What is the "pOH" of a solution containing 7.5 xx 10^(-2) "M H"^(+)?
1 Answer
Aug 10, 2017
Well, at
"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