What is [H_3O^+][H3O+] in an aqueous solution whose pOH=8.26pOH=8.26?

1 Answer
anor277
May 6, 2017

By definition pOH=-log_(10)[HO^-]pOH=−log10[HO−]..........and

Explanation:

And also, we KNOW that in aqueous solution,

pH+pOH=14pH+pOH=14

a.a. And thus, for pOH=8.26pOH=8.26, [HO^-]=10^(-8.26)*mol*L^-1[HO−]=10−8.26⋅mol⋅L−1

=5xx10^-9*mol*L^-1=5×10−9⋅mol⋅L−1

And thus for pH=5.74pH=5.74,............................................

[H_3O^+]=10^(-5.74)*mol*L^-1=1.82xx10^-6*mol*L^-1[H3O+]=10−5.74⋅mol⋅L−1=1.82×10−6⋅mol⋅L−1.

I will let you do b.b., and c.c.. Post the values back in thread if you want them checked.

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