What is #[H^+]# if #pH=4.18#?

1 Answer
Jul 20, 2017

#[H^+]=10^(-4.18)*mol*L^-1#

Explanation:

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

We take antilogs..........

#[H_3O^+]=10^(-4.18)*mol*L^-1=6.61xx10^-5*mol*L^-1#.

See https://socratic.org/questions/does-flat-coke-go-ph-neutral for background......and also [here.](https://socratic.org/questions/what-is-the-ph-of-a-1-00-x-10-4-m-solution-of-lithium-hydroxide-solution#306871)