$310$ $K$ તાપમાને પાણીનો આયનીય ગુણાકાર $2.7 \times 10^{-14}$ છે. આ તાપમાને તટસ્થ પાણીના દ્રાવણની $\mathrm{pH}$ કેટલી હશે ?

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 $6.78$