physical chemistry - Find the pH of an amphiprotic salt dissolved in water at room temperature

Let's say I have a polyprotic acid $\ce{H2A}$ with the following properties:

$$\begin{align} \ce{H2A &<=> HA- + H+} & K_1 &= 1\cdot 10^{-1}\\ \ce{HA- &<=> A^2-+ H+} & K_2 &= 1\cdot 10^{-4} \end{align}$$

Now a 0.5 M solution of $\ce{HA-}$ is prepared; what will be the $\mathrm{pH}$ of the solution?

My textbook (and many other sources) says it should be

$$\frac{\mathrm{p}K_1 + \mathrm{p}K_2}{2}$$

which sounds a little counterintuitive to me.

First of all, it implies that the $\mathrm{pH}$ is independent of the concentration of $\ce{HA-}$ present.

Also, if $1/\mathrm{p}K_1 > \mathrm{p}K_2$, then $\ce{HA-}$ should have the tendency to accept a $\ce{H+}$ more than it has of losing one, then why is the $\mathrm{pH}$ exactly partway between $\mathrm{p}K_1$ and $\mathrm{p}K_2$?