NaOH + HC2H3O2 ==> C2H3O2Na + H2O

moles NaOH used = 0.020 L x 0.10 mol/L = 0.0020 moles NaOH

moles HC2H3O2 used = 0.04 L x 0.20 mol/L = 0.008 moles HC2H3O2

This forms a buffer of the weak acid (HC2H3O2) and the salt of that acid C3H3O2Na)

Final volume = 20 ml + 40 ml = 60 ml = 0.06 L

Final [salt] = 0.002 moles/0.06 L = 0.033 M

Final [acid] = 0.008-0.002 moles/0.06L = 0.006 mol/0.06L = 0.1 M

pH = pKa + log [salt]/[acid] = 4.74 + log (0.033/0.1) ... note:pKa = -log Ka and -log 1.8x10^-5 = 4.74

pH = 4.74 + (-0.48)

pH = 4.3