23.2 Enthalpy of solution & Hydration
Candidates should be able to:
- Define and use the term enthalpy change with reference to hydration, ΔHhyd, and solution, ΔHsol
- Construct and use an energy cycle involving the enthalpy change of solution, lattice energy, and enthalpy change of hydration
- Carry out calculations involving these energy cycles
- Explain, in qualitative terms, the effect of ionic charge and ionic radius on the numerical magnitude of an enthalpy change of hydration
- Describe and explain qualitatively the variation in solubility and the enthalpy change of solution of the hydroxides and sulfates of Group 2 metals, in terms of the relative magnitudes of the enthalpy change of hydration and the lattice energy
When an ionic solid dissolves in water, it undergoes a fascinating energetic transformation. This process boils down to two main competing forces: breaking an ionic lattice apart and bonding the free ions to polar water molecules.
1Key Definitions: Hydration & Solution
Enthalpy Change of Hydration
The energy released when gaseous ions dissolve in water is called the standard enthalpy change of hydration.
The enthalpy change of hydration, ΔH°hyd, is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution.
Cl−(g) + aq → Cl−(aq) ΔH°hyd = −364 kJ mol−1
The word “gaseous” is essential in your definition. The ion must start as a gaseous ion. Many students lose marks by writing “ions dissolve in water” without specifying gaseous.
Enthalpy Change of Solution
The standard enthalpy change of solution, ΔH°sol, is the energy absorbed or released when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution.
NaCl(s) + aq → Na+(aq) + Cl−(aq) ΔH°sol = +3.9 kJ mol−1
Note that:
- The symbol ‘aq’ represents the very large amount of water used
- Enthalpy changes of solution can be positive (endothermic) or negative (exothermic)
- A compound is likely to be soluble only if ΔH°sol is negative or has a small positive value; large positive values mean the substance is relatively insoluble
Water molecules are polar: the oxygen end carries a partial negative charge (δ−) and the hydrogen ends carry partial positive charges (δ+). This polarity is what makes water such an effective solvent for ionic compounds:
- Cations are surrounded by the δ− oxygen atoms of water molecules
- Anions are surrounded by the δ+ hydrogen atoms of water molecules
These attractive forces are called ion–dipole interactions, and the energy they release is the hydration enthalpy.
Summary of the Three Key Terms
| Term | Symbol | Always exo/endo? | Key phrase |
|---|---|---|---|
| Enthalpy change of hydration | ΔH°hyd | Always exothermic (−) | 1 mol of gaseous ions → very dilute solution |
| Enthalpy change of solution | ΔH°sol | Can be either | 1 mol of ionic solid → very dilute solution |
| Lattice energy (formation) | ΔH°latt | Always exothermic (−) | 1 mol of ionic solid formed from gaseous ions |
2Constructing Energy Cycles
By Hess’s Law, we can link the three quantities above in an energy cycle. Dissolving an ionic solid can be thought of as two steps:
- Step 1 — Lattice breaking: The ionic lattice is broken apart to give gaseous ions. This is the reverse of lattice energy formation, so the value is +ΔH°latt
- Step 2 — Hydration: The gaseous ions are hydrated by water molecules. The energy released is the sum of all hydration enthalpies
ΔH°hyd is the sum of both individual ions. Always pay close attention to the stoichiometry of the salt formula! For example, MgCl2 produces 2 moles of Cl− ions, so you must multiply the hydration enthalpy of Cl− by 2 in your calculations.
Given: ΔH°latt(NaCl) = −787 kJ mol−1, ΔH°hyd(Na+) = −406 kJ mol−1, ΔH°hyd(Cl−) = −364 kJ mol−1
Step 1 — Apply Hess’s Law:
ΔH°sol = +787 + (−770)
ΔH°sol = +17 kJ mol−1
NaCl is slightly endothermic to dissolve, but the value is small – NaCl is soluble.
Given: ΔH°latt(MgCl2) = −2526 kJ mol−1, ΔH°hyd(Mg2+) = −1920 kJ mol−1, ΔH°hyd(Cl−) = −364 kJ mol−1
MgCl2 gives 1 mol Mg2+ and 2 mol Cl− — so multiply ΔH°hyd(Cl−) by 2!
ΔH°sol = +2526 + (−1920) + (−728)
ΔH°sol = −122 kJ mol−1
Always check the formula! For salts such as MgCl2, AlCl3, and CaCl2, you must multiply ΔH°hyd of the anion by the number of anions produced from one formula unit. Forgetting this is the most common calculation error in this topic.
3Ionic Charge & Radius Effects
ΔH°hyd becomes more exothermic (more negative) when:
| Factor | Why? |
|---|---|
| Smaller ionic radius | Higher charge density → stronger ion–dipole attractions with water → more energy released |
| Higher ionic charge | Greater electrostatic attraction to polar water molecules → more energy released |
Hydration enthalpy is a measure of the strength of ion–dipole interactions between ions and water. It’s always exothermic because bonds form.
Charge density = ionic charge ÷ ionic radius. A small, highly charged ion has a very high charge density, and therefore a very large (very exothermic) hydration enthalpy.
Comparison: Mg2+ (ΔH°hyd = −1920 kJ mol−1) vs Ba2+ (ΔH°hyd = −1305 kJ mol−1). Both have charge 2+, but Mg2+ is much smaller → much more negative ΔH°hyd.
4Solubility Trends: Group 2 Compounds
Sulfates — Solubility Decreases Down the Group
| Compound | Solubility / mol dm−³ (25°C) |
|---|---|
| Magnesium sulfate, MgSO4 | 1.83 |
| Calcium sulfate, CaSO4 | 4.66 × 10−2 |
| Strontium sulfate, SrSO4 | 7.11 × 10−4 |
| Barium sulfate, BaSO4 | 9.43 × 10−6 |
We can explain this variation using the relative values of ΔH°hyd and ΔH°latt:
- Hydration enthalpy decreases (less exothermic) down the group: Mg2+ > Ca2+ > Sr2+ > Ba2+. This is a large decrease, depending entirely on cation size
- Lattice energy also decreases down the group, but by a smaller amount — because the large sulfate ion dominates the lattice energy term, so changing cation size has less impact
ΔH°sol = −ΔH°latt + ΔH°hyd. Since ΔH°hyd falls faster than ΔH°latt, ΔH°sol becomes increasingly endothermic (more positive) down the group → solubility decreases.
The solubility of the Group 2 sulfates decreases down the group. Explain this trend.
Show mark scheme
M1 ΔHlatt and ΔHhyd decrease / both become less exothermic
M2 ΔHlatt changes less / becomes less exothermic by a smaller extent OR ΔHhyd changes more / is the dominant factor
M3 ΔHsol becomes less exothermic / more endothermic OR ΔHsol = ΔHhyd − ΔHlatt expression AND reaction becomes less exothermic
Hydroxides — Solubility Increases Down the Group
The hydroxide ion (OH−) is much smaller than the sulfate ion. This changes the analysis:
- Lattice energy decreases as cation size increases, and because OH− is small, this decrease is large
- Hydration enthalpy also decreases, but by a smaller amount (it is constant for OH− across all four hydroxides)
- Since ΔH°latt falls faster than ΔH°hyd, ΔH°sol becomes less endothermic (more negative) down the group → solubility increases
The trends are opposite — the key is the size of the anion:
Large anion (SO42−)
ΔH°latt is insensitive to cation size → ΔH°hyd falls faster → solubility decreases down the group
Small anion (OH−)
ΔH°latt is sensitive to cation size → ΔH°latt falls faster → solubility increases down the group
The solubility of the Group 2 hydroxides increases down the group. Explain this trend.
Show mark scheme
M1 ΔHlatt and ΔHhyd both become less exothermic / less negative
M2 ΔHhyd changes less / becomes less exothermic by a smaller extent OR ΔHlatt changes more / is the dominant factor / changes faster
M3 ΔHsol becomes more exothermic / more negative OR ΔHsol becomes less endothermic / less positive
Core Definitions (Must Know Exactly)
| Term | Symbol | Definition | Sign |
|---|---|---|---|
| Standard enthalpy change of solution | ΔH°sol | Enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form a very dilute solution under standard conditions | Exo (−) or endo (+) |
| Standard enthalpy change of hydration | ΔH°hyd | Enthalpy change when 1 mole of specified gaseous ions dissolves in sufficient water to form a very dilute solution under standard conditions | Always exo (−) |
| Lattice energy (formation) | ΔH°latt | Enthalpy change when 1 mole of an ionic lattice is formed from its gaseous ions under standard conditions | Always exo (−) |
- Always state standard conditions (298 K, 100 kPa, 1 mol dm−3 solution)
- Draw the energy cycle clearly with arrows and labels – examiners award marks for the cycle
- Check signs: lattice formation is (−), breaking it is (+)
- Watch out for MgCl2, AlCl3, etc. – multiply hydration enthalpy by the number of ions
- Define hydration as “gaseous ions → aqueous ions” – many students miss “gaseous”
- Explain factors using “charge density” and “ion–dipole attractions” – not just “size” or “charge” alone
Download PDF 23.2 Enthalpy Change of Hydration and Solution – Notes
Download PDF 23.2 Enthalpy Change of Hydration and Solution – Worksheet