When a solid ionic compound dissolves in water, the giant ionic lattice breaks apart and the separated ions interact with water molecules. To understand why some substances dissolve easily while others do not, we must examine the enthalpy changes that occur during this process.
Key Principle
The dissolving of an ionic lattice is a two-step process: first, the lattice is broken down into gaseous ions (lattice dissociation, which is highly endothermic), and second, the gaseous ions are surrounded by water molecules (hydration, which is highly exothermic).
Important Enthalpy Definitions
The enthalpy change when 1 mole of an ionic solid is completely dissolved in water under standard conditions to form an infinitely dilute solution. It can be endothermic or exothermic:
\( \text{NaCl}(s) \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(aq) + \text{Cl}^-(aq) \quad \Delta H_{\text{sol}}^\theta = +4\text{ kJ mol}^{-1} \)
The enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions under standard conditions. This process is always exothermic because attractions are formed between the ions and polar water molecules:
\( \text{Na}^+(g) \xrightarrow{\text{aq}} \text{Na}^+(aq) \quad \Delta H_{\text{hyd}}^\theta = -406\text{ kJ mol}^{-1} \)
\( \text{Cl}^-(g) \xrightarrow{\text{aq}} \text{Cl}^-(aq) \quad \Delta H_{\text{hyd}}^\theta = -378\text{ kJ mol}^{-1} \)
Ensure you use correct state symbols in these equations. For hydration, the reactants are gaseous ions, and the products are aqueous. For solution, the reactant is the solid lattice, and the products are aqueous ions. Writing the incorrect state symbols can cost you full marks in written questions.
The Dissolving Enthalpy Cycle
Hess's law allows us to construct an enthalpy cycle to calculate the enthalpy of solution. We connect the solid ionic lattice to the aqueous ions by passing through a gaseous ion intermediate state:
From the cycle above, we can write the equation:
Or if using the lattice enthalpy of formation (which is the negative of the lattice dissociation enthalpy):
ΔHL-form⦵(NaCl) = −788 kJ mol⁻¹
ΔHhyd⦵(Na⁺) = −406 kJ mol⁻¹
ΔHhyd⦵(Cl⁻) = −378 kJ mol⁻¹
Step 1: Set up the cycle equation. We know that:
ΔHsol⦵ = −ΔHL-form⦵+ΔHhyd⦵(Na⁺) + ΔHhyd⦵(Cl⁻)
Note that we change the sign of the lattice formation enthalpy to get the lattice dissociation enthalpy (+788 kJ mol⁻¹).
Step 2: Substitute the values:
ΔHsol⦵ = −(−788) + (−406) + (−378)
ΔHsol⦵ = 788 − 784 = +4 kJ mol⁻¹
The standard enthalpy of solution for sodium chloride is +4 kJ mol⁻¹.
ΔHsol⦵(MgCl₂) = −155 kJ mol⁻¹
ΔHL-diss⦵(MgCl₂) = +2526 kJ mol⁻¹
ΔHhyd⦵(Cl⁻) = −378 kJ mol⁻¹
Step 1: Express the cycle equation for MgCl₂. Note that there are 2 moles of chloride ions per mole of magnesium chloride:
ΔHsol⦵ = ΔHL-diss⦵+ΔHhyd⦵(Mg²⁺) + 2ΔHhyd⦵(Cl⁻)
Step 2: Rearrange the equation to solve for ΔHhyd⦵(Mg²⁺):
ΔHhyd⦵(Mg²⁺) = ΔHsol⦵−ΔHL-diss⦵−2ΔHhyd⦵(Cl⁻)
Step 3: Substitute the known values:
ΔHhyd⦵(Mg²⁺) = −155 − 2526 − 2(−378)
ΔHhyd⦵(Mg²⁺) = −155 − 2526 + 756
ΔHhyd⦵(Mg²⁺) = −1925 kJ mol⁻¹
The standard enthalpy of hydration of magnesium ions is −1925 kJ mol⁻¹.
Factors Affecting Lattice and Hydration Enthalpies
The values of both lattice enthalpy and hydration enthalpy depend on the strength of electrostatic attractions. Two key factors determine this strength: ionic radius and ionic charge.
1. Ionic Radius (Size)
As the ionic radius increases:
- The distance between the centres of opposite charges increases.
- The electrostatic attractions become weaker.
- Lattice Enthalpy: Becomes less exothermic (less energy is released when forming the lattice).
- Hydration Enthalpy: Becomes less exothermic (water molecules are less strongly attracted to the larger ion, which has a lower charge density).
2. Ionic Charge
As the charge on the ions increases:
- The electrostatic attractions between oppositely charged ions in the lattice, and between ions and polar water molecules, become much stronger.
- Lattice Enthalpy: Becomes significantly more exothermic.
- Hydration Enthalpy: Becomes significantly more exothermic due to the increased attraction to the water dipoles.
Comparing Sodium and Magnesium Ions
The magnesium ion (Mg²⁺) has a higher charge (+2 vs +1) and a smaller ionic radius (0.072 nm vs 0.095 nm) than the sodium ion (Na⁺). Therefore, Mg²⁺ has a much higher charge density.
This explains why the hydration enthalpy of Mg²⁺ (−1925 kJ mol⁻¹) is far more exothermic than that of Na⁺ (−406 kJ mol⁻¹).
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