We interrogate the equilibrium:
#2H_2O(l) rightleftharpoons H_3O^(+) + HO^(-)#
Now this equilibrium has been carefully measured under standard conditions, and at #298*K#:
#[H_3O^+][HO^-]=10^(-14)#
This is a mathematical relationship, we can divide, multiply or otherwise manipulate. One thing we can do is to take #-log_10# OF BOTH SIDES,
#-log_10[H_3O^+]-log_10[HO^-]=-log_(10)10^-14#
But the RHS, #-log_(10)10^-14=-(-14)=14#, by the very definition of the #"logarithmic function"#. And thus,
#+14=-log_10[H_3O^+]-log_10[HO^-]#
And, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH#. This is simply by definition of the #p# function, and thus,
#+14=pH+pOH#
And thus we are given that #[H_3O^+]=1.55xx10^-3*mol*L^-1#, and so #pH=-(-2.81)=2.81.#
And so #pOH=11.19#
Capisce?
