# What is the "pH" of a solution with ["OH"^(-)] = 3.9 xx 10^(-8) "M"?

Aug 10, 2017

Well, at ${25}^{\circ} \text{C}$ and $\text{1 atm}$...

${\text{pH" + "pOH" = 14 = "pK}}_{w}$

and you ought to memorize that

"pOH" = -log["OH"^(-)].

Thus,

$\text{pH" = 14 - "pOH}$

= 14 + log(3.9 xx 10^(-8) "M") = ???

Do you expect this number to be less than $7$? Why?