I was wondering which is more soluble in water, hydrogen fluoride or sulfur dioxide? Since $\ce{HF}$ will hydrogen bond to other HF particles (as opposed to $\ce{SO2}$ dipole-dipole attraction).

  • Wouldn't $\ce{SO2}$ dissolve better as it would be easier for the $\ce{H2O}$ particles to dissociate?
  • Or would $\ce{HF}$ dissolve more readily as it will form hydrogen bonds with the $\ce{H2O}$ particles?
  • 1
    $\begingroup$ You're on the right track. I'm not quite sure I understand your reason for SO2 being more soluble though...could you expand or restate that? $\endgroup$
    – airhuff
    Mar 25 '17 at 22:54

There are a couple different forces at play here. One big difference between the compounds is that $\ce{SO2}$ is much more volatile, with a boiling point of $\pu{-10 ^\circ C}$; in other words it's a gas. It does strongly hydrogen bond with water however, and is pretty soluble at $\pu{94 g/L}$.

$\ce{HF}$ is also pretty volatile, but much less so than $\ce{SO2}$ with a boiling point of $\pu{20 ^\circ C}$. It also is a more polar molecule than $\ce{SO2}$ and hydrogen bonds even stronger with water. In fact, $\ce{HF}$ is fully miscible with water, so clearly that would be the more water soluble compound.

To summarize, the water solubility of $\ce{SO2}$ is $\pu{94 g/L}$, whereas $\ce{HF}$ is fully miscible, thus $\ce{HF}$ is the more soluble compound in water.

Reference for all solubility and melting point data:
Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.


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