Chapter 18, Problem 111A

Interpretation Introduction

Interpretation:

The effect of change in temperature on $pH$ value of water is to be explained.

Concept introduction:

Ionic product of water is given by product of hydrogen and hydroxyl ion concentration.

$k_w=\left[H^{+}\right]\times \left[O{H}^{-}\right]$

Its value is fixed at fixed temperature.

Answer to Problem 111A

It is incorrect to say that $pH$ of pure water is always $7$. It is $7$ at $298K$

Explanation of Solution

In pure water,

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

So,

$\begin{array}{l}{k}_w={\left[H^{+}\right]}^2\\ \left[H^{+}\right]=\sqrt{k_w}\end{array}$

At $10°C$

$k_w=2.92\times {10}^{-15}$

$\begin{array}{c}\left[H^{+}\right]=\sqrt{2.92\times {10}^{-15}}\\ =5.40\times {10}^{-8}M\\ pH=-\log (5.40\times {10}^{-8})\\ =7.268\end{array}$

At $25°C$

$k_w=1\times {10}^{-14}$

$\begin{array}{c}\left[H^{+}\right]=\sqrt{1\times {10}^{-14}}\\ =1\times {10}^{-7}M\\ pH=-\log (1\times {10}^{-7})\\ =7\end{array}$

At $40°C$

$k_w=2.92\times {10}^{-14}$

$\begin{array}{c}\left[H^{+}\right]=\sqrt{2.92\times {10}^{-14}}\\ =1.71\times {10}^{-7}M\\ pH=-\log (1.71\times {10}^{-7})\\ =6.767\end{array}$

Conclusion

Thus, $pH$ of pure water is not always $7$. It is $7$ at $298K$