# Question #897fe

Jul 20, 2015

Here's why that happens.

#### Explanation:

As you know, the strength of an acid is determined by the amount of protons it realeases in aqueous solution.

This means that an acid's degree of dissociation will determine its strength

• Complete or significant dissociation $\to$ stronger acids;
• Partial dissociation $\to$ weaker acid;

So, hydrofluoric acid is indeed weaker than hydroiodic acid because it does not dissociate to the same extent that $H I$ does.

There are actually two points of view to take into consideration here

• how tight is the proton being held
• how stable the resulting anion is

Let's take the first one. The difference in electronegativity between fluorine, which is the most electronegative element in the periodic table, and hydrogen causes the bond that forms between the two atoms to be very polar.

This means that the bonding electrons will spend most of their time around fluorine, leaving fluorine with a partial negative charge and hydrogen with a partial positive charge.

By comparison, the bond between hydrogen and iodine is much less polar.

This is important because the more polar a covalent bond is, the easier it is for the hydrogen to be plucked off by the water molecules.

If you go by this factor alone, $H F$ would be the stronger acid. However, this is where the stability of the resulting anion comes into play.

Fluorine is a very small atom. That means that the fluoride anion, ${F}^{-}$, which results when $H F$ is deprotonated, will be very unstable.

That happens because the negative charge must be distributed on a very small surface, i.e. it will be much more concentrated.

By comparison, iodine is a much larger atom than fluorine is. This means that the negaative charge that the iodide anion, ${I}^{-}$, develops when the $H I$ molecule is deprotonated can be distributed on a much larger surface area, i.e. it will be less concentrated.

This in turn increases the stability of the anion considerably.

As it turns out, these two factors are competing against each other. When it comes to the ease with which the proton can be lost, $H F$ wins hands-down.

But the significant stability that the iodide anion has makes $H I$ a much stronger acid than $H F$.

In fact, this is why hydrofluoric acid is the only hydrohalic acid (compounds that have hydrogen bonded to a halogen) that's not considered a strong acid.