# Chemistry - Why stepwise dissociation for acids but not hydroxide bases?

## Solution 1:

This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.

In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.

As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)

## Solution 2:

The $\ce{H+}$ is bonded covalently to the rest of the molecule. So, when you take off the first $\ce{H+}$, the molecule fundamentally changes. It becomes another molecule. The second $\ce{H+}$ is being taken away from a different species and so it is governed by a different equilibrium.

In $\ce{Ca(OH)2}$, $\ce{Ca^2+}$ and $\ce{OH-}$ are held together via ionic bonds. When you dissolve that in water, all the ionic bonds are going to break at the same time. You cannot break some of the ionic bonds without also breaking the rest.