The former would be all you needed to write in an exam. But as to background, we KNOW from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.
We could represent this reaction by #(i)#:
#2H_2O(l) rightleftharpoonsH_3O^+ +HO^-#
OR by #(ii)#:
#H_2O(l) rightleftharpoonsH^+ +HO^-#
Note that #(i)# and #(ii)# ARE EQUIVALENT REPRESENTATIONS, and it really is a matter of preference which equation you decide to use. As far as anyone knows, the actual acidium ion in solution is
#H_5O_2^+# or #H_7O_3^+#, i.e. a cluster of 2 or 3 or 4 (or more) water molecules with an EXTRA #H^+# tacked on. We can use #H^+#, #"protium ion"# (as is used here) or #H_3O^+#, #"hydronium ion"# equivalently to represent this species.
The equilibrium constant for the reaction, under standard conditions, is..........#K_w=[H_3O^+][""^(-)OH]=10^-14#.
And to make the arithmetic a bit easier we can use the #pH# function, where #pH=-log_10[H_3O^+]#, and #pOH=-log_10[HO^-]#. And thus in aqueous solution under the given standard conditions, #pH+pOH=14#.
And since we are directly given #[H^+]# or #[H_3O^+]#.....
#pH=-log_10[H_3O^+]=-log_10(2.3xx10^-3)=2.64#
How do you think #pH# and #pOH# would evolve under non-standard conditions, i.e. at a temperature of say #373*K#? Would #K_w# decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?
