Enthalpy of Solution and Hydration — AQA A-level Chemistry

SPEC 3.1.8.1 - THERMODYNAMICS

Enthalpies of Solution & Hydration

Understanding what happens when ionic compounds dissolve in water

Introduction

In Lesson 8.01, we explored lattice enthalpy — the energy involved when gaseous ions come together to form an ionic lattice. But what happens when that lattice dissolves in water?

When an ionic compound dissolves, two key processes occur:

Process 1

Breaking the lattice

The ionic bonds holding the lattice together must be broken. This requires energy (endothermic).

Process 2

Hydrating the ions

Water molecules surround and bond to the ions. This releases energy (exothermic).

The enthalpy of solution is the overall energy change — the balance between these two competing processes.

Key Insight

Whether dissolving is exothermic or endothermic overall depends on whether more energy is released by hydration than is needed to break the lattice.

Enthalpy of Solution

Definition

The standard enthalpy of solution (ΔH_sol) is the enthalpy change when one mole of an ionic solid dissolves in sufficient water to form an infinitely dilute solution, under standard conditions.

"Infinitely dilute" means the solution is so dilute that adding more water causes no further enthalpy change — the ions are completely separated and only interact with water molecules, not with each other.

The symbol (aq) is used to show that the solid is dissolved in sufficient water.

Example Equations

For sodium chloride:NaCl(s) → Na⁺(aq) + Cl⁻(aq)For magnesium chloride:MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq)

SymbolMeaningSign

ΔH_solEnthalpy of solutionCan be +ve or −ve

(s)Solid ionic compound—

(aq)Aqueous (dissolved in water)—

Exothermic vs Endothermic

The enthalpy of solution can be exothermic (negative) or endothermic (positive), depending on the balance between lattice breaking and ion hydration.

Exothermic

ΔH_sol < 0

More energy is released by hydration than is needed to break the lattice.

Example: NaOH, LiCl

Solution feels warm

Endothermic

ΔH_sol > 0

More energy is needed to break the lattice than is released by hydration.

Example: NH₄NO₃, KNO₃

Solution feels cold

Important

Even if ΔH_sol is endothermic, dissolving can still occur spontaneously. This is because entropy increases when a solid dissolves (more disorder), which can make the overall process favourable. We'll explore this in Lesson 8.03 on Gibbs free energy.

Solubility and Enthalpy of Solution

For a substance to dissolve readily, the energy released from hydration must be similar to or greater than the energy required to break up the lattice.

Therefore, soluble substances usually have exothermic enthalpies of solution.

Substances with highly positive ΔH_sol values (e.g. AgCl, +77 kJ mol⁻¹) tend to be insoluble because the hydration energy cannot compensate for the lattice energy.

Enthalpy of Hydration

Definition

The standard enthalpy of hydration (ΔH_hyd) is the enthalpy change when one mole of gaseous ions dissolves in sufficient water to form an infinitely dilute solution, under standard conditions.

Example Equations

For sodium ions:Na⁺(g) → Na⁺(aq) ΔH_hyd = −406 kJ mol⁻¹For chloride ions:Cl⁻(g) → Cl⁻(aq) ΔH_hyd = −364 kJ mol⁻¹For magnesium ions:Mg²⁺(g) → Mg²⁺(aq) ΔH_hyd = −1891 kJ mol⁻¹

Key Point

Hydration enthalpies are always exothermic (negative values) because energy is released when the ions become surrounded by water molecules, forming ion-dipole interactions.

Alternative Wording

You may also see this definition phrased as: "The enthalpy change when 1 mole of aqueous ions is formed from 1 mole of gaseous ions."

Both phrasings mean the same thing — gaseous ions becoming hydrated (aqueous).

Understanding Hydration

Hydration enthalpies are the measure of the energy that is released when there is an attraction formed between the ions and water molecules.

Each ion has its own individual ΔH_hyd value — you hydrate the cation and anion separately, which must be taken into account during calculations.

Why is Hydration Exothermic?

Water is a polar molecule. The oxygen atom carries a partial negative charge (δ−) and the hydrogen atoms carry partial positive charges (δ+). This allows water to form ion-dipole interactions with both positive and negative ions.

Ion-Dipole Interactions

Cations (positive ions) attract the δ− oxygen atoms of water molecules.

Anions (negative ions) attract the δ+ hydrogen atoms of water molecules.

These attractions are called ion-dipole interactions. Forming them releases energy.

The Hydration Process

The polar water molecules will form ion-dipole bonds with the ions in solution, causing the ions to become hydrated. When an ionic solid dissolves in water, positive and negative ions are formed, and water molecules arrange themselves around each ion according to their charge.

Factors Affecting Hydration Enthalpy

The magnitude of hydration enthalpy depends on charge density — how concentrated the ionic charge is. Higher charge density = stronger ion-dipole interactions = more exothermic hydration enthalpy.

Factor 1: Ionic Charge

Higher charge → more exothermic ΔH_hyd

A 2+ ion attracts water molecules more strongly than a 1+ ion of similar size.

Example:

Mg²⁺: −1891 kJ mol⁻¹

Na⁺: −406 kJ mol⁻¹

Factor 2: Ionic Radius

Smaller radius → more exothermic ΔH_hyd

Smaller ions have higher charge density, so attract water more strongly.

Example:

F⁻: −506 kJ mol⁻¹

I⁻: −307 kJ mol⁻¹

Charge Density

Charge density = charge ÷ volume. Ions with high charge and small radius have the highest charge density and the most exothermic hydration enthalpies.

IonChargeRelative SizeΔH_hyd / kJ mol⁻¹

Li⁺+1Smallest Group 1−519

Na⁺+1Medium−406

K⁺+1Larger−322

Mg²⁺+2Small−1891

Ca²⁺+2Larger−1579

Trends in Hydration Enthalpy

Key Trends

Down a group: Hydration enthalpy becomes less exothermic (less negative) as ionic radius increases.

Across groups: Group 2 ions have much more exothermic hydration enthalpies than Group 1 ions due to their higher charge (2+ vs 1+).

The Energy Cycle

We can link lattice enthalpy, hydration enthalpies, and enthalpy of solution using an energy cycle based on Hess's Law.

Hess's Law Relationship

There are two routes from solid MX to aqueous ions:

Route 1 (direct): MX(s) → M⁺(aq) + X⁻(aq) [ΔH_sol]

Route 2 (indirect): MX(s) → M⁺(g) + X⁻(g) → M⁺(aq) + X⁻(aq) [ΔH_LE + ΣΔH_hyd]

By Hess's Law: ΔH_sol = ΔH_LE(dissociation) + ΣΔH_hyd

According to Hess's Law

The enthalpy change for both routes is the same. This means we can use this energy cycle to calculate unknown enthalpy values if we know the other values.

The total ΔH_hyd is found by adding the ΔH_hyd values of both anions and cations together.

Calculating Lattice Enthalpy Indirectly

These enthalpy cycles can be used as another way of measuring lattice enthalpies indirectly.

By rearranging the formula, we can find lattice enthalpy from solution and hydration data:ΔH_LE = ΣΔH_hyd − ΔH_sol

Alternative: Energy Level Diagram

You can also draw a Born-Haber style energy level diagram as an alternative approach. Both methods give the same answer — use whichever you prefer.

Examiner Tip

It doesn't matter whether you use Hess cycles or Born-Haber style cycles to solve these problems, as long as the information is correctly labelled and the direction of the arrows matches the definitions.

Exam problems in this topic often show diagrams with missing labels which you have to complete and find unknown values. The key to success is having a careful step-by-step approach.

Sign Conventions

Be careful with sign conventions! The key formula depends on which definition of lattice enthalpy you use:

DefinitionSignFormula for ΔH_sol

Lattice dissociation enthalpy MX(s) → M⁺(g) + X⁻(g)+ve (endothermic)ΔH_sol = ΔH_LE + ΣΔH_hyd

Lattice formation enthalpy M⁺(g) + X⁻(g) → MX(s)−ve (exothermic)ΔH_sol = −ΔH_LE + ΣΔH_hyd

AQA Convention

AQA typically uses lattice dissociation enthalpy (positive values) for these calculations. Always check which convention the question uses by looking at the sign of the lattice enthalpy given.

Remember

Hydration enthalpies are always negative (exothermic) — when you add them, you're adding negative numbers, which makes the total more negative.

For compounds like MgCl₂, don't forget to multiply the chloride hydration enthalpy by 2!

01Worked Example

Calculate the enthalpy of solution of sodium chloride given the following data:

Lattice dissociation enthalpy of NaCl = +787 kJ mol⁻¹

Enthalpy of hydration of Na⁺ = −406 kJ mol⁻¹

Enthalpy of hydration of Cl⁻ = −364 kJ mol⁻¹

1

Write the relationship: ΔH_sol = ΔH_LE + ΣΔH_hyd

2

Identify values: ΔH_LE = +787 kJ mol⁻¹

3

Sum hydration enthalpies: ΣΔH_hyd = (−406) + (−364) = −770 kJ mol⁻¹

4

Calculate: ΔH_sol = (+787) + (−770) = +17 kJ mol⁻¹

Answer: ΔH_sol = +17 kJ mol⁻¹

The enthalpy of solution is slightly endothermic — more energy is needed to break the lattice than is released by hydration.

02Worked Example

Calculate the enthalpy of solution of potassium bromide given:

Lattice dissociation enthalpy of KBr = +689 kJ mol⁻¹

Enthalpy of hydration of K⁺ = −322 kJ mol⁻¹

Enthalpy of hydration of Br⁻ = −351 kJ mol⁻¹

1

Formula: ΔH_sol = ΔH_LE + ΣΔH_hyd

2

Sum hydration enthalpies: ΣΔH_hyd = (−322) + (−351) = −673 kJ mol⁻¹

3

Calculate: ΔH_sol = (+689) + (−673) = +16 kJ mol⁻¹

Answer: ΔH_sol = +16 kJ mol⁻¹

Compounds with Multiple Ions

For compounds like MgCl₂ or CaCl₂, you must account for the stoichiometry — there are two chloride ions per formula unit.

If there is more than one cation or anion, then you must multiply by the appropriate coefficient for that ion. Since there are 2 moles of Cl⁻, the hydration enthalpy (ΔH_hyd) for chloride will be doubled.

For MgCl₂

ΔH_sol = ΔH_LE + ΔH_hyd(Mg²⁺) + 2×ΔH_hyd(Cl⁻)

Common Mistake

Forgetting to multiply the hydration enthalpy by the number of ions is one of the most common errors in these calculations. Always check the formula of the compound!

MgCl₂: Energy Level Diagram

Here is the same information shown as a Born-Haber style energy level diagram. Notice how the 2×ΔH_hyd(Cl⁻) appears because there are 2 moles of chloride ions.

Stoichiometry Reminder

Each ion has its own individual ΔH_hyd value. For MgCl₂, you hydrate one Mg²⁺ and two Cl⁻ ions separately. The total hydration enthalpy is:ΣΔH_hyd = ΔH_hyd(Mg²⁺) + 2×ΔH_hyd(Cl⁻)

03Worked Example

Calculate the enthalpy of solution of magnesium chloride given:

Lattice dissociation enthalpy of MgCl₂ = +2523 kJ mol⁻¹

Enthalpy of hydration of Mg²⁺ = −1891 kJ mol⁻¹

Enthalpy of hydration of Cl⁻ = −364 kJ mol⁻¹

1

Write formula: ΔH_sol = ΔH_LE + ΔH_hyd(Mg²⁺) + 2×ΔH_hyd(Cl⁻)

2

Note: We need 2× the chloride hydration enthalpy because there are 2 Cl⁻ ions

3

Substitute: ΔH_sol = (+2523) + (−1891) + 2×(−364)

4

Calculate: ΔH_sol = +2523 − 1891 − 728 = −96 kJ mol⁻¹

Answer: ΔH_sol = −96 kJ mol⁻¹

This is exothermic — the very large hydration enthalpy of Mg²⁺ (due to its high charge density) makes dissolving favourable.

Finding Unknown Values

Exam questions often give you three values and ask you to calculate the fourth. You need to rearrange the formula.

Rearrangements

Starting from: ΔH_sol = ΔH_LE + ΣΔH_hyd

To find lattice enthalpy:ΔH_LE = ΔH_sol − ΣΔH_hydTo find sum of hydration enthalpies:ΣΔH_hyd = ΔH_sol − ΔH_LE

Strategy

If asked to find the hydration enthalpy of one specific ion, first find ΣΔH_hyd, then subtract the known hydration enthalpy of the other ion(s).

04Worked Example

Calculate the enthalpy of hydration of the bromide ion given:

Enthalpy of solution of NaBr = +1 kJ mol⁻¹

Lattice dissociation enthalpy of NaBr = +742 kJ mol⁻¹

Enthalpy of hydration of Na⁺ = −406 kJ mol⁻¹

1

Start with: ΔH_sol = ΔH_LE + ΔH_hyd(Na⁺) + ΔH_hyd(Br⁻)

2

Rearrange: ΔH_hyd(Br⁻) = ΔH_sol − ΔH_LE − ΔH_hyd(Na⁺)

3

Substitute: ΔH_hyd(Br⁻) = (+1) − (+742) − (−406)

4

Calculate: ΔH_hyd(Br⁻) = +1 − 742 + 406 = −335 kJ mol⁻¹

Answer: ΔH_hyd(Br⁻) = −335 kJ mol⁻¹

05Worked Example

The lattice enthalpy of formation of CaCl₂ is −2254 kJ mol⁻¹. Calculate the enthalpy of solution given:

Enthalpy of hydration of Ca²⁺ = −1579 kJ mol⁻¹

Enthalpy of hydration of Cl⁻ = −364 kJ mol⁻¹

1

Note: This is lattice formation enthalpy (negative), so: ΔH_sol = −ΔH_LE(form) + ΣΔH_hyd

2

Sum hydration enthalpies: ΣΔH_hyd = (−1579) + 2×(−364) = −1579 − 728 = −2307 kJ mol⁻¹

3

Calculate: ΔH_sol = −(−2254) + (−2307) = +2254 − 2307 = −53 kJ mol⁻¹

Answer: ΔH_sol = −53 kJ mol⁻¹

Note: When given lattice formation enthalpy (negative), reverse its sign to get dissociation enthalpy.

Practice

Question 1:

Lattice dissociation enthalpy of LiCl = +853 kJ mol⁻¹

Enthalpy of hydration of Li⁺ = −519 kJ mol⁻¹

Enthalpy of hydration of Cl⁻ = −364 kJ mol⁻¹

1

ΔH_sol = ΔH_LE + ΔH_hyd(Li⁺) + ΔH_hyd(Cl⁻)

2

ΔH_sol = (+853) + (−519) + (−364)

3

ΔH_sol = 853 − 519 − 364 = −30 kJ mol⁻¹

Practice

Question 2:

Enthalpy of solution of KI = +21 kJ mol⁻¹

Lattice dissociation enthalpy of KI = +649 kJ mol⁻¹

Enthalpy of hydration of K⁺ = −322 kJ mol⁻¹

1

Rearrange: ΔH_hyd(I⁻) = ΔH_sol − ΔH_LE − ΔH_hyd(K⁺)

2

ΔH_hyd(I⁻) = (+21) − (+649) − (−322)

3

ΔH_hyd(I⁻) = 21 − 649 + 322 = −306 kJ mol⁻¹

Practice

Question 3:

Lattice dissociation enthalpy of BaCl₂ = +2056 kJ mol⁻¹

Enthalpy of hydration of Ba²⁺ = −1360 kJ mol⁻¹

Enthalpy of hydration of Cl⁻ = −364 kJ mol⁻¹

1

ΔH_sol = ΔH_LE + ΔH_hyd(Ba²⁺) + 2×ΔH_hyd(Cl⁻)

2

ΔH_sol = (+2056) + (−1360) + 2×(−364)

3

ΔH_sol = 2056 − 1360 − 728 = −32 kJ mol⁻¹

Common Mistakes

Mistake 1

Forgetting to multiply for multiple ions

For MgCl₂, you need 2× the chloride hydration enthalpy. For Al₂O₃, you'd need 2× for Al³⁺ and 3× for O²⁻.

Mistake 2

Confusing lattice formation and dissociation

Dissociation (breaking lattice) is endothermic (+ve). Formation (making lattice) is exothermic (−ve). They are opposite signs!

Mistake 3

Sign errors with negative numbers

When subtracting a negative: −(−406) = +406. When adding negatives: (−519) + (−364) = −883.

Mistake 4

Thinking hydration can be endothermic

Hydration enthalpies are always exothermic (negative) because forming ion-dipole bonds releases energy.

Summary

Key Definitions

TermDefinition

ΔH_solEnthalpy change when 1 mole of solid dissolves to form infinitely dilute solution

ΔH_hydEnthalpy change when 1 mole of gaseous ions forms infinitely dilute solution

Key Formula

ΔH_sol = ΔH_LE(dissociation) + ΣΔH_hyd

Key Exam Skills

Hydration enthalpies are always exothermic (negative)

Higher charge density → more exothermic hydration enthalpy

Multiply hydration enthalpies by the number of ions (2× for MgCl₂, etc.)

Check whether lattice enthalpy is formation or dissociation

Draw energy cycles to visualise the relationships

Show your workings and use brackets to separate mathematical operations from the enthalpy changes