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