Explanation:
[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The dissociation constant of the water is given as :
[tex]K=\frac{[H_3O^+][OH^-]}{[H_2O][H_2O]}[/tex]
[tex]K_w=K\times [H_2O][H_2O]=[H_3O^+][OH^-][/tex]
The value of [tex]K_w[/tex] st 25°C = [tex]1\times 10^{-14}[/tex]
[tex]K_w=[H_3O^+][OH^-][/tex]
[tex]1\times 10^{-14}=[H_3O^+][OH^-][/tex]
Water is neutral in nature with equal amount of hydronium and hydroxide ions.
[tex]1\times 10^{-14}=[H_3O^+][H_3O^+][/tex]
[tex][H_3O^+]=10^{-7} M[/tex]
[tex]pH=-\log[H_3O^+][/tex]
[tex]pH=-\log[10^{-7} M]=7[/tex]
The pH of the water at 25°C is 7.
With increase in temperature dissociation of water will increase with which concentration of hydronium and hydroxide ions will also increase which will in return will lower the down the pH of the water at higher temperature.
