What is the #"pH"# of a solution with #["OH"^(-)] = 3.9 xx 10^(-8) "M"#?
1 Answer
Aug 10, 2017
Well, at
#"pH" + "pOH" = 14 = "pK"_w#
and you ought to memorize that
#"pOH" = -log["OH"^(-)]# .
Thus,
#"pH" = 14 - "pOH"#
#= 14 + log(3.9 xx 10^(-8) "M") = ???#
Do you expect this number to be less than