# Why is the ionic product of water $K_w$ the same in acid, water, and alkali? [duplicate]

Seen on the web:

The concentration of [H+] and [OH−] both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When [H+] goes up, [OH−] goes down, in the same proportion.

The word "remarkable" was used to describe the ionic product of water (Kw). Ideally, it should not seem more remarkable than anything else, there should be simple explanations that make sense. Why is Kw constant?

• You seem to be implying that something that has an explanation that makes sense can't be remarkable? – Aaron Stevens Mar 21 '20 at 14:01
• Also you should give where you found the quote. "On the web" isn't helpful nor a valid way to give credit. – Aaron Stevens Mar 21 '20 at 14:06
• "Bohr was inconsistent, unclear, willfully obscure and right. Einstein was consistent, clear, down-to-earth and wrong. " John Bell on the Bohr-Einstein debates. I'm afraid in real physics "simple explanations that make sense" don't always work, even for genius. – StephenG Mar 21 '20 at 14:27
• @AaronStevens I'm implying that explanations and descriptions are different (but related) things. Keppler observed, for example, his third law, empirically. Surface tension on water, another empirical observation. That it is a result of the gel-phase of water, or "ionized ice", still not very broad consensus, socially. – Rwi1 Mar 21 '20 at 17:18
• Another way to put that, people tend to conflate cause and effect. They'll observe an effect, and, a bit lazily, ascribe it as the cause. Lazy thinking. Common in the history of ideas, "generatio spontanea". – Rwi1 Mar 21 '20 at 17:20

The equilibrium reaction for the auto-dissociation of water is:

$$2\text{ H}_2 \text{O}(l) \leftarrow \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$ The associated equilibrium constant $$K_w$$ is: $$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}$$ (Strictly speaking the expression is: $$\frac{[\text{H}_3\text{O}^+]\times [\text{OH}^-]}{[\text{H}_2\text{O}]}$$

But because $$[\text{H}_2\text{O}]=\text{constant}$$ for dilute solutions we can use the first expression.)

The reaction rates for the reactions are, where $$k$$ are the rate constants:

Forward reaction:

$$\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$

$$v_f=k_f[\text{H}_2\text{O}]$$

Reverse reaction:

$$\text{H}_2\text{O}(l) \leftarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$

$$v_r=k_r[\text{H}_3\text{O}^+][\text{OH}^-]$$ We can now show easily that:

$$\boxed{K_w=\frac{k_f}{k_r}}$$

Because in our case the rate constants $$k$$ are only dependent on temperature (see Arrhenius) that means that $$K_w$$ is a temperature (only) dependent constant.

One can also say that any equilibrium constant $$K$$ is related to the change of free enthalpy $${\Delta G°}$$ of the particular reaction :

$${\Delta G° = RT lnK}$$ And this $${\Delta G°}$$ does not depend on any other compound present in the solution.