Question

What is the `[H_3O^+]` in a solution with `[OH^-] = 2.3 * 10^-12` `M`?

Answer 1

The ion product `[HO^-][H_3O^+]=10^(-14)` at `1` `atm` and `298*K`.

Explanation:

Thus `[H_3O^+]=([10^-14])/([HO^-])`

Chemists typically use `pH` and `pOH`, where `pH=-log_10([H_3O^+])` and `pOH=-log_10([HO^-])`.

The use of logarithms allows us to use the relationship:

`pH+pOH=14` under standard conditions.

Under non-standard conditions, for instance at `373*K`, how do you think `K_w` would evolve?