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

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Answer 1 · 2016-06-04T18:32:25

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

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?

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