Answer
$CH_3COO^-(aq) + H_2O(l) \lt -- \gt CH_3COOH(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}$
Work Step by Step
1. Write the ionization chemical equation:
- Since $CH_3COO^-$ is a base, write the reaction where it takes a proton from a water molecule:
$CH_3COO^-(aq) + H_2O(l) \lt -- \gt CH_3COOH(aq) + OH^-(aq)$
2. Now, write the $K_b$ expression:
- The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_b = \frac{[Products]}{[Reactants]}$
$K_b = \frac{[OH^-][CH_3COOH]}{[CH_3COO^-]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
