[NathanielZhu]

Calculate the pOH after addition of 13.46 mL of 0.1525 M LiOH to 25.00 mL of 0.1503 M C9H7O4H. Ka(C9H7O4H) = 3.000e-4
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Ok, so I've been working on a ton of these types of questions, and I've been getting them wrong over and over again.

So, for this, I get the new concentrations from the given concentrations and volumes.
Then I create an ICE chart and plug in the new values for initial. Then from there, I set the equilibrium values to Kb which I find by dividing Kw/Ka.
I solve for x, and finally plug in my value for x into the expression for [OH]. Then I take the -log of it to get pOH.

[JustinCh3m]

well, in the balanced eq'n all mole ratios are 1:1 so I don't see that being your problem ("2x" for example).  Can you post your work?  That might let me/us help you better.