Question

$K_w$ for water is $9.62\times {10}^{-14}$ at ${60}^{o}C$. Calculate $pH$ of water.

Answer 1

As we know that,

$K_w=\left[H^{+}\right]\left[{OH}^{-}\right]$

As the water is neutral, and for a neutral solution-

$\left[H^{+}\right]=\left[{OH}^{-}\right]$

$\therefore K_w={\left[H^{+}\right]}^2$

$\because K_w=9.62\times {10}^{-14}\left(\text{Given}\right)$

$\Rightarrow {\left[H^{+}\right]}^2=9.62\times {10}^{-14}$

$\Rightarrow \left[H^{+}\right]=\sqrt{9.62\times {10}^{-14}}=31\times {10}^{-8}$

Now, as we know that,

$pH=-\log \left[H^{+}\right]$

$\therefore pH=-\log (3.1\times {10}^{-7})=6.5$

Hence the pH of water at $60℃$ is $6.5$.