Hydrogen fluoride $(HF)$ is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride $(AlF)$ does not dissolve in it. Sodium fluoride $(NaF)$ is an ionic compound and when it is added to the mixture, $AlF$ dissolves. This is because of the availability of free $F^{-}$. The reaction involved in the process is:

$Al{F_3} + 3NaF \to \mathop {N{a_3}[Al{F_6}]}\limits_{Sodium\,hexafluroalu\min ate\,(III)} $

When boron trifluoride $\left( BF _{3}\right)$ is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when $BF _{3}$ is added to the solution, $B$ replaces $Al$ from the complexes according to the following reaction:

$N{a_3}[Al{F_6}] + 3B{F_3} \to 3Na[B{F_4}] + Al{F_3}$