# Does deprotonation of a hexaaquaion continue until neutral charge

I learned that deprotonation is the process in which a proton is lost from ligand $\ce{H2O}$ molecules surrounding a central metal ion (I am referring to hexaaqua ions). Also, the process continues until the charge of the complex ion is neutralized.

My question: Why does it only continue until the charge is neutralized, why don't all the six ligand water molecules lose a proton?

Let's look at an example: Hexaaqua aluminium, or $\ce{[Al(OH2)6]^{3+}}$

It is acidic and will undergo deprotonation of a ligand water molecule spontaneously: $$\ce{[Al(OH2)6]^{3+} <=> H+ + [Al(OH)(OH2)5]^{2+}}$$

You can even get it to deprotonate twice: $$\ce{[Al(OH)(OH2)5]^{2+} <=> H+ + [Al(OH)2(OH2)4]^{+}}$$

But, as indicated, both reactions are equilibrium reactions. So if the concentration of protons in solution is high enough, the hydroxo ligands will be reprotonated and become neutral aqua ligands.

If you add some hydroxide to the solution, you can precipitate aluminium hydroxide, since it is not soluble in water. While this might be a driving force for the deprotonation, it doesn't seem to be the case here.

So we have shown that the process does not necessarily continue until the charge of the complex ion is neutralized.

Now, concerning your question why this doesn't happen until all aqua ligands have lost a proton:

Again, a part of the solution is the equilibrium idea brought forward in the upper part. But there's another way to think about it: If you have a negatively charged ion, you also need a positively charged counterion. The only counterion that exists in the solution of water and the aluminium compounds is the freshly abstracted proton, which happily will bind back to the hydroxoligand.

There are also $\pi$-donor effects that are stronger in the hydroxo than the aqua ligand. So it might be favourable to have no more than three hydroxo ligands in an octahedral complex so no two $\pi$-donors are across each other.