What is the #"pOH"# for a solution of #4.8 xx 10^(-10) "M"# #"H"^(+)#?

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Answer 1 · 2017-08-10T18:28:27

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(4.8 xx 10^(-10) "M") = ???#

Do you expect this number to be less than #7#? Why?