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If #pH-=8.25#, what is #[H_3O^+]#?

1 Answer

Apr 27, 2017

By definition, #pH=-log_10[H_3O^+]#.........

Explanation:

And thus #[H_3O^+]=10^(-8.25)*mol*L^-1#

#=5.0xx10^-9*mol*L^-1#

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