Calculate solution enthalpy from an ionic cycle
Combine the reversed lattice-formation term with hydration enthalpies to calculate the energy change for dissolving an ionic solid.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 23. Use the syllabus sign convention for lattice enthalpy (gaseous ions forming the ionic solid); state phases, signs, units and temperature. Written practice is self-checked
Before you begin
- Use AS enthalpy changes and Hess's law, formulae for ions, balanced equations, and temperature in kelvins.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the separation contribution if ΔHₗₐₜ = −700 kJ mol⁻¹ for gas ions → solid?
What you will learn
- Calculate solution enthalpy from lattice and hydration values
- Apply formula-unit coefficients to ion hydration
- Check the sign against the net dissolution equation
Use the lattice-formation sign convention explicitly
For this course's definition, ΔHₗₐₜ is gaseous ions → solid. Dissolution therefore uses the reverse separation step: ΔHsol = −ΔHₗₐₜ + ΣΔHhyd.
For a formula unit containing two identical anions, include two hydration enthalpies. Keep all terms in the same units and per mole of ionic solid dissolved.
Interpret the result only after checking the sum
If ΔHsol is positive, the dissolution is endothermic; if negative, it is exothermic. A positive result does not alone show that dissolution cannot occur, because entropy also affects feasibility through ΔG.
Substitute your answer back into the Hess cycle and check that it equals the sum of the physical steps: lattice separation plus hydration of every ion.
Pause and explain
If ΔHsol is positive, what does that say about the dissolution enthalpy?
Worked example
For a hypothetical 1:1 salt MX, ΔHₗₐₜ = −780, ΔHhyd(M⁺) = −390 and ΔHhyd(X⁻) = −360 kJ mol⁻¹. Calculate ΔHsol.
Show the worked solution
- Reverse lattice formation: −(−780) = +780.
- Add both hydration terms: −390 − 360 = −750.
- Sum the separation and hydration steps.
Answer ΔHsol = 780 − 390 − 360 = +30 kJ mol⁻¹; dissolving is endothermic for these illustrative values.
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Test this topic →From guided practice to a new situation
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.
For MX₂, ΔHₗₐₜ = −2100, ΔHhyd(M²⁺) = −1900 and ΔHhyd(X⁻) = −350 kJ mol⁻¹. Calculate ΔHsol.
Give me a hint
Reverse the lattice term and include two X⁻ hydration steps.
Compare my reasoning
- Separation term = +2100.
- Hydration total = −1900 + 2(−350) = −2600.
- Add the terms.
ΔHsol = 2100 − 1900 − 700 = −500 kJ mol⁻¹.
Look for these in your work
- Lattice reversed
- Two anion terms included
- Signed total correct
A 1:1 salt has ΔHₗₐₜ = −650, ΔHhyd(M⁺) = −300 and ΔHhyd(X⁻) = −330 kJ mol⁻¹. Find ΔHsol and classify it.
Give me a hint
Use one hydration term for each ion and reverse lattice formation.
Compare my reasoning
- Reverse lattice: +650.
- Add hydration: −300 − 330 = −630.
- Combine the signed values.
ΔHsol = +650 − 300 − 330 = +20 kJ mol⁻¹; dissolution is endothermic.
Look for these in your work
- Correct lattice direction
- Both ions counted
- Sign and classification correct
Two salts have the same ΔHsol, but salt A has a much larger positive lattice-separation term. What must be true about its hydration sum compared with salt B?
Give me a hint
Use the cycle equation and compare the terms that balance the same total.
Compare my reasoning
- ΔHsol equals separation plus hydration.
- A's larger separation would make its total larger if hydration were unchanged.
- A's hydration sum must be more negative to reach the same ΔHsol.
Salt A must have a more negative total hydration enthalpy to offset its larger lattice-separation contribution and give the same ΔHsol.
Look for these in your work
- Equation applied
- Unchanged-term comparison made
- Hydration comparison stated
Common mistakes
- Adding the negative lattice-formation value instead of reversing it
- Omitting an ion or its formula coefficient
- Mixing kJ mol⁻¹ and J mol⁻¹
- Treating positive ΔHsol as proof that dissolution is impossible
What can you explain now?
What is the separation contribution if ΔHₗₐₜ = −700 kJ mol⁻¹ for gas ions → solid?
Compare with the explanation
+700 kJ mol⁻¹
The solid must be separated, so reverse the negative formation step.
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Make it stick
How do you calculate dissolution enthalpy when the supplied lattice value describes formation of the solid?
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.