pKa and pH graphically
You should state the charge balance or the proton balance.

[H3O+] = [OH-] + [CH3COO-]

We take the logarithm of this and get:
log [H3O+] = log([OH-] + [CH3COO-]).

Now, you can plot this in your nice diagram.

log [H3O+] will follow the straight line in your diagram, i.e. log [H+].

The line for log([OH-] + [CH3COO-]) will at very high pH follow the line for log([OH-], but quite soon it will bend off and only follow the line for log [CH3COO-]. This is because [OH+] &lt;&lt; [CH3COO-] at lower pH.

When [OH+] &lt;&lt; [CH3COO-], you get log [H3O+] = log [CH3COO-]). When the line for log [H3O+] crosses the line for log [CH3COO-], you can calculate the pH of the solution from the diagram.
The lines are crossing each other at right angel: pH = (1.00 + 4.65)/2 = 2.82(5).
Very good, I am encouraging you to use logarithmic diagrams for solving acid-base equilibria.