$\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{\mathrm{Acid}}$
$\mathrm{pH}=4.5+\log \frac{(\mathrm{Salt}]}{\mathrm{Acid}}$
As HA is $50 \%$ ionized so $[$ Salt $]=[$ Acid $]$
$\mathrm{pH}=4.5+\log (1)=4.5$
Now, $\mathrm{pH}+\mathrm{pOH}=14$
$\therefore \mathrm{pOH}=14-4.5=9.5$
(આપેલ છે : ${Zn}({OH})_{2}$નો દ્રાવ્યતા ગુણાકાર $2 \times 10^{-20}$ છે.)