Chapter 14.6, Problem 14.44QAP

Interpretation Introduction

Interpretation: ”The $pH$ of solution can be determined by $\left[O{H}^{-}\right]$”should be explained

Concept Introduction:

► $pH$ of a solution is defined as the negative logarithm of the hydrogen ion concentration of the aqueous solution.

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

► Water is a very weak conductor of electricity. This suggests that it undergoes self-ionization to a very small extent. Due to self ionization, the following equilibrium is set up between water and its ions.

$H_2{O}_{(l)}+H_2{O}_{(l)}\leftrightarrow H^{+}{}_{(aq)}+O{H}^{-}{}_{(aq)}$

Applying law of mass action, the equilibrium constant $K_c$ for the self ionization of water can be written as

$\begin{array}{l}{K}_c=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{{\left[H_{2}O\right]}^2}\\ K_c{\left[H_{2}O\right]}^2=\left[H^{+}\right]\left[O{H}^{-}\right]\end{array}$

Since degree of ionization of water is very low, its ionization does not appreciably changes the concentration of water molecules. Therefore ${\left[H_{2}O\right]}^2$ may be regarded as constant.

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

where $K$ is a constant representing ${\left[H_{2}O\right]}^2$

Now,

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

Where $K_w$ is the product of two constant $K_{c}andK$

$K_w$ is called the ionic product of water and its value is $1\times 10^{-14}$