Q.Define ionic product of water. Give its value at room temperature.

Pure water has a slight tendency to auto-ionise: one water molecule donates a proton to another, $H_2O+H_2O \rightleftharpoons H_3O^++OH^-$, with one molecule acting as the acid and the other as the base. Its equilibrium constant is $K=\dfrac{[H_3O^+][OH^-]}{[H_2O]^2}$; since the concentration of pure liquid water is essentially constant (taken as 1), this simplifies to 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 $25^\circ C$.

Temperature dependence. Like every equilibrium constant, $K_w$ varies with temperature; since water's auto-ionisation is endothermic, $K_w$ increases with rising temperature -- from $1.14\times10^{-15}$ at $0^\circ C$ to $5.30\times10^{-14}$ at $50^\circ C$. …