The value of the ion product constant for wat | vedclass

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(D) For a neutral solution at $60\,^{\circ}C$,$[H_3O^{+}] = [OH^{-}]$.Since $K_w = [H_3O^{+}][OH^{-}] = 9.6 	imes 10^{-14}$,we have $[H_3O^{+}]^2 = 9

Vedclass · Accepted Answer

$3.1 	imes 10^{-7}$,basic

Vedclass · Answer

(D) For a neutral solution at $60\,^{\circ}C$,$[H_3O^{+}] = [OH^{-}]$.Since $K_w = [H_3O^{+}][OH^{-}] = 9.6 	imes 10^{-14}$,we have $[H_3O^{+}]^2 = 9.6 	imes 10^{-14}$.Therefore,$[H_3O^{+}] = \sqrt{9.6 	imes 10^{-14}} \approx 3.1 	imes 10^{-7} \,M$.The neutral $pH$ at $60\,^{\circ}C$ is $pH = -\log(3.1 	imes 10^{-7}) \approx 6.51$.Since the neutral $pH$ is $6.51$,a solution with $pH = 7.0$ (which is greater than $6.51$) is basic in nature.

The value of the ion product constant for water,$(K_w)$ at $60\,^{\circ}C$ is $9.6 \times 10^{-14} \,M^2$. What are the $[H_3O^{+}]$ of a neutral aqueous solution at $60\,^{\circ}C$ and the nature of an aqueous solution with a $pH = 7.0$ at $60\,^{\circ}C$,respectively?

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