What is the pH when 100ml of 0.1 M NaOH is added to 150ml of 0.2 M HAc if pKa for acetic acid is 4.76?

The reaction that takes place when NaOH is added to acetic
acid is:

NaOH + HAc --> NaAc +
H2O

100 mL of 0.1 M NaOH contains 0.01 moles of NaOH. 150
mL of 0.2 M HAc contains 0.03 M of HAc. After the reaction we are left with 0.03 - 0.01
= 0.02 moles of acetic acid.

Ka for acetic acid is
[H+][CH3COO-]/[CH3COOH] where [] indicates the
concentration.

pKa = -log(10)[Ka] = log(10)[CH3COOH] -
log(10)[H+] - log(10)[CH3COO-]

as [H+] =
[CH3COO-]

pKa = log(10)[CH3COOH] -
2*log(10)[H+]

For acetic acid pKa = 4.76 and -log(10)[H+] =
pH

4.76 = log(10)(0.2) +
2*pH

=> 2*pH = 4.76 +
0.698

=> 5.458

pH =
2.729

The pH when 100ml of 0.1 M NaOH is added to 150ml of
0.2 M HAc is 2.729