n[H+] = .02 x .08
= 1.6 x 10^-3
n[OH-] = .05 x .03
= 1.5 x 10^-3
Therefore, H+ is in excess, and 1 x 10^-4 is left over after mix.
New volume = .05 L
Therefore [H+] = 1 x 10^-4 / .05
= 2.69
Well I'm an idiot.
I skipped a step there somewhere but that's the gist of it.