The dissociation constant of pure water at 18 | vedclass

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(B) The dissociation of water is given by: $H_2O \rightleftharpoons H^+ + OH^-$. The dissociation constant $K_w$ is defined as $[H^+][OH^-]$. In pure

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$1.8 	imes 10^{-16}$

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(B) The dissociation of water is given by: $H_2O ightleftharpoons H^+ + OH^-$. The dissociation constant $K_w$ is defined as $[H^+][OH^-]$. In pure water,$[H^+] = [OH^-] = \alpha 	imes [H_2O]$,where $\alpha$ is the degree of dissociation and $[H_2O]$ is the molar concentration of water. The molar concentration of water is $[H_2O] = \frac{1000 \, g/L}{18 \, g/mol} \approx 55.55 \, M$. Given $\alpha = 1.8 	imes 10^{-9}$,then $[H^+] = 1.8 	imes 10^{-9} 	imes 55.55 \approx 10^{-7} \, M$. Thus,$K_w = [H^+][OH^-] = (10^{-7}) 	imes (10^{-7}) = 10^{-14}$. However,using the provided $\alpha$ directly in the formula $K_w = \alpha^2 	imes [H_2O]^2$ is not standard; usually,$K_w$ is calculated as $[H^+][OH^-]$. Given the options provided and the standard value of $K_w$ at $25\,^oC$ being $10^{-14}$,the closest value at $18\,^oC$ is $1.8 	imes 10^{-16}$ based on the specific $\alpha$ provided.

The dissociation constant of pure water at $18\,^oC$ will be .......... (Degree of dissociation of pure water at $18\,^oC = 1.8 \times 10^{-9}$)

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