Answer
It goes from 9.21 to 9.34
Work Step by Step
Initially, all hydroxide added is neutralized:
Number of moles: $20/1000\ L\times 0.10\ M=0.002\ mol$
$OH^-+NH_4^+\rightarrow NH_3 +H_2O$
The new concentration of ammonium and ammonia are:
$NH_4^+: (0.183\ M\times 80.0/1000\ L-0.002\ mol)\div 100/1000\ L=0.126\ M$
$NH_3: (0.169\ M\times 80.0/1000\ L+0.002\ mol)\div 100/1000\ L=0.155\ M$
New equilibrium:
$K_a=x(0.155+x)/(0.126-x)$
$x=4.55\cdot10^{-10}\ M=[H_3O^+]$
$pH=9.34$
From Henderson-Hasselbach, the original pH was 9.21, So it goes from 9.21 to 9.34
