Question

If `pH=2` upon dilution what is `[HClO_4(aq)]`?

Answer 1

`[HClO_4]=0.01*mol*L^-1...................`

Explanation:

By definition, `pH=-log_(10)[H_3O^+]`

And thus, if `pH=2`, then.......................................

`[H_3O^+]=10^(-pH)=10^-2*mol*L^-1=0.01*mol*L^-1`.

And of course, when you dilute conc. acid, you always add acid to water. Why?

`"Because if you spit in acid, it spits backs!"`