HELP! Use the Henderson-Hasselbalch equation to calculate the pH of a solution of 0.3 M acetic acid?

You didn't say, but in order to do this calculation you must know the volume of acetic acid solution to which you are adding the 50 mL of NaOH to. I'm going to assume an initial volume of 100 mL. Your answer will be completely dependent on that volume, though.

OK...So, the initial solution of acetic acid contains 0.3 mol/L X 0.1 L = 0.030 mol of HAc (acetic acid). You are adding to that solution 0.05 L X 0.12 mol/L = 0.006 mol of NaOH to it. That NaOH will react with HAc forming 0.006 mol of Ac-, and leaving you with 0.024 mol HAc in the soltuion. (Even though the volume of the solution has changed, these two components are in the same solution, so its volume is really unimportant.)

So, using the H-H equation for the solution:

pH = pKa + log ([Ac-]/[HAc]

pH = 4.76 + log (0.006 / 0.024)
pH = 4.16