Answer
$K_a * K_b = K_w$
Work Step by Step
$NH_3$ Kb formula:
$K_b = \frac{[OH^-][NH_4^+]}{[NH_3]}$
$NH_4^+$: Ka formula:
$K_a = \frac{[H^+][NH_3]}{[NH_4^+]}$
- Isolating $[NH_4^+]$ in both equations:
$[NH_4^+] = \frac{K_b * [NH_3]}{[OH^-]}$
$[NH_4^+] = \frac{[H^+][NH_3]}{K_a}$
Therefore:
$\frac{K_b * [NH_3]}{[OH^-]} = \frac{[H^+][NH_3]}{K_a}$
We can eliminate the $[NH_3]$.
$\frac{K_b}{[OH^-]} = \frac{[H^+]}{K_a}$
$K_a * K_b = [H^+][OH^-]$
So:
$K_a * K_b = K_w$
