We can write the equilibrium as follows:
#2H_2OrightleftharpoonsH_3O^+ + HO^-#
And, #K_w# #=# #[H_3O^+][HO^-]#. This equilibrium has been carefully measured (by conductivity experiments) over a range of temperatures. #[H_2O]# does not appear in the equilibrium because its concentration is effectively constant.
At #298*K# we know that #K_w = 10^-14.# How do we know this? By careful and repeated measurements of the conductivity of the water solvent.
So #K_w = [H_3O^+][HO^-]=10^-14.#
Now this is an arithmetic expression, that we can divide, multiply, add to, subtract from, provided that we do it to both sides of the equation. We can take #log_10# of both sides to give:
#log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=log_10(10^-14)#
But, by definition of logarithms, if #a^b=c#, #log_(a)c=b#, then #log_10(10^-14)=-14#, and thus,
#log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=-14#
But by definition, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH#
#pK_w = pH+ pOH=14# as required.
And after all that we learn that #pOH+pH=14# in water under standard conditions....
And so here #pOH=-log_10(0.005)=2.30#..
#pH=14-2.30=11.70#
