How do you find the pH of a solution whose #[H^+] = 3.44 times 10^-12?#?

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Answer 1 · 2017-05-27T08:41:33

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

So here #pH=-log_10(3.44xx10^-12).............#

#=-(-11.46)=11.5#

For a few more details see this old answer.

Answer 2 · 2017-05-27T08:41:54

pH = 11.5

My own work (Mert METİN)