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

1 Answer
anor277
Jun 22, 2018

Well, [H^+]-=[H_3O^+]=0.0025*mol*L^-1[H+]≡[H3O+]=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^(-)HClO4(aq)+H2O(l)→H3O++ClO−4

Now pH=-log_10[H_3O^+]=2.60pH=−log10[H3O+]=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...

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