What is pOH of 0.1M H2SO4?

If you are in beginning chemistry, the easy answer is that it is 0.2M in (H^+) so pH = -log(0.2). Then pH + pOH = pKw = 14 and you get pOH from that. However, that isn't quite right since the second H^+ is not a strong acid and more advanced classes will do it slightly different. (Actually, a 0.1M solution will be about 1.2 or so for pH.

H2SO4 ==> H^+ + HSO4^- and (H^+) from this first ionization is 0.1.
Then the second one comes off like this.
.......HSO4^- ==> H^+ + SO4^2-
I......0.1.......0.1.....0
C......-x.........+x.....x
E.....0.1-x.....0.1+x....x

Then k2 = (H^+)(SO4^2-)/(HSO4^-)
k2 = (0.1+x)(x)/(0.1-x)
Look up k2, solve for x and total H^+ then is 0.1x. Note that you MUST solve the quadratic equation. Then pH = -log(H^+) and pOH = pKw-pH