The pH of a solution is 8.7, what is the pOH?

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Answer 1 · 2016-07-05T01:37:11

pOH = 5.3

You can answer this question in one of two ways:

Take the anti-log of the pH to obtain the concentration of $H^+$ ions in solution. After that, use the self-ionization of water formula:
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Where $K_w$ has a value of $1.0xx10^-14$ Then you can rearrange the equation to solve for [ $OH^-$ ]. Take the -log of that value to obtain the pOH.

I'll show you both ways using these equations:

![www.slideshare.net]( useruploads.socratic.org )

THE PROCESS FOR METHOD 1:

$[H^+]= 10^(-8.7) = 1.995xx10^-9 M$
$K_w = ["H"_3"O"^(+)]["OH"^(-)] = 1.0xx10^(-14)$

Kw / [H+] = [OH-]

$(1.0xx10^(-14))/(1.995xx10^(-9) "M") = 5.01xx10^(-6)"$
$[OH^(-)] = 5.01xx10^(-6)M$
$pOH = -log(5.01xx10^(-6))M$
$pOH = 5.3$

THE PROCESS FOR METHOD 2:

$14-8.7 = 5.3$