# Ionic product of water at 310 K is 2.7 × 10–14.

Question:

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

Solution:

Ionic product,

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

Let $\left[\mathrm{H}^{+}\right]=x$

Since $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right], K_{\mathrm{w}}=x^{2} .$

$\Rightarrow K_{w}$ at $310 \mathrm{~K}$ is $2.7 \times 10^{-14}$

$\therefore 2.7 \times 10^{-14}=x^{2}$

$\Rightarrow x=1.64 \times 10^{-7}$

$\Rightarrow\left[\mathrm{H}^{+}\right]=1.64 \times 10^{-7}$

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

$=-\log \left[1.64 \times 10^{-7}\right]$

$=6.78$

Hence, the pH of neutral water is 6.78.