We assess the autoprotolysis reaction of water:
#H_2O(l) rightleftharpoonsH^(+) +HO^-#
Equivalently we could write this is as...........
#2H_2O(l) rightleftharpoonsH_3O^(+) +HO^-#
And #K_w=[H_3O^+][HO^-]=10^-14# at #298*K#
We can take #log_10# of BOTH sides to give........
#log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14#
#log_10[H_3O^+]+log_10[HO^-]=-14#, or, multiplying each side by #-1#
#-log_10[H_3O^+]-log_10[HO^-]=+14#,
But by definition, #-log_10[H_3O^+]=pH# and #-log_10[HO^-]=pOH#. And so after all that..........
#pH+pOH=14#, which is our defining relationship.
At LOW #pH#, #[H_3O^+]# is HIGH and #[HO^-]# is low; and at high #pH#, #[H_3O^+]# is LOW and #[HO^-]# is HIGH.
When #pH=pOH#, #pH=7#. Why?
Remember the definition of the #"log function"#. If #log_(a)b=c#, then #a^c=b#, and thus #log_(10)1=0#, #log_(10)10=1#, #log_(10)10^-1=-1#,#log_(10)100=2#.
