Question

What are the [H+] and [OH–] for a 0.0025 M solution of perchloric acid?

Answer 1

Well, `[H^+]-=[H_3O^+]=0.0025*mol*L^-1`

Explanation:

Perchloric acid is a STRONG acid, that undergoes more or less complete protonolysis…

`HClO_4(aq) + H_2O(l) rarrH_3O^+ + ClO_4^(-)`

Now `pH=-log_10[H_3O^+]=2.60`...

Given that in water...`pH+pOH=14...` `pOH=14-2.60=11.40`..

`[HO^-]=10^(-11.40)*mol*L^-1=4.0xx10^-12*mol*L^-1`...