# What is the "pOH" for a solution of 4.8 xx 10^(-10) "M" "H"^(+)?

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

"pH" = -log["H"^(+)].

Thus,

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

= 14 + log(4.8 xx 10^(-10) "M") = ???

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