water acts as both a weak acid and a weak base. it is more stable as H2O, but one one molecule can act as a nucleophile and take another’s H+. the result is H3O and OH. the H2O molecule is about 550,000,000 x more common than the hydronium ion or hydrogen ion, but the ions are highly reactive and therefore significant.
Hydronium (H3O) is the strongest acid that can exist in (aq) without dissociating. Hydroxide is a super strong base. As quickly as they dissociate, they react with each other to return to H2O form, unless they collide with something else reactive first.
If H3O donates a proton to a species other than OH, it returns to H2O. the [H] has decreased and [OH-] increased; the solution has become more basic.
OH and H3O are not in equilibrium with each other; this was a hard one for me to grasp. they are in equilibrium with H2O, although their formation never puts a dent in [H2O]. as one is lost, the concentration of the other goes up. the product (rather than sum) of their respective concentrations always equals 1e-14 (this being their equilibrium constant). this means that their concentrations increase and reciprocally decrease exponentially, which is still too much for my brain to wrap around.
edit: nope. 1e-14 is water’s ionization constant (=dissociation constant, equilibrium constant since water is a weak acid/base). this means that if ions are removed, add’l water molecules dissociate, and if ions are added, add’l water molecules are formed (unless one exceeds the other…). as a result, (H)(OH)=1e-14, and just because this is a tiny number and you can’t find an online calculator that provides that many significant figures doesn’t mean that the relationship does anything wacky. [H] = 1e-14/[OH]. it’s linear.
If OH acts as a nucleophile and reacts with a molecule other than H3O (such as CO2), the [OH] decreases, [H+] increases, and the solution becomes more acidic.