Enthalpies of Hydration and Solution

When an ionic compound dissolves in water, two processes occur: the ionic lattice must be broken apart, and the separated ions must become surrounded by water molecules. Understanding the enthalpy changes associated with these processes allows us to predict whether dissolving will be exothermic or endothermic, and even to rationalise why some salts are more soluble than others. This lesson covers hydration enthalpies, solution enthalpies, the energy cycle linking them, and detailed worked calculations.

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Enthalpy of Hydration (ΔH°hyd)

The standard enthalpy of hydration is the enthalpy change when one mole of gaseous ions becomes surrounded by water molecules to form an aqueous solution (at infinite dilution).

Na⁺(g) → Na⁺(aq) ΔH°hyd = −406 kJ mol⁻¹

Cl⁻(g) → Cl⁻(aq) ΔH°hyd = −363 kJ mol⁻¹

Key points:

• Hydration is always exothermic (negative ΔH°hyd) because ion-dipole attractions form between the ion and surrounding water molecules, releasing energy.
• Hydration enthalpies are more exothermic for ions with higher charge and smaller radius (higher charge density), because these ions attract water molecules more strongly.
• Cations are typically more strongly hydrated than anions of similar size because cations are generally smaller.

Trends in Hydration Enthalpy: Group 1 Cations

Ion	Radius / pm	ΔH°hyd / kJ mol⁻¹
Li⁺	76	−520
Na⁺	102	−406
K⁺	138	−322
Rb⁺	152	−293
Cs⁺	167	−264

As the cation gets larger down the group, the charge density decreases, water molecules are attracted less strongly, and the hydration enthalpy becomes less exothermic.

Trends in Hydration Enthalpy: Halide Anions

Ion	Radius / pm	ΔH°hyd / kJ mol⁻¹
F⁻	133	−506
Cl⁻	181	−363
Br⁻	196	−336
I⁻	220	−295

The same trend: larger ions have less exothermic hydration enthalpies.

Trends in Hydration Enthalpy: Effect of Charge

Ion	Charge	Radius / pm	ΔH°hyd / kJ mol⁻¹
Na⁺	1+	102	−406
Mg²⁺	2+	72	−1920
Al³⁺	3+	54	−4690

Moving from Na⁺ to Al³⁺, both the charge increases and the radius decreases, leading to a dramatically more exothermic hydration enthalpy. The Mg²⁺ value is roughly five times that of Na⁺, reflecting both the doubled charge and the smaller radius.

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Enthalpy of Solution (ΔH°sol)

The standard enthalpy of solution is the enthalpy change when one mole of a solute dissolves in a sufficient amount of water to form an infinitely dilute solution.

NaCl(s) → Na⁺(aq) + Cl⁻(aq) ΔH°sol = +3.9 kJ mol⁻¹

Key points:

• Enthalpy of solution can be positive or negative depending on the balance between lattice breaking (endothermic) and hydration (exothermic).
• A small positive ΔH°sol means the compound still dissolves — the entropy increase on dissolving drives the process.

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The Energy Cycle: Linking Lattice Energy, Hydration, and Solution

The enthalpy of solution can be calculated from the lattice energy (dissociation convention) and the hydration enthalpies of the individual ions:

ΔH°sol = −ΔH°latt(formation) + Σ ΔH°hyd(ions)

Or equivalently, using the dissociation convention for lattice energy:

ΔH°sol = ΔH°latt(dissociation) + Σ ΔH°hyd(ions)

graph TD
    A["NaCl(s)"] -->|"ΔH°sol"| C["Na⁺(aq) + Cl⁻(aq)"]
    A -->|"−ΔH°latt = +787"| B["Na⁺(g) + Cl⁻(g)"]
    B -->|"ΔH°hyd(Na⁺) + ΔH°hyd(Cl⁻)"| C

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Worked Example 1: Enthalpy of Solution of NaCl

Given:

• ΔH°latt(NaCl, formation convention) = −787 kJ mol⁻¹
• ΔH°hyd(Na⁺) = −406 kJ mol⁻¹
• ΔH°hyd(Cl⁻) = −363 kJ mol⁻¹

ΔH°sol = −(−787) + (−406) + (−363)

ΔH°sol = +787 − 406 − 363

ΔH°sol = +18 kJ mol⁻¹

The accepted value is about +3.9 kJ mol⁻¹ — the discrepancy arises from using rounded lattice energy and hydration data. The key point is that the value is slightly positive (endothermic), yet NaCl still dissolves because the entropy increase (ΔS > 0) makes ΔG negative at room temperature.

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Worked Example 2: Enthalpy of Solution of MgCl₂

Given:

• ΔH°latt(MgCl₂, formation convention) = −2523 kJ mol⁻¹
• ΔH°hyd(Mg²⁺) = −1920 kJ mol⁻¹
• ΔH°hyd(Cl⁻) = −363 kJ mol⁻¹

Note: There are two Cl⁻ ions per formula unit.

ΔH°sol = −(−2523) + (−1920) + 2(−363)

ΔH°sol = +2523 − 1920 − 726

ΔH°sol = −123 kJ mol⁻¹

This is significantly exothermic, which explains why MgCl₂ dissolves very readily and the solution gets warm. The very exothermic hydration of the small, doubly-charged Mg²⁺ ion more than compensates for the large lattice energy.

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Worked Example 3: Calculating a Hydration Enthalpy

The enthalpy of solution of KBr is +19.9 kJ mol⁻¹. The lattice energy (formation convention) is −679 kJ mol⁻¹. The hydration enthalpy of K⁺ is −322 kJ mol⁻¹. Calculate the hydration enthalpy of Br⁻.

ΔH°sol = −ΔH°latt + ΔH°hyd(K⁺) + ΔH°hyd(Br⁻)

+19.9 = +679 + (−322) + ΔH°hyd(Br⁻)

+19.9 = +357 + ΔH°hyd(Br⁻)

ΔH°hyd(Br⁻) = 19.9 − 357 = −337 kJ mol⁻¹

This is close to the data book value of −336 kJ mol⁻¹.

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Predicting Whether Dissolving Is Exothermic or Endothermic

The key question is: does the hydration of the ions release enough energy to compensate for breaking the lattice?

Scenario	Result	Example
ΣΔH°hyd > ΔH°latt(dissociation)	Exothermic dissolving (ΔH°sol < 0)	MgCl₂, LiCl
ΣΔH°hyd < ΔH°latt(dissociation)	Endothermic dissolving (ΔH°sol > 0)	NaCl, KBr, NH₄NO₃
ΣΔH°hyd ≈ ΔH°latt(dissociation)	ΔH°sol ≈ 0	NaCl (only slightly positive)

Why do endothermic salts still dissolve? Because dissolving increases the entropy of the system (ions become dispersed in solution). If TΔS > ΔH, then ΔG is negative and the process is feasible. This is why ammonium nitrate dissolves despite being endothermic — the entropy gain is large enough.

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Real-World Applications

• Instant cold packs: Ammonium nitrate dissolving in water is endothermic (ΔH°sol = +25.7 kJ mol⁻¹). When the inner pouch is broken, the salt dissolves and absorbs heat, cooling the pack for treating injuries.
• Self-heating meals: Calcium chloride dissolving is exothermic (ΔH°sol = −82 kJ mol⁻¹). Military ration packs and some ready meals use this reaction to heat food without an external heat source.
• Road de-icing: Sodium chloride and calcium chloride are spread on roads. The dissolution process (especially for CaCl₂) releases heat, helping to melt ice. CaCl₂ is preferred in very cold conditions because its more exothermic dissolution provides additional heat.

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Common Mistakes

Mistake	Correction
Forgetting to reverse lattice energy sign	If given the formation convention (negative), reverse to positive for the dissociation step
Not doubling hydration enthalpy for compounds like MgCl₂	Two Cl⁻ ions means 2 × ΔH°hyd(Cl⁻)
Confusing enthalpy of hydration with enthalpy of solution	Hydration: gaseous ion → aqueous ion. Solution: solid compound → aqueous ions. They are different quantities.
Assuming endothermic dissolving means insoluble	Many endothermic salts dissolve because entropy drives the process

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Summary

• Enthalpy of hydration: gaseous ion → aqueous ion; always exothermic; depends on charge density.
• Enthalpy of solution = −lattice energy (formation) + sum of hydration enthalpies.
• Small, highly charged ions have the most exothermic hydration enthalpies.
• Whether dissolving is exothermic or endothermic depends on the balance between lattice energy and hydration enthalpies.
• Entropy changes explain why endothermic dissolving processes still occur.

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A-Level Deep Dive: Enthalpies of Hydration and Solution

Spec mapping

Edexcel 9CH0 specification Topic 13 requires use of the relationship ΔH°_solution = ΔH°_lattice(dissociation) + Σ ΔH°_hydration to analyse the dissolution of ionic solids in water (refer to the official specification document for exact wording). Candidates must construct enthalpy cycles linking the solid lattice, separate gaseous ions, and aqueous ions; compute one quantity given the other two; and use the data to predict whether a salt is likely to be soluble. Trends in solubility — particularly the decrease in group 2 sulfate solubility down the group, and the increase in group 2 hydroxide solubility down the group — are explained by competing changes in lattice-dissociation enthalpy and total hydration enthalpy. These questions appear on Paper 1 (cycle calculations) and Paper 3 (synoptic with group 2 chemistry, ionic bonding, and entropy). The Edexcel data booklet provides hydration enthalpies for common cations and anions.

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Worked example with full mark scheme

Question (8 marks):

(a) Construct an enthalpy cycle linking ΔH°_solution of NaCl to its lattice dissociation enthalpy and the hydration enthalpies of Na⁺(g) and Cl⁻(g). (2)

(b) Calculate ΔH°_sol[NaCl] given: ΔH°_lat(diss) = +787; ΔH°_hyd[Na⁺] = −406; ΔH°_hyd[Cl⁻] = −364 kJ mol⁻¹. (3)

(c) NaCl is moderately soluble in water at 298 K. Comment on the magnitude and sign of your answer in light of this observation, and explain why entropy is also relevant. (3)

Solution with mark scheme:

(a) Cycle:

NaCl(s) ──ΔH°_sol──→ Na⁺(aq) + Cl⁻(aq) │ ↑ │ ΔH°_lat(diss) │ ΔH°_hyd[Na⁺] + ΔH°_hyd[Cl⁻] ↓ │ Na⁺(g) + Cl⁻(g) ───────────────────┘

M1 — three-corner cycle drawn with arrows. A1 — gaseous ions identified as the common intermediate.

(b) ΔH°_sol = ΔH°_lat(diss) + ΔH°_hyd[Na⁺] + ΔH°_hyd[Cl⁻] = 787 + (−406) + (−364) = 787 − 770 = +17 kJ mol⁻¹

M1 — correct cycle equation. M1 — correct substitution. A1 — answer with sign.

(c) B1 — ΔH°_sol is small and slightly positive: dissolving NaCl is mildly endothermic (consistent with NaCl solutions feeling slightly cool on dissolution).

B1 — despite the positive ΔH, NaCl dissolves readily because dissolution increases entropy substantially: 1 mole of solid → 2 moles of free, hydrated ions in solution.

B1 — Gibbs free energy ΔG°_sol = ΔH°_sol − TΔS°_sol becomes negative at 298 K because the −TΔS term outweighs the modest +ΔH; the process is entropy-driven.

Total: 8 marks (M3 A1 B4).

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Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):

The solubility of group 2 sulfates in water decreases down the group (MgSO₄ very soluble; BaSO₄ essentially insoluble). The lattice dissociation enthalpies are MgSO₄ +2962, CaSO₄ +2641, SrSO₄ +2477, BaSO₄ +2374 kJ mol⁻¹. The hydration enthalpies of M²⁺(g) ions are Mg²⁺ −1920, Ca²⁺ −1650, Sr²⁺ −1480, Ba²⁺ −1360 kJ mol⁻¹. The hydration enthalpy of SO₄²⁻(g) is −1099 kJ mol⁻¹.

(a) Compute ΔH°_sol for MgSO₄ and BaSO₄. (3) (b) Explain the trend in solubility down the group using your answers. (3)

Mark scheme decomposition by AO: