Answer
$K_a \times K_b = K_w$
Work Step by Step
- Let's create a imaginary acid, and its conjugate base:
$HA$ and $A^-$.
- The $K_a$ expression for $HA$ will be:
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
- The $K_b$ expression for $A^-$ will be:
$K_b = \frac{[OH^-][HA]}{[A^-]}$
- So, if we multiply $K_a$ and $K_b$:
$K_a * K_b$
$\frac{[H_3O^+][A^-]}{[HA]} * \frac{[OH^-][HA]}{[A^-]}$
- We can cancel $[H_3O^+]$ and $[OH^-]$ using algebra.
$[H_3O^+] * [OH^-]$, which is equal to $K_w$
- This is the expression for $K_w$
- Therefore:
$K_a * K_b = K_w$
