The $\mathrm{pH}$ discussions only make sense in the way you are using them for free $\ce{edta}$. Free $\ce{edta^4-}$ can indeed accept up to six protons depending on the acidity of the solvent; the most common forms are probably the neutral compound $\ce{H4edta}$ and the disodium salt $\ce{Na2H2edta}$. For these species, a $\mathrm{pH}$ equilibrium exists as you know it for polyprotic acids and only for these species is the discussion you’re beginning your post with correct.
Once $\ce{edta}$ forms a coordination compound, e.g. with calcium, things change. Six atoms of $\ce{edta}$ — four oxygens and to nitrogens — are now octahedrally surrounding the central metal ion. The complex will look something like this:
The calcium ion is of course calcium(II), the $\ce{edta}$ ligand carries four negative charges resulting in the complex anion being a dianion $\ce{[Ca(edta)]^2-}$. Technically, this species can still accept up to four protons, but only the carbonyl oxygens can now be protonated to some extent. If you do, you will encounter a very acidic oxygen-centred cation which will want to get rid of the extraneous proton again. Note that the two sites which I intuitively expect to be the most basic, the two nitrogen atoms, cannot accept any more protons since they are already tetracoordinated.
Thus, once you form a complex from free $\ce{edta}$, the equilibrium situation changes completely. We now no longer may consider $\ce{edta}$ by itself but must consider the complex as a whole; the remaining acid-base equilibrium is
$$\ce{[Ca(edta)]^2- + H+ <<=> [Ca(Hedta)]-}\tag{1}$$
As I mentioned, protonating the ligand will lead to a very acidic species (akin to protonating a protonated carboxylic acid) and therefore the equilibrium is shifted to the left.
So even though the $\ce{pH}$ values may tell you that you are dealing with a tri- or tetraanion of free $\ce{edta}$ in solution, the complex has quite different acidity constants.
