# Ionic Product, Kw and lonisation Constant, K of water

Though water is regarded as a non-conductor, yet even the purest water has a small conductivity as it ionises to a very slight extent. $H_2O \leftrightharpoons H^+ + OH^-$

Since it is a reversible process, the law of mass action can be applied to the equilibrium, therefore, $K = \dfrac{[H^+][OH^-]}{[H_2O]}$

where K is called the Dissociation constant or ionisation constant of water.

Since the degree of ionisation of water is very low (nearly one molecule ionises out of 550 molecules), the concentration of ionised water may be assumed as constant. $\therefore [H^+][OH^-] = K[H_2O] K_w$ $K_w$  is a constant at a given temperature and is known as Ionic product of water and may be defined as: “The product of I-1+ and OH” ions concentration is constant at a constant temperature and is called the ionic product of water”. The value of ionic product of water is approximately equal to $10^{-14} at 25^0 Cor 298 K$. $\text{Therefore}K_w = [H^+][OH^-] \\[3mm] 1 \times 10^{-14}$

And $[H_2O]$ = no. of water moles per liter $= \dfrac{1000}{18} = 55.66$

Now substituting both the values in above equation, we get $K = \dfrac{[H^+][OH^-]}{[H_2O]} \\[3mm] = \dfrac{K_w}{[H_2O]} \\[3mm] 1 \times 10^{-14}{55.55} \\[3mm] = 1.8 \times 10^{-16}$

Hence the value of ionisation constant of water is $1.8 \times 10^{-16} \text{and} \dfrac{1}{55.55}$ times that of $K_w$, the ionic product of water.

It is evident from the above discussion that, if $[H^+] = [OH^-]:$ the solution is neutral $[H^+] > [OH^-]:$ the solution is acidic $[H^+] < [OH^-]:$ the solution is basic or alkaline. $pK_w = -\log_{10} K_w \\[3mm] = -\log_{10} 1 \times 10^{-14} \\[3mm] pK_w = 14$ At 298 K

When temperature is increased the value of $K_w$ also increases. For e.g., at $100^0C$ the value of $K_w \text{is} 5.5 \times 10^{-13}$

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