Answer
$HC{O_3}^-(aq) + H_2O(aq) \lt -- \gt H_2CO_3(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][H_2CO_3]}{[HC{O_3}^-]}$
Work Step by Step
1. Write the ionization chemical equation:
- Write the reaction where $HC{O_3}^-$ takes a proton from a water molecule:
$HC{O_3}^-(aq) + H_2O(aq) \lt -- \gt H_2CO_3(aq) + OH^-(aq)$
2. Now, write the $K_b$ expression:
- The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_b = \frac{[Products]}{[Reactants]}$
$K_b = \frac{[OH^-][H_2CO_3]}{[HC{O_3}^-]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
