Answer
a) $NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$
b) $C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][C_5H_5NH^+]}{[C_5H_5N]}$
Work Step by Step
1. Write the ionization chemical equation:
- Write the reactions where $NH_3$ and $C_5H_5N$ take a proton from a water molecule:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
$C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq)$
2. Now, write the $K_b$ expression:
- The $K_b$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_b = \frac{[Products]}{[Reactants]}$
a) $K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$
b) $K_b = \frac{[OH^-][C_5H_5NH^+]}{[C_5H_5N]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
