23.2 Enthalpies of solution and hydration
- Syllabus
- 9701–2028–2029
- Topic
- 23.2
- Level
- A2
Learning objectives
23.2.1Hydration and solution enthalpies• Define/use: enthalpy change using hydration, ΔHhyd, and solution, ΔHsol23.2.2An energy cycle involving enthalpy change• Construct/use: an energy cycle involving enthalpy change of solution, lattice energy and enthalpy change of hydration23.2.3Calculations: the energy cycles in 23.2.2• Calculate: the energy cycles in 23.2.223.2.4Explain, in qualitative terms, the effect• Explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of an enthalpy change of hydration
Define hydration and solution enthalpy from exact state changes
| Term | Process for one mole | Sign |
|---|---|---|
| hydration enthalpy, ΔHₕyd | one mole of a specified gaseous ion → that aqueous ion | always exothermic / negative |
| solution enthalpy, ΔHₛₒₗ | one mole of ionic solid → its stoichiometric aqueous ions | may be positive or negative |
NaX+(g)NaX+(aq)
ClX−(g)ClX−(aq)
NaCl(s)NaX+(aq)+ClX−(aq)
Hydration releases energy when ion–dipole attractions form between ions and water. Dissolution also requires lattice separation, so the balance between endothermic separation and exothermic hydration determines the sign of ΔHₛₒₗ.
A hydration value belongs to one ion, not a whole salt. State charge and phase: Na⁺(g) → Na⁺(aq) is hydration, whereas NaCl(s) → aqueous ions is solution.
Construct the solution cycle through the same gaseous ions
Connect the solid directly to its aqueous ions by ΔHₛₒₗ, and indirectly through the stoichiometric gaseous ions. The indirect route reverses lattice formation, then hydrates each gaseous ion.
| MgCl₂ path | State equation | Enthalpy contribution |
|---|---|---|
| direct dissolution | MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq) | ΔHₛₒₗ |
| lattice separation | MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g) | −ΔHₗₐₜₜ (formation) |
| hydrate cation | Mg²⁺(g) → Mg²⁺(aq) | ΔHₕyd(Mg²⁺) |
| hydrate anions | 2Cl⁻(g) → 2Cl⁻(aq) | 2ΔHₕyd(Cl⁻) |
ΔHsol=−ΔHlatt+∑ΔHhyd
The summation includes stoichiometric coefficients. Do not use one hydration value for MgCl₂ as a whole, omit the factor 2 for chloride, or insert negative lattice formation without reversing its direction.
Calculate a missing cycle term without changing its defined sign
Use one convention throughout. With lattice formation, the gaseous-ion-to-aqueous route gives ΣΔHₕyd = ΔHₗₐₜₜ + ΔHₛₒₗ. Apply coefficients first, substitute signed data second, and rearrange last.
| KCl datum | Value / kJ mol⁻¹ |
|---|---|
| ΔHₗₐₜₜ[KCl] (formation) | −711 |
| ΔHₛₒₗ[KCl] | +26 |
| ΔHₕyd[K⁺] | −322 |
ΔHhyd(K+)+ΔHhyd(Cl−)=ΔHlatt+ΔHsol
ΔHhyd(Cl−)=−711+26−(−322)=−363 kJ mol−1
Hydration must be exothermic, so the negative answer is physically consistent. A positive result here signals a likely direction, coefficient or rearrangement error; do not flip a sign merely to force the expectation.
Higher charge density makes hydration more exothermic
∣ΔHhyd∣ increases as ionic charge density r∣q∣ increases
| Ion change | Ion–dipole attraction to water | ΔHₕyd |
|---|---|---|
| higher charge at similar radius | stronger | more negative; larger magnitude |
| smaller radius at same charge | charge is closer to water dipoles; stronger | more negative; larger magnitude |
| larger radius at same charge | weaker | less negative; smaller magnitude |
Water's Oδ− end points towards a cation, while its Hδ+ ends point towards an anion. Stronger attraction releases more energy as the hydration shell forms.
Mg²⁺ has a much more negative hydration enthalpy than Na⁺ because it has higher charge and high charge density. Within Group 1, Na⁺ is more negative than K⁺ because Na⁺ is smaller at the same +1 charge.
Say more negative or larger magnitude; “larger hydration enthalpy” is ambiguous. Compare charge first and radius among equal-charge ions rather than using radius alone across unlike charges.