Answer
See the answer below.
Work Step by Step
The number of moles of acetic acid:
$0.15\ M\times 22/1000\ L=3.3\cdot10^{-3}\ mol$
New molarity: $3.3\cdot10^{-3}\ mol\div 44/1000\ L=0.075\ M$
Net ionic reaction:
$CH_3CO_2^-+H_2O\leftrightarrow CH_3CO_2H+OH^-$
$K_b=[ CH_3CO_2H][OH^-]/[CH_3CO_2^-]$
$5.6\cdot10^{-10}=x^2/(0.075-x)$
$x=6.48\cdot10^{-6}\ M=[OH^-]$
$pH=14+log([OH^-])=8.81$
$[H_3O^+]=10^{-pH}=1.54\cdot10^{-9}\ M$
