What is Autoprotolysis, ionic product, pH?

Water undergoes autoprotolysis, $\ce{2H2O <=> H3O+ + OH-}$, of constant $K_e = [\ce{H3O+}][\ce{OH-}]$, the ionic product of water; at $25\,{}^{\circ}\mathrm{C}$, $K_e = 10^{-14.00}$, so $\mathrm{p}K_e = 14.00$. The pH of a solution is $\mathrm{pH} = -\log a(\ce{H3O+}) \approx -\log[\ce{H3O+}]$. A solution is neutral when $[\ce{H3O+}] = [\ce{OH-}]$, that is at $\mathrm{pH} = \mathrm{p}K_e/2 = 7.00$ at $25\,{}^{\circ}\mathrm{C}$.