Water is known to undergo an autoprotolysis reaction as shown, and this has been accurately measured under standard conditions.

#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#

As for any equilibrium, we can write the equilibrium expression:

#K'_w=([H_3O^+][HO^-])/([H_2O(l)])#

Now because #[H_2O]# is effectively a constant, so this can be included on the left had side of the equation,

#K_w=[H_3O^+][HO^-]# #=# #10^-14# at #298*K#

We can take #log_(10)# of both sides to get:

#log_(10)K_w=log_(10)[H_3O^+] + log_(10)[HO^-]#

OR,

#-log_(10)K_w=-log_(10)[H_3O^+] - log_(10)[HO^-]#

But by definition, #- log_(10)[HO^-]=pOH#, and #- log_(10)[H_3O^+]=pH#, and #-log_(10)K_w=-log_10(10^-14)=14#.

Thus #pH + pOH =14#

This equation tells us that in acid solutions, #pH# is low, and in alkaline solutions, #pOH# is low, but the equilibrium relationship between #H_3O^+# and #HO^-# is maintained.