The $p{K}_a$ of a weak acid ($HA$) is $4.5$. The $pOH$ of an aqueous buffered solution of $HA$ in which $50\%$ of the acid is ionized is:

For an acidic buffer, the Henderson-Hasselbalch equation is:
$$pH=p{K}_a+\log \frac{[\text{Salt}]}{[\text{Acid}]}$$
When the acid $HA$ is $50\%$ ionized, the concentration of the salt (formed from the conjugate base $A^{-}$) is equal to the concentration of the remaining unionized acid.
$$[\text{Salt}]=[\text{Acid}]$$
$$pH=4.5+\log (1)=4.5+0=4.5$$
We know that at $298\text{K}$:
$$pH+pOH=14$$
$$pOH=14-pH=14-4.5=9.5$$