Why is the ionic product for water $10^{-14}$?
$$\ce{2H2O &lt;=&gt; H3O+ + OH-}$$

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

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.