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

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Answer 1 · 2017-05-06T18:20:05

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

And also, we KNOW that in aqueous solution,

#pH+pOH=14#

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

#=5xx10^-9*mol*L^-1#

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

#[H_3O^+]=10^(-5.74)*mol*L^-1=1.82xx10^-6*mol*L^-1#.

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