A-Level Chemistry 9701 / Topic 23 / A2

Pull the lattice apart, let water wrap the ions.

Dissolving an ionic solid is a contest: the lattice energy holds the ions together, the hydration of each ion releases energy as water molecules cluster round it. A simple Hess cycle links the two to the enthalpy change of solution.

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The key idea

The enthalpy change of hydration is the enthalpy change when 1 mol of gaseous ions dissolves in water, e.g. Na⁺(g) → Na⁺(aq); it is always exothermic because water dipoles are attracted to the ion. The enthalpy change of solution is the enthalpy change when 1 mol of an ionic solid dissolves in enough water to give an infinitely dilute solution, e.g. NaCl(s) → Na⁺(aq) + Cl⁻(aq). A Hess cycle through the gaseous ions gives LE + ΔH_sol = ΣΔH_hyd, so for NaCl ΔH_sol = (−406 − 377) − (−787) = +4 kJ mol⁻¹. Hydration is more exothermic for ions of higher charge and smaller radius (higher charge density).

Fig. 1: energy cycle for dissolving sodium chloride (values in kJ mol−1)

Section 01

The solution cycle.

Set up the triangle in the simulator: solid to gaseous ions (reverse of lattice energy), then gaseous ions to aqueous ions (hydration).

Salt	LE	ΔH_hyd(cation)	ΔH_hyd(anion)	ΔH_sol = ΣΔH_hyd − LE
NaCl	−787	−406	−377	−783 + 787 = +4
LiF	−1037	−519	−506	−1025 + 1037 = +12
MgCl₂	−2521	−1920	2 × (−377)	−2674 + 2521 = −153

All values in " + KJ + ". Count every ion: MgCl₂ releases two moles of Cl⁻, so its anion hydration is doubled.
Rearranging: the same cycle finds a missing hydration enthalpy. For KCl, ΔH_sol = +17 and LE = −718, so ΔH_hyd(K⁺) = 17 − 718 + 377 = −324 " + KJ + ".
Sign of ΔH_sol is small and can be either sign, because it is the difference between two large numbers.

Stage 1 · Learn

Check what the sim just showed you

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Which equation represents the enthalpy change of hydration of the chloride ion?

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Why are enthalpy changes of hydration always exothermic?

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Which cation has the most exothermic enthalpy change of hydration?

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What is ΔH_sol(NaCl) if LE = −787, ΔH_hyd(Na⁺) = −406 and ΔH_hyd(Cl⁻) = −377 kJ mol⁻¹?

Section 02

Charge density and solubility.

Switch the solution cycle in the simulator between the Group 2 sulfates and hydroxides, then explain the sulfate trend with the two competing energy terms.

Ion	Charge	Relative size	ΔH_hyd / kJ mol⁻¹
Li⁺	1+	smallest of the three 1+ ions	−519
Na⁺	1+	larger	−406
K⁺	1+	largest	−324
Mg²⁺	2+	small	−1920

Charge density: a small, highly charged ion attracts the δ− oxygen of water molecules (or the δ+ hydrogen, for anions) more strongly, so more energy is released.
Group 2 sulfates: the sulfate ion is large, so the lattice energy changes only a little down the group, but the cation hydration enthalpy becomes much less exothermic. ΔH_sol becomes less exothermic, and solubility falls from MgSO₄ to BaSO₄.

Examiner trap

Hydration starts from gaseous ions, not the solid. In the cycle, the solid goes to gaseous ions by the reverse of lattice energy, so ΔH_sol = ΣΔH_hyd − LE; adding LE instead gives a hugely negative answer. Multiply each hydration enthalpy by the number of ions in the formula.

Stage 2 · Exam

Exam-style questions

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LE(MgCl₂) = −2521, ΔH_hyd(Mg²⁺) = −1920, ΔH_hyd(Cl⁻) = −377 kJ mol⁻¹. What is ΔH_sol(MgCl₂)?

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For KCl, ΔH_sol = +17 and LE = −718 kJ mol⁻¹. ΔH_hyd(Cl⁻) = −377 kJ mol⁻¹. What is ΔH_hyd(K⁺)?

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Why does the solubility of the Group 2 sulfates decrease down the group?

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Why is ΔH_hyd(Li⁺) more exothermic than ΔH_hyd(K⁺)?

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Six original Cambridge-style questions

Defining enthalpy changes of solution and hydration, using the cycle LE + ΔH(sol) = sum of ΔH(hyd), and explaining trends in hydration enthalpy and Group 2 sulfate solubility.