Enthalpy of Solution Revision Notes | A Level AQA Chemistry
Enthalpy of solution = lattice enthalpy (dissociation) + the sum of the hydration enthalpies of the ions. Free AQA notes and 14 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Enthalpy of Solution is part of Thermodynamics (A-level only) in A Level AQA Chemistry.
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What this lesson covers
Enthalpy of solution = lattice enthalpy (dissociation) + the sum of the hydration enthalpies of the ions. Free AQA notes and 14 questions answered.
Describe the enthalpy of solution in terms of lattice enthalpy and enthalpy of hydration for ionic compounds
Construct Born-Haber-style energy cycles linking lattice enthalpy, enthalpies of hydration, and enthalpy of solution
Calculate the enthalpy of solution of an ionic compound using a Hess's Law cycle
Calculate an unknown lattice enthalpy or enthalpy of hydration from a given enthalpy of solution
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Use this lesson to stay inside Thermodynamics (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Enthalpy of solution: The enthalpy change when one mole of an ionic compound dissolves completely in sufficient water to form an infinitely dilute solution, under standard conditions. Symbol: $\Delta_{sol}H^\circ$. Can be exothermic or endothermic.
Lattice enthalpy (dissociation): The enthalpy change when one mole of an ionic compound is completely dissociated into its constituent gaseous ions under standard conditions. Always endothermic (positive). Symbol: $\Delta_{latt}H^\circ$.
Enthalpy of hydration: The enthalpy change when one mole of gaseous ions is completely surrounded by water molecules to form an aqueous solution under standard conditions. Always exothermic (negative). Symbol: $\Delta_{hyd}H^\circ$.
Revision notes
Read the core explanations from Enthalpy of Solution before testing yourself from memory.
1. Why Does Ionic Dissolving Have an Energy Cost?
Picture a crystal of sodium chloride sitting in water. The ions are locked in a rigid lattice, held together by strong electrostatic forces. For dissolving to happen at all, those forces must be overcome — and that costs energy.
But water molecules are not passive bystanders. They are polar, and they cluster around the freed ions, releasing energy as they do so. The overall enthalpy of solution is the net result of these two competing processes: energy in to break the lattice, energy out as ions become hydrated.
This is the central idea of the topic: $\Delta_{sol}H^\circ$ is not measured directly in a Born-Haber-style cycle — it is calculated from lattice enthalpy and enthalpies of hydration using Hess's Law. That calculation is what AQA expects you to perform fluently.
2. The Three Key Quantities
Two further quantities feed into the cycle. The first is lattice enthalpy. AQA uses the dissociation convention: the energy required to convert one mole of solid ionic compound into its gaseous ions. Because you are breaking electrostatic attractions, this is always endothermic.
Notice the pattern: charge density governs hydration enthalpy. A small, highly charged ion such as $\text{Mg}^{2+}$ has a far more exothermic hydration enthalpy than a large, singly charged ion such as $\text{Cs}^+$. This will matter when you evaluate why some compounds dissolve readily and others do not.
3. Constructing the Energy Cycle
The cycle connects three energy levels. At the top sits the ionic solid plus water. At the bottom sit the aqueous ions. There are two routes between these two levels — and by Hess's Law, the total enthalpy change is the same regardless of route.
Route 1 (direct): The solid dissolves directly in water. The enthalpy change for this route is $\Delta_{sol}H^\circ$.
Route 2 (indirect, via gaseous ions): First, the lattice is broken to give gaseous ions — this requires $+\Delta_{latt}H^\circ$ (dissociation). Then each type of gaseous ion is hydrated — this releases $\Delta_{hyd}H^\circ$ for the cation and $\Delta_{hyd}H^\circ$ for the anion. The sum of these two steps equals the enthalpy change for Route 2.
For a 1:1 salt such as NaCl, the sum of hydration enthalpies is $\Delta_{hyd}H^\circ(\text{Na}^+) + \Delta_{hyd}H^\circ(\text{Cl}^-)$. For a salt such as MgCl$_2$, you must include two moles of $\Delta_{hyd}H^\circ(\text{Cl}^-)$ because one formula unit produces two chloride ions.
4. Worked Calculation: Enthalpy of Solution of NaCl
Theory is one thing; exam questions demand you apply it with numbers. Here is a full worked example using sodium chloride, the most commonly examined 1:1 ionic compound. Work through each step before revealing the next — this mirrors how you should approach a calculation in an exam.
Worked example: Calculating the Enthalpy of Solution of NaCl
Use the following data to calculate the enthalpy of solution of sodium chloride, NaCl(s).
$\Delta_{latt}H^\circ$(NaCl, dissociation) = +787 kJ mol$^{-1}$
$\Delta_{hyd}H^\circ$(Na$^+$) = −406 kJ mol$^{-1}$
$\Delta_{hyd}H^\circ$(Cl$^-$) = −363 kJ mol$^{-1}$
$\Delta_{sol}H^\circ = \Delta_{latt}H^\circ(\text{dissociation}) + \Delta_{hyd}H^\circ(\text{Na}^+) + \Delta_{hyd}H^\circ(\text{Cl}^-)$ — NaCl is a 1:1 salt, so one mole of each ion is produced per mole of NaCl dissolved. The cycle has two steps in Route 2: lattice dissociation, then hydration of both ions.
$\Delta_{sol}H^\circ = (+787) + (-406) + (-363)$ kJ mol$^{-1}$ — Keep the signs as given. The lattice enthalpy is positive (endothermic); both hydration enthalpies are negative (exothermic). Carry units through every step.
$(-406) + (-363) = -769$ kJ mol$^{-1}$ — Adding two negative numbers: $406 + 363 = 769$, so the sum is $-769$ kJ mol$^{-1}$.
$\Delta_{sol}H^\circ = +787 + (-769) = +787 - 769 = +18$ kJ mol$^{-1}$ — The lattice enthalpy (+787) is larger in magnitude than the total hydration enthalpy (−769), so the overall process is slightly endothermic. This matches the known behaviour of NaCl, which dissolves endothermically but readily because entropy drives the process.
$\Delta_{sol}H^\circ(\text{NaCl}) = +18$ kJ mol$^{-1}$ — The positive sign indicates an endothermic enthalpy of solution. Always state the sign explicitly — omitting it costs marks on AQA papers.
5. Worked Calculation: Finding an Unknown Hydration Enthalpy
AQA frequently reverses the calculation: you are given the enthalpy of solution and asked to find a missing lattice enthalpy or hydration enthalpy. The algebra is straightforward — rearrange the same Hess's Law expression. Here is a worked example using magnesium chloride, which also tests your handling of stoichiometry.
Worked example: Finding the Enthalpy of Hydration of Mg$^{2+}$
Use the following data to calculate the enthalpy of hydration of the Mg$^{2+}$ ion.
$\Delta_{sol}H^\circ$(MgCl$_2$) = −155 kJ mol$^{-1}$
$\Delta_{latt}H^\circ$(MgCl$_2$, dissociation) = +2526 kJ mol$^{-1}$
$\Delta_{hyd}H^\circ$(Cl$^-$) = −363 kJ mol$^{-1}$
$\Delta_{sol}H^\circ = \Delta_{latt}H^\circ + \Delta_{hyd}H^\circ(\text{Mg}^{2+}) + 2\Delta_{hyd}H^\circ(\text{Cl}^-)$ — MgCl$_2$ produces one Mg$^{2+}$ and two Cl$^-$ per formula unit, so the Cl$^-$ hydration enthalpy is multiplied by 2. This stoichiometry step is a common source of errors.
$-155 = +2526 + \Delta_{hyd}H^\circ(\text{Mg}^{2+}) + (-726)$ — Replace each symbol with its numerical value, keeping signs. Leave $\Delta_{hyd}H^\circ(\text{Mg}^{2+})$ as the unknown.
$\Delta_{hyd}H^\circ(\text{Mg}^{2+}) = -155 - 1800 = -1955$ kJ mol$^{-1}$ — Subtract 1800 from both sides: $-155 - 1800 = -1955$. The large negative value is consistent with Mg$^{2+}$ being a small, doubly charged ion with very high charge density.
$\Delta_{hyd}H^\circ(\text{Mg}^{2+}) = -1955$ kJ mol$^{-1}$ — The strongly negative value confirms that Mg$^{2+}$ is very strongly hydrated — its high charge density attracts water molecules powerfully. Always check: hydration enthalpies must be negative. If you get a positive value, recheck your signs.
6. Interpreting the Sign and Magnitude of $\Delta_{sol}H^\circ$
Once you can calculate $\Delta_{sol}H^\circ$, the next level of demand is to explain its sign and compare values across compounds. This is where charge density reasoning becomes essential.
When the total hydration enthalpy is more negative than the lattice enthalpy is positive, the enthalpy of solution is negative — dissolving is exothermic. When the lattice enthalpy dominates, dissolving is endothermic. The balance between these two quantities determines the sign.
Compound
$\Delta_{latt}H^\circ$ / kJ mol$^{-1}$
$\sum \Delta_{hyd}H^\circ$ / kJ mol$^{-1}$
$\Delta_{sol}H^\circ$ / kJ mol$^{-1}$
Sign
NaCl
+787
−769
+18
Endothermic
NaOH
+900
−932
−32
Exothermic
KBr
+682
−658
+24
Endothermic
LiCl
+853
−912
−59
Exothermic
Notice that LiCl has a more exothermic enthalpy of solution than NaCl. The Li$^+$ ion is smaller than Na$^+$, giving it higher charge density and a more exothermic hydration enthalpy. Although LiCl also has a larger lattice enthalpy than NaCl, the hydration gain outweighs the extra lattice energy cost.
7. Exam Technique and Common Mistakes
AQA calculation questions on this topic carry marks for method as well as accuracy. Even if you make an arithmetic slip, you can still earn method marks by showing the correct cycle expression and substitution. Always write the Hess's Law relationship before substituting numbers.
Explain-style questions ask you to account for differences in enthalpy of solution between compounds. The mark-winning structure is: state the relevant quantity (lattice enthalpy or hydration enthalpy), link it to ion size or charge, and then connect to the effect on $\Delta_{sol}H^\circ$. Vague answers that say only "stronger forces" without specifying which forces and why they differ will not score full marks.
The enthalpy of solution cycle is a direct application of Hess's Law to ionic dissolving: lattice enthalpy quantifies the energy cost of separating the ions, hydration enthalpies quantify the energy recovered as water surrounds them, and the algebraic sum gives $\Delta_{sol}H^\circ$ — covering both AQA specification requirements for this topic.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
Enthalpy of solution = lattice enthalpy (dissociation) + the sum of the hydration enthalpies of the ions. For NaCl, (+787) + (−406) + (−363) = +18 kJ mol⁻¹, so dissolving is slightly endothermic. For MgCl₂, the chloride hydration enthalpy is multiplied by 2 because there are two chloride ions.
What is enthalpy of hydration?
Enthalpy of hydration is the enthalpy change when one mole of a specified gaseous ion is completely hydrated by water molecules under standard conditions. Ion–dipole attractions form, so it is always exothermic, and smaller, more highly charged ions such as Mg²⁺ have more exothermic values than ions such as Cs⁺.
What is enthalpy of solution?
Enthalpy of solution is the enthalpy change when one mole of an ionic compound dissolves completely in sufficient water to produce an infinitely dilute solution under standard conditions. It can be exothermic or endothermic depending on the relative magnitudes of the lattice enthalpy and the hydration enthalpies.
How do you draw an enthalpy of solution cycle?
An enthalpy of solution cycle has the ionic solid plus water at the top and the aqueous ions at the bottom. Route 1 is the direct dissolving, ΔsolH. Route 2 breaks the lattice into gaseous ions, then hydrates each ion; by Hess's Law both routes give the same enthalpy change.
Why is the chloride hydration enthalpy doubled for MgCl₂?
The chloride hydration enthalpy is doubled for MgCl₂ because one formula unit produces one Mg²⁺ ion and two Cl⁻ ions. The cycle is ΔsolH = ΔlattH + ΔhydH(Mg²⁺) + 2 × ΔhydH(Cl⁻), and missing this stoichiometry step is a common source of errors.
What is the enthalpy of solution of NaCl?
The enthalpy of solution of NaCl is +18 kJ mol⁻¹. The lattice enthalpy of dissociation is +787 kJ mol⁻¹, and the hydration enthalpies of Na⁺ (−406) and Cl⁻ (−363) sum to −769 kJ mol⁻¹, so +787 − 769 = +18 kJ mol⁻¹, slightly endothermic.
Why is the enthalpy of solution of LiCl more exothermic than NaCl?
LiCl has a more exothermic enthalpy of solution than NaCl, −59 compared with +18 kJ mol⁻¹, because the Li⁺ ion is smaller than Na⁺, giving it higher charge density and a more exothermic hydration enthalpy. This hydration gain outweighs the larger lattice enthalpy of LiCl.
What decides whether dissolving is exothermic or endothermic?
Dissolving is exothermic when the total hydration enthalpy is larger in magnitude than the lattice enthalpy, and endothermic when the lattice enthalpy dominates. NaOH, with a lattice enthalpy of +900 and total hydration of −932 kJ mol⁻¹, is exothermic overall, while KBr, with +682 and −658, is endothermic overall.
How do you calculate the enthalpy of hydration of Mg²⁺?
The enthalpy of hydration of Mg²⁺ is found by rearranging the enthalpy of solution cycle for MgCl₂. With ΔsolH = −155, lattice enthalpy +2526 and two Cl⁻ at −363 each (−726), −155 = 1800 + ΔhydH(Mg²⁺), so ΔhydH(Mg²⁺) = −1955 kJ mol⁻¹.
Why does NaCl dissolve if its enthalpy of solution is endothermic?
NaCl dissolves readily even though its enthalpy of solution is endothermic, +18 kJ mol⁻¹, because entropy drives the process. The lattice enthalpy is only slightly larger in magnitude than the total hydration enthalpy, so the energy cost is small.
Why is enthalpy of hydration always negative?
Enthalpy of hydration is always negative because ion–dipole attractions form between the gaseous ion and polar water molecules, releasing energy. If a calculated hydration enthalpy comes out positive, the signs in the working should be rechecked.
What is the enthalpy of solution of NaOH?
The enthalpy of solution of NaOH is −32 kJ mol⁻¹, so dissolving it is exothermic. Its lattice enthalpy is +900 kJ mol⁻¹ and its total hydration enthalpy is −932 kJ mol⁻¹; the hydration term is larger in magnitude, so the overall value is negative.
Which lattice enthalpy convention is used in enthalpy of solution cycles?
Enthalpy of solution cycles use the dissociation convention for lattice enthalpy: the energy required to convert one mole of solid ionic compound into its gaseous ions. Because electrostatic attractions are broken, this value is always endothermic, or positive.
How does charge density affect hydration enthalpy?
Higher charge density gives a more exothermic hydration enthalpy. A small, highly charged ion such as Mg²⁺ has a far more exothermic hydration enthalpy than a large, singly charged ion such as Cs⁺, because it attracts water molecules more powerfully.