Ionic product of water at 310 , K is 2.7 time | vedclass

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Question

(A) Ionic product,$K_{w} = [H^{+}] [OH^{-}]$. Let $[H^{+}] = x$. Since $[H^{+}] = [OH^{-}]$ for neutral water,$K_{w} = x^{2}$. Given $K_{w}$ at $310 \

Vedclass · Accepted Answer

$6.78$

Vedclass · Answer

(A) Ionic product,$K_{w} = [H^{+}] [OH^{-}]$. Let $[H^{+}] = x$. Since $[H^{+}] = [OH^{-}]$ for neutral water,$K_{w} = x^{2}$. Given $K_{w}$ at $310 \, K$ is $2.7 	imes 10^{-14}$. Therefore,$2.7 	imes 10^{-14} = x^{2}$. $x = \sqrt{2.7 	imes 10^{-14}} = 1.64 	imes 10^{-7}$. So,$[H^{+}] = 1.64 	imes 10^{-7} \, M$. $pH = -\log [H^{+}] = -\log [1.64 	imes 10^{-7}]$. $pH = 7 - \log(1.64) = 7 - 0.215 = 6.785 \approx 6.78$. Hence,the $pH$ of neutral water at $310 \, K$ is $6.78$.

Ionic product of water at $310 \, K$ is $2.7 \times 10^{-14}$. What is the $pH$ of neutral water at this temperature?

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