Question

What is the `"pOH"` of a solution if the `["OH"^-] = 33.5 xx 10^-2` `"M"`?

Answer 1

`"pOH"=0.475`

Explanation:

The chemical definition of `"pOH"` or in fact, the `"p"` of any value/constant is given as the cologarithm of that thing. That is the negative (or reciprocal) logarithm of that thing.

`therefore` `"pOH"=-log["OH"^-]=log(1/(["OH"^-]))`

`"pOH"=-log(3.35xx10^-1)=0.475`