What is the $pOH$ $"pOH"$ of a solution containing $7.5 \times 10^{- 2} {M H}^{+}$ $7.5 xx 10^(-2) "M H"^(+)$ ?

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Answer 1 · 2017-08-10T18:28:56

Well, at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ ...

$pH + pOH = 14 = {pK}_{w}$ $"pH" + "pOH" = 14 = "pK"_w$

and you ought to memorize that

$pH = - log [H^{+}]$ $"pH" = -log["H"^(+)]$ .

Thus,

$pOH = 14 - pH$ $"pOH" = 14 - "pH"$

$= 14 + log (7.5 \times 10^{- 2} M) = ? ? ?$ $= 14 + log(7.5 xx 10^(-2) "M") = ???$

Do you expect this number to be greater than $7$ $7$ ? Why?

What is the pOH"pOH" of a solution containing 7.5×10−2M H+7.5 xx 10^(-2) "M H"^(+)?