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

1 Answer
anor277
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.

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