Question

What is `[H_3O^+]` for a solution whose `pH-=2.85`?

Answer 1

Well, first we had better specify what `pH` is..........

Explanation:

And I assume that `2.85=-log_10[H^+]`

And taking antilogs..........

`10^(-2.85)=[H^+]`

So `[H^+]=1.41xx10^-3*mol*L^-1.............`

See here for an example...........