# Chapter 16 - Acid-Base Equilibria - Exercises - Page 717: 16.19b

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.

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