Question

What is the pOH if the [OH⁻] is 4.1 x 10⁻¹¹?

Answer 1

`10.39`

Explanation:

`"pOH" = -log["OH"^-]`

`"pOH" = -log(4.1 × 10^-11) = 11 - log(4.1) = 10.39`