Answer
$$C_6H_5-NH_2(aq) + H_2O(l) \leftrightharpoons C_6H_5-NH{_3}^+(aq) + OH^-(aq)$$
$$K_b = \frac{[OH^-][ C_6H_5-NH{_3}^+]}{[ C_6H_5-NH_2]} = 4.0 \times 10^{-10}$$
Work Step by Step
1. Write the reactants and products:
Aniline: $C_6H_5-NH_2$
Water: $H_2O$
Conjugate acid: $C_6H_5-NH{_3}^+$
Hydroxide ion: $OH^-$
2. Write the equation:
$$C_6H_5-NH_2(aq) + H_2O(l) \leftrightharpoons C_6H_5-NH{_3}^+(aq) + OH^-(aq)$$
3. The $K_b$ expression follows this pattern:
$$K_b = \frac{[OH^-][Conj. \space Acid]}{[Base]}$$
Substituting the compounds:
$$K_b = \frac{[OH^-][ C_6H_5-NH{_3}^+]}{[ C_6H_5-NH_2]} = 4.0 \times 10^{-10}$$
