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

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Question

Ionic product,  $K_{w}=\left[ H ^{+}\right]\left[ OH ^{-}\right]$

Let $\left[ H ^{+}\right]=x$

Since $\left[ H ^{+}\right]=\left[ OH ^{-}\r

Vedclass · Accepted Answer

Vedclass · Answer

Ionic product,  $K_{w}=\left[ H ^{+}ight]\left[ OH ^{-}ight]$

Let $\left[ H ^{+}ight]=x$

Since $\left[ H ^{+}ight]=\left[ OH ^{-}ight], K_{ w }=x^{2}$

$\Rightarrow K_{ w }$ at $310 \,K$ is $2.7 	imes 10^{-14}$.

$	herefore 2.7 	imes 10^{-14}=x^{2}$

$\Rightarrow x=1.64 	imes 10^{-7}$

$\Rightarrow\left[ H ^{+}ight]=1.64 	imes 10^{-7}$

$\Rightarrow pH =-\log \left[ H ^{+}ight]$

$=-\log \left[1.64 	imes 10^{-7}ight]$

$=6.78$

Hence, the $pH$ of neutral water is $6.78$

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

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