Question

If `[H_3O^+]=7.30xx10^-4*mol*L^-1`, what is `pH`?

Answer 1

`pH=3.14`

Explanation:

By definition, `pH=-log_10[H_3O^+]`...

...we are given here that..........

`[H^+]"/"[H_3O^+]=7.30xx10^-4*mol*L^-1`

And so `pH=-log_10[7.30xx10^-4]=-(-3.14)=??`