Answer
a. $$C{O_3}^{2-}(aq) + H_2O(l) \leftrightharpoons HC{O_3}^-(aq) + OH^-(aq)$$
$$K_b = \frac{[HC{O_3}^-][OH^-]}{[C{O_3}^{2-}]}$$
b. $$C_6H_5NH_2(aq) + H_2O(l) \leftrightharpoons C_6H_5NH{_3}^+(aq) + OH^-(aq)$$
$$K_b = \frac{[C_6H_5NH{_3}^+][OH^-]}{[C_6H_5NH_2]}$$
c. $$C_2H_5NH_2(aq) + H_2O(l) \leftrightharpoons C_2H_5N{H_3}^+(aq) + OH^-(aq)$$
$$K_b = \frac{[C_2H_5N{H_3}^+][OH^-]}{[C_2H_5NH_2]}$$
Work Step by Step
A weak base ionizes, receiving one proton $(H^+)$ from a water molecule, producing its conjugate acid and hydroxide ion.
$$K_b = \frac{[Products]}{[Reactants]}$$
* Water does not appear in the Kb expression since it is a liquid.
