For a full understanding of why ionic compounds are soluble we need to consider both the enthalpy and entropy changes involved …
NaCl(s) + (aq) ⇾ Na+(aq) + Cl–(aq)
We can think of dissolving in 2 steps: breaking up the lattice to form gaseous ions and then hydrating the ions to form an infinitely dilute solution where we assume that there is negligible electrical attraction between oppositely charged ions.
Hydrating ions involves a number of water molecules (4-8) directly coordinating themselves around the ion to form a primary hydration shell. These water molecules also affect their neighbours, binding them more tightly than they would be in the free solvent and so forming a secondary hydration shell.
As result ions in solution appear much larger than might be expected.
| Ion | Li+ | Na+ | K+ | Cs+ |
| Hydration number | 22 | 13 | 7 | 6 |
Although the charge on the ion remains the same as we move down the group, the ionic radius increases so the charge density of the ion decreases. Li+ has the highest charge density as the +1 charge is spread out over a small ion and so attracts water molecules more strongly in solution.
ΔhydH⦵ becomes more negative (exothermic) as the radius decreases and the charge increases because as the charge density of the ion increases, more ion-dipole bonds are formed.
However, hydrating ions isn’t just about enthalpy changes – we must also consider the entropy changes to the solvent.
1. For most ions, the entropy change for X+(g) ⇾ X+(aq) is negative (not favourable) as water molecules become more ordered / less random around the ion. Smaller, highly charged ions have a greater effect on more surrounding water molecules and so the entropy change is more negative.
2. But the entropy change of hydration for some large, singly charged ions is positive – somehow it seems that there is an increase in the chaos / random nature of the surrounding water molecules. How can we explain this?
- There is an extensive network of strong hydrogen bonds between the molecules in pure water which restricts their movement and is disrupted when some of these water molecules from hydration shells around the ions.
- Large, singly charged ions with a low charge density only coordinate with a few water molecules. The decrease in entropy accompanying the formation of a hydration shell must be outweighed the overall increase in entropy arising from the widespread disruption of hydrogen bonding allowing for greater movement of water molecules in the solution.
Practice questions
Use the information below to answer the questions that follow.
| Compound | Lattice enthalpy / kJ mol-1 | Ion (g) | Enthalpy of hydration / kJ mol-1 |
| AgF | -958 | Ag+ | -446 |
| AgCl | -905 | F– | -506 |
| Cl– | -364 |
(a) Explain the difference is lattice enthalpy for the two silver halides.
(b) Explain the difference in the enthalpy change of hydration of the fluoride and chloride ions.
(c) Draw an enthalpy cycle / Hess cycle linking ΔsolH⦵, ΔhydH⦵ and ΔLEH⦵ together for silver fluoride.
(d) Calculate ΔsolH⦵ for both silver fluoride and silver chloride.
(e) What do the enthalpy changes of solution suggest about the relative solubilities of the silver halides?
Answers
(a) AgCl has a less exothermic ΔLEH⦵ which suggests weaker ionic bonding between Ag+ and Cl– than between Ag+ and F–. This is because the Cl– ion is larger than F– and so electrostatic attractions between the oppositely charged ions are weaker.
(b) ΔhydH⦵ is more exothermic for the fluoride ion as it has a higher charge density so attracts more water molecules to its hydration shell = stronger ion-dipole bonding.
(d) ΔsolH⦵ (AgF) = – (-958) + (-446) + (-506) = +6 kJ mol-1
ΔsolH⦵ (AgCl) = – (-905) + (-446) + (-364) = +95 kJ mol-1
(e) Both ΔsolH⦵ are endothermic but solubility also depends on entropy changes. AgF may be soluble, AgCl is likely to be insoluble.
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