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

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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.](https://socratic.org/questions/what-is-the-result-of-dissociation-of-water)

Answer 2 · 2017-05-27T08:41:54

pH = 11.5

My own work (Mert METİN)

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