The pH of pure water at 80^{circ} C is | vedclass

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(D) At $25^{\circ} C$,the ionic product of water $(K_w)$ is $1.0 	imes 10^{-14}$.For pure water,$[H^{ }] = [OH^{-}] = \sqrt{K_w} = 10^{-7} \ M$,which

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$< 7.0$

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(D) At $25^{\circ} C$,the ionic product of water $(K_w)$ is $1.0 	imes 10^{-14}$.For pure water,$[H^{ }] = [OH^{-}] = \sqrt{K_w} = 10^{-7} \ M$,which gives $pH = 7$.The auto-ionization of water is an endothermic process $(\Delta H > 0)$.According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of $K_w$.At $80^{\circ} C$,$K_w > 10^{-14}$,which implies $[H^{ }] > 10^{-7} \ M$.Since $pH = -\log[H^{ }]$,an increase in $[H^{ }]$ results in a decrease in $pH$.Therefore,at $80^{\circ} C$,the $pH$ of pure water is $< 7.0$.

The $pH$ of pure water at $80^{\circ} C$ is

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