Answer
pOH is the negative logarithm of the $OH^-$ concentration ($M$).
$pOH = -log[OH^-]$
The equation that relates pOH and pH is:
$pH + pOH = 14$
Work Step by Step
This equation is the derivation of the $K_w$ equation:
$[H^+] \times [OH^-] = 10^{-14}$
$log ([H^+] \times [OH^-]) = log(10^{-14})$
$log[H^+] + log[OH^-] = -14$
$-pH + (-pOH) = -14$
$pH + pOH = 14$
