Question

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

Answer 1

Ionic product,

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
Let $\left[H^{+}\right]=x$
Since $\left[H^{+}\right]=\left[O{H}^{-}\right],K_w=x^2$
$\Rightarrow K_{w}is=2.7\times {10}^{-14}\therefore 2.7\times {10}^{-14}=x^2\Rightarrow x=1.64\times {10}^{-7}\Rightarrow \left[H^{+}\right]=1.64\times {10}^{-7}\Rightarrow pH=-log\left[H^{+}\right]=-log[1.64\times {10}^{-7}]=6.78$
Hence, the pH of neutral water is 6.78.