The pH of D_2O and H_2O at 298 K is | vedclass

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(B) At $298 \ K$,the ionic product of water $(K_w)$ is $1.0 	imes 10^{-14}$,so $pH = -\log[H^+] = 7.0$. For heavy water $(D_2O)$,the dissociation con

Vedclass · Accepted Answer

$7.35, 7.0$

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(B) At $298 \ K$,the ionic product of water $(K_w)$ is $1.0 	imes 10^{-14}$,so $pH = -\log[H^+] = 7.0$. For heavy water $(D_2O)$,the dissociation constant $(K_w')$ is approximately $1.6 	imes 10^{-15}$ at $298 \ K$. Since $pD = -\log[D^+]$,and $[D^+] = \sqrt{K_w'} = \sqrt{1.6 	imes 10^{-15}} \approx 4.0 	imes 10^{-8} \ M$. Therefore,$pD = -\log(4.0 	imes 10^{-8}) \approx 7.4$. Thus,the $pH$ (or $pD$) of $D_2O$ is approximately $7.35$ and for $H_2O$ it is $7.0$.

The $pH$ of $D_2O$ and $H_2O$ at $298 \ K$ is

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