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In para-aminobenzoic acid, how can I know whether the amino group loses a proton or not when $\mathrm{pH}$ exceeds $8.5$? I have only found two values for $\mathrm{p}K_\mathrm{a}$ but I never found info about the pKa (in this case it would be the third $\mathrm{p}K_\mathrm{a}$ at $\mathrm{pH}>8.5$) of the amino group.
In any case I guess at basic $\mathrm{pH}$, negative charge over $\ce{NH2}$ would be stabilized by aromatic ring and $\ce{COO-}$.
The two acidity constants in 4-aminobenzoic acid are due to the loss of $\ce{H^+}$ from the protonated form of 4-aminobenzoic acid (the 4-carboxyphenylammonium cation), transforming it into 4-aminobenzoic acid, and then a further loss of $\ce{H^+}$ turning it into the 4-aminobenzoate anion.
The first two ionizations have a $\mathrm{p}K_\mathrm{a}$ of around 5, while the next ionization quite likely has a $\mathrm{p}K_\mathrm{a}$ well below ~25 (compare with other substituted anilines), so you won't find any appreciable amount of deprotonated nitrogen atoms in an aqueous solution. Groups with N−H bonds are far less acidic than their O−H counterparts due to the lower electronegativity of the nitrogen atom, which is less capable of stabilizing the extra negative charge.
The third $\mathrm{p}K_\mathrm{a}$ would refer to the deprotonation of the amino group of the $\ce{NH2-Ph-COO-}$ (singly deprotonated) species.
This would be very difficult to deprotonate since the molecule is already negatively charged. Deprotonation would make a doubly charged species which is normally unfavorable.
Contrary to your assertion, the $\ce{COO-}$ provides no stabilization for the negative charge on amine, and is likely to modestly inhibit delocalization into the benzene ring as well.
Very strong bases $(\ce{BuLi}$ or similar) might be able to deprotonate this compound.
