show chem step plz

Ka=[H+][ethanoate]/[ethanoic acid]

assume sodium ethanoate had 100% dissociation

in buffer, concentration of ethanoate and ethanoic acid is equal and is cancelled out
Ka=[H+]
[H+]= 10^-4.74=Ka
number of mole of NaOH added= 4x5/1000=0.02mol
CH3COOH + OH- --> CH3COO- +H2O
number of mole of substance after reaction:
ethanoate: 0.32
ethanoic acid: 0.28
(0.32)/(0.28)= 1.142857
[H+]=10^-4.74/1.142857 = 1.592238x10^-5M
pH = 4.797992=4.80
NaOH adding into water, negate water's self dissociation
[OH-]= 0.02/1.005=0.0199
10^-14 = [H+][OH-]
[H+]=10^-14/0.0199
[H+]=5.025x10^-13
pH= 12.3