Lattice enthalpy, hydration and dissolving

Introduction: The Chemistry of Dissolving

Have you ever wondered why table salt, \(\text{NaCl}\), dissolves effortlessly in a pot of boiling water or in the vast oceans, while limestone, \(\text{CaCO}_3\), can sit in the sea for thousands of years without disappearing? Dissolving seems like a simple everyday event, but at the particle level, it is a finely tuned energetic "tug-of-war".

In this chapter from the Oceans storyline of OCR Chemistry B (Salters), we explore the energetics that govern the dissolving of ionic compounds. We will break down the process into three core enthalpy changes: lattice enthalpy, enthalpy change of hydration, and enthalpy change of solution. We will also learn how to calculate these values using an enthalpy cycle, examine how ionic size and charge control these forces, and discover how to quantify the solubility of sparingly soluble salts using the solubility product, \(K_{\text{sp}}\).

Key Enthalpy Definitions

Don't worry if these terms look similar at first glance. Once you picture what the particles are actually doing, the definitions become completely logical.

1. Lattice Enthalpy (\(\Delta_{\text{latt}}H^\ominus\))

An ionic solid consists of a giant three-dimensional lattice of alternating positive and negative ions held firmly together by strong electrostatic attractions.

Standard lattice enthalpy is defined as the enthalpy change when one mole of an ionic lattice is formed from its constituent gaseous ions under standard conditions.

For example, for sodium chloride:
\(\text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)} \rightarrow \text{NaCl(s)}\)   \(\Delta_{\text{latt}}H^\ominus = -787\text{ kJ mol}^{-1}\)

Exam Note: Because bonds/attractions are being formed between gaseous ions to build the crystal lattice, lattice formation is always strongly exothermic (the value of \(\Delta_{\text{latt}}H\) has a negative sign). If you are asked to break the lattice apart into gaseous ions (lattice dissociation), it requires the exact same amount of energy, but with a positive sign: \(+787\text{ kJ mol}^{-1}\).

2. Enthalpy Change of Hydration (\(\Delta_{\text{hyd}}H^\ominus\))

When gaseous ions enter water, the polar water molecules surround them. The partially negative oxygen atoms of water are attracted to positive cations, and the partially positive hydrogen atoms are attracted to negative anions. These ion-dipole attractions release energy.

Standard enthalpy change of hydration is the enthalpy change when one mole of specified gaseous ions dissolves in sufficient water to form an infinitely dilute solution under standard conditions.

For sodium ions:
\(\text{Na}^+\text{(g)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)}\)   \(\Delta_{\text{hyd}}H^\ominus = -406\text{ kJ mol}^{-1}\)

For chloride ions:
\(\text{Cl}^-\text{(g)} + \text{aq} \rightarrow \text{Cl}^-\text{(aq)}\)   \(\Delta_{\text{hyd}}H^\ominus = -378\text{ kJ mol}^{-1}\)

Memory Trick: Hydration is an attraction-forming process, so \(\Delta_{\text{hyd}}H^\ominus\) is always negative (exothermic).

3. Enthalpy Change of Solution (\(\Delta_{\text{sol}}H^\ominus\))

This is the overall heat change that occurs when you physically dissolve a solid in water.

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

For sodium chloride:
\(\text{NaCl(s)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)}\)   \(\Delta_{\text{sol}}H^\ominus = +4\text{ kJ mol}^{-1}\)

Depending on the substance, \(\Delta_{\text{sol}}H^\ominus\) can be either slightly endothermic (cooling the beaker down, like \(\text{NH}_4\text{NO}_3\)) or exothermic (warming the beaker up, like \(\text{CaCl}_2\)).

Quick Review — Key Definitions:

• Lattice enthalpy: Gaseous ions \(\rightarrow\) 1 mole of solid lattice (always exothermic, negative).
• Hydration enthalpy: 1 mole of gaseous ions \(\rightarrow\) aqueous ions (always exothermic, negative).
• Solution enthalpy: 1 mole of solid \(\rightarrow\) aqueous ions (can be positive or negative).

The Dissolving Enthalpy Cycle

To understand the energetic balance of dissolving, we can think of dissolving in two imaginary steps using Hess's Law:

Step 1: Pull the solid crystal lattice apart into separated gaseous ions. This is the reverse of lattice formation, requiring the input of energy equal to \(-\Delta_{\text{latt}}H^\ominus\) (or \(+\text{lattice dissociation energy}\)).
Step 2: Allow water molecules to surround these separated gaseous ions, releasing hydration enthalpy for each ion: \(\sum \Delta_{\text{hyd}}H^\ominus\).

Combining these two steps gives us our master relationship:

\(\Delta_{\text{sol}}H^\ominus = \sum \Delta_{\text{hyd}}H^\ominus - \Delta_{\text{latt}}H^\ominus\)

(where \(\Delta_{\text{latt}}H^\ominus\) is defined as the exothermic value for lattice formation)

Alternatively, in terms of words:

\(\Delta_{\text{sol}}H^\ominus = [\text{Hydration of Cations} + \text{Hydration of Anions}] - [\text{Lattice Enthalpy of Formation}]\)

Worked Example: Calculating \(\Delta_{\text{sol}}H^\ominus\)

Calculate the standard enthalpy change of solution for magnesium chloride, \(\text{MgCl}_2\), using the following data:
• Lattice enthalpy of formation of \(\text{MgCl}_2\): \(-2526\text{ kJ mol}^{-1}\)
• Enthalpy of hydration of \(\text{Mg}^{2+}\): \(-1920\text{ kJ mol}^{-1}\)
• Enthalpy of hydration of \(\text{Cl}^-\): \(-378\text{ kJ mol}^{-1}\)

Step 1: Write down the equation showing the stoichiometry
\(\text{MgCl}_2\text{(s)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}\)
Notice that there are two moles of chloride ions per mole of magnesium chloride!

Step 2: Calculate total hydration enthalpy
\(\sum \Delta_{\text{hyd}}H^\ominus = \Delta_{\text{hyd}}H(\text{Mg}^{2+}) + 2 \times \Delta_{\text{hyd}}H(\text{Cl}^-)\)
\(\sum \Delta_{\text{hyd}}H^\ominus = -1920 + 2(-378) = -1920 - 756 = -2676\text{ kJ mol}^{-1}\)

Step 3: Apply the enthalpy cycle formula
\(\Delta_{\text{sol}}H^\ominus = \sum \Delta_{\text{hyd}}H^\ominus - \Delta_{\text{latt}}H^\ominus\)
\(\Delta_{\text{sol}}H^\ominus = -2676 - (-2526)\)
\(\Delta_{\text{sol}}H^\ominus = -2676 + 2526 = -150\text{ kJ mol}^{-1}\)

Conclusion: Dissolving anhydrous \(\text{MgCl}_2\) is an exothermic process; the solution heats up appreciably!

Factors Affecting Lattice and Hydration Enthalpies

Both lattice enthalpy and hydration enthalpy depend directly on the strength of electrostatic attraction between electric charges. Electrostatic attraction is governed by two main factors:

1. Ionic Charge: Greater charge leads to much stronger electrostatic attractions.
2. Ionic Radius: Smaller ions can get closer together (or closer to water dipoles), which creates much stronger electrostatic forces. Together, these determine the charge density of the ion.

Trends in Lattice Enthalpy

• Effect of Charge: Compare sodium chloride (\(\text{Na}^+\) and \(\text{Cl}^-\), \(\Delta_{\text{latt}}H = -787\text{ kJ mol}^{-1}\)) with magnesium oxide (\(\text{Mg}^{2+}\) and \(\text{O}^{2-}\), \(\Delta_{\text{latt}}H = -3791\text{ kJ mol}^{-1}\)). Doubling the charge on both ions leads to nearly a four-fold increase in lattice enthalpy!
• Effect of Size: Down Group 1 from \(\text{LiF}\) to \(\text{CsI}\), the ionic radii increase. The centres of positive and negative charge are further apart, so electrostatic attraction weakens, and lattice enthalpy becomes less negative (less exothermic).

Trends in Hydration Enthalpy

• Effect of Charge: A \(\text{Mg}^{2+}\) ion (\(-1920\text{ kJ mol}^{-1}\)) has a far more exothermic hydration enthalpy than a \(\text{Na}^+\) ion (\(-406\text{ kJ mol}^{-1}\)) because its \(+2\) charge attracts the lone pairs on the oxygen atom of water much more strongly.
• Effect of Size: Down Group 2, as we move from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\), the ionic radius increases while the charge stays \(+2\). The charge is spread over a larger volume (lower charge density), water molecules cannot get as close to the nucleus, and so hydration enthalpy becomes less negative.

Why Do Salts Dissolve? Enthalpy and Entropy

You might expect that only compounds with exothermic enthalpies of solution (\(\Delta_{\text{sol}}H < 0\)) would dissolve. However, common salt (\(\text{NaCl}\)) has an endothermic \(\Delta_{\text{sol}}H\) (\(+4\text{ kJ mol}^{-1}\)), yet it dissolves very easily.

Why is this possible? The answer lies in entropy (\(\Delta S\)):

• An ionic solid is an ordered, rigid crystalline lattice with low entropy.
• In solution, ions are free to move throughout the liquid, resulting in a large increase in disorder, so \(\Delta S_{\text{system}}\) is usually positive.
• For a process to be feasible, the total entropy change must be positive (\(\Delta S_{\text{total}} > 0\)). As covered in the chapter on entropy and feasibility, when \(\Delta_{\text{sol}}H^\ominus\) is only slightly positive, the large positive \(\Delta S_{\text{system}}\) easily outweighs any small negative \(\Delta S_{\text{surroundings}}\), making dissolving spontaneous.

However, if the lattice enthalpy is enormously large compared to hydration enthalpy (as in \(\text{CaCO}_3\) or \(\text{BaSO}_4\)), \(\Delta_{\text{sol}}H^\ominus\) becomes too endothermic. The unfavorable enthalpy change cannot be overcome by the entropy increase, rendering the solid virtually insoluble.

Sparingly Soluble Salts and Solubility Product (\(K_{\text{sp}}\))

When an ionic compound has very low solubility, we call it sparingly soluble. When placed in water, only a minute fraction dissolves before the solution becomes saturated.

The Dynamic Equilibrium

In a saturated solution of a sparingly soluble salt in contact with undissolved solid, a dynamic equilibrium exists between the solid lattice and the dissolved ions:

\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}\)

Since the concentration of the pure solid \([\text{AgCl(s)}]\) is essentially constant, it is incorporated into the equilibrium constant to define the solubility product, \(K_{\text{sp}}\):

\(K_{\text{sp}} = [\text{Ag}^+\text{(aq)}][\text{Cl}^-\text{(aq)}]\)

General Definition: The solubility product, \(K_{\text{sp}}\), is the product of the concentrations of each dissolved ion in a saturated solution at a given temperature, with each concentration raised to the power of its stoichiometric coefficient in the balanced dissociation equation.

Examples of \(K_{\text{sp}}\) Expressions and Units

1. A 1:1 Salt: Silver chloride, \(\text{AgCl}\)
\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}\)
\(K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-]\)
Units: \((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3}) = \text{mol}^2\text{ dm}^{-6}\)

2. A 1:2 Salt: Lead(II) iodide, \(\text{PbI}_2\)
\(\text{PbI}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)}\)
\(K_{\text{sp}} = [\text{Pb}^{2+}][\text{I}^-]^2\)
Units: \((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2 = \text{mol}^3\text{ dm}^{-9}\)

3. A 1:2 Salt: Calcium hydroxide, \(\text{Ca(OH)}_2\)
\(\text{Ca(OH)}_2\text{(s)} \rightleftharpoons \text{Ca}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}\)
\(K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2\)
Units: \(\text{mol}^3\text{ dm}^{-9}\)

Calculations Involving \(K_{\text{sp}}\)

Type 1: Calculating \(K_{\text{sp}}\) from Solubility

Problem: The solubility of \(\text{CaSO}_4\) in pure water at \(298\text{ K}\) is \(4.90 \times 10^{-3}\text{ mol dm}^{-3}\). Calculate its solubility product, \(K_{\text{sp}}\), and give the units.

Solution:
The equilibrium equation is:
\(\text{CaSO}_4\text{(s)} \rightleftharpoons \text{Ca}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}\)
Since 1 mole of \(\text{CaSO}_4\) gives 1 mole of \(\text{Ca}^{2+}\) and 1 mole of \(\text{SO}_4^{2-}\):
\([\text{Ca}^{2+}] = 4.90 \times 10^{-3}\text{ mol dm}^{-3}\)
\([\text{SO}_4^{2-}] = 4.90 \times 10^{-3}\text{ mol dm}^{-3}\)
\(K_{\text{sp}} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (4.90 \times 10^{-3}) \times (4.90 \times 10^{-3}) = 2.40 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}\)

Type 2: Calculating Solubility from \(K_{\text{sp}}\) (1:2 stoichiometry)

Problem: The solubility product of magnesium hydroxide, \(\text{Mg(OH)}_2\), is \(1.80 \times 10^{-11}\text{ mol}^3\text{ dm}^{-9}\) at \(298\text{ K}\). Calculate the solubility of \(\text{Mg(OH)}_2\) in \(\text{mol dm}^{-3}\).

Solution:
1. Let the solubility of \(\text{Mg(OH)}_2\) be \(s\text{ mol dm}^{-3}\).
2. Dissociation equation:
\(\text{Mg(OH)}_2\text{(s)} \rightleftharpoons \text{Mg}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)}\)
From the stoichiometry:
\([\text{Mg}^{2+}] = s\)
\([\text{OH}^-] = 2s\)
Common Pitfall Alert: Remember that the concentration of hydroxide is \(2s\), AND in the \(K_{\text{sp}}\) expression, this concentration must be squared!
3. Set up the expression:
\(K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^-]^2\)
\(K_{\text{sp}} = (s) \times (2s)^2 = s \times 4s^2 = 4s^3\)
4. Solve for \(s\):
\(4s^3 = 1.80 \times 10^{-11}\)
\(s^3 = \frac{1.80 \times 10^{-11}}{4} = 4.50 \times 10^{-12}\)
\(s = \sqrt[3]{4.50 \times 10^{-12}} = 1.65 \times 10^{-4}\text{ mol dm}^{-3}\)

Type 3: Predicting Precipitation (The Ion Product)

We can use \(K_{\text{sp}}\) to predict whether a precipitate will form when two solutions containing relevant ions are mixed.

We calculate the ion product using the concentrations of ions in the mixed solution (just like a \(K_{\text{sp}}\) expression):

• If Ion Product \(< K_{\text{sp}}\): The solution is unsaturated. No precipitate forms, and more solid could dissolve.
• If Ion Product \(= K_{\text{sp}}\): The solution is exactly saturated. It is in dynamic equilibrium.
• If Ion Product \(> K_{\text{sp}}\): The solution cannot hold that many ions! A precipitate forms until the concentrations fall back so that the ion product equals \(K_{\text{sp}}\).

Exam Tip for Mixing Problems: When two equal volumes of solutions are mixed, the volume doubles, which means each initial ion concentration is halved before you calculate the ion product!

Common Mistakes to Avoid

• Forgetting stoichiometry in \(K_{\text{sp}}\): For a salt like \(\text{PbCl}_2\), if solubility is \(s\), then \([\text{Cl}^-] = 2s\). The expression is \([s][2s]^2 = 4s^3\), not \(2s^2\) or \(2s^3\).
• Sign errors with \(\Delta_{\text{latt}}H^\ominus\): Remember that lattice formation is exothermic (\(-\)). Breaking the lattice requires positive energy (\(-\Delta_{\text{latt}}H^\ominus\)). Watch the double negatives in your calculation!
• Missing state symbols: Always include state symbols when defining enthalpy terms: lattice enthalpy must have \(\text{(g)}\) on the left and \(\text{(s)}\) on the right; hydration must have \(\text{(g)}\) on the left and \(\text{(aq)}\) on the right.
• Units of \(K_{\text{sp}}\): These change depending on the salt stoichiometry. Always work them out explicitly from the ion concentrations.

Summary & Key Takeaways

• Lattice enthalpy measures the electrostatic attraction within the crystal lattice; higher charges and smaller ionic radii produce a more exothermic lattice enthalpy.
• Hydration enthalpy measures the attraction between gaseous ions and polar water molecules; smaller ions with higher charges have more exothermic hydration enthalpies.
• Enthalpy of solution is the net balance: \(\Delta_{\text{sol}}H^\ominus = \sum \Delta_{\text{hyd}}H^\ominus - \Delta_{\text{latt}}H^\ominus\).
• Solubility depends on the interplay between enthalpy of solution and the favorable entropy change of dispersing ions into solution.
• Solubility product (\(K_{\text{sp}}\)) quantifies the solubility of sparingly soluble salts in saturated solutions. If the ion product exceeds \(K_{\text{sp}}\), a precipitate forms.