The concentration of hydrogen ion in pure wat | vedclass

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(B) In pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^+ + OH^-$. Since the concentration of $H^+$ ions equals the concentration

Vedclass · Accepted Answer

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

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(B) In pure water,the dissociation equilibrium is $H_2O ightleftharpoons H^+ + OH^-$. Since the concentration of $H^+$ ions equals the concentration of $OH^-$ ions,let $[H^+] = [OH^-] = x$. The ionic product of water $(K_w)$ at $298 \ K$ is $1 	imes 10^{-14}$. Therefore,$K_w = [H^+][OH^-] = x^2 = 1 	imes 10^{-14}$. Solving for $x$,we get $x = \sqrt{1 	imes 10^{-14}} = 1 	imes 10^{-7} \ M$.

The concentration of hydrogen ion in pure water at $298 \ K$ is:

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