Ionisation of Water

Pure water itself has a slight tendency to dissociate -- one water molecule donates a proton to another, called the auto-ionisation of water: $H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$, in which one molecule acts as the acid and the other as the base. Its dissociation constant is $K = \dfrac{[H_3O^+][OH^-]}{[H_2O]^2}$; treating the concentration of pure liquid water as effectively constant (taken as 1) gives the ionic product of water, $K_w = [H_3O^+][OH^-]$. Experimentally, $[H_3O^+]$ in pure water is $1\times10^{-7}$ M at $25^\circ C$, and since auto-ionisation produces equal numbers of $H_3O^+$ and $OH^-$, $[OH^-]$ is also $1\times10^{-7}$ M, giving $K_w = (1\times10^{-7})(1\times10^{-7}) = 1\times10^{-14}$ at $25^\circ C$. Like every equilibrium constant, $K_w$ is temperature-dependent -- since water's dissociation is endothermic, $K_w$ increases as temperature rises. In a neutral solution such as aqueous $NaCl$, $[H_3O^+]=[OH^-]$; but once an acidic or basic solute is added, that balance shifts -- e.g. in aqueous $HCl$, the dissociation of $HCl$ adds extra $H_3O^+$ on top of water's own auto-ionisation, so $[H_3O^+] > [OH^-]$ and the solution is acidic, while in aqueous $NH_3$ or $NaOH$, $[OH^-] > [H_3O^+]$ and the solution is basic.

Example 8.1 – [OH⁻] in a fruit juice. Calculate $[OH^-]$ in a fruit juice with $[H_3O^+]=2\times10^{-3}$ M. From $K_w=[H_3O^+][OH^-]$: $[OH^-]=\dfrac{K_w}{[H_3O^+]}=\dfrac{1\times10^{-14}}{2\times10^{-3}}=5\times10^{-12}$ M. Since $[H_3O^+] (2\times10^{-3}) \gg [OH^-] (5\times10^{-12})$, the juice is acidic in nature. …