In neutral water under standard conditions, what are [H_3O^+][H3O+], and [HO^-][HO]?

1 Answer
Apr 19, 2017

[H_3O^+]=[HO^(-)]=10^(-7)*mol*L^-1...................

Explanation:

By careful measurement at 298*K, K_w,"the ion product" -=

[H_3O^+][HO^-]=10^(-14). Since pure water is neutral, [HO^-]=[H_3O^+], and thus [HO^-]=[H_3O^+]=sqrt(K_w).

=sqrt(10^-14)=10^-7*mol*L^-1.

At higher temperatures, how do you think K_w would evolve? Remember that the given equation is a BOND BREAKING reaction.