Why is para-aminobenzoic acid more acidic than ortho-aminobenzoic acid?

Question

Compare the acidic strength of o-, m-, p-aminobenzoic acids.

I got that meta will be the most acidic as it won't be able to show R+ effect of $\ce{NH2}.$

But among the other two, i.e. ortho and para, shouldn't ortho be more acidic as it will cause ortho effect? But according to my book and the data I found online:

$$ \begin{array}{lc} \hline \text{Compound} & \mathrm{p}K_\mathrm{a}\text{(amino)} \\ \hline \textit{o}\text{-Aminobenzoic acid} & 4.89 \\ \textit{m}\text{-Aminobenzoic acid} & 4.79 \\ \textit{p}\text{-Aminobenzoic acid} & 4.77 \\ \hline \end{array} $$

Answer 1

The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compared to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the distance between the two groups.

I would like to add that since the hydrogen of the $\ce{-COOH}$ group is more acidic and $\ce{N}$ has a lone pair. This is why H-bond occurs with $\ce{-NH2}$ group as electron donor and not the other way round.

Also, I presume that by ortho effect you intended to tell that due to repulsion of $\ce{-NH2}$ group $\ce{-COOH}$ group would go out of the plane. However, as we have seen, it is difficult as it is already H-bonded with the nitrogen atom and moving out of the plane would put strain into the H-bond. Moreover, the repulsion by $\ce{-NH2}$ group is not too intense so it would (in most part) avoid the strain.