## Chemistry 12th Edition

$K_a * K_b = K_w$
$NH_3$ 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: $K_a * K_b = K_w$