Deprotonated bases

[Jeffrey Hablewitz 2I]

I am confused by this statement:
"When pOH is less than pK_b, the acid is deprotonated."
What does it mean for a base to be deprotonated?

[IanWheeler3F]

For this I guess it helps to look at the base version of the H-H equation, pOH = pKb + log(BH+/B)
So if pOH < pKb then pOH - pKb is negative, so the equation becomes: -N = log(BH+/B), where -N is just some random negative number
then if you get rid of the log by doing 10^ to both sides you get 10^-N = BH+/B
But we know bc of power rules that that 10^-N is 1/(10^N) so the denominator is greater than the numerator, and using the equation that means that B is greater than BH+, since B is the deprotonated version this means when pOH < pKb there is more B (deprotonated) than BH+ (protonated)

[Brian Acevedo 2E]

The way I like to think of it is that for a strong base, for instance, deprotonation would imply that the base in its current state would likely resemble its conjugate acid. In the calculations for pk_b values, deproponated would mean the concentration term of the "B" species that is not "BH".