WE know from classic experiments that water undergoes autoprotolysis, i.e. self-ionization.
We could represent this reaction by (i):
2H2O(l)⇌H3O++HO−
OR by (ii):
H2O(l)⇌H++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
H5O+2 or H7O+3, i.e. a cluster of 2 or 3 or 4 water molecules with an EXTRA H+ tacked on.
We can use H+, protium ion or H3O+, hydronium ion equivalently to represent this species.
The equilibrium constant for the reaction, under standard conditions, is..........Kw=[H3O+][−OH]=10−14.
And so Kw=[H3O+]2 because [HO−]=[H3O+] at neutrality, and thus..........
[H3O+]=[HO−]=√10−14⋅mol2⋅L−2=10−7⋅mol⋅L−1
And to make the arithmetic a bit easier we can use the pH function, where pH=−log10[H3O+], and pOH=−log10[HO−]
And thus in aqueous solution under the given standard conditions, pH+pOH=14. And when pH is BELOW 7, NECESSARILY [H3O+]>[HO−]. If we got conc, HCl, [HCl]=10.6⋅mol⋅L−1; pH=−log10(10.6)=−(1.02)=−1.02.
How do you think pH and pOH would evolve under non-standard conditions, i.e. at a temperature of say 373⋅K? Would Kw decrease, remain constant, increase? Remember that the reaction as written is A BOND BREAKING reaction. How would this inform your choice?
