Enthalpy of Solution Revision Notes | A Level AQA Chemistry

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$^{-1}$

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.
2. $\Delta_{sol}H^\circ = (+787) + (-406) + (-363)$ 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.
3. $(-406) + (-363) = -769$ kJ mol$^{-1}$ — Adding two negative numbers: $406 + 363 = 769$, so the sum is $-769$ kJ mol$^{-1}$.
4. $\Delta_{sol}H^\circ = +787 + (-769) = +787 - 769 = +18$ 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.
5. $\Delta_{sol}H^\circ(\text{NaCl}) = +18$ kJ mol$^{-1}$ — The positive sign indicates an endothermic enthalpy of solution. Always state the sign explicitly — omitting it costs marks on AQA papers.

Answer: $\Delta_{sol}H^\circ(\text{NaCl}) = +18$ kJ mol$^{-1}$ (endothermic)

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$^{-1}$

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.
2. $2 \times (-363) = -726$ kJ mol$^{-1}$ — $2 \times 363 = 726$, so $2 \times (-363) = -726$ kJ mol$^{-1}$.
3. $-155 = +2526 + \Delta_{hyd}H^\circ(\text{Mg}^{2+}) + (-726)$ — Replace each symbol with its numerical value, keeping signs. Leave $\Delta_{hyd}H^\circ(\text{Mg}^{2+})$ as the unknown.
4. $-155 = 2526 - 726 + \Delta_{hyd}H^\circ(\text{Mg}^{2+}) = 1800 + \Delta_{hyd}H^\circ(\text{Mg}^{2+})$ — $2526 + (-726) = 2526 - 726 = 1800$ kJ mol$^{-1}$.
5. $\Delta_{hyd}H^\circ(\text{Mg}^{2+}) = -155 - 1800 = -1955$ 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.
6. $\Delta_{hyd}H^\circ(\text{Mg}^{2+}) = -1955$ 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.

Answer: $\Delta_{hyd}H^\circ(\text{Mg}^{2+}) = -1955$ kJ mol$^{-1}$

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.

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.