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6-2.Equilibrium-II (Ionic Equilibrium)
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Ionic product of water at $310 \,K$ is $2.7 \times 10^{-14}$. What is the $\mathrm{pH}$ of neutral water at this temperature?
Option A
Option B
Option C
Option D
Solution
Ionic product, $K_{w}=\left[ H ^{+}\right]\left[ OH ^{-}\right]$
Let $\left[ H ^{+}\right]=x$
Since $\left[ H ^{+}\right]=\left[ OH ^{-}\right], K_{ w }=x^{2}$
$\Rightarrow K_{ w }$ at $310 \,K$ is $2.7 \times 10^{-14}$.
$\therefore 2.7 \times 10^{-14}=x^{2}$
$\Rightarrow x=1.64 \times 10^{-7}$
$\Rightarrow\left[ H ^{+}\right]=1.64 \times 10^{-7}$
$\Rightarrow pH =-\log \left[ H ^{+}\right]$
$=-\log \left[1.64 \times 10^{-7}\right]$
$=6.78$
Hence, the $pH$ of neutral water is $6.78$
Standard 11
Chemistry
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