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?

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