Enthalpy of Solution and Hydration

Dissolving an ionic solid in water is an everyday chemical process — table salt in soup, calcium chloride on icy pavements, ammonium nitrate in instant cold packs — yet the underlying thermochemistry exposes one of the most elegant balances in physical chemistry. The standard enthalpy of solution (ΔsolH⦵) is the enthalpy change when one mole of an ionic compound dissolves completely in a large excess of water to form aqueous ions at infinite dilution under standard conditions (100 kPa, 298 K, solid in its standard state). The standard enthalpy of hydration (ΔhydH⦵) is the enthalpy change when one mole of gas-phase ions becomes one mole of aqueous ions, again at infinite dilution. These two quantities are linked to the lattice enthalpy through a single Hess-cycle relationship: ΔsolH = ΔlattH(dissociation) + Σ ΔhydH(ions). In this lesson we will use that cycle to compute solution enthalpies, to extract ionic hydration enthalpies from experimental data, and — most importantly — to rationalise the famous opposite solubility trends of Group 2 sulfates (which decrease in solubility down the group) and Group 2 hydroxides (which increase). We will close by noting that enthalpy alone is not the whole story: entropy completes the picture, anticipating lesson 5.

Spec mapping (AQA 7405): This lesson maps to §3.1.8 of the AQA A-Level Chemistry specification, the A2 thermodynamics topic dealing with Born-Haber cycles, enthalpies of solution and hydration, entropy and Gibbs free energy. It builds directly on lesson 3 of this course (Born-Haber cycles, which deliver lattice enthalpies of formation and dissociation), and on lesson 2 (Hess's law as the general framework for thermochemical cycles). It also threads forward to lesson 5 (entropy and Gibbs free energy — the missing piece in deciding solubility). The qualitative solubility patterns discussed here connect to the inorganic course §3.2 (trends in Group 1 and Group 2), and the molecular basis of hydration — ion-dipole interactions — connects to lesson 4 of the §3.1.3 intermolecular forces topic. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: Definitions of ΔsolH and ΔhydH (with the precise stipulations about aqueous ions at infinite dilution and the stated standard state for the solid) are AO1 recall. Computing one quantity in the cycle when the other two are given — for example, finding ΔsolH(NaCl) from ΔlattH(dissoc) and the two hydration enthalpies, or extracting ΔhydH(Cl⁻) from ΔsolH, ΔlattH and ΔhydH(Na⁺) — is core AO2. Rationalising the solubility trends in Group 2 sulfates (decreasing down the group) and Group 2 hydroxides (increasing down the group), using the competing trends in ΔlattH and ΔhydH and the effect of anion size, tests AO3 synthesis and is the topic-defining application of the specification.

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Defining ΔsolH and ΔhydH

The two key definitions must be stated with care, because both involve hidden stipulations that the mark scheme rewards.

Key Definition — Enthalpy of solution: The standard enthalpy of solution, ΔsolH⦵, is the enthalpy change when one mole of an ionic compound dissolves completely in a large excess of water to form aqueous ions at infinite dilution, under standard conditions (100 kPa, 298 K), with the solid initially in its standard state.

Key Definition — Enthalpy of hydration: The standard enthalpy of hydration, ΔhydH⦵, is the enthalpy change when one mole of gas-phase ions becomes one mole of aqueous ions at infinite dilution, under standard conditions.

Two stipulations matter. First, "infinite dilution" means we ignore ion-ion interactions in solution — every aqueous ion is surrounded only by water molecules, never by other solutes. In a real 1 mol dm⁻³ solution, neighbouring ions do interact (the basis of the Debye-Hückel theory mentioned in the Going Further section), so tabulated ΔhydH values are extrapolations to zero concentration. Second, the symbol Δ_hyd_H refers to a single ionic species — for example, ΔhydH(Na⁺) or ΔhydH(Cl⁻) — not to an ion pair. When we dissolve NaCl we sum the contributions from both ions separately.

Both definitions, by convention, are exothermic for spontaneous hydration of typical small ions: gas-phase ions attract polar water molecules through ion-dipole forces, releasing energy. ΔsolH, on the other hand, can be exothermic or endothermic — it is the small net difference between two large numbers (the endothermic ΔlattH(dissoc) and the exothermic ΔhydH sum), and its sign depends on which dominates.

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The Thermochemical Cycle

Because enthalpy is a state function (Hess's law, lesson 2), we can construct a two-step alternative route from solid ionic compound to aqueous ions:

Step 1. Solid → gas-phase ions, governed by the lattice enthalpy of dissociation ΔlattH(dissoc), which is endothermic and equal in magnitude but opposite in sign to the lattice enthalpy of formation given by a Born-Haber cycle.

Step 2. Gas-phase ions → aqueous ions, governed by the sum of the hydration enthalpies of all the ions produced, which is exothermic.

The cycle for a generic 1:1 ionic compound MX is:

                ΔlattH(dissoc)
   MX(s)  ─────────────────────►  M⁺(g) + X⁻(g)
     │                                    │
     │                                    │ ΔhydH(M⁺) + ΔhydH(X⁻)
     │           ΔsolH                    │
     ▼                                    ▼
   M⁺(aq) + X⁻(aq)  ◄──────────────  M⁺(aq) + X⁻(aq)

By Hess's law:

ΔsolH = ΔlattH(dissoc) + Σ ΔhydH(ions)

For a 1:1 salt MX: ΔsolH = ΔlattH(dissoc) + ΔhydH(M⁺) + ΔhydH(X⁻).

For a 1:2 salt MX₂ (such as MgCl₂ or CaCl₂): ΔsolH = ΔlattH(dissoc) + ΔhydH(M²⁺) + 2 × ΔhydH(X⁻). The factor of two is a frequent exam trap (see Common Errors).

For a 2:1 salt M₂X (such as Na₂SO₄): ΔsolH = ΔlattH(dissoc) + 2 × ΔhydH(M⁺) + ΔhydH(X²⁻).

Key Point — Sign of ΔlattH: Many data tables list the lattice enthalpy of formation (gas-phase ions → solid), which is exothermic and negative. In the solution cycle we need the dissociation value, which is the reverse process and therefore positive: ΔlattH(dissoc) = −ΔlattH(form). Reversing the sign is the single commonest error in this topic.

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Worked Example 1: ΔsolH of Sodium Chloride

Calculate ΔsolH for NaCl, given:

• ΔlattH(dissoc, NaCl) = +788 kJ mol⁻¹
• ΔhydH(Na⁺) = −406 kJ mol⁻¹
• ΔhydH(Cl⁻) = −364 kJ mol⁻¹

Applying the cycle:

ΔsolH(NaCl) = ΔlattH(dissoc) + ΔhydH(Na⁺) + ΔhydH(Cl⁻) = (+788) + (−406) + (−364) = +788 − 770 = +18 kJ mol⁻¹

The slightly endothermic value is consistent with NaCl being soluble but with no detectable warming or cooling on dissolution — a household observation: salt dissolves in cold water without a measurable temperature change. The fact that ΔsolH is positive at all yet NaCl dissolves readily is a foretaste of the entropy argument (see lesson 5): the entropy gain on dissolving the ordered crystal is large enough to make ΔG negative despite the slightly unfavourable enthalpy.

Worked Example 2: ΔsolH of Magnesium Chloride

For MgCl₂:

• ΔlattH(dissoc, MgCl₂) = +2493 kJ mol⁻¹
• ΔhydH(Mg²⁺) = −1920 kJ mol⁻¹
• ΔhydH(Cl⁻) = −364 kJ mol⁻¹

ΔsolH(MgCl₂) = ΔlattH(dissoc) + ΔhydH(Mg²⁺) + 2 × ΔhydH(Cl⁻) = (+2493) + (−1920) + 2 × (−364) = +2493 − 1920 − 728 = −155 kJ mol⁻¹

The very negative ΔsolH is consistent with the observable behaviour of MgCl₂: it dissolves with appreciable warming of the water. The dominant factor is the very large exothermic hydration enthalpy of the small, doubly-charged Mg²⁺ ion (−1920 kJ mol⁻¹) compared with the singly-charged Na⁺ (−406 kJ mol⁻¹) — even though the lattice enthalpy of dissociation is also much larger for the 2+/1− compound. The factor of 2 on the chloride hydration enthalpy is essential — leave it out and you would get ΔsolH = +209 kJ mol⁻¹, which is wrong both in sign and in magnitude.

Worked Example 3: Extracting ΔhydH(Cl⁻)

Given ΔsolH(NaCl) = +4 kJ mol⁻¹ (an experimentally measured value), ΔlattH(dissoc, NaCl) = +780 kJ mol⁻¹, and ΔhydH(Na⁺) = −406 kJ mol⁻¹, calculate ΔhydH(Cl⁻).

Rearrange the cycle:

ΔhydH(Cl⁻) = ΔsolH − ΔlattH(dissoc) − ΔhydH(Na⁺) = (+4) − (+780) − (−406) = 4 − 780 + 406 = −370 kJ mol⁻¹

This is the standard route by which single-ion hydration enthalpies are obtained: by combining measurable bulk quantities (ΔsolH from calorimetry, ΔlattH from a Born-Haber cycle) with a reference ion whose hydration enthalpy is fixed by convention. The conventional reference is ΔhydH(H⁺) = −1110 kJ mol⁻¹, but for A-Level we treat tabulated single-ion ΔhydH values as given.

Worked Example 4: ΔsolH of Potassium Chloride

For KCl, with ΔlattH(dissoc) = +711 kJ mol⁻¹, ΔhydH(K⁺) = −322 kJ mol⁻¹ and ΔhydH(Cl⁻) = −364 kJ mol⁻¹:

ΔsolH(KCl) = +711 − 322 − 364 = +25 kJ mol⁻¹

Slightly more endothermic than NaCl, consistent with the small cooling observed when KCl dissolves. The larger K⁺ ion has a weaker (less exothermic) hydration enthalpy than Na⁺, but the lattice enthalpy of dissociation is also smaller. The two effects partly cancel, leaving a marginally more endothermic ΔsolH than NaCl. This near-cancellation is typical of Group 1 chlorides, all of which are soluble but with small ΔsolH values that hover near zero.

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Factors Affecting ΔhydH

The magnitude of the (exothermic, negative) enthalpy of hydration depends on the charge density of the ion — defined qualitatively as charge per unit volume, often approximated as charge / radius:

|ΔhydH| ∝ charge² / radius

The two trends that follow are:

1. Smaller ions are more strongly hydrated. For a fixed charge, decreasing ionic radius increases the electric field at the ion's surface, polarising nearby water molecules more strongly and increasing the ion-dipole attraction. Comparing the Group 1 cations:

   Ion	Ionic radius / pm	ΔhydH / kJ mol⁻¹
   Li⁺	76	−519
   Na⁺	102	−406
   K⁺	138	−322
   Rb⁺	152	−293
   Cs⁺	167	−264

   The magnitude decreases smoothly down the group, exactly as the charge-density argument predicts.

2. Higher charges hydrate much more strongly. Doubling the charge quadruples the charge-squared term, so ΔhydH becomes vastly more exothermic for divalent and trivalent ions:

   Ion	Ionic radius / pm	ΔhydH / kJ mol⁻¹
   Na⁺	102	−406
   Mg²⁺	72	−1920
   Al³⁺	54	−4690

   Mg²⁺ is hydrated nearly five times as strongly as Na⁺ despite being only slightly smaller. Al³⁺ is so strongly hydrated that the aqua complex [Al(H₂O)₆]³⁺ is sufficiently acidic to hydrolyse water — a topic developed in §3.2.5 (acid-base behaviour of aqua ions).

For anions the same rules apply: F⁻ (small) is hydrated more strongly than I⁻ (large); O²⁻ would be hydrated extraordinarily strongly were it not that O²⁻(g) is itself thermodynamically unstable and is not isolable. SO₄²⁻ and CO₃²⁻ are weakly hydrated despite their 2− charge, because the negative charge is delocalised over several oxygen atoms — the effective charge density at any one point on the surface is modest.

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Factors Affecting ΔlattH(dissoc)

The lattice enthalpy of dissociation (positive, the energy required to break the ionic solid into gas-phase ions) follows essentially the same charge-density rule, but applied to the ion pair rather than the single ion:

ΔlattH ∝ (charge of cation × charge of anion) / (sum of ionic radii)

Comparing some examples (all values are lattice enthalpies of dissociation, in kJ mol⁻¹):

Compound	ΔlattH(dissoc)
NaCl	+788
NaI	+704
KCl	+711
MgCl₂	+2493
MgO	+3791
CaO	+3401
Al₂O₃	+15916

The pattern is clear: small ions and high charges give very large lattice enthalpies. MgO has both small ions and 2+/2− charges, hence its huge ΔlattH and very high melting point (2852 °C). Al₂O₃ combines 3+ ions with 2− ions and is correspondingly even more refractory.

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Solubility Trends in Group 2 Sulfates

This is the topic-defining application. Going down Group 2 from Mg to Ba:

Compound	Solubility / mol per 100 g H₂O at 20 °C
MgSO₄	~0.36 (very soluble)
CaSO₄	~2 × 10⁻³ (sparingly soluble)
SrSO₄	~5 × 10⁻⁴
BaSO₄	~1 × 10⁻⁵ (insoluble)

Solubility decreases sharply down the group. The thermochemical explanation:

• The sulfate ion SO₄²⁻ is large (effective radius ≈ 230 pm). Therefore, in ΔlattH ∝ (q⁺ × q⁻)/(r₊ + r₋), the denominator (r₊ + r₋) is dominated by the sulfate radius. Changing the cation from Mg²⁺ to Ba²⁺ alters r₊ from 72 pm to 135 pm — but since the sum (r₊ + r₋) goes only from ~300 pm to ~365 pm (a ~22% increase), ΔlattH decreases only modestly down the group.
• ΔhydH(M²⁺), in contrast, depends only on the cation. Going from Mg²⁺ (−1920) to Ba²⁺ (−1360) is a ~30% decrease in magnitude — much larger than the change in ΔlattH.
• Therefore the hydration term shrinks faster than the lattice term: the cycle becomes progressively more endothermic down the group, and solubility decreases.

A typical mark-scheme phrasing: "The lattice enthalpy of dissociation changes by less than the sum of the hydration enthalpies down the group, because the sulfate ion is large and dominates the cation-anion distance. The hydration enthalpy of the cation decreases more steeply than the lattice enthalpy, so ΔsolH becomes more endothermic and solubility decreases."

Solubility Trends in Group 2 Hydroxides

The pattern is opposite to the sulfates:

Compound	Solubility / mol per 100 g H₂O at 20 °C
Mg(OH)₂	~2 × 10⁻⁵ (insoluble)
Ca(OH)₂	~2 × 10⁻³
Sr(OH)₂	~3 × 10⁻²
Ba(OH)₂	~1 × 10⁻¹ (soluble)

Solubility increases down the group. The explanation inverts the sulfate argument:

• The hydroxide ion OH⁻ is small (radius ≈ 153 pm), comparable in size to the cations themselves. Therefore the denominator (r₊ + r₋) in the lattice enthalpy expression changes substantially when we swap Mg²⁺ for Ba²⁺ — the sum goes from ~225 pm to ~288 pm (a ~28% increase). ΔlattH(dissoc) decreases sharply down the group.
• ΔhydH(M²⁺) still decreases by ~30% going Mg²⁺ → Ba²⁺.
• The two effects are now comparable in size, and the lattice term — which is endothermic in the cycle — dominates the change. ΔsolH becomes less endothermic (more favourable) down the group, and solubility increases.

The key contrast is therefore: anion size determines which trend dominates. Large anion (sulfate) → lattice enthalpy changes little, hydration dominates, solubility decreases down the group. Small anion (hydroxide) → lattice enthalpy changes a lot, lattice dominates, solubility increases down the group.

Solubility Trends in Group 1 Carbonates

The carbonate ion CO₃²⁻ is large (radius ≈ 178 pm), so the pattern is similar to the sulfates: solubility of Group 2 carbonates decreases down the group (MgCO₃ slightly soluble, BaCO₃ effectively insoluble). For Group 1 carbonates, all are soluble in water, but lithium carbonate (Li₂CO₃) is much less soluble than the others — an anomaly that reflects Li⁺'s unusually high charge density and large hydration enthalpy stabilising the dissolved ion only modestly relative to the small-ion lattice. This anomaly is industrially relevant: Li₂CO₃ is precipitated from brines as the standard route to lithium feedstock for lithium-ion batteries.

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Enthalpy Is Not the Whole Story — Entropy

A central caveat: ΔsolH alone does not determine solubility. The thermodynamic criterion for a spontaneous process is ΔG < 0, where ΔG = ΔH − TΔS. Dissolving an ionic solid almost always produces a large positive ΔS — the ordered crystal lattice is replaced by free ions moving around in solution, and disorder increases substantially. Hence even reactions with ΔsolH > 0 (slightly endothermic dissolution) can be spontaneous if TΔS > ΔH.

NaCl is the textbook example: ΔsolH = +4 kJ mol⁻¹, ΔS = +43 J K⁻¹ mol⁻¹, so at 298 K, TΔS = +12.8 kJ mol⁻¹, giving ΔG = +4 − 12.8 = −8.8 kJ mol⁻¹ — comfortably negative, hence soluble. Ammonium nitrate is the dramatic example: ΔsolH = +26 kJ mol⁻¹ (strongly endothermic — the basis of instant cold packs) yet ΔS = +109 J K⁻¹ mol⁻¹, so TΔS = +32.5 kJ mol⁻¹ and ΔG = −6.5 kJ mol⁻¹.

Conversely, "insoluble" salts (BaSO₄, AgCl) have very unfavourable ΔG of solution, dominated by the small or negative ΔS associated with strong ion-pair attractions in concentrated brine — but they are never literally insoluble; they have small but finite solubility-product constants K_sp. We will develop ΔG and the Gibbs equation fully in lesson 5.

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