Which is more soluble, sodium iodide, or sodium fluoride?

Jul 9, 2017

I would presume $N a I$ would be the more soluble in aqueous solution.......

Explanation:

So before we interrogate the data, why should I so presume? The iodide ion is MUCH larger than the fluoride ion. The degree of electrostatic interaction between the gegenions is $\text{a priori}$ greater for the sodium cation, and fluoride anion, with the result that this salt should be the LEAST soluble.

But chemistry is an experimental science, first and foremost. Are the data consistent with my (potentially) whack reasoning? Let's see.......

This site quotes a solubility of $40.4 \cdot g \cdot {L}^{-} 1$ at $20$ ""^@C for $N a F$; on t'other hand this site quotes $184.2 \cdot g \cdot {L}^{-} 1$ at $25$ ""^@C for $N a I$ (the temperatures are close enuff!). This translates to molar solubilities of $0.962 \cdot m o l \cdot {L}^{-} 1$ for $N a F$, versus $1.229 \cdot m o l \cdot {L}^{-} 1$ for $N a I$ (assuming little volume change). And thus, reasonably, $N a I$ is the more soluble salt, and we predict this on the basis of the reduced electrostatic force between the larger anion, and the metal cation.

The smaller fluoride anion in solution is also more polarizing, and potentially induces more solvent order upon solvation. Certainly, the reduced acidity of $H F$, a weak acid, versus strong acids $H C l$, $H B r$, and $H I$, is well-known to be an entropy effect. Confused yet?

Anyway $\Delta {G}^{\circ} \text{_"dissolution} \left(N a I\right)$ is thus LESS (i.e. more negative) than that for $N a F$ both on grounds of $\Delta {H}^{\circ}$ and $\Delta {S}^{\circ}$, and thus the dissolution reaction is more favourable for sodium iodide.

Note that because of the size mis-match you can also get iodide salts up into solvents such as acetone. This is something that you cannot do with the lower halides......