You have 325 mL of an 0.11 M acetic acid solution. What volume (V) of 2.00 M NaOH solution must you add in order to prepare an acetate buffer of pH = 5.07? (The pKa of acetic acid is 4.76.)
millimols HAc = 325 x 0.11 = about 36 but you need to do it more accurately.
........HAc + NaOH ==> NaAc + H2O
I.......36.....0........0
add............x...........
C.......-x....-x........+x
E......36-x....0........+x
Redo so you have good numbers for the E line, then substitute the revised E line into the HH equation and solve for x which gives you millimols NaOH.
Then M = millimols/M. You know M and mmols, solve for mL NaOH needed. The answer should be close to 12 mL of the 2.00 M NaOH.