At 60,^{circ}C,the concentration of [H_3O^{+} | vedclass

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(D) In pure water,the dissociation reaction is: $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.Since the water is pure,the concentration of $

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$10^{-13.4}$

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(D) In pure water,the dissociation reaction is: $2H_2O(l) ightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.Since the water is pure,the concentration of $[H_3O^{+}]$ must be equal to the concentration of $[OH^{-}]$.Given: $[H_3O^{+}] = 10^{-6.7} \, 	ext{mol/L}$.Therefore,$[OH^{-}] = 10^{-6.7} \, 	ext{mol/L}$.The ionic product of water $(K_W)$ is defined as: $K_W = [H_3O^{+}][OH^{-}]$.Substituting the values: $K_W = (10^{-6.7}) 	imes (10^{-6.7}) = 10^{-6.7 - 6.7} = 10^{-13.4}$.

At $60\,^{\circ}C$,the concentration of $[H_3O^{+}]$ in pure water is $10^{-6.7} \, \text{mol/L}$. What is the value of $K_W$ at $60\,^{\circ}C$?

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