In neutral water at #298*K#, the following equilibrium operates:
#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#
Careful measurement has established that at #298*K#, the ion product, #[H_3O^+][""^(-)OH]=10^(-14)#.
If the solution is neutral then, #[H_3O^+]=[""^(-)OH]=10^(-7)mol*L^-1#.
If #[H_3O^+]<[""^(-)OH]#, then the solution is alkaline, and if #[H_3O^+]>[""^(-)OH]#, then the solution is acidic.
We can go a step farther than this, and define #pH# and #pOH#. If we take #-log_10# of both sides of the equation,#[H_3O^+][""^(-)OH]=10^(-14)#, we get #-log_10[H_3O^+]-log_10[""^(-)OH]=-14#.
If we define #pH=-log_10[H_3O^+]# and #pOH=-log_10[""^(-)OH]#, then,
#pH + pOH =14.#
From above, at neutrality, #pH = pOH =7#; i.e. #[H_3O^+]# and #[HO^-]# are EQUAL.
