# Question #f6709

Jun 4, 2016

See explanation below:

#### Explanation:

When comparing two acids , we look at their conjugate bases stabilities and compare them.

A strong acid will give a stable conjugate base , so the reaction is favoured in the forward direction (acid dissociation).

For $H F$:

$H F \to {H}^{+} + {\underbrace{{F}^{-}}}_{\text{conjugate base}}$

For $C {H}_{4}$:

$C {H}_{4} \to {H}^{+} + {\underbrace{C {H}_{3}^{-}}}_{\text{conjugate base}}$

Looking at both conjugate bases, we can see that the negative charge is carried by fluorine for ${F}^{-}$ and by the carbon atom of $C {H}_{3}^{-}$.

Since fluorine is a more electronegative than carbon atom, the negative charge will be more stabilized on fluorine and thus ${F}^{-}$ is a more stable conjugate base than $C {H}_{3}^{-}$.

Therefore, $H F$ is a stronger acid than $C {H}_{4}$.