Answer
$S{O_3}^{2-}(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][HS{O_3}^-]}{[S{O_3}^{2-}]}$
Work Step by Step
1. Write the ionization chemical equation:
- Since $S{O_3}^{2-}$ is a base, write the reaction where it takes a proton from a water molecule:
$S{O_3}^{2-}(aq) + H_2O(l) \lt -- \gt HS{O_3}^-(aq) + OH^-(aq)$
2. Now, write the $K_b$ expression:
- The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_b = \frac{[Products]}{[Reactants]}$
$K_b = \frac{[OH^-][HS{O_3}^-]}{[S{O_3}^{2-}]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
