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