $\ce{Cu(OH)2}$. I just saw a video on this, posted by NileRed, but he doesn't go into detail. Does anyone have an idea of what's going on?

As noted in your question and in one of the comments, copper forms many different complexes having a variety of colors from red to green to blue to black and probably more. Of course the situation here is even more "complex" (sorry) as you can end up with mixtures of the different copper complexes, as appears to be the case over the course of your experiment.

The formation of some cupric oxide ($\ce{CuO}$) would account for the appearance of an insoluble black precipitate. This could happen from the decomposition of some of the $\ce{Cu(OH)2}$.

$\ce{Cu(OH)2}$ itself should be a greenish to pale-blue insoluble precipitate.

It is possible that the original $\ce{Cu(OH)2}$ precipitate appeared more blue than green due to the blue $\ce{CuSO4}$ solution. As the reaction proceeded, $\ce{CuO}$ and more $\ce{Cu(OH)2}$ precipitated to give the black-green precipitate you observed.

• The wiki link shows Cu(OH)2 as a "blue" solid. It looks blusih-green to me. Sort of turquoise color. I think you also have to consider "wet" vs "dry" Cu(OH)2. Cu(SO4)*(H2O)5 is much darker blue than Cu(SO4). – MaxW Jan 26 '17 at 20:14
• Good point about the hydration @MaxW. I also tried to give a better description of the turquoise-ish color. It seems to me to change with the angle of my monitor, so I included the description given in the text;) – airhuff Jan 26 '17 at 20:19
• Hydroxide ppts are notorious for absorbing other ions. In a strong hydroxide solution I'd expect the ppt to drag extra $\ce{OH^-}$ out of solution. I'd guess that the decomposition of $\ce{Cu(OH)2}$ is base catalyzed. So $$\ce{Cu(OH)2 + OH^- -> CuO(OH)^- + H2O}$$ $$\ce{CuO(OH)^- -> CuO + OH^-}$$ Out of an ammonia solution the ppt must form slowly enough so as to absorb very little extra $\ce{OH^-}$. – MaxW Jan 27 '17 at 0:04
• If I remember correctly, anhydrous copper sulfate is white, but that should hardly be an issue here. – Martin - マーチン Jan 27 '17 at 6:46

Yes, as noted by @airhuff, there is not one but two reactions involving, one is formation of greenish-pale blue copper sulfate, $\ce{CuSO4}$ and other one is formation of black-brown copper(II) oxide $\ce{CuO}$. Also, there are also solvation reactions occurring involving formation of solvated complex but we are going to neglect that. \begin{align} \ce{CuSO4 + 2NaOH &-> Cu(OH)2 + Na2SO4}\tag{1}\label{one}\\ \ce{2CuSO4 + 2NaOH &-> [CuO + H2O] + Na2SO4}\tag{2}\label{two} \end{align}

Some copper(II) hydroxide decomposes to form copper(II) oxide and water, $$\ce{Cu(OH)2 -> CuO + H2O},$$ and that's how reaction $\eqref{two}$ proceeds.