HI is a stronger acid than HF. Why? Because when dissolved in water, the bigger iodide ion handles the negative charge way better than the small fluoride ion. So Iodide ion is a weak conjugate base making HI a strong acid.

Which one is more soluble in water, AgF or AgI? Answer is AgF. Why? It is an ionic compound and by Fajans' rules, an ionic bond shows more covalent character if it is more polarised. The smaller the anion, the less polarisation there will be, thus more ionic character, hence more solubility. But what if I said that AgI is more soluble in water, because what happens when AgI dissolves in water? It turns into Ag+ ion and I- ion. And I- handles the negative charge better than F-. What's wrong with that?

  • $\begingroup$ Ever heard of HSAB? $\endgroup$
    – Mithoron
    Jun 4, 2018 at 20:46
  • $\begingroup$ nope...but had a look in the wikipedia few moments ago. $\endgroup$
    – Sami
    Jun 6, 2018 at 14:27

1 Answer 1


When pondering solubility, you can't just think about what happens to the compound, you have to think about what happens to the solvent, too. Solubility is a balance between the loss of interactions between the particles of the compound and the gain of interactions between those particles and the solvent particles.

In this case, the small highly charged F- anion will have much stronger ion-dipole interactions with water, as reflected in its much higher heat of hydration (-524 kJ/mol for F- versus -308 kJ/mol for I-). That suggests to me that even if the lost Ag+/F- interaction (whether defined as "ionic" or "covalent") is greater than the equivalent lost Ag+/I- interaction when AgI dissolves, the much strong interaction of the F- with water more than compensates, giving AgF the greater solubility.

  • $\begingroup$ can you please describe the words "lost Ag+/F- interaction is greater than the lost Ag+/I- interaction". What I am thinking is maybe lost Ag+/F- interaction means the breakdown of the AgF bonds giving Ag+, F- ions. If that's correct then I don't get "the much strong interaction of the F- with water" compensates what. Because the previous line would itself mean AgF is more soluble. $\endgroup$
    – Sami
    Jun 4, 2018 at 18:02
  • 1
    $\begingroup$ Yes, that's about it. But recall the free energy change for the process is sum of dG for each step, so dG_dissolve = dG_ion_breakup + dG_ion_solvation. It may well be that breaking up the Ag/F interaction in AgF has a dG that is more positive than breaking up the AgI interaction in AgI. But if dG_ion_solvation is much more negative for Ag+/F- due to the strong new attractions between F- and water, then the overall process may be more favorable for AgF than AgI. $\endgroup$ Jun 4, 2018 at 23:21

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