What is the #"pH"# of a solution with #["OH"^(-)] = 3.9 xx 10^(-8) "M"#?

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Answer 1 · 2017-08-10T18:31:28

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

#"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 #7#? Why?