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

1 Answer
Aug 10, 2017

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?