I did a lab in my Chemistry 12 class, but I don't really understand equilibrium and cannot explain why it happened. CuSO4Cu 2+ +SO4 2- We added NH3 into CuSO4 solution, and it turned lighter blue then turned to darker blue as we added more NH3 into it. Then, we added 1M HCl which made the solution clear, and there were blue specks sitting on the bottom of the test tube. I have to show that what was the stress to this equilibrium and which way the equilibrium system shifted (either left or right). And stress would be increased/decreased temperature, increased/decreased product, increased/decreased reactant, increased/decreased pressure.

From your question, I believe that the primary purpose of this lab is to demonstrate and understand Le Chatelier's principle (which is essentially equilibrium) and the concept of complex ion formation.
In your case, you are starting out with a copper sulfate (CuSO4) solution and adding ammonia (NH3) to it. Remember that NH3 is a weak alkaline/base which produces hydroxide ion in water:

NH3 + H2O <==> NH4(+) + OH(-)


Also, copper sulfate is an ionic solid that dissociates in water:


CuSO4 + H2O ==> Cu^2+ (aq) + SO4 ^2+ (aq)



Adding the NH3 will give you precipitate of copper (II) hydroxide as the following reaction will occur:


CuSO4(aq) + 2NH3(aq) + 2H2O(l) <==>Cu(OH)2(s) + (NH4)2SO4(aq)


Also the copper(II) reacts with ammonia to form the complex ion.


Cu^+2 + 4NH3 ==> Cu(NH3)42+


The solution turning darker blue is the Cu(NH3)42+ and the precipitate is the Cu(OH)2(s) being formed. Hence, in terms of understanding the stress to this equilibrium, adding NH3 is causing the reaction to shift towards the right, resulting in the formation of the Cu(OH)2(s).


Later on, you added 1M of HCl, which is a strong acid. The HCl will react with the Cu(OH)2(s), which shifts the equilibrium to the left. This means that less of the solid will be present, and the blue specks that you see at the bottom of the test tube should just be the remaining Cu(OH)2(s) that has not reacted with the HCl.

When thinking about stress including temperature, reactant/prodct, pressure, it is helpful to think which "side" of the equation these stressors belong. For instance, if you are adding something from the left side, equilibrium will favor the reaction proceeding to the right in order to try to balance both sides.

Hope this is helpful!

If I understand correctly, you’re asking what the stress is to the system of copper II sulphate in solution, when first adding ammonia (NH3) and then hydrochloric acid (HCl). Here’s the equation for the system we’re looking at, which you’ve already written:
a)CuS04(aq)-->Cu2++SO42-
When you add NH3to this solution, remember that there’s also water( H20) present. The NH3 will react with H20 to form the ammonium ion (NH4+) and the hydroxide ion (OH-) in the same solution:
b)NH3(aq)-->NH4++ OH-
So now, there are 4 ions in solution: Cu2+, SO42-, NH4+and OH-. The Cu2+and the OH- will then react with each other to form the blue precipitate you observed, which is copper II hydroxide, Cu(OH)2
c) Cu2++ OH- -->Cu(OH)2(s)
Therefore, adding the ammonia stresses the system in reaction (a) to the right by increasing the product, because adding ammonia effectively takes the copper ions out of the solution.
In the second part of the experiment, when HCl is added to the solution, it dissolves the precipitate of Cu(OH)2, returning the solution to a clear colour (with remaining blue specs, which is leftover precipitate that has not dissolved). This stresses the system in reaction (a) to the left, increasing the reactant since it releases the copper ions. The copper ions will then be in solution with chloride ions and water since copper II chloride (shown below) does not form a precipitate.
d) Cu(OH)2(s)+2HCl-(aq)-->Cu(Cl)2(aq) +2H20(l)