Topic 7.14: Free Energy of Dissolution Revision Notes

What ΔG° of Dissolution Means

Every dissolution has a standard free energy change, ΔG°, which tells you whether the solid wants to dissolve under standard conditions. A negative ΔG° means dissolution is favoured; a positive ΔG° means it is unfavoured, so very little dissolves.

For sparingly soluble salts ΔG° is positive, which is exactly why so little dissolves: AgCl has ΔG° = +55.6 kJ/mol and Ksp = 1.8 × 10⁻¹⁰, while Ca(OH)₂ has ΔG° = +30.0 kJ/mol and Ksp = 5.5 × 10⁻⁶.

This connection is the whole topic: if you know Ksp, you can find ΔG°, and if you know ΔG°, you can find Ksp. If you know how ΔG° changes with temperature, you can predict whether warming raises or lowers the solubility.

Standard conditions. The ° in ΔG° means standard conditions: all dissolved species at 1 mol/L, the solid pure, and the temperature usually 298 K (25 °C). A positive ΔG° does not mean nothing dissolves: the equilibrium lies far to the solid side.

ΔG° = −RT ln Ksp

The link between the free energy and the equilibrium constant is

ΔG° = −RT ln Ksp

where R = 8.314 J mol⁻¹ K⁻¹ and T is the temperature in kelvin. At 298 K,

RT = 8.314 × 298 = 2,478 J/mol

or 2.48 kJ/mol. The natural logarithm, ln, is the logarithm to base e. Ksp values are far below 1 for sparingly soluble salts, so ln Ksp is negative and ΔG° comes out positive.

For AgCl, Ksp = 1.8 × 10⁻¹⁰, so ln Ksp = −22.44, and ΔG° = −(2,478)(−22.44) = +55.6 kJ/mol. For Mg(OH)₂, Ksp = 5.6 × 10⁻¹², so ln Ksp = −25.91, and ΔG° = +64.2 kJ/mol. For PbI₂, Ksp = 9.8 × 10⁻⁹, so ln Ksp = −18.44, and ΔG° = +45.7 kJ/mol.

The pattern: the larger the Ksp, the less negative the ln Ksp, and the smaller the ΔG°.

Reading the Pattern: Ksp and ΔG° Side by Side

Lead(II) iodide, with Ksp = 9.8 × 10⁻⁹, has ΔG° = +45.7 kJ/mol, while silver chloride, with Ksp = 1.8 × 10⁻¹⁰, has ΔG° = +55.6 kJ/mol.

The extreme case is iron(III) hydroxide: Ksp = 2.8 × 10⁻³⁹ gives ln Ksp = −88.77 and ΔG° = +220 kJ/mol. At the other end, calcium hydroxide has Ksp = 5.5 × 10⁻⁶, ln Ksp = −12.11 and ΔG° = +30.0 kJ/mol.

Barium sulfate sits near silver chloride: Ksp = 1.1 × 10⁻¹⁰, ln Ksp = −22.93 and ΔG° = +56.8 kJ/mol. Calcium fluoride falls between: Ksp = 3.9 × 10⁻¹¹, ln Ksp = −23.97 and ΔG° = +59.4 kJ/mol. Silver chromate is the least favourable of this group: Ksp = 1.1 × 10⁻¹² and ΔG° = +68.2 kJ/mol.

The three landmark values. If Ksp = 1, then ln Ksp = 0 and ΔG° = 0. If Ksp is greater than 1, ln Ksp is positive and ΔG° is negative: dissolution is favoured. If Ksp is less than 1, ln Ksp is negative and ΔG° is positive: dissolution is unfavoured, which is the case for every sparingly soluble salt in this unit.

A useful benchmark: Ksp = 1.0 × 10⁻³ gives ln Ksp = −6.91 and ΔG° = +17.1 kJ/mol at 298 K.

Finding Ksp from ΔG°

The equation works in reverse:

ln Ksp = −ΔG°/RT

and then Ksp = e^(−ΔG°/RT).

Take calcium fluoride with ΔG° = +59.4 kJ/mol at 298 K: ln Ksp = −59,400/2,478 = −23.97, and Ksp = 3.9 × 10⁻¹¹. Take lead(II) iodide with ΔG° = +45.7 kJ/mol: ln Ksp = −18.44, and Ksp = 9.8 × 10⁻⁹. Take silver chloride with ΔG° = +55.6 kJ/mol: ln Ksp = −22.44, and Ksp = 1.8 × 10⁻¹⁰.

Each calculation returns the Ksp you started from.

Temperature and ΔG° = ΔH° − TΔS°

Free energy also splits into enthalpy and entropy:

ΔG° = ΔH° − TΔS°

Dissolving a salt usually increases disorder, so ΔS° is usually positive. When ΔS° is positive, warming makes the −TΔS° term more negative, so ΔG° falls and Ksp rises.

Take a salt with ΔH° = +30.0 kJ/mol and ΔS° = +60 J mol⁻¹ K⁻¹. At 298 K, ΔG° = 30,000 − (298)(60) = +12.1 kJ/mol, and Ksp = e^(−12,120/2,478) = 0.0075. At 350 K, ΔG° = 30,000 − (350)(60) = +9.0 kJ/mol, and Ksp = e^(−9,000/2,910) = 0.045. Warming by 52 K raised Ksp by a factor of six.

The dissolution is endothermic, so warming shifts the equilibrium toward the dissolved ions.

When Warming Lowers Solubility

If dissolution is exothermic, ΔH° is negative, and warming can lower the solubility. Take a salt with ΔH° = −10.0 kJ/mol and ΔS° = +40 J mol⁻¹ K⁻¹. At 298 K, ΔG° = −10,000 − (298)(40) = −21.9 kJ/mol, and Ksp = e^(21,920/2,478) = 7.0 × 10³, a very soluble salt.

Warming lowers solubility when dissolution is exothermic, whatever the sign of ΔS°. The general rule: for an endothermic dissolution with positive entropy, solubility rises with temperature.

This is why hot water dissolves more solid than cold water: as T grows, the −TΔS° term drags ΔG° down and Ksp up.

Worked Examples

Example 1: ΔG° for AgCl. AgCl has Ksp = 1.8 × 10⁻¹⁰ at 298 K. First find ln Ksp = −22.44. Then ΔG° = −RT ln Ksp = −(2,478)(−22.44) = +55.6 kJ/mol. The positive sign is expected: Ksp is far below 1, so ln Ksp is negative.

Example 2: ΔG° for Mg(OH)₂, compared with PbI₂. Mg(OH)₂ has Ksp = 5.6 × 10⁻¹², so ln Ksp = −25.91 and ΔG° = −(2,478)(−25.91) = +64.2 kJ/mol. PbI₂ has Ksp = 9.8 × 10⁻⁹, ln Ksp = −18.44 and ΔG° = +45.7 kJ/mol. The difference is 64.2 − 45.7 = 18.5 kJ/mol.

Example 3: Ksp from ΔG° for CaF₂. CaF₂ has ΔG° = +59.4 kJ/mol at 298 K. Then ln Ksp = −ΔG°/RT = −59,400/2,478 = −23.97, and Ksp = e^(−23.97) = 3.9 × 10⁻¹¹.

Example 4: warming an endothermic dissolution. A salt has ΔH° = +30.0 kJ/mol and ΔS° = +60 J mol⁻¹ K⁻¹. At 298 K, ΔG° = 30,000 − (298)(60) = +12.1 kJ/mol, and Ksp = 0.0075. At 350 K, ΔG° = 30,000 − (350)(60) = +9.0 kJ/mol. At 350 K, RT = 8.314 × 350 = 2,910 J/mol, so Ksp = e^(−9,000/2,910) = 0.045. Warming raised Ksp from 0.0075 to 0.045, a factor of six.

Traps That Cost Marks in Topic 7.14

“ΔG° is negative because the salt dissolves a little.” A sparingly soluble salt has Ksp below 1, so ln Ksp is negative and ΔG° is positive.

“log₁₀ Ksp works in ΔG° = −RT ln Ksp.” The equation needs the natural logarithm: for AgCl, log₁₀ Ksp is −9.74 while ln Ksp is −22.44, and mixing them up changes ΔG° by a factor of 2.303.

“ΔH° = +30 kJ/mol, so ΔG° = 30,000 − 298 × 60 in kJ.” The units must match: convert ΔH° from kJ/mol to J/mol before subtracting TΔS° in J/mol.

“T = 25 °C in ΔG° = −RT ln Ksp.” T must be in kelvin: at 25 °C, T = 298 K.

“A larger Ksp means a larger ΔG°.” It is the reverse: a larger Ksp means a less negative ln Ksp and a smaller ΔG°.

“Warming always raises solubility.” Warming raises solubility when the dissolution is endothermic, and lowers it when the dissolution is exothermic.

Key Relationships

• ΔG° = −RT ln Ksp, with R = 8.314 J mol⁻¹ K⁻¹ and T in kelvin; at 298 K, RT = 2,478 J/mol
• Ksp below 1 gives a positive ΔG°; Ksp above 1 gives a negative ΔG°; Ksp = 1 gives ΔG° = 0
• AgCl, Ksp = 1.8 × 10⁻¹⁰: ln Ksp = −22.44, ΔG° = +55.6 kJ/mol
• Mg(OH)₂, Ksp = 5.6 × 10⁻¹²: ln Ksp = −25.91, ΔG° = +64.2 kJ/mol
• PbI₂, Ksp = 9.8 × 10⁻⁹: ln Ksp = −18.44, ΔG° = +45.7 kJ/mol
• Fe(OH)₃, Ksp = 2.8 × 10⁻³⁹: ln Ksp = −88.77, ΔG° = +220 kJ/mol
• Ca(OH)₂, Ksp = 5.5 × 10⁻⁶: ln Ksp = −12.11, ΔG° = +30.0 kJ/mol
• ln Ksp = −ΔG°/RT, so Ksp = e^(−ΔG°/RT)
• ΔG° = ΔH° − TΔS°; with ΔS° positive, warming lowers ΔG° and raises Ksp
• Use ln, not log₁₀; convert kJ to J; use kelvin, not Celsius