# Which is the stronger acid, "HF" or "CH"_4? Why does "HF" ionize but "CH"_4 does not? What is the difference between ionization and dissociation?

Jul 20, 2016

To ionize is to dissociate, so that aside...

$\text{pKa}$ is a nice relative measure of acidity. The higher the $\text{pKa}$, the weaker the acid (and the stronger its bonds with its ${\text{H}}^{+}$).

$\text{HF}$ is an acid; it has an $\setminus m a t h b f \left({\text{H}}^{+}\right)$ on it, since the anion is ${\text{F}}^{-}$. Therefore, it can dissociate to donate an ${\text{H}}^{+}$ to solution, making it follow (at least) the definition of a Bronsted acid and that of an Arrhenius acid.

Its $\text{pKa}$ (the $- {\log}_{10}$ of the acid dissociation constant ${K}_{a}$) is about $3.17$, and if you place it in water, whose $\text{pKa}$ is higher, $\text{HF}$ will be the stronger acid.

The lower $p K a$ indicates the stronger acid.

${\text{CH}}_{4}$ on the other hand has a $\text{pKa}$ of about $\setminus m a t h b f \left(50\right)$, so it's very basic (or very poorly acidic). That means its $\text{H}$ will never in a million years want to dissociate in water as ${\text{H}}^{+}$; its $\text{pKa}$ is much higher than that of water.

Its $\text{C"-"H}$ bonds are too strong (thermodynamically stable), because the $\text{pKa}$ of ${\text{CH}}_{4}$ is too high, relative to water, meaning that ${\text{CH}}_{4}$ is a much weaker acid.

The equilibrium lies on the side of the weaker acid, so $C {H}_{4}$ will want to remain as $C {H}_{4}$.

But, this largely depends on the solvent. If these are placed into water, then yes, $\text{HF}$ dissociates partially, giving an acidic solution, and ${\text{CH}}_{4}$ won't do anything at all.