Answer
$pOH = pKb +log\frac{[X^+]}{[XOH]}$
Work Step by Step
1. Write the Kb formula:
$Kb = \frac{[X^+][OH^-]}{[XOH]}$
2. Isolate the $[OH^-]$ in one side of the equation:
$Kb \times \frac{[XOH]}{[X^+]} = [OH^-]$
3. Put a $(-log)$ in each side of the equation:
$-log(Kb \times \frac{[XOH]}{[X^+]}) = -log[OH^-]$
$-log(Kb) - log( \frac{[XOH]}{[X^+]}) = -log[OH^-]$
$pKb - log( \frac{[XOH]}{[X^+]}) = pOH$
4. Invert the $- log( \frac{[XOH]}{[X^+]}) $:
$pKb + log( \frac{[X^+]}{[XOH]}) = pOH$
