Topic 12 · page 14 of 17

Enthalpy of solution and hydration

Dissolving an ionic solid is lattice energy reversed, followed by hydration of the gaseous ions. The enthalpy of solution is the small difference between two large numbers.

14 min read12.1612.17

Read first: Covalent character in ionic bonds

Two definitions 12.16

When an ionic solid dissolves in water, the lattice breaks up and the separated ions are surrounded by water molecules. Two enthalpy changes describe this.

NaCl(s)+aq→NaX+(aq)+ClX−(aq) \ce{NaCl(s) + aq -> Na+(aq) + Cl-(aq)}

ΔsolH\Delta_\text{sol}H can be positive or negative. "aq" means an excess of water.

NaX+(g)+aq→NaX+(aq)ΔhydH=−406 kJ mol−1 \ce{Na+(g) + aq -> Na+(aq)} \qquad \Delta_\text{hyd}H = -406\ \text{kJ mol}^{-1}

Hydration is always exothermic. Bonds form between the ion and the water molecules and nothing is broken. Water is a polar molecule, with a δ− oxygen and δ+ hydrogens, so it is attracted to ions in a particular way: the oxygen points towards a cation and a hydrogen points towards an anion.

Two panels. On the left, a sodium ion with four water molecules around it, each with its oxygen pointing in towards the ion. On the right, a chloride ion with four water molecules around it, each with one hydrogen atom pointing in towards the ion.
Ion-dipole attractions between an ion and the surrounding water molecules. Making these bonds is what releases the hydration enthalpy.

"Infinitely dilute" means that adding more water produces no further enthalpy change.

The solution cycle 12.17

Dissolving can be split into two steps that you can follow with a cycle, as with Born-Haber.

  1. Break up the lattice into gaseous ions. This is the reverse of the lattice energy, so it is endothermic: its value is −(lattice energy).
  2. Hydrate each gaseous ion. This is exothermic.

The direct route (dissolving the solid) and this two-step route start and end at the same places, so by Hess's law:

ΔsolH=−(lattice energy)+Σ ΔhydH \Delta_\text{sol}H = -(\text{lattice energy}) + \Sigma\,\Delta_\text{hyd}H

Worked exampleSodium chloride

Lattice energy = −787; ΔhydH\Delta_\text{hyd}H(Na⁺) = −406; ΔhydH\Delta_\text{hyd}H(Cl⁻) = −364 (kJ mol⁻¹).

  1. Break the lattice: +787
  2. Hydrate the ions: −406 + (−364) = −770
  3. ΔsolH\Delta_\text{sol}H = +787 − 770 = +17 kJ mol⁻¹
An energy level diagram for dissolving NaCl, drawn to scale. NaCl solid is at zero. An arrow up of plus 787 reaches gaseous Na plus and Cl minus ions. An arrow down of minus 406 reaches aqueous Na plus with gaseous Cl minus, and a second arrow down of minus 364 reaches the aqueous ions, which are 17 above the solid. A small bracket marks delta H sol, plus 17.
Solution cycle for NaCl, drawn to scale. Two large steps, +787 up and −770 down, leave a small difference: ΔsolH = +17 kJ mol⁻¹.

The measured value for NaCl is about +4 kJ mol⁻¹. The calculation gives +17, because a small difference between two large numbers is very sensitive to the data used. The point of the diagram is the method, and the sign: slightly endothermic.

Do the same for a salt with a 2+ ion. Here the cycle needs two chloride ions.

Worked exampleMagnesium chloride

Lattice energy = −2524; ΔhydH\Delta_\text{hyd}H(Mg²⁺) = −1920; ΔhydH\Delta_\text{hyd}H(Cl⁻) = −364 (kJ mol⁻¹).

  1. Break the lattice: +2524
  2. Hydrate the ions: −1920 + 2 × (−364) = −1920 − 728 = −2648
  3. ΔsolH\Delta_\text{sol}H = +2524 − 2648 = −124 kJ mol⁻¹
An energy level diagram for dissolving MgCl2, drawn to scale. MgCl2 solid is at zero. An arrow up of plus 2524 reaches gaseous Mg 2 plus and two Cl minus ions. An arrow down of minus 1920 reaches aqueous Mg 2 plus with gaseous Cl minus, and a second arrow of minus 728 reaches the aqueous ions, which are 124 below the solid. A small bracket marks delta H sol, minus 124.
Solution cycle for MgCl₂, drawn to scale. The hydration steps outweigh the lattice by 124 kJ mol⁻¹, so dissolving is exothermic.

The measured value is about −155 kJ mol⁻¹. The calculation reproduces the sign and the size to within a few tens of kJ.

MethodSolving a solution-enthalpy problem

  1. Draw the cycle: solid at the bottom left, gaseous ions above it, aqueous ions to the right.
  2. The arrow from solid to gaseous ions is −(lattice energy), drawn up.
  3. Hydration arrows come down from the gaseous ions. Use one for each ion and multiply by the number of ions in the formula.
  4. ΔsolH\Delta_\text{sol}H goes directly from the solid to the aqueous ions. Add along the route through the gaseous ions.
  5. Check that the units are kJ mol⁻¹ and that the sign makes sense.

Check yourself

All values in kJ mol⁻¹.

  1. Write an equation, with state symbols, for the enthalpy change of solution of sodium bromide, and another for the enthalpy change of hydration of the bromide ion.
  2. Explain why enthalpy changes of hydration are always exothermic.
  3. Calculate ΔsolH\Delta_\text{sol}H of KCl: lattice energy −718; ΔhydH\Delta_\text{hyd}H(K⁺) = −322; ΔhydH\Delta_\text{hyd}H(Cl⁻) = −364.
  4. For NaBr, ΔsolH\Delta_\text{sol}H = +6 (illustrative), lattice energy = −747 and ΔhydH\Delta_\text{hyd}H(Na⁺) = −406. Calculate ΔhydH\Delta_\text{hyd}H of the bromide ion.
  5. Describe, with a diagram, how water molecules are arranged round a sodium ion and round a chloride ion.
  6. Why is the calculated ΔsolH\Delta_\text{sol}H sensitive to small changes in the data?

Answers cover these until you have tried the questions

  1. NaBr(s)+aq→NaX+(aq)+BrX−(aq)\ce{NaBr(s) + aq -> Na+(aq) + Br-(aq)} and BrX−(g)+aq→BrX−(aq)\ce{Br-(g) + aq -> Br-(aq)}.
  2. Hydration forms attractions (ion-dipole bonds) between the ions and water molecules, and no bonds are broken, so energy is released.
  3. ΔsolH\Delta_\text{sol}H = +718 + (−322) + (−364) = 718 − 686 = +32 kJ mol⁻¹.
  4. ΔsolH\Delta_\text{sol}H = −(lattice energy) + ΔhydH\Delta_\text{hyd}H(Na⁺) + ΔhydH\Delta_\text{hyd}H(Br⁻), so +6 = +747 − 406 + ΔhydH\Delta_\text{hyd}H(Br⁻), giving ΔhydH\Delta_\text{hyd}H(Br⁻) = 6 − 747 + 406 = −335 kJ mol⁻¹.
  5. Water has a δ− oxygen and δ+ hydrogens. Round a sodium ion the water molecules point their oxygen atoms towards the positive ion. Round a chloride ion they point a hydrogen atom towards the negative ion. A diagram shows several water molecules around each ion with these orientations (see the figure above).
  6. It is the difference between two very large numbers (for example +787 and −770), so a small error or rounding in either number is a large fraction of the answer.