Question

Determine the [OH−] in a 0.22 M solution of HClO4. Answer in units of M?

I tried using 10^14 but that didnt work.

Answer 1

Well, `14=pH+pOH`...we get `[HO^-]=4.55xx10^-14*mol*L^-1`

Explanation:

...are we clear on this? And `"perchloric acid"` is an exceptionally strong acid...

And so we got the stoichiometric reaction in water..

`HClO_4(aq) + H_2O(l) rarr H_3O^+ + ClO_4^(-)`

...and `pH=-log_10[H_3O^+]=-log_10(0.22)=+0.658`

`pOH=14-pH=14-0.658=13.34`...a high `pH` as it consistent with a largish hydrogen ion concentration....

And finally,

`[HO^-]=10^(-13.34)*mol*L^-1=4.55xx10^-14*mol*L^-1`