Answer
The electron deficiency in $ \mathrm{AlF_3}$ explains its high insolubility, but it forms $\left[\mathrm{AlF}_{4}\right]^{-}$ in the presence of $\mathrm{F}^{-}$ from $\mathrm{KF}$, which facilitates the solubility. On the other hand, $\mathrm{BF}_{3}$ is a stronger Lewis acid than $ \mathrm{AlF_3}$ and abstracts a $\mathrm{F}^{-}$ ion from $\left[\mathrm{AlF}_{4}\right]^{-}$, yielding the formation of the insoluble $ \mathrm{AlF_3}$ once again as foolows:
$\left[\mathrm{AIF}_{4}\right]^{-}+\mathrm{BF}_{3} \longrightarrow \mathrm{AlF}_{3}(\mathrm{s})+\left[\mathrm{BF}_{4}\right]^{-}$
Work Step by Step
See answer.