In neutral water under standard conditions, what are $[H_{3} O^{+}]$ $[H_3O^+]$ , and $[H O^{-}]$ $[HO^-]$ ?

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Answer 1 · 2017-04-19T18:49:25

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

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.

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