Answer
Conjugate Acid: $H_2SO_3$
Conjugate Base: ${SO_3}^{2-}$
Work Step by Step
As we found in 16.19a these are the reactions for:
$HS{O_3}^-$ acting like an acid:
$HS{O_3}^-(aq) + H_2O(l) --> {SO_3}^{2-}(aq) + H_3O^+(aq)$
Since $S{O_3}^{2-}$ is the $HS{O_3}^-$ after donating one proton, it is the conjugate base.
$HS{O_3}^-$ acting like a base:
$H{SO}_3^-(aq) + H_2O(l) --> H_2SO_3(aq) + OH^-(aq)$
Since $H_2SO_3$ is the $HS{O_3}^-$ after receiving one proton, it is the conjugate base.
