Answer
a. $$NH_3(aq) + H_2O(l) \leftrightharpoons N{H_4}^+(aq) + OH^-(aq)$$
$$K_b = \frac{[NH{_4}^{+}][OH^-]}{[NH_3]}$$
b. $$HC{O_3}^-(aq) + H_2O(l) \leftrightharpoons H_2CO_3(aq) + OH^-(aq)$$
$$K_b = \frac{[H_2CO_3][OH^-]}{[HC{O_3}^-]}$$
c. $$CH_3NH_2(aq) + H_2O(l) \leftrightharpoons CH_3N{H_3}^+(aq) + OH^-(aq)$$
$$K_b = \frac{[CH_3N{H_3}^+][OH^-]}{[CH_3NH_2]}$$
Work Step by Step
A weak base ionizes receiving one proton $(H^+)$ from a water molecule, producing its conjugate acid and a hydroxide ion.
$$K_b = \frac{[Products]}{[Reactants]}$$
* Water does not appear in the Kb expression since it is a liquid.
