As you know, in aqueous solution at #298*K#, #pH# #=# #-log_10[H_3O^+]#, #pOH# #=# #-log_10[HO^-]#, AND #pH+pOH=14#. Back in the day, before the advent of electronic calculators, students and engineers would routinely use log tables to enable complex multiplication and division. If I have a basic solution, I can thus calculate #[H_3O^+]# on the basis of #pOH# by means of simple addition or subtraction rather than by more complex multiplication/division.
If you are still unsure, come back here, and someone will help you.
In the meantime, #K_w# #=([H_3O^+][HO^-])=10^(14))# under standard conditions, i.e. #298*K#, and #1*atm#. If we were to increase the temperature, say to water's normal boiling point, #373*K#, how would you predict #K_w# and #pK_w# to evolve?
Remember that the autoprotolysis of water is an example of a bond-breaking reaction.
#2H_2O(l) rarr H_3O^+ + HO^-#
