Mr Cole Chemistry

Trends in Enthalpy of Solution Questions

Solubility of Group 2 Compounds

Thermodynamics Worksheet

Solubility Trends

1. Use enthalpy of solution calculations to demonstrate the trend in solubility of the group 2 hydroxides.
Group 2 Hydroxide Data
Compound Lattice Enthalpy / kJmol-1 Enthalpy of hydration of metal / kJmol-1
Mg(OH)2-3006-1921
Ca(OH)2-2645-1577
Sr(OH)2-2470-1443
Ba(OH)2-2339-1305

The enthalpy of hydration for OH– is -479.5 kJmol-1

Show Calculations & Answer
Calculation Diagram for Hydroxides
Calculations (ΔHsol = ΣΔHhyd – ΔHLE): Mg(OH)2: [-1921 + 2(-479.5)] – (-3006) = -2880 + 3006 = +126 kJ mol-1 Ca(OH)2: [-1577 + 2(-479.5)] – (-2645) = -2536 + 2645 = +109 kJ mol-1 Sr(OH)2: [-1443 + 2(-479.5)] – (-2470) = -2402 + 2470 = +68 kJ mol-1 Ba(OH)2: [-1305 + 2(-479.5)] – (-2339) = -2264 + 2339 = +75 kJ mol-1

Conclusion: You can see that as you go down the group, the enthalpy of solution generally becomes less endothermic (more exothermic in trend), indicating increasing solubility.

2. Explain the trend in enthalpy of hydration of group 2 ions.

As you go down the group, the enthalpy of hydration gets lower (less exothermic). This is because the charge remains the same (+2) but the ionic radii increases. This results in a lower charge density, and therefore a weaker electrostatic attraction between the aqueous ion and the partially negative oxygen atom in the water molecules.

3. Use enthalpy of solution calculations to demonstrate the trend in solubility of the group 2 sulfates.
Group 2 Sulfate Data
Compound Lattice Enthalpy / kJmol-1 Enthalpy of hydration of metal / kJmol-1
MgSO4-2874-1921
CaSO4-2653-1577
SrSO4-2540-1443
BaSO4-2423-1305

The enthalpy of hydration for SO42- is -973.3 kJmol-1

Show Calculations & Answer
Calculation Diagram for Sulfates
Calculations (ΔHsol = ΣΔHhyd – ΔHLE): MgSO4: [-1921 + (-973.3)] – (-2874) = -2894.3 + 2874 = -20.3 kJ mol-1 CaSO4: [-1577 + (-973.3)] – (-2653) = -2550.3 + 2653 = +102.7 kJ mol-1 SrSO4: [-1443 + (-973.3)] – (-2540) = -2416.3 + 2540 = +123.7 kJ mol-1 BaSO4: [-1305 + (-973.3)] – (-2423) = -2278.3 + 2423 = +144.7 kJ mol-1

Conclusion: You can see that as you go down the group, the enthalpy of solution becomes significantly more endothermic, indicating decreasing solubility.

4. Discuss the difference in enthalpy of solution trends with group 2 sulfates and group 2 hydroxides and suggest why that is the case.

Group 2 Hydroxides (OH–): Solubility increases down the group (Mg to Ba).

Group 2 Sulfates (SO42-): Solubility decreases down the group (Mg to Ba).

The Reason

It comes down to the size of the anion and how it affects the balance between Lattice Enthalpy (breaking the solid) and Hydration Enthalpy (forming the solution). Both enthalpies decrease down the group, but at different rates.

1. Hydroxides (Small Anion)

Because OH– is small, the increasing size of the cation (Mg2+ → Ba2+) causes a massive disruption to the lattice structure.

Result: The Lattice Enthalpy falls faster than the Hydration Enthalpy. It becomes much easier to break the lattice than to hydrate the ions, so solubility increases.

2. Sulfates (Large Anion)

Because SO42- is large, the increasing size of the cation makes very little difference to the overall lattice spacing.

Result: The Hydration Enthalpy falls faster than the Lattice Enthalpy. The energy released by water binding to the ions drops significantly, making it energetically unfavorable to dissolve, so solubility decreases.

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