Enthalpy of Hydration and Solution

Spec Mapping — OCR H432 Module 5.2.1 — Enthalpy of hydration and enthalpy of solution, covering definitions of $\Delta_{hyd} H^\ominus$Δhyd​H⊖ (gaseous ion to aqueous ion) and $\Delta_{sol} H^\ominus$Δsol​H⊖ (one mole of ionic solid dissolving to give an infinitely dilute solution); construction of the energy-cycle linking $\Delta_{LE} H^\ominus$ΔLE​H⊖, individual $\Delta_{hyd} H^\ominus$Δhyd​H⊖ for cation and anion, and $\Delta_{sol} H^\ominus$Δsol​H⊖ via the Hess equation $\Delta_{sol} H^\ominus = -\Delta_{LE} H^\ominus + \sum \nu_i \Delta_{hyd} H_i^\ominus$Δsol​H⊖=−ΔLE​H⊖+∑νi​Δhyd​Hi⊖​; numerical calculations for Group 1 halides (NaCl, KBr, LiF), Group 2 halides (CaCl$_2$2​, MgSO$_4$4​), and ammonium nitrate; qualitative justification of hydration-enthalpy trends from ionic charge and radius; explanation of Group 2 sulphate and hydroxide solubility trends as a lattice-versus-hydration balance; identification of insolubility cases (AgCl) where hydration is too small to overcome the lattice; and recognition that small-positive $\Delta_{sol} H^\ominus$Δsol​H⊖ values can still give spontaneous dissolution when the accompanying entropy term is favourable (Lesson 7) (refer to the official OCR H432 specification document for exact wording).

Lesson 4 closed by hinting that lattice-enthalpy magnitude controls macroscopic observables such as solubility — but only through a competing energetic process: hydration. When an ionic crystal is dropped into water, the lattice must be torn apart (endothermic, equal in magnitude to $|\Delta_{LE} H^\ominus|$∣ΔLE​H⊖∣) and the gaseous ions must be enveloped by water molecules (exothermic, $\Delta_{hyd} H^\ominus$Δhyd​H⊖ per ion). The net is the enthalpy of solution, $\Delta_{sol} H^\ominus$Δsol​H⊖ — a difference between two large opposing quantities of $\sim 1000$∼1000 kJ mol$^{-1}$−1 each. This single number decides whether a salt feels warm (CaCl$_2$2​ road de-icer, MgSO$_4$4​ hand-warmer), cold (NH$_4$4​NO$_3$3​ sports cold-pack), or neutral (NaCl seasoning) in your hand. This lesson develops the Hess cycle linking the three enthalpies, walks through five worked examples (NaCl, CaCl$_2$2​, MgSO$_4$4​, NH$_4$4​NO$_3$3​, AgCl), explains Group 2 sulphate vs hydroxide solubility trends, and closes with a tiered specimen question. Lesson 6 adds entropy, which is what saves "endothermic-but-soluble" cases like KBr and NH$_4$4​NO$_3$3​.

Key Equation: for any soluble ionic compound the Hess cycle gives $\Delta_{sol} H^\ominus = -\Delta_{LE} H^\ominus + \sum_{\text{ions}} \nu_i \,\Delta_{hyd} H_i^\ominus$Δsol​H⊖=−ΔLE​H⊖+∑ions​νi​Δhyd​Hi⊖​ The leading $-\Delta_{LE} H^\ominus$−ΔLE​H⊖ is positive (the lattice step in the cycle is breaking the lattice, opposite in sign to the OCR formation convention). Stoichiometric coefficients $\nu_i$νi​ apply to every ion: for CaCl$_2$2​ the chloride hydration appears twice, for MgSO$_4$4​ both ions appear once.

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Learning Objectives (OCR Spec 5.2.1)

By the end of this lesson you should be able to:

• Define enthalpy of hydration ($\Delta_{hyd} H^\ominus$Δhyd​H⊖) and enthalpy of solution ($\Delta_{sol} H^\ominus$Δsol​H⊖) with full OCR precision
• Construct a Hess-law cycle linking $\Delta_{LE} H^\ominus$ΔLE​H⊖, the individual $\Delta_{hyd} H^\ominus$Δhyd​H⊖ values, and $\Delta_{sol} H^\ominus$Δsol​H⊖
• Use the cycle algebraically to calculate any missing enthalpy from the other three
• Predict and justify the trend in hydration enthalpy with ionic charge and radius (down Group 1, down the halides, across Period 3 cations)
• Explain why some salts dissolve exothermically (CaCl$_2$2​), endothermically (NH$_4$4​NO$_3$3​, NaCl), or are essentially insoluble (AgCl, BaSO$_4$4​) in terms of the lattice-vs-hydration balance
• Apply the lattice-vs-hydration balance to the Group 2 sulphate trend (solubility decreases down the group) and the Group 2 hydroxide trend (solubility increases down the group)
• Recognise that thermodynamic feasibility of dissolution also depends on the entropy term (Lesson 6–7) — a small positive $\Delta_{sol} H^\ominus$Δsol​H⊖ does not forbid dissolving

The Dissolving Process at the Particulate Level

When an ionic crystal is placed in water, two opposing energetic steps occur: (1) the lattice breaks, costing $|\Delta_{LE} H^\ominus|$∣ΔLE​H⊖∣ — under OCR's convention $\Delta_{LE} H^\ominus$ΔLE​H⊖ is negative for formation, so the breaking step is positive and equal to $-\Delta_{LE} H^\ominus$−ΔLE​H⊖; (2) the gaseous ions become aqueous, each surrounded by an oriented shell of water dipoles (oxygen $\to$→ cation; hydrogen $\to$→ anion) — the always-exothermic hydration step. The net is the experimentally observable $\Delta_{sol} H^\ominus$Δsol​H⊖. For NaCl, $+787 - 406 - 364 = +17$+787−406−364=+17 kJ mol$^{-1}$−1 — a 17 kJ difference between summed terms of $\sim 800$∼800 kJ mol$^{-1}$−1 each; a 2 % error on any input flips the sign.

Key Definitions (OCR):

• Enthalpy of hydration ($\Delta_{hyd} H^\ominus$Δhyd​H⊖) — the enthalpy change when one mole of gaseous ions dissolves in water to form one mole of aqueous ions under standard conditions (298 K, 100 kPa, infinite dilution).
• Enthalpy of solution ($\Delta_{sol} H^\ominus$Δsol​H⊖) — the enthalpy change when one mole of an ionic solid dissolves in sufficient water to give an "infinitely dilute" solution under standard conditions.
• Standard conditions for solution data: 298 K, 100 kPa, and the convention that the dissolved species is at infinite dilution so that ion–ion interactions in the solution are negligible.

Enthalpy of Hydration: the Microscopic Picture

A water molecule is a permanent dipole — oxygen $\delta^-$δ−, hydrogens $\delta^+$δ+. Cations are surrounded by water with oxygens pointing inward (Mg$^{2+}$2+ in an octahedral first shell of six waters); anions are surrounded by water with one H hydrogen-bonded to the anion. Hydration is always exothermic — ion–dipole interactions have the same Coulombic skeleton as ion–ion ($q/r$q/r scaling), giving two dominant factors:

1. Ionic charge ($q$q): higher charge means stronger ion–dipole attraction. Mg$^{2+}$2+ ($-1920$−1920 kJ mol$^{-1}$−1) is over four times more exothermic to hydrate than Na$^+$+ ($-406$−406 kJ mol$^{-1}$−1), even though their radii are similar (72 vs 102 pm).
2. Ionic radius ($r$r): smaller ions have higher charge density (charge per unit surface area), giving stronger attraction to water dipoles. Down Group 1: Li$^+$+ ($-519$−519) > Na$^+$+ ($-406$−406) > K$^+$+ ($-322$−322) > Rb$^+$+ ($-301$−301) > Cs$^+$+ ($-276$−276). Down the halides: F$^-$− ($-506$−506) > Cl$^-$− ($-364$−364) > Br$^-$− ($-335$−335) > I$^-$− ($-293$−293).

Ion	Charge	Radius / pm	Charge density (q/r)	$\Delta_{hyd} H^\ominus$Δhyd​H⊖ / kJ mol$^{-1}$−1
Li$^+$+	+1	76	0.013	$-519$−519
Na$^+$+	+1	102	0.010	$-406$−406
K$^+$+	+1	138	0.007	$-322$−322
Rb$^+$+	+1	152	0.007	$-301$−301
Cs$^+$+	+1	167	0.006	$-276$−276
Mg$^{2+}$2+	+2	72	0.028	$-1920$−1920
Ca$^{2+}$2+	+2	100	0.020	$-1650$−1650
Ba$^{2+}$2+	+2	135	0.015	$-1360$−1360
F$^-$−	$-1$−1	133	0.008	$-506$−506
Cl$^-$−	$-1$−1	181	0.006	$-364$−364
Br$^-$−	$-1$−1	196	0.005	$-335$−335
I$^-$−	$-1$−1	220	0.005	$-293$−293

Note the Group 2 cations: doubling the charge from Group 1 to Group 2 makes hydration enthalpy roughly four times more exothermic at comparable radii, exactly as the $q/r$q/r scaling predicts.

graph TD
  A["Gaseous ion (Na+ or Cl-)"] -->|water dipoles orient and pack around the ion| B[First hydration shell forms]
  B -->|"ion-dipole electrostatic stabilisation: ΔhydH negative"| C[Aqueous ion]
  C --> D[Outer hydration shells: weaker, exchangeable]

Enthalpy of Solution: the Macroscopic Number

The enthalpy of solution is what a school calorimeter measures (PAG 3): dissolve a known mass of salt in water, record $\Delta T$ΔT, calculate $\Delta_{sol} H^\ominus$Δsol​H⊖ from $q = mc\Delta T$q=mcΔT. Selected data-book examples:

Compound	$\Delta_{sol} H^\ominus$Δsol​H⊖ / kJ mol$^{-1}$−1	Practical observation
NaCl	$+3.9$+3.9	barely any temperature change
NH$_4$4​NO$_3$3​	$+26$+26	cools noticeably (cold pack)
CaCl$_2$2​	$-82$−82	warms strongly (road de-icer)
MgSO$_4$4​ (anhydrous)	$-91$−91	warms strongly (hand-warmer)
AgCl	$+65$+65	essentially insoluble
BaSO$_4$4​	very positive	essentially insoluble

The Hess-Law Cycle

Energy is a state function, so the enthalpy change of going from solid ionic compound to aqueous ions must be the same whether you take the direct route (measurable in a calorimeter) or the indirect route via gaseous ions:

graph TD
  A["MX(s) solid"] -->|"ΔsolH (direct, calorimeter)"| B["M+(aq) + X-(aq)"]
  A -->|"-ΔLEH (break lattice into gaseous ions)"| C["M+(g) + X-(g) gaseous"]
  C -->|"ΔhydH(M+) + ΔhydH(X-)"| B

By Hess's law the two routes are equal:

$\Delta_{sol} H^\ominus = -\Delta_{LE} H^\ominus + \Delta_{hyd} H^\ominus(\text{cation}) + \Delta_{hyd} H^\ominus(\text{anion})$Δsol​H⊖=−ΔLE​H⊖+Δhyd​H⊖(cation)+Δhyd​H⊖(anion)

The key bookkeeping points:

• The lattice step in the cycle goes in the reverse direction to lattice enthalpy. Under the OCR convention $\Delta_{LE} H^\ominus$ΔLE​H⊖ is the enthalpy of formation of the lattice (negative); to break it you flip the sign, getting $-\Delta_{LE} H^\ominus$−ΔLE​H⊖ (positive).
• For ionic compounds with stoichiometry M$_a$a​X$_b$b​ both hydration terms carry the stoichiometric coefficient. CaCl$_2$2​ needs $\Delta_{hyd} H^\ominus(\text{Ca}^{2+}) + 2 \,\Delta_{hyd} H^\ominus(\text{Cl}^-)$Δhyd​H⊖(Ca2+)+2Δhyd​H⊖(Cl−). Al$_2$2​(SO$_4$4​)$_3$3​ needs $2 \,\Delta_{hyd} H^\ominus(\text{Al}^{3+}) + 3 \,\Delta_{hyd} H^\ominus(\text{SO}_4^{2-})$2Δhyd​H⊖(Al3+)+3Δhyd​H⊖(SO42−​).

Worked Example 1: NaCl (slightly endothermic, entropy-driven)

Given $\Delta_{LE} H^\ominus(\text{NaCl}) = -787$ΔLE​H⊖(NaCl)=−787, $\Delta_{hyd} H^\ominus(\text{Na}^+) = -406$Δhyd​H⊖(Na+)=−406, $\Delta_{hyd} H^\ominus(\text{Cl}^-) = -364$Δhyd​H⊖(Cl−)=−364 kJ mol$^{-1}$−1, calculate $\Delta_{sol} H^\ominus$Δsol​H⊖.

Hess equation: $\Delta_{sol} H^\ominus = -\Delta_{LE} H^\ominus + \Delta_{hyd} H^\ominus(\text{Na}^+) + \Delta_{hyd} H^\ominus(\text{Cl}^-)$Δsol​H⊖=−ΔLE​H⊖+Δhyd​H⊖(Na+)+Δhyd​H⊖(Cl−)

Substitute: $\Delta_{sol} H^\ominus = -(-787) + (-406) + (-364) = +787 - 406 - 364 = +17\;\text{kJ mol}^{-1}$Δsol​H⊖=−(−787)+(−406)+(−364)=+787−406−364=+17kJ mol−1

The 13 kJ discrepancy from the data-booklet value of $+3.9$+3.9 kJ mol$^{-1}$−1 comes from rounding in the input data — when subtracting two nearly-equal large quantities, 1 % rounding errors on inputs become very visible in the answer. The qualitative message — NaCl dissolves only slightly endothermically — is correct. The reason NaCl is freely soluble despite this positive enthalpy of solution is entropy: the dispersion of two moles of ions into water provides a large positive $\Delta_{sol} S^\ominus$Δsol​S⊖ (about $+43$+43 J K$^{-1}$−1 mol$^{-1}$−1), so the Gibbs free energy of solution $\Delta_{sol} G^\ominus = \Delta_{sol} H^\ominus - T \Delta_{sol} S^\ominus = +3.9 - 298 \times 0.043 = -9$Δsol​G⊖=Δsol​H⊖−TΔsol​S⊖=+3.9−298×0.043=−9 kJ mol$^{-1}$−1 is comfortably negative (Lessons 6 and 7).

Worked Example 2: CaCl$_2$2​ (the exothermic hand-warmer case)

Given $\Delta_{LE} H^\ominus(\text{CaCl}_2) = -2258$ΔLE​H⊖(CaCl2​)=−2258, $\Delta_{hyd} H^\ominus(\text{Ca}^{2+}) = -1650$Δhyd​H⊖(Ca2+)=−1650, $\Delta_{hyd} H^\ominus(\text{Cl}^-) = -364$Δhyd​H⊖(Cl−)=−364 kJ mol$^{-1}$−1. Note the $\nu = 2$ν=2 on chloride:

$\Delta_{sol} H^\ominus = -(-2258) + (-1650) + 2(-364) = +2258 - 1650 - 728 = -120\;\text{kJ mol}^{-1}$Δsol​H⊖=−(−2258)+(−1650)+2(−364)=+2258−1650−728=−120kJ mol−1

Strongly exothermic — the basis of anhydrous-CaCl$_2$2​ hand-warmers and self-heating ready-meal cans. Dissolving 100 g in 100 g water releases $\sim 108$∼108 kJ, warming from 5 °C to over 70 °C. (Literature: $-82$−82 kJ mol$^{-1}$−1; our 38 kJ deficit is typical Hess-cycle rounding.) The reverse logic — exothermic dissolving in any film of water on an icy road — is why CaCl$_2$2​ is spread as a winter road de-icer, with the dissolved chloride also lowering the freezing point.

Worked Example 3: MgSO$_4$4​ (hand-warmer chemistry, double-Group-2 case)

Given $\Delta_{LE} H^\ominus(\text{MgSO}_4) = -2833$ΔLE​H⊖(MgSO4​)=−2833, $\Delta_{hyd} H^\ominus(\text{Mg}^{2+}) = -1920$Δhyd​H⊖(Mg2+)=−1920, $\Delta_{hyd} H^\ominus(\text{SO}_4^{2-}) = -1004$Δhyd​H⊖(SO42−​)=−1004 kJ mol$^{-1}$−1:

$\Delta_{sol} H^\ominus = -(-2833) + (-1920) + (-1004) = +2833 - 1920 - 1004 = -91\;\text{kJ mol}^{-1}$Δsol​H⊖=−(−2833)+(−1920)+(−1004)=+2833−1920−1004=−91kJ mol−1

Excellent agreement with the literature value of $-91$−91 kJ mol$^{-1}$−1. This is the "Epsom salt warming compress" chemistry: anhydrous MgSO$_4$4​ encountering skin moisture releases heat. (Hydrated MgSO_4$$\cdot7H$_2$2​O has $\Delta_{sol} H^\ominus$Δsol​H⊖ close to zero because the hydration has effectively already happened when the crystal formed.)

Worked Example 4: NH$_4$4​NO$_3$3​ (the cold-pack case)

Given $\Delta_{LE} H^\ominus(\text{NH}_4\text{NO}_3) = -649$ΔLE​H⊖(NH4​NO3​)=−649, $\Delta_{hyd} H^\ominus(\text{NH}_4^+) = -307$Δhyd​H⊖(NH4+​)=−307, $\Delta_{hyd} H^\ominus(\text{NO}_3^-) = -314$Δhyd​H⊖(NO3−​)=−314 kJ mol$^{-1}$−1:

$\Delta_{sol} H^\ominus = -(-649) + (-307) + (-314) = +649 - 307 - 314 = +28\;\text{kJ mol}^{-1}$Δsol​H⊖=−(−649)+(−307)+(−314)=+649−307−314=+28kJ mol−1

Strongly endothermic (lit. $+26$+26). Dissolving 100 g cools a beaker of water from 25 °C to about $-3$−3 °C — the disposable sports-injury cold-pack. The dissolving still proceeds spontaneously because of the large positive entropy of solution ($+108$+108 J K$^{-1}$−1 mol$^{-1}$−1), giving $\Delta_{sol} G^\ominus \approx +26 - 298 \times 0.108 = -6$Δsol​G⊖≈+26−298×0.108=−6 kJ mol$^{-1}$−1.

Worked Example 5: AgCl (insoluble — hydration too small to break the lattice)

Given $\Delta_{LE} H^\ominus(\text{AgCl}) = -905$ΔLE​H⊖(AgCl)=−905, $\Delta_{hyd} H^\ominus(\text{Ag}^+) = -464$Δhyd​H⊖(Ag+)=−464, $\Delta_{hyd} H^\ominus(\text{Cl}^-) = -364$Δhyd​H⊖(Cl−)=−364 kJ mol$^{-1}$−1:

$\Delta_{sol} H^\ominus = -(-905) + (-464) + (-364) = +905 - 464 - 364 = +77\;\text{kJ mol}^{-1}$Δsol​H⊖=−(−905)+(−464)+(−364)=+905−464−364=+77kJ mol−1

Strongly positive. With $\Delta_{sol} S^\ominus$Δsol​S⊖ only $+33$+33 J K$^{-1}$−1 mol$^{-1}$−1, $T\Delta S$TΔS at 298 K is just $\sim 10$∼10 kJ mol$^{-1}$−1 — nowhere near enough to bridge the $+77$+77 enthalpy gap. So $\Delta_{sol} G^\ominus \approx +67$Δsol​G⊖≈+67 kJ mol$^{-1}$−1 and $K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}$Ksp​(AgCl)=1.8×10−10. The lattice term wins because of Fajans-rules covalent character (Lesson 4): Ag$^+$+'s polarising d$^{10}$10 configuration makes $\Delta_{LE} H^\ominus$ΔLE​H⊖ about $135$135 kJ mol$^{-1}$−1 more exothermic than point-charge predicts; the hydration term cannot compensate.

Worked Example 6: Finding an Unknown Hydration Enthalpy (LiF)

The cycle can be rearranged to solve for any unknown term. Given $\Delta_{sol} H^\ominus(\text{LiF}) = +5$Δsol​H⊖(LiF)=+5 kJ mol$^{-1}$−1, $\Delta_{LE} H^\ominus(\text{LiF}) = -1031$ΔLE​H⊖(LiF)=−1031 kJ mol$^{-1}$−1, $\Delta_{hyd} H^\ominus(\text{F}^-) = -506$Δhyd​H⊖(F−)=−506 kJ mol$^{-1}$−1, find $\Delta_{hyd} H^\ominus(\text{Li}^+)$Δhyd​H⊖(Li+).

Hess equation: $\Delta_{sol} H^\ominus = -\Delta_{LE} H^\ominus + \Delta_{hyd} H^\ominus(\text{Li}^+) + \Delta_{hyd} H^\ominus(\text{F}^-)$Δsol​H⊖=−ΔLE​H⊖+Δhyd​H⊖(Li+)+Δhyd​H⊖(F−)

Rearrange:

$\Delta_{hyd} H^\ominus(\text{Li}^+) = \Delta_{sol} H^\ominus + \Delta_{LE} H^\ominus - \Delta_{hyd} H^\ominus(\text{F}^-)$Δhyd​H⊖(Li+)=Δsol​H⊖+ΔLE​H⊖−Δhyd​H⊖(F−)

Substitute: $\Delta_{hyd} H^\ominus(\text{Li}^+) = (+5) + (-1031) - (-506) = +5 - 1031 + 506 = -520\;\text{kJ mol}^{-1}$Δhyd​H⊖(Li+)=(+5)+(−1031)−(−506)=+5−1031+506=−520kJ mol−1

Close to the accepted value of $-519$−519 kJ mol$^{-1}$−1 (and the highest of the Group 1 cations, as expected — Li$^+$+ has the smallest radius and therefore highest charge density).

Group 2 Sulphate Trend: Solubility Decreases Down the Group

Group 2 sulphates show one of the cleanest solubility trends in the A-Level syllabus: MgSO$_4$4​ is freely soluble (35 g per 100 g water at 20 °C), CaSO$_4$4​ is sparingly soluble (0.2 g per 100 g water — gypsum, plaster of Paris), SrSO$_4$4​ is very poorly soluble (0.014 g per 100 g), BaSO$_4$4​ is essentially insoluble (0.0002 g per 100 g — clinical barium meal). The trend is entirely explained by the lattice-vs-hydration balance.