What is #[H_3O^+]# for a solution whose #pH-=2.85#?

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1 Answer
anor277
Mar 26, 2017

Well, first we had better specify what #pH# is..........

Explanation:

And I assume that #2.85=-log_10[H^+]#

And taking antilogs..........

#10^(-2.85)=[H^+]#

So #[H^+]=1.41xx10^-3*mol*L^-1.............#

See here for an example...........

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