"Superficially this is a bewildering field." W.E. Dasent (1).
Solubility and Hesss Law: enthalpies.
The standard argument in many A level texts (including mine (2), but I try to show the limitations) concerning the solubility of ionic compounds relates only to enthalpy changes. Essentially, if the hydration enthalpies of the gaseous ions are more exothermic than the negative of the lattice enthalpy, then the enthalpy of solution will be negative and the substance will dissolve. The hydration enthalpy offsets the enthalpy change in breaking up the lattice. The more negative the hydration enthalpy, then the more soluble the substance will be.
This is fine as far as it goes, But like all thermodynamic arguments based on enthalpy changes alone, and especially ones concerning equilibrium, it doesnt go far enough and gives a misleading impression if the entropy changes of the process are ignored.
Solubility is an aspect of equilibrium; we are concerned with the concentrations of the ions in a saturated solution, that is one that has excess solid at the bottom of the beaker. If the salt is, say, sodium chloride, the equilibrium of interest is
NaCl(s) Û Na+(aq) + Cl-(aq).
We can write an equilibrium constant for this process, and the larger it is the greater the solubility. Since the equilibrium constant is related to the free energy change DG by
D
G = -RT ln K,and the free energy change is related to the enthalpy and entropy changes by the fundamental equation
D
G = DH - TDS,its clear that the entropy must figure in the considerations of the solubility of substances.
Why, then, is it often ignored? I find it amazing that such knowledge is not required, but then I am often regarded as difficult The examples are chosen carefully in A level schemes that do not require a knowledge of entropy.
The Gibbs Free Energy DG.
The spontaneity of a chemical reaction or process is governed by DG; it is not, whatever syllabuses might wish, governed by anything else.
If DG is negative, the process is spontaneous.
This does not necessarily mean that it is fast. Be careful to separate thermodynamics from kinetics.
With many reactions, perhaps even most, the enthalpy change is dominant. Such reactions or processes are enthalpy-controlled. This is why many people ignore the entropy changes in elementary work, and why misleading impressions are then imparted. However, there are some reactions or processes where the feasibility of the process depends on the entropy contributions. These are entropy-controlled reactions or processes. They are found in endothermic reactions where DH is positive but where DG can still be negative if TDS is large and positive.
The result of this is that it is perfectly possible for substances to dissolve endothermically and so apparently contravene our criteria from the Hesss Law argument. Some substances that dissolve very endothermically are also very soluble; ammonium salts are the best examples of this.
The true thermodynamics of solubility.
In the case of silver halides, the fluoride is the most soluble (which is why the test using silver nitrate in nitric acid cannot be used to detect fluoride ion) and the iodide is the least; for the case of calcium halides the reverse trend is seen. The solubility of group 2 hydroxides increases down the group, whereas the group 2 sulphates show an opposing trend.
The arguments concerning the magnitudes of the contribution of enthalpy and entropy changes can be illustrated using the solubility of the sodium and the silver halides. These are given in Table 1; the values are at 298K. The standard free energy changes DGo for the solution of each of these salts, and the contributory values of DH o and TDSo, all at 298K, are also given. The more negative DGo is, the more soluble the salt will be since the larger K will be.
Salt |
Solid phase |
Solubility, mol/100g H2O |
D Go/ kJ mol-1 |
D H o/ kJ mol-1 |
TDSo/kJ mol-1 |
NaF | NaF | 0.0987 | +2.6 | +0.2 | -2.4 |
NaCl | NaCl | 0.614 | -9.0 | +3.9 | +12.9 |
NaBr | NaBr.2H2O | 0.919 | -17.0 | -0.6 | +16.4 |
NaI | NaI.2H2O | 1.226 | -31.1 | -7.6 | +23.5 |
AgF | AgF.4H2O | 1.397 | -14.4 | -20.3 | -5.9 |
AgCl | AgCl | 1.35 x 10-6 | +55.6 | +65.4 | +9.8 |
AgBr | AgBr | 7.19 x 10-8 | +70.2 | +84.4 | +14.2 |
AgI | AgI | 1.11 x 10-9 | +91.7 | +112.3 | +20.6 |
If you look at the values of DH o and compare them with the solubility values, then you will see that the solubility of the silver halides parallels these values, but that the solubility of the sodium halides does not.
The reasons are not hard to find. Look at all of the values for the silver halides. In this case the values for DH o are dominant, being roughly five times as large as the contribution from TDS o. The dissolution of the silver halides is enthalpy-controlled, and so the elementary picture works well for this series of salts. For the sodium halides, however, DH o does not parallel the solubility, though (of course) DG o does. The solubility of the sodium halides is entropy-controlled, the contribution of TDS o being much larger than that of DH o.
The contribution from the enthalpy, DH o.
The Hesss Law approach divides the solution of the salt into two steps, the breaking up of the lattice and the hydration of the resulting gaseous ions:
I | MX(s) à M+(g) + X- (g) | - DHlatt | |
II | M+(g) + X- (g) à M+(aq) + X- (aq) | S ( DHhydration) | |
III | MX(s) à M+(aq) + X- (aq) | DHsolution |
(This is where the standard dont involve the entropy treatment stops.)
Compare now the enthalpy contributions for processes I and II for the two series of halides, and the overall result for III all values are at 298 K.
NaF |
NaCl |
NaBr |
NaI |
|
step I | +919.2 | +787 | +752 | +703 |
step II | - 918.8 | - 783 | - 753 | - 711 |
total for III | + 0.4 | + 4.0 | - 1 | - 8 |
AgF |
AgCl |
AgBr |
AgI |
|
step I | +966 | + 917 | + 905 | + 891 |
step II | - 986 | - 851 | - 820 | - 778 |
total for III | - 20 | + 66 | + 85 | + 113 |
From this it is clear that the differences between the lattice enthalpy and the hydration enthalpies of the gaseous ions are very small in the case of the sodium halides, but much larger in the case of the silver halides. It is reasonable to ask why this is.
For a given halide ion X the hydration enthalpy of the silver salt is always greater than that of the corresponding sodium salt by 67 kJ mol-1, so this must be the difference between the hydration enthalpy of the sodium ion and the silver ion. However, the differences in the lattice enthalpies increase from fluoride to iodide. The AgX lattice becomes more stable than the NaX lattice in passing from fluoride to iodide; this corresponds to the larger increase in covalence in going from AgF to AgI compared with that in going from NaF to NaI. This is shown by a comparison of the calculated (theoretical) values of DHlatt from an ionic model with the experimental values from a Born-Haber cycle. All of the latter are more exothermic:
NaF | NaCl | NaBr | NaI | |
- DHlatt (calc) | 912 | 770 | 735 | 687 |
- DHlatt (B-H) | 918 | 780 | 742 | 705 |
Difference | - 6 | - 10 | - 7 | - 18 |
AgF | AgCl | AgBr | AgI | |
- DHlatt (calc) | 920 | 833 | 816 | 778 |
- DHlatt (B-H) | 958 | 905 | 891 | 889 |
Difference | - 38 | - 72 | - 75 | - 111 |
The entropy contribution T DS o
The values of TDS o (3) at 298 K are given below for steps I and II which, for convenience, are given again here:
I | MX(s) à M+(g) + X- (g) | - DHlatt | |
II | M+(g) + X- (g) à M+(aq) + X- (aq) | S ( DHhydration) | |
III | MX(s) à M+(aq) + X- (aq) | DHsolution |
NaF |
NaCl |
NaBr |
NaI |
|
step I | + 70 | + 68 | + 67 | + 67 |
step II | - 72 | - 55 | - 51 | - 44 |
total for III | - 2 | + 13 | + 16 | + 23 |
AgF |
AgCl |
AgBr |
AgI |
|
step I | +68 | + 67 | + 67 | + 66 |
step II | - 74 | - 57 | - 52 | - 46 |
total for III | - 6 | + 10 | + 15 | + 20 |
In the case of all of the salts apart from the fluorides the entropy term for the overall process III is positive, favouring solution since this tends to make DG o negative. In these cases the entropy increase in forming the gaseous ions outweighs the entropy decrease in turning these into the hydrated ions. For the fluorides the converse is true; this is due to the hydrogen bonding of the water solvent with the fluoride ion, a process that increases order and thus decreases the entropy.
And finally:
These results, taken in conjunction with the enthalpy analysis already discussed, illustrate the difficulties encountered in seeking explanations of solubility in terms of a handful of simple rules; in general each case must be individually subjected to the above sort of analysis before any considerable progress can be made towards understanding the factors which determine solubility. (4)
Bibliography:
1 Dasent, W.E., Inorganic Energetics, Penguin Books, 1970, p135ff.
2 Beavon, J.R.G. Structure, Bonding, and the Periodic Table, Thomas Nelson, 1997.
3 Noyes, R. M., Journal of the American Chemical Society, 40, (1), 2-10, 1962.
4 Dasent, op. cit., p 139.