# If liquid hydrogen fluoride autoionises, how can it be an insulator?

According to Wikipedia, one of the last major hurdles in the isolation of elemental fluorine was the need to mix hydrogen fluoride with another substance in order for its electrolysis to be possible (Henri Moissan eventually hit upon potassium bifluoride, which is still used for this purpose today).

From "Fluorine Isolation":

Edmond Frémy postulated that electrolysis of pure hydrogen fluoride to generate fluorine was feasible and devised a method to produce anhydrous samples from acidified potassium bifluoride; instead, he discovered that the resulting (dry) hydrogen fluoride did not conduct electricity.

And from "Henri Moissan Preparation of elemental fluorine":

Moissan eventually succeeded in preparing fluorine in 1886 by the electrolysis of a solution of potassium hydrogen difluoride ($\ce{KHF2}$) in liquid hydrogen fluoride ($\ce{HF}$). The mixture was needed because hydrogen fluoride is a nonconductor.

However, hydrogen fluoride is a protic solvent, and undergoes autoionisation via the following reaction:

$$\ce{3 HF <=> H2F^+ + HF2^-}$$

Pure liquid $\ce{HF}$ should, therefore, conduct electricity at least weakly, like liquid water does.

What gives?

• PURE Water like HF is an excellent insulator - For a simplified version scienceabc.com/pure-sciences/… . Basically the concentration of the ions in the pure substance is too low to allow a reasonable amount of conduction – Ian Bush Jul 8 '18 at 21:12

Unlike metals which can conduct with free electrons, conduction in non-metallic liquids (ionic and covalent) is dependent on the ion concentration to carry charges. Pure water has a autoionization constant of $\rm K_w = 1\!\times\!10^{-14}$, which corresponds to a total ion concentration of $\rm 2\!\times\!10^{-7} \pu M$. At this concentration the resistivity of water is $\pu{18.2 M\Omega~ cm}$ which corresponds to a conductivity of $\pu{5.5 \mu S/cm}$ and is considered to be a poor conductor. Sea water on the other hand has about $\pu{3.4\%}$ sodium chloride and a conductivity of about $\pu{55,000 \mu S /cm}$ which is a million times improvement in conductivity which is considered conductive.

Hydrogen fluoride has an autoionization constant of $\rm 10^{-12.5} ~~(3.162\!\times\!10^-13)$ which corresponds to a total ion concentration of $\rm 1.125\!\times\!10^{-6}$ which is about five-times that of water. Five times the ions gives us five times the conductivity of $\pu{275\mu S/cm}$ or a resitivity of $\pu{2.28M\Omega ~cm}$ which is still a quite poor conductor. Thus it is necessary to add an ionic substance to increase conductivity to have appreciable conduction of the hydrogen fluoride similar to sea water.