\(\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{\mathrm{Acid}}\)
\(\mathrm{pH}=4.5+\log \frac{(\mathrm{Salt}]}{\mathrm{Acid}}\)
As HA is \(50 \%\) ionized so \([\) Salt \(]=[\) Acid \(]\)
\(\mathrm{pH}=4.5+\log (1)=4.5\)
Now, \(\mathrm{pH}+\mathrm{pOH}=14\)
\(\therefore \mathrm{pOH}=14-4.5=9.5\)