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

May 27, 2017

Well, by definition $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] \ldots \ldots \ldots \ldots .$

#### Explanation:

So here $p H = - {\log}_{10} \left(3.44 \times {10}^{-} 12\right) \ldots \ldots \ldots \ldots .$

$= - \left(- 11.46\right) = 11.5$

For a few more details see this old answer.

May 27, 2017

pH = 11.5