What is the "pOH" of a solution containing 7.5 xx 10^(-2) "M H"^(+)?

1 Answer
Aug 10, 2017

Well, at 25^@ "C" and "1 atm"...

"pH" + "pOH" = 14 = "pK"_w

and you ought to memorize that

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

Thus,

"pOH" = 14 - "pH"

= 14 + log(7.5 xx 10^(-2) "M") = ???

Do you expect this number to be greater than 7? Why?