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

1 Answer
Jun 4, 2016

Answer:

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?