# Why is the ionic product for water $10^{-14}$? [closed]

Why is the reference value of $K_w=10^{-14}\ \mathrm{M}^2$, such that $[\ce{H+}]= 10^{-7}\ \mathrm{M}$ and $[\ce{OH^-}]= 10^{-7}\ \mathrm{M}$ for pure water at $20\ ^\circ \mathrm{C}$?

For example, why isn't $K_w=10^{-18}\ \mathrm{M}^2$, and thus $[\ce{H+}]= 10^{-9}\ \mathrm{M}$ and $[\ce{OH-}]= 10^{-9}\ \mathrm{M}$?

## closed as too broad by hBy2Py, jerepierre, ringo, M.A.R. ಠ_ಠ, orthocresol♦Jan 14 '16 at 11:25

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• Questions of the "why" of specific numerical values of physical constants exit the realm of chemistry and enter the sphere of scientific theology. One might well ask in the same vein why Planck's constant is $6.626\times 10^{−34}\ \mathrm{J\cdot s}$ instead of $7.358 \dots$, or why the universal gravitational constant is $6.674\times 10^{−11}\ \frac{\mathrm{N\cdot m^2}}{\mathrm{kg^2}}$ instead of $5.123\dots$. – hBy2Py Dec 15 '15 at 17:02
• That being said, the mechanistic (if not explanatory) answer is that the quantum and statistical mechanics of the interactions of $\ce{H2O}$, $\ce{H3O+}$, and $\ce{OH-}$ in the condensed phase just work out that way. – hBy2Py Dec 15 '15 at 17:08
• @Brian see section 15.3.6 "Computation of the Water Dissociation Constant" in Bridging the Time Scales: Molecular Simulations for the Next Decade and see if still think it is the same vein as Planck's constant books.google.com/… – DavePhD Dec 15 '15 at 17:09
• @DavePhD It's not fundamental in the same way that Planck's constant is, I agree, but I'd argue the fact that it bears the value it is observed to have at a given set of conditions is scientifically unexplainable in the same fashion as is the fact of the value of $h$. – hBy2Py Dec 15 '15 at 17:11
• $K_w$ is actually measured for pure water. The value of $K_w$ does change with temperature. The value is $1 \times 10^{-14}$ at room temperature. – MaxW Dec 15 '15 at 17:16

$$\ce{2H2O <=> H3O+ + OH-}$$

$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$

where $a$ is "activity of".

Then you can approximate activity of water as 1 if appropriate and activity of the ions as concentration of the ions.

Just like any other equilibrium constant, the relative chemical potential (Gibbs energy) of the products and reactant determine the constant.

Consider the relative strengths of the bonds of $\ce{H2O}$, compare to those of $\ce{OH-}$ and $\ce{H3O+}$, as well as the number and strength of intermolecular hydrogen bonds to each of these.

For $\ce{D2O}$, the equilibrium constant is 8 times small.

• But why does the $\Delta G$ have the value such that $K_w=10^{-14}$? – hBy2Py Dec 15 '15 at 17:23
• @Brian see Ab Initio QM/MM Simulation with Proper Sampling:  “First Principle” Calculations of the Free Energy of the Autodissociation of Water in Aqueous Solution . Phys. Chem. B, 2002, 106 (51), pp 13333–13343 pubs.acs.org/doi/abs/10.1021/jp021625h – DavePhD Dec 15 '15 at 17:33
• But why does this highly accurate QM/MM simulation with proper sampling provide this particular $\Delta G$ value? Why not a different one? Why do the quantum and statistical mechanics give this numerical value and not another? – hBy2Py Dec 15 '15 at 17:38
• @Brian yes, you can keep asking way, but we should get to a reasonable set of constants that QED and QCD are based upon. – DavePhD Dec 15 '15 at 17:42
• Agreed! And then we reach the parallel to Planck's constant: the values of the QED and QCD constants, of which $h$ is one, are "scientifically arbitrary" -- science cannot explain why they hold the values they do. Thus, values derived from these constants are similarly "arbitrary," in this sense. Depending on how fundamental an explanation Mehdi is looking for, your answer is either spot-on, or no answer can be given. – hBy2Py Dec 15 '15 at 17:49