What is #pOH# for a solution that is #0.565*mol*L^-1# in #HClO_4(aq)#?

1 Answer
anor277
Apr 26, 2017

#pH+pOH=14#

Explanation:

#pH+pOH=14# in aqueous solution under standard conditions.

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

Now #"perchloric acid"# is thus stoichiometric in #H_3O^+#, #"hydronium ion"#, and since #pH=-log_10[H_3O^+]#, #pH=-log_10(0.272)=0.565#, and...................

#pOH=14-0.565=13.4#

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