Question

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

Answer 1

`[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.