[hoihoi]

Greetings, and a fine forum you all have here. Appreciate the knowledge.

I am struggling to fully understand why there is such a stark water solubility difference between (NH₄)₂SO₄ and CaSO₄. I think part of the answer is that the ionic bonds that are involved are stronger for the CaSO₄, and I believe that has to do with the charges of the cation (and shape?) involved. Perhaps also the ionic radius has something to do with it?

Any help in understanding the bonding and solubility differences would be greatly appreciated. Thanks.

[thetada]

Hello and welcome. You've got some good ideas about the explanation for this. How might ionic radius be involved?

[hoihoi]

Hello and welcome. You've got some good ideas about the explanation for this. How might ionic radius be involved?

Assuming that the ionic radius will be variable for the ammonium since it's polyatomic, and therefore weaker, and since the Ca will be more of a true radius (i may be imagining this wrong), it will form a stronger ionic bond with the sulfate?

[mjc123]

You may find this thread of interest: http://www.chemicalforums.com/index.php?topic=81896.msg297960
Solubility is complicated because it involves several competing factors that depend on charge and ionic radius, and you are generally dealing with a relatively small difference between large quantities. The above thread considers enthalpy of solution, but to explain solubility you also need the entropy.

[Corribus]

(This is maybe superfluous because of the above responses, but I wrote it so figured I might as well post it.)

Rationalizing solubility trends is not always straightforward because of the number of factors involved. You have to consider not only the thermodynamic effects (both entropy and enthalpy) of hydrating the ions, but also the thermodynamic effects of breaking apart the crystal lattice. Which form wins out ultimately determines the thermodynamics of dissolution and, ultimately, solubility at equilibrium. You are right that generally the different chemistry for the two compounds results in the relative strength of interactions possible when comparing the small and very positively charged calcium ion versus the large and less-positively-charged ammonium ion. Note that it's not quite this straightforward because when you dissolve ammonium sulfate, you release two ammonium ions, but dissolving calcium sulfate releases only a single calcium ion.

You can get a rough idea of what is going on by looking at the relative heats of formation and standard entropies of the various compounds involved.

For example, the heats of formation for aqueous calcium ion and ammonium ion are -543 kJ/mol and -133 kJ/mol, respectively, and the standard entropies are -56.2 J/mol K and 111.17 J/mol K, respectively. Can you think about what causes these values to be what they are? Likewise, the heats of formation and standard entropy for solid calcium sulfate are -1434.5 kJ/mol and 106.5 J/mol K, respectively; the heat of formation and standard entropy for ammonium sulfate (s) are -1180.9 kJ/mol and 220.1 J/mol K, respectively. Together with the values for aqueous sulfate ion (enthalpy is -909.3 kJ/mol and entropy is 18.5 J/mol K), you can calculate Gibbs energies for the dissolution and show that the dissolution of calcium sulfate is expected to not be spontaneous at 298 K but for ammonium sulfate it is spontaneous at 298 K. This agrees with the solubility observation.

(All thermodynamic values were from the CRC 96th Edition)

Also: sometimes you get unexpected results. Another interesting comparison is between calcium and magnesium sulfate. Based on a simple explanation that the differences in solubilities is due to the cation size, we might predict magnesium sulfate to be even less soluble than calcium sulfate because magnesium is an even smaller ion (with same charge), and therefore even less favorable to be stabilized in water. But, magnesium sulfate is far more soluble than calcium sulfate - 26.9 g/100 mL versus ~0.21 g/199 mL. What is going on? If you compare the Mg2+ ion to Ca2+ ion, the former has heat of formation of -467 kJ/mol and standard entropy of -137 J/mol K. So compared to calcium, the strength of intermolecular interactions with smaller water molecules is actually weaker (would seem to favor less dissolution). Also, because it's a smaller ion, the entropy cost is greater to dissolve, because there is more ordering of water molecules around it. This would also seem to favor less dissolution. So this can't explain the better solubility of magnesium sulfate. We must turn to the crystal lattice. MgSO4 has a heat of formation of -1284.9 kJ/mol and a standard entropy of 91.6 J/mol. Based on these numbers, calcium sulfate is anticipated to be a more stable salt. So evidently although the magnesium ion is less favorable to be stabilized by water than calcium, the stability of the calcium sulfate lattice wins out in the equation, meaning his is the less soluble material. By my calculations, the heats of dissolution for calcium sulfate and magnesium sulfate are -17.8 kJ/mol and -91.4 kJ/mol, respectively, and the entropies of dissolution are -144.2 J/K mol and -210.1 J/K mol, respectively. Translating into Gibbs energies of reaction of +25.19 kJ/mol and -28.75 kJ/mol at 298 K for calcium sulfate and magnesium sulfate, respectively: consistent with the observation that magnesium sulfate is the more soluble salt at room temperature.

It is not easy to explicitly say why calcium sulfate is a more stable lattice than magnesium sulfate. Crystal lattice stability is a complex function of the relative sizes of the ions involved and their crystal structure (how the ions are located relative to each other). From what I can tell, calcium sulfate adopts an orthorhombic crystal and magnesium sulfate a monoclinic structure, so we're not even comparing apples to apples any more. Ammonium sulfate, having a completely different stoichiometry, will be even less of an apples-to-applies comparison.

Note that I think this method of using thermodynamic values to estimate solution thermochemistry is useful only for qualitative comparisons. But, maybe it helps you see that it's not a simple question to answer. There are a lot of complexities involved.