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Question

(A) The dissociation constant $K$ for water is given by the expression $K = \frac{[H^+][OH^-]}{[H_2O]}$.At $25^{\circ}C$,the ionic product of water $K

Vedclass · Accepted Answer

$10^{-14} 	imes (55.4)^{-1}$

Vedclass · Answer

(A) The dissociation constant $K$ for water is given by the expression $K = \frac{[H^+][OH^-]}{[H_2O]}$.At $25^{\circ}C$,the ionic product of water $K_w = [H^+][OH^-] = 10^{-14}$.The molar concentration of water $[H_2O]$ is calculated as $\frac{1000 \ g/L}{18 \ g/mol} \approx 55.4 \ mol/L$.Therefore,$K = \frac{10^{-14}}{55.4} = 10^{-14} 	imes (55.4)^{-1}$.

What is the dissociation constant of water at $25^{\circ}C$?

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