A-Level Chemistry Study Notes: Enthalpies of Solution and Hydration (9701, Topic 23.2)

Hello future Chemists! This chapter takes us deeper into energetics, specifically looking at what happens, energetically speaking, when an ionic solid dissolves in water. This concept links directly to Lattice Energy from the previous section and is crucial for understanding solubility and the feasibility of processes.

Don't worry if all the Greek letters ($\Delta H$) seem intimidating—we are simply applying Hess’s Law (which you already know!) to dissolution processes. Let's dive in!

1. Defining Key Enthalpy Terms

To understand how a salt dissolves, we need to look at two fundamental energetic steps: breaking the solid apart, and cuddling the resulting ions with water molecules.

1.1 Enthalpy Change of Solution ($\Delta H_{\text{sol}}$)

The Enthalpy Change of Solution ($\Delta H_{\text{sol}}$) is the enthalpy change when one mole of an ionic solid dissolves completely in a large enough amount of solvent (usually water) to form an infinitely dilute solution.

This process occurs under standard conditions (298 K and 101 kPa).

  • Equation Example (for NaCl):
    $$ \text{NaCl}(s) + \text{aq} \xrightarrow{\Delta H_{\text{sol}}} \text{Na}^+(aq) + \text{Cl}^-(aq) $$
Did you know? (Endothermic vs Exothermic Dissolution)

You experience $\Delta H_{\text{sol}}$ every day!

  • If $\Delta H_{\text{sol}}$ is negative (exothermic), the solution gets warmer. (Example: Dissolving anhydrous Calcium Chloride used in self-heating packs.)
  • If $\Delta H_{\text{sol}}$ is positive (endothermic), the solution gets colder. (Example: Dissolving Ammonium Nitrate used in instant cold packs for sports injuries.)

1.2 Enthalpy Change of Hydration ($\Delta H_{\text{hyd}}$)

The Enthalpy Change of Hydration ($\Delta H_{\text{hyd}}$) is the enthalpy change when one mole of gaseous ions dissolves completely in a large amount of water to form an infinitely dilute aqueous solution.

This process is always exothermic (releases energy) because it involves the attraction between the polar water molecules and the charged ions, which forms bonds (electrostatic attractions).

  • Equation Example (for separate ions):
    $$ \text{Na}^+(g) + \text{aq} \xrightarrow{\Delta H_{\text{hyd}}(\text{Na}^+)} \text{Na}^+(aq) $$ $$ \text{Cl}^-(g) + \text{aq} \xrightarrow{\Delta H_{\text{hyd}}(\text{Cl}^-)} \text{Cl}^-(aq) $$
Analogy: The Hydration Hug

Think of the gaseous ion as a lonely, very hot tourist who just arrived at the airport. Water molecules are like helpful friends rushing in to give the tourist a cooling hug. This "hug" (the electrostatic attraction) releases energy, making the process exothermic. The stronger the attraction, the bigger the negative value for $\Delta H_{\text{hyd}}$.

Key Takeaway: $\Delta H_{\text{sol}}$ is the overall heat change when a solid dissolves. $\Delta H_{\text{hyd}}$ is the heat released when separated gaseous ions are surrounded by water.

2. The Energy Cycle: Linking Solution, Hydration, and Lattice Energy

Dissolving an ionic solid ($\Delta H_{\text{sol}}$) can be imagined as two separate steps, which allows us to construct an energy cycle based on Hess's Law.

Hess's Law states that the total enthalpy change for a reaction is independent of the path taken.

2.1 The Two Hypothetical Steps of Dissolution

The cycle considers two main paths between the starting point (solid crystal) and the end point (aqueous ions):

  1. Path 1 (Direct Dissolution): Solid ionic compound to aqueous ions ($\Delta H_{\text{sol}}$).
  2. Path 2 (Hypothetical Route):

    Step A: Lattice Dissociation

    We must break the bonds holding the solid lattice together to form separate gaseous ions. This step is highly endothermic (requires energy input). This enthalpy change is the negative of the Lattice Energy ($\Delta H_{\text{latt}}$), as defined in the syllabus (gas ions $\to$ solid lattice is $\Delta H_{\text{latt}}$, which is negative).

    $$ \text{MX}(s) \xrightarrow{\text{Lattice Dissociation Energy}} \text{M}^+(g) + \text{X}^-(g) $$

    Step B: Hydration

    The separated gaseous ions are surrounded by water molecules. This step is always exothermic ($\Delta H_{\text{hyd}}$).

    $$ \text{M}^+(g) + \text{X}^-(g) + \text{aq} \xrightarrow{\Delta H_{\text{hyd}}} \text{M}^+(aq) + \text{X}^-(aq) $$

2.2 Constructing the Solution Cycle

We can use a simple Hess’s Law cycle to relate these terms:

$$ \text{MX}(s) \xrightarrow{\Delta H_{\text{sol}}} \text{M}^+(aq) + \text{X}^-(aq) $$ $$ \text{MX}(s) \xleftarrow{\Delta H_{\text{latt}}} \text{M}^+(g) + \text{X}^-(g) $$

The gaseous ions, $\text{M}^+(g) + \text{X}^-(g)$, also go down to the aqueous ions:

$$ \text{M}^+(g) + \text{X}^-(g) \xrightarrow{\Delta H_{\text{hyd}}} \text{M}^+(aq) + \text{X}^-(aq) $$

Applying Hess's Law (going down the sides must equal going straight across):

$$ \Delta H_{\text{sol}} = (-\Delta H_{\text{latt}}) + \Delta H_{\text{hyd}} $$

Warning on Signs: Be extremely careful with the sign of the Lattice Energy ($\Delta H_{\text{latt}}$)!

  • The syllabus defines $\Delta H_{\text{latt}}$ as the formation of the solid from gaseous ions, which is always exothermic (negative value).
  • However, in the solution cycle, we need the energy to break apart the solid lattice, which is the reverse reaction (Lattice Dissociation Energy, always positive).
  • Therefore, if you are given a negative $\Delta H_{\text{latt}}$ value, you must use the positive equivalent (or use $-\Delta H_{\text{latt}}$) in the equation for dissolution.

Quick Review Box:
The energy required to dissolve a salt is the balance between the energy needed to pull the ions apart ($\text{Lattice Energy Dissociation}$) and the energy released when water molecules surround the ions ($\Delta H_{\text{hyd}}$).

$ \Delta H_{\text{sol}} = \text{(Energy Absorbed)} + \text{(Energy Released)} $

3. Calculations Involving the Energy Cycle

Since we have a robust Hess's Law equation, we can calculate any one of the three terms ($\Delta H_{\text{sol}}$, $\Delta H_{\text{latt}}$, or $\Delta H_{\text{hyd}}$) if the other two are known.

3.1 The Combined Hydration Enthalpy

Note that the Enthalpy of Hydration ($\Delta H_{\text{hyd}}$) is always the sum of the individual hydration enthalpies for the cation and the anion:

$$ \Delta H_{\text{hyd}} = \Delta H_{\text{hyd}}(\text{cation}) + \Delta H_{\text{hyd}}(\text{anion}) $$

3.2 Calculation Practice: Finding Enthalpy of Solution

Problem: Calculate the $\Delta H_{\text{sol}}$ for sodium chloride, given the following standard values:

(1) Lattice Energy ($\Delta H_{\text{latt}}$) for $\text{NaCl}(g) \to \text{NaCl}(s)$: $ -787 \, \text{kJ} \, \text{mol}^{-1} $
(2) Hydration Enthalpy for $\text{Na}^+(g)$: $ -406 \, \text{kJ} \, \text{mol}^{-1} $
(3) Hydration Enthalpy for $\text{Cl}^-(g)$: $ -378 \, \text{kJ} \, \text{mol}^{-1} $

Step 1: Calculate Total Enthalpy of Hydration ($\Delta H_{\text{hyd}}$)
$$ \Delta H_{\text{hyd}} = (-406) + (-378) = -784 \, \text{kJ} \, \text{mol}^{-1} $$

Step 2: Determine Lattice Dissociation Energy
Since $\Delta H_{\text{latt}}$ (formation) is $-787 \, \text{kJ} \, \text{mol}^{-1}$, the Lattice Dissociation Energy (breaking the lattice) is the opposite sign:
$$ \text{Lattice Dissociation Energy} = - \Delta H_{\text{latt}} = +787 \, \text{kJ} \, \text{mol}^{-1} $$

Step 3: Calculate Enthalpy of Solution ($\Delta H_{\text{sol}}$)
$$ \Delta H_{\text{sol}} = \text{(Lattice Dissociation Energy)} + \Delta H_{\text{hyd}} $$ $$ \Delta H_{\text{sol}} = (+787) + (-784) = +3 \, \text{kJ} \, \text{mol}^{-1} $$

Since $\Delta H_{\text{sol}}$ is small and positive, dissolving NaCl is slightly endothermic (it makes the water slightly cooler).

Common Mistake to Avoid: Forgetting to reverse the sign of the Lattice Energy when calculating $\Delta H_{\text{sol}}$. The reaction $\text{MX}(s) \to \text{M}^+(g) + \text{X}^-(g)$ always costs energy!

Key Takeaway: Use the relationship $\Delta H_{\text{sol}} = (-\Delta H_{\text{latt}}) + \Delta H_{\text{hyd}}$ and remember that $\Delta H_{\text{hyd}}$ is the sum of the cation and anion hydration enthalpies.

4. Qualitative Explanation of Hydration Enthalpy

As specified in the syllabus, we need to understand how the physical properties of the ion affect the numerical magnitude of the hydration enthalpy ($\Delta H_{\text{hyd}}$).

Remember, hydration is an electrostatic attraction between the ion and the water dipole. The stronger this attraction, the more energy is released, and the larger the negative numerical value of $\Delta H_{\text{hyd}}$.

4.1 Effect of Ionic Charge ($Q$)

The attraction is directly proportional to the charge on the ion.

  • Higher Ionic Charge ($Q$): Leads to a stronger electrostatic attraction to the polar water molecules.
  • Result: The magnitude of $\Delta H_{\text{hyd}}$ increases significantly (becomes much more negative).

Example: $\text{Mg}^{2+}$ has a much more negative $\Delta H_{\text{hyd}}$ than $\text{Na}^{+}$ because $\text{Mg}^{2+}$ has twice the positive charge, leading to a much stronger "hydration hug."

4.2 Effect of Ionic Radius ($r$)

The attraction is inversely proportional to the distance between the nucleus and the water molecules. We measure this distance via the ionic radius.

  • Smaller Ionic Radius ($r$): This results in a higher charge density (charge concentrated in a smaller volume).
  • Result: The ion can get closer to the water molecules, leading to a stronger attraction. The magnitude of $\Delta H_{\text{hyd}}$ increases (becomes more negative).

Example: Comparing ions in the same group (e.g., $\text{Li}^{+}$ vs $\text{Cs}^{+}$). $\text{Li}^{+}$ is much smaller than $\text{Cs}^{+}$, so it has a much higher charge density and a much more negative $\Delta H_{\text{hyd}}$.

Summary of Factors (The "Charge Density Rule")

The magnitude of the hydration enthalpy depends heavily on the Charge Density of the ion.

For an ion, the hydration enthalpy becomes more negative (larger magnitude) when:

  1. The ionic charge ($Q$) increases.
  2. The ionic radius ($r$) decreases.

Key Takeaway: Small, highly charged ions ($\text{e.g., Al}^{3+}$ or $\text{Mg}^{2+}$) interact very strongly with water, resulting in large, highly exothermic (very negative) hydration enthalpies.

5. Bringing it Together: Solubility and Enthalpy

The enthalpy of solution ($\Delta H_{\text{sol}}$) helps predict if dissolving is energetically favourable.

5.1 Comparing Magnitudes

Whether a salt dissolves exothermically or endothermically depends entirely on the balance of the two energy terms:

  • For Exothermic Dissolution ($\Delta H_{\text{sol}}$ is negative):
    The energy released by hydration ($\Delta H_{\text{hyd}}$, negative value) must be greater than the energy required to break the lattice (Lattice Dissociation Energy, positive value).
  • For Endothermic Dissolution ($\Delta H_{\text{sol}}$ is positive):
    The energy required to break the lattice must be greater than the energy released by hydration.

5.2 Example: Group 2 Trends

If we look down Group 2, the ions get larger (e.g., $\text{Mg}^{2+} \to \text{Ba}^{2+}$).

  • As the ion gets larger, its $\Delta H_{\text{hyd}}$ becomes less negative (weaker attraction to water).
  • The lattice energy (the "difficulty" of breaking the crystal) also generally becomes less exothermic (easier to break) as ions get bigger, but this decrease is often offset by the decrease in hydration energy.

The final solubility trend (e.g., why Group 2 sulfates decrease in solubility down the group) is determined by how the difference between the hydration energy and the lattice energy changes as the ionic size increases. (A small drop in hydration energy often outweighs a small drop in lattice energy, making $\Delta H_{\text{sol}}$ more positive, thus less soluble).

Don't worry if this seems tricky at first! The main goal is to remember: Dissolution is a competition between the energy holding the ions together ($\Delta H_{\text{latt}}$) and the energy pulling the ions into water ($\Delta H_{\text{hyd}}$).

If the "water hug" energy released is strong enough, the compound will dissolve easily!