Buffer help please?

Question

So I am doing practice problems in my book and I got the first part to this bt I can't get the second part.....

calculate the pH of the 0.30M NH3/0.36M NH4Cl buffer system. What is the pH after the addition of 20.0ml of o.050M NaOH to 80.0ml of the buffer solution?

Answer 1

pOH = pKb + log [NH4+] / [NH3]

pOH = 4.74 + log 0.36/0.30 = 4.82

pH = 14 - 4.82 =9.18

Moles NH3 = 0.0800 L x 0.30 M =0.024

moles NH4+ = 0.080 L x 0.36 M = 0.0288

Moles OH- = 0.0200 L x 0.050 M = 0.0010

NH4+ + OH- >> NH3 + H2O

moles NH4+ = 0.0288 - 0.0010 =0.0278

moles NH3 = 0.024 + 0.0010 =0.0250

total volume = 80.0 + 20.0 = 100.0 mL => 0.100 L

concentration NH4+ = 0.0278 / 0.100 = 0.278 M

concentration NH3 = 0.0250 / 0.100 = 0.250

pOH = 4.74 + log 0.278 / 0.250 = 4.79