Let's call CH3COOH HA and the acetate ion will be A-
millimoles HA = 500 mL x 0.30M = 150.
mmoles A- = 500 mL x 0.20M = 100
We will add 93mL x 1.00M = 93 mmoles NaOH.
............HA + OH- ==> A- + H2O
initial....150...0.......100......
add..............93................
change.....-93...-93.....+93...+93
equil.......57....0......193
pH = pKa + log(base)/(acid)
Plug and chug.
My quickie answer is 5.3 but you should confirm that and don't estimate as I've done.
You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 93 mL of 1.00 M NaOH solution?
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