What is the concentration of [H_3O^+] in dist | vedclass

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(C) In pure distilled water at $298 \; K$,the water undergoes auto-ionization according to the equation: $2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^

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$1 	imes 10^{-7} \; M$

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(C) In pure distilled water at $298 \; K$,the water undergoes auto-ionization according to the equation: $2H_2O(l) ightleftharpoons H_3O^+(aq) + OH^-(aq)$.For pure water,the concentration of hydronium ions $[H_3O^+]$ is equal to the concentration of hydroxide ions $[OH^-]$.The ionic product of water $(K_w)$ at $298 \; K$ is $1 	imes 10^{-14}$.Since $K_w = [H_3O^+][OH^-]$,and $[H_3O^+] = [OH^-]$,we can write $K_w = [H_3O^+]^2$.Therefore,$[H_3O^+] = \sqrt{K_w} = \sqrt{1 	imes 10^{-14}} = 1 	imes 10^{-7} \; M$.

What is the concentration of $[H_3O^+]$ in distilled water at $298 \; K$?

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