We assess the autoprotolysis reaction of water:
H2O(l)⇌H++HO−
Equivalently we could write this is as...........
2H2O(l)⇌H3O++HO−
And Kw=[H3O+][HO−]=10−14 at 298⋅K
We can take log10 of BOTH sides to give........
log10[H3O+]+log10[HO−]=log1010−14
log10[H3O+]+log10[HO−]=−14, or, multiplying each side by −1
−log10[H3O+]−log10[HO−]=+14,
But by definition, −log10[H3O+]=pH and −log10[HO−]=pOH. And so after all that..........
pH+pOH=14, which is our defining relationship.
At LOW pH, [H3O+] is HIGH and [HO−] is low; and at high pH, [H3O+] is LOW and [HO−] is HIGH.
When pH=pOH, pH=7. Why?
Remember the definition of the log function. If logab=c, then ac=b, and thus log101=0, log1010=1, log1010−1=−1,log10100=2.
