# Why does copper sulfate disassociate into Cu+2 and SO4-2 when dissolved in water?

I'm trying to understand why exactly copper sulfate disassociates into $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$ ions? How is it calculated? I'm trying to find information but it seems as if it is treated as a given. I'm trying to understand what happens exactly? What happens to the electrons in the process? Where do they come from? How to arrive to the charge of the ions?

Thank you.

• chemistry.stackexchange.com/questions/41694/… – bonCodigo Aug 17 '17 at 7:56
• Those are the ways of salts: metals are cations, and acid residues are anions. "Calculated" is not the right word; the information has been around for quite a while now, and everybody just got used to it. So will you. – Ivan Neretin Aug 17 '17 at 7:59
• But what happens in the process? how do we arrive to it? I want to know. I don't like "just treated as a given". I want to know. – user50190 Aug 17 '17 at 8:00
• Put two wires in the solution and connect them to a battery. The one which is connected to (-) will get covered with metal. So there must have been positive ions of Cu. – Ivan Neretin Aug 17 '17 at 8:02
• They were there from the very beginning. By current understanding, solid CuSO4 consists of ions. – Ivan Neretin Aug 17 '17 at 8:09

Basically $\ce{CuSO4}$ already exists as $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$. It's anhydride form exists as an orthorhombic arrangement of ions. The opposite charge attraction balances the like-charge repulsions and the crystal stays bound.
When you dissolve it into water, a few things happen. First of all, with water being quite a polar molecule, tends to surround ions. This is called hydration of ions. Basically this just separates the ions from the tightly packed lattice and spreads them evenly along the water sample. Hydration is the main reason why the $\ce{Cu^2+}$ ion is quite stable in aqueous medium.