Build a lattice–hydration cycle for solution enthalpy
Separate lattice dissociation, ion hydration and net solution enthalpy before explaining Group 2 solubility patterns.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 27. Compare Group 2 ions at the same charge and distinguish lattice formation, lattice dissociation, hydration and solution enthalpy. Trend explanations are qualitative unless data are supplied. Written practice is self-checked
Before you begin
- Use AS Group 2 reactivity trends, ionic charge and radius, Hess cycles, and balanced equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens in lattice dissociation?
What you will learn
- Distinguish lattice dissociation enthalpy from lattice formation enthalpy
- Relate hydration enthalpies to gaseous and aqueous ions
- Use a Hess cycle to express solution enthalpy
Dissolution has an energy cost and an energy release
For an ionic solid dissolving, one step separates the crystal into gaseous ions; this lattice-dissociation step is endothermic. Hydration then transfers each gaseous ion into water, releasing energy and giving negative hydration enthalpies.
Write the stoichiometric number of every ion. For M(OH)₂, hydration includes one M²⁺ and two OH⁻ ions; for MSO₄, it includes one M²⁺ and one SO₄²⁻ ion.
Add signed steps to obtain ΔHsol
Using lattice dissociation enthalpy, ΔHsol = ΔH_latt(diss) + ΣΔH_hyd. If a source instead gives lattice formation enthalpy for gaseous ions forming the solid, reverse that step: ΔHsol = −ΔH_latt(form) + ΣΔH_hyd.
Hydration becomes less exothermic for a larger ion of the same charge because charge density falls. Lattice dissociation also becomes less endothermic as the cation radius increases, since the solid's electrostatic attractions are weaker. The solubility and ΔHsol trends depend on the balance of these changing contributions.
Pause and explain
When ΔH_latt is defined as formation from gaseous ions, what sign is used in the dissolution cycle?
Worked example
Write the Hess relation for dissolving Ba(OH)₂(s) using lattice-dissociation and hydration enthalpies.
Show the worked solution
- Break the solid into one Ba²⁺ and two OH⁻ gaseous ions.
- Hydrate each ion in the formula ratio.
- Add the signed lattice and hydration contributions.
Answer Ba(OH)₂(s) → Ba²⁺(aq) + 2OH⁻(aq); ΔHsol = ΔH_latt(diss) + ΔH_hyd(Ba²⁺) + 2ΔH_hyd(OH⁻).
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the enthalpy-cycle relation for MgSO₄(s) dissolving, using lattice-dissociation enthalpy.
Give me a hint
Count one Mg²⁺ and one SO₄²⁻ in each formula unit.
Compare my reasoning
- Separate MgSO₄(s) into gaseous ions.
- Hydrate Mg²⁺ once.
- Hydrate SO₄²⁻ once and sum signed terms.
ΔHsol = ΔH_latt(diss) + ΔH_hyd(Mg²⁺) + ΔH_hyd(SO₄²⁻).
Look for these in your work
- Correct ion pair
- Stoichiometry retained
- Signed contributions combined
A table gives ΔH_latt(form) = −2200 kJ mol⁻¹ and total hydration enthalpy = −2050 kJ mol⁻¹ for a hypothetical salt. Calculate ΔHsol.
Give me a hint
Reverse the formation enthalpy, then add the hydration contribution.
Compare my reasoning
- Dissociation contribution is +2200 kJ mol⁻¹.
- Add −2050 kJ mol⁻¹.
- Interpret the sign of the sum.
ΔHsol = −(−2200) − 2050 = +150 kJ mol⁻¹; the illustrative dissolution is endothermic.
Look for these in your work
- Lattice direction reversed
- Hydration sign retained
- Sum and sign interpreted
Why can ΔHsol not, by itself, establish whether a salt is soluble?
Give me a hint
Consider that solubility is an equilibrium property and dissolution also changes particle dispersal.
Compare my reasoning
- ΔHsol describes the enthalpy change, not the full free-energy change.
- Entropy and temperature also affect feasibility.
- Solubility should be supported by equilibrium evidence or the stated syllabus trend.
ΔHsol is only the enthalpy contribution. Entropy and temperature also affect Gibbs energy and the dissolution equilibrium, so the enthalpy sign alone does not establish solubility.
Look for these in your work
- Enthalpy scope stated
- Other thermodynamic contribution recognised
- Equilibrium limitation explained
Common mistakes
- Adding formation lattice enthalpy without reversing its sign
- Omitting the coefficient 2 for hydroxide ions
- Calling hydration endothermic for the stated gas-to-aqueous direction
- Using ΔHsol alone as a complete proof of solubility
What can you explain now?
What happens in lattice dissociation?
Compare with the explanation
An ionic solid separates into gaseous ions
This step overcomes the crystal's electrostatic attractions.
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Make it stick
How do lattice dissociation and hydration combine, with signs, to give solution enthalpy?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.