[Roddilla]

Is Kw reliant on how much water dissociates or on the Ka and Kb of an acid and its conjugate base pair? Or are they interdependent because the OH- formed by conjugate base must react with the H produced by the acid though not all because there is a lot of H ions?

[Unknown008]

Kw is a constant. It doesn't change at all, unless temperature is changed.

I'm sure you know by now how to get Kw.

Ka and Kb are completely different things. They refer to the dissociation constant of an acid and a base respectively; not to water.

[Roddilla]

first and foremost thanks a lot for your help. I always find you when I come up with a stupid idea but which is at the same time intriguing.

secondly, so why do we say that Ka * Kb = Kw if you said that Kw depends solely on water? According to the equation is seems as if Kw depends on the dissociation of the acid and the conjugate base.

[Unknown008]

Kw is a constant, as I told you. It is equal to:



Which makes:

And as you probably know:

pH + pOH = 14

That said, I prefer not to mess up the equilibrium constants.

The Ka can be anything, yet you will still get Kw when you multiply the Ka of the acid with the Kb of the conjugate base.

Thus, Kw does not depend on Ka, since there will always be a Kb which makes the product become Kw.

If the world somehow got different where the concentrations of H ions and OH ions in water change, then Kw would change (actually, this happens when you change the temperature of water. It doesn't change much, but it does).

[Roddilla]

To conclude:
If I have ethanoic acid + water <-----> ethanoate + H3O+
The ethanoate + water <-----> ethanoic acid + OH-

Some of the H3O + ions react with OH - ions react, though not all since there are much more H30+ ions than OH- ions or else we wouldn't say that ethanoic acid is a bronsted-lowry acid.
The H3O+ and OH- are at equilibrium with water, thus Kw.
Kw cannot change at a temperature since it depends on the reaction between H3O+ and OH- ions.

If you have a very strong acid the second equilibrium hardly occurs and the very few OH- which form react immediately with H+ to form water. Thus pOH is nearly O. The OH- ions which form from the auto dissociation of water (1 x 10^-7) also end up reacting with some H+ thus making pOH nearly 0.

If you have a weaker acid the second equilibrium forms more since the conjugate base is a stronger base and accepts more H+ from water. If that is so, more OH- are present when compared to strong acid and pOH is lower (i.e. more alkaline) while pH is a higher than in strong acid.


Ss shown the Kw of water is created because of the auto dissociation of water. This already existing equilibrium is changed a bit because of different amounts of H+ and OH- formed from acid/conjugate base or base/conjugate acid. The stronger the acid the more H+ ions are added to equilibrium as compared to OH- while the more OH- added as compared to H+ shifts the equilibrium in such a way that still OH- predominates. H2O + H2O <-----> H3O+ + OH-
Without this pre-existing equilibrium things would be different.