Answer
$H_2SO_3(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + H_3O^+(aq)$
$HS{O_3}^-(aq) + H_2O(l) \lt -- \gt S{O_3}^{2-}(aq) + H_3O^+(aq)$
Work Step by Step
1. Write the reaction where $H_2SO_3$ behaves as an acid in water.
$H_2SO_3(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + H_3O^+(aq)$
- It will donate one proton to the water molecule, producing $H_3O^+$.
2. Write the reaction where the conjugate base acts as an acid:
$HS{O_3}^-(aq) + H_2O(l) \lt -- \gt S{O_3}^{2-}(aq) + H_3O^+(aq)$
- Since $S{O_3}^{2-}$ can't donate protons, the dissociation stops there.
