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

1 Answer
May 6, 2017

Answer:

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

Explanation:

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.