The pH of pure water or neutral solution at 5 | vedclass

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(D) For a neutral solution,$[H^{+}] = [OH^{-}]$.Given $pK_w = 13.26$,we know that $K_w = [H^{+}] [OH^{-}] = 10^{-pK_w} = 10^{-13.26}$.Since $[H^{+}] =

Vedclass · Accepted Answer

$6.63$

Vedclass · Answer

(D) For a neutral solution,$[H^{+}] = [OH^{-}]$.Given $pK_w = 13.26$,we know that $K_w = [H^{+}] [OH^{-}] = 10^{-pK_w} = 10^{-13.26}$.Since $[H^{+}] = [OH^{-}]$,we have $[H^{+}]^2 = 10^{-13.26}$.Taking the square root,$[H^{+}] = 10^{-13.26 / 2} = 10^{-6.63}$.By definition,$pH = -\log[H^{+}] = -\log(10^{-6.63}) = 6.63$.Therefore,the correct option is $(d)$.

The $pH$ of pure water or neutral solution at $50\,^{\circ}C$ is .... ($pK_w = 13.26$ at $50\,^{\circ}C$)

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