Chemistry - Why is silver chloride less soluble than silver nitrate?
Solution 1:
In the comment to my previous answer, you asked for a theoretical reason for the solubilities, not considering energy data. Since I know from energy considerations that the issue is not the solvation of the anions, I can present a reason based on the strength of the ionic bond in the two compounds. This reference (as well as others) states the bonding in $\ce{AgCl}$ has an unusually high covalent character which makes it a tighter bond. The $\ce{Ag+}$ ion and the $\ce{Cl-}$ ion are close to the same size (with the silver ion being smaller), so they can approach each other quite closely. In silver nitrate, the $\ce{NO3-}$ ion is larger and does not allow as close an approach as the chloride ion, so the bond is weaker, easier to break up, and the salt is more soluble.
Solution 2:
I'm afraid this is rather a non-answer (or why is it so difficult to answer this)...
Pretty much all nitrates are soluble. This is often explained by the exceptionally good delocalization of the negative charge.
On the other hand, silver salts in general aren't well soluble (I recall only fluoride, nitrate and perchlorate as soluble. Sulfate, carbonate, oxide, sulfide (of course), even the stochiometric cyanide (if I remember correctly) isn't soluble.
So in the case of the nitrate, usually good solubility of nitrates seem to win over the usually not so good solubility of silver salts.
For the Chloride, it is maybe similar: the Chloride stands between the good solubility of AgF and the low solubility of AgBr. Just here, "unsoluble" wins.
(There's also - related to Fajan I think - the HSAB concept which tries to extend also to hard-hard unsoluble salts [e.g. $\ce{CaF2}$] - However, for the case here it is about equally unhelpful as Fajan. I think those two concepts work better in a row of similar ions, like Ag-halogenides)
I think the difficulty in explaining/predicting from theory may come from the fact that there are contrary but strong effects to be considered, so it is hard to predict what exactly the result will be (loss of significance: subtracting two almost equal large numbers, you may not even be able to be sure about the sign of the result).
substance solubility / (mol/l) AgF 1.4 ⋅ 10¹ = 1.4e1 AgNO3 1 - 5 ⋅ 10⁰ = 1e0 - 5e0 AgCl 5 ⋅ 10⁻⁴ = 5e-4 AgBr 7 ⋅ 10⁻⁷ = 5e-7 AgI 9 ⋅ 10⁻⁹ = 9e-9 Ag2S 1.7 ⋅ 10⁻¹⁷ = 1.7e-17
The solubility between different halogenides varies 10 orders of magnitude between fluoride and iodide. That corresponds to a 10fold change in $\Delta_RG^0$. A factor 10 in energy is like bond strength covalent vs. H-bond or like rotational vs. vibrational vs. electronic transition of a molecule.
- Have a look at the paper that I linked from the other question. It is actually focused on $\ce{AgCl}$.
Solution 3:
In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve.
I found data that the crystal lattice energy of $\ce{AgCl}$ is -916.3 kJ/mol (experimental), while the lattice energy of $\ce{AgNO_3}$ is -820 kJ/mol. (Since lattice energy is defined as the energy released when the ions join to form a solid, it is always negative. Reverse the sign for the amount of energy you need to put in to break the crystal apart into separate ions in the gas phase.) From this data you can see that it takes quite a bit more energy to break apart the $\ce{Ag+}$ and $\ce{Cl-}$ ions than it does the $\ce{Ag+}$ and $\ce{NO3-}$ ions, presumably because the chloride ion is smaller and more tightly held. If the enthalpy of hydration value is about the same for the two cases (I couldn't find actual values), then it may not be enough to overcome the extra energy needed to break up the ions.
So, both enthalpy of hydration and crystal lattice energy are important considerations for solubility.There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference.
Added this paragraph to the answer after putting it in the comments below first. Calculation from enthalpy of hydration data (-850.7 for $\ce{AgCl}$ and -794.4 for $\ce{AgNO3}$ says that $\ce{Cl−}$ is the more soluble ion, by -61.3 kJ/mol (since the contribution from solvation of $\ce{Ag+}$ would be the same in each case). Nitrate would be the less soluble ion (from energy concerns) in agreement with Fajan's rules. The enthalpy of hydration data was obtained from problems in a chemistry book.
Solution 4:
Look at the crystalline structure of $\ce{AgCl}$ vs $\ce{AgNO3}$.
Chlorides usually form face centred cubic crystals while nitrates form trigonal planar crystals. Nitrate is a triangular molecule with a positive charge on the nitrogen and a (2–) charge on the oxygens.
In FCC each ion interacts with six ions of opposing charge in an octahedral arrangement.
But with a nitrate the nitrate ions are only interacting with 2 ions of opposing charge above and below the 2/3 negative charge on the oxygens point to neighboring positively charged nitrogen. This arrangment is far more unstable and in silver nitrate ions are far more likely to be whisked away by ion-dipole interactions.
Solution 5:
I'm not much sure about this. I too was much curious about precipitates and this is similar to what I had read somewhere on the net (can't find the site, got it from Wikipedia)
Solubility occurs under dynamic equilibrium, which means that solubility results from the simultaneous and opposing processes of dissolution and phase joining (e.g., precipitation of solids). The solubility equilibrium occurs when the two processes proceed at a constant rate.
So according to this the process of phase joining is very very more than dissolution in $\ce{AgCl}$ thus it is insoluble in water, but in $\ce{AgNO3}$ it is just the opposite.