Question

If `pH=5.25`, what is `[H_3O^+]`?

Answer 1

`[H_3O^+]` `=` `10^(-5.25)*mol*L^-1`.

Explanation:

`pH` `=` `-log_10[H_3O^+]`.

It is worthwhile reviewing your grasp of the logarithmic function.

When we write, `log_ab`, we ask to what power we raise the base `a` to get `b`. Thus `log_(10)100=2`, and `log_(10)0.1=-1`.