Answer
a) $C_6H_5NH_2(aq) + H_2O(l) \lt -- \gt C_6H_5N{H_3}^+(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][C_6H_5N{H_3}^+]}{[C_6H_5NH_2]}$
b) $(CH_3)_2NH(aq) + H_2O(l) \lt -- \gt (CH_3)_2N{H_2}^+(aq) + OH^-(aq)$
Work Step by Step
1. Write the ionization chemical equation:
- Write the reaction where $C_6H_5NH_2$ and $(CH_3)_2NH$ take a proton from a water molecule:
a) $C_6H_5NH_2(aq) + H_2O(l) \lt -- \gt C_6H_5N{H_3}^+(aq) + OH^-(aq)$
b) $(CH_3)_2NH(aq) + H_2O(l) \lt -- \gt (CH_3)_2N{H_2}^+(aq) + OH^-(aq)$
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) $K_b = \frac{[OH^-][C_6H_5N{H_3}^+]}{[C_6H_5NH_2]}$
b) $K_b = \frac{[OH^-][(CH_3)_2N{H_2}^+]}{[(CH_3)_2NH]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
