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.
Lattice energies – breaking the lattice
The lattice energy for an ionic compounds is calculated theoretically from experimental data put into a Born-Haber cycle. In the context of dissolving an ionic compound, we are actually thinking about breaking up the ionic lattice rather than forming one. This is going to be a highly endothermic process as we need to find the energy to overcome the strong ionic bonds between oppositely charged ions in the lattice.
As the ionic radius of the cation or anion in the lattice increases, the less exothermic the lattice energy becomes (assuming we are comparing ions with the same charge). Breaking up a caesium iodide lattice will be far easier (less endothermic) than breaking up a lithium fluoride lattice.
LiF(s) ⇾ Li+(g) + F–(g) – ΔLEH⦵ = +1049 kJ mol-1
CsI(s) ⇾ Cs+(g) + I–(g) – ΔLEH⦵ = +604 kJ mol-1
The lattice energy increases with increasing charge on the anion or cation. The greater the charge on the ion, the higher the charge density and the stronger the ionic bonds / electrostatic attractions in the lattice. It takes more energy to break up a lattice of magnesium oxide than a lattice of lithium fluoride.
LiF(s) ⇾ Li+(g) + F–(g) – ΔLEH⦵ = +1049 kJ mol-1
MgO(s) ⇾ Mg2+(g) + O2-(g) – ΔLEH⦵ = +3923 kJ mol-1
Hydrating gaseous ions
Clearly we need to find a lot of energy to break up an ionic lattice, and that energy comes from hydrating the gaseous ions. The attractions between cations or anions and water molecules are called ion-dipole bonds. Forming ion-dipole bonds is a highly exothermic process.
Li+(g) + (aq) ⇾ Li+(aq) ΔhydH⦵ = -519 kJ mol-1
F–(g) + (aq) ⇾ F–(aq) ΔhydH⦵ = -506 kJ mol-1
The enthalpy change of hydration, ΔhydH⦵, is defined as the enthalpy change when 1 mole of a gaseous ion dissolves in water to form an infinitely dilute solution under standard conditions (298K, 101kPa).
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 a 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.
Notes:
1. For most ions, the entropy change for X+(g) ⇾ X+(aq) is negative (not favourable) as the hydration shell forms around the ion and the water molecules become more ordered / less random. 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 water molecules in pure water which restricts their movement and is disrupted when some of these water molecules form hydration shells around an ion.
- 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.
Enthalpy changes of solution
The enthalpy change of solution, ΔsolH⦵, is the enthalpy change when 1 mole of an ionic compound dissolves in water to form an infinitely dilute solution under standard conditions (in which we assume that there is negligible electrical attraction between oppositely charged ions).
ΔsolH⦵ may be exothermic or endothermic. If it is only just endothermic the ionic compound may be still be soluble due to a positive total entropy change e.g. sodium chloride, but if ΔsolH⦵ has a large endothermic value, the ionic compound will be insoluble e.g. silver bromide.
NaCl(s) + (aq) ⇾ Na+(aq) + Cl–(aq) ΔsolH⦵ = +3.9 kJ mol-1
AgBr(s) + (aq) ⇾ Ag+(aq) + Br–(aq) ΔsolH⦵ = +84.5 kJ mol-1
The enthalpy change of solution is calculated by constructing a Hess / enthalpy cycle with ΔLEH⦵ for the ionic compound and ΔhydH⦵ for both the anion and the cation (alternatively you might be asked to construct as energy level diagram with the same information – see Q.2 below).
The sharp-eyed amongst you will notice that the value I get for ΔsolH⦵ for NaCl using the cycle is not +3.9 kJ mol-1 as given in the example above! Every data source has slightly different values for lattice enthalpies and enthalpies of hydration, none of which give us a value of +3.9, and yet all the data sources insist this is the correct value for ΔsolH⦵ for NaCl, citing ‘experimental differences’ as the explanation. I’ll leave it with you…
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?
2. (a) Use the following information to draw an energy level diagram to determine the enthalpy change of solution of calcium hydroxide. Calculate a value for ΔsolH⦵.
| Compound | Lattice enthalpy / kJ mol-1 | Ion (g) | Enthalpy of hydration / kJ mol-1 |
| Ca(OH)2 | -2506 | Ca2+ | -1579 |
| OH– | -460 |
(b) The ΔsolH⦵ for Mg(OH)2 is +152 kJ mol-1. Suggest reasons for the difference between the enthalpy change of solution of magnesium hydroxide and calcium hydroxide.
3. Potassium reacts vigorously with water to give potassium hydroxide and hydrogen.
(a) Write an equation, with state symbols, for the reaction of 1 mole of potassium with water.
(b) Define enthalpy change of hydration, ΔhydH⦵, for a potassium ion.
(c) Use the information in the table to determine the enthalpy change for
K(s) ⇾ K+(aq) + e–
| Element | ΔatH⦵/ kJ mol-1 | ΔhydH⦵/ kJ mol-1 | ΔIEH⦵/ kJ mol-1 |
| Potassium | +90 | -322 | +418 |
(d) The enthalpy change for the reaction of 1 mole of potassium with water is -196 kJ mol-1. Use this information, in addition to your answers to parts (a) and (c), to calculate the enthalpy change of reaction per mole of hydrogen when water forms hydroxide ions:
2H2O(l) + 2e– ⇾ 2OH–(aq) + H2(g)
(e) Draw an energy level diagram to illustrate your answer to part (d).
(f) Calculate the enthalpy change of reaction of potassium with water per gram of potassium.
4. Magnesium sulphate is considerably more soluble in water than strontium sulphate, despite magnesium sulphate having a more exothermic lattice enthalpy. Explain why, in terms of the enthalpy changes involved when 1 mole of an ionic compound dissolves in water to form an aqueous solution.
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 / ionic bonds 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.
2.
(b) Mg2+ is a smaller ion than Ca2+ but it has the same charge, and so a higher charge density.
This means that the strength of the attractions within the lattice between oppositely charges ions will be stronger in Mg(OH)2 .
The more exothermic lattice enthalpy of Mg(OH)2 must outweigh the more exothermic ΔhydH⦵ that would be expected because ions with a higher charge density attract more water molecules into hydration shells , forming stronger ion-dipole bonds.
3. (a) K(s) + H2O(l) ⇾ KOH(aq) + ½H2(g) (equation must be for 1 mole of potassium so we can’t double up)
(b) ΔhydH⦵ is the enthalpy change when one mole of gaseous potassium ions form one mole of aqueous ions.
(f) ΔrH⦵ for the reaction of potassium with water = -196 kJ mol-1 ; Mr = 39.1 g mol-1
ΔrH⦵ per (g) = -196 / 39.1 = -5.01 kJ g-1
4. ΔsolH⦵ = -ΔLEH⦵ + ΔhydH⦵(x+) + ΔhydH⦵(y–)
ΔsolH⦵is the enthalpy change when one mole of an ionic compound dissolves to form an aqueous solution and it can be broken down into two steps. Firstly, breaking up the lattice to form gaseous ions (endothermic) and secondly, hydrating the cation and anion (exothermic).
MgSO4 is more soluble than SrSO4 because the Mg2+ ion has a higher charge density and so attracts more water molecules to its hydration shell (stronger ion-dipole bonds ). ΔhydH⦵(Mg2+) must be considerably more exothermic than ΔLEH⦵(Mg2+) to compensate for the energy needed to overcome the lattice enthalpy.