Free energy of dissolution — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Gibbs free energy change of dissolution ($\Delta G_{soln}^{\circ }$, $\Delta G_{soln}$), the relationship between $\Delta G_{soln}^{\circ }$ and $K_{sp}$, solubility prediction from enthalpy and entropy of dissolution, and $\Delta G$ analysis of unsaturated/saturated/supersaturated solutions.

You should already know: Gibbs free energy relation $\Delta G=\Delta G^{\circ }+RT\ln Q$, the definition of the solubility product constant $K_{sp}$, enthalpy and entropy changes for solution formation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Free energy of dissolution?

Free energy of dissolution refers to the Gibbs free energy change that occurs when 1 mole of solute dissolves in a solvent to form a solution at constant temperature and pressure. For AP Chemistry, standard state conditions are 1 atm pressure, 1 M solute concentration, and 298 K unless stated otherwise. The standard free energy of dissolution is denoted $\Delta G_{soln}^{\circ }$, while the non-standard free energy change for dissolution at non-standard concentrations is called $\Delta G_{soln}$.

This topic is part of Unit 7: Equilibrium, which accounts for 7–9% of the total AP Chemistry exam weight. Free energy of dissolution appears in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically test conceptual understanding of spontaneity and solubility trends, while FRQ often ask for calculations connecting $\Delta G_{soln}^{\circ }$ to $K_{sp}$ or prediction of temperature-dependent solubility. It bridges core thermodynamics and equilibrium, two of the big ideas in the AP Chemistry curriculum.

2. Standard Free Energy of Dissolution and the $\Delta G^{\circ }$-$K_{sp}$ Relationship

Dissolution is a homogeneous equilibrium process, so the general relationship between standard Gibbs free energy change and the equilibrium constant applies directly. For the dissolution of a generic ionic solid $M_x{A}_y(s)$: $$M_x{A}_y(s)\rightleftharpoons x{M}^{y+}(aq)+y{A}^{x-}(aq)$$ The equilibrium constant for this reaction is the solubility product constant $K_{sp}$. The core formula relating $\Delta G_{soln}^{\circ }$ to $K_{sp}$ is: $$\Delta G_{soln}^{\circ }=-RT\ln K_{sp}$$ Where $R=8.314\text{ J/(molcdotpK)}=0.008314\text{ kJ/(molcdotpK)}$, and $T$ is absolute temperature in Kelvin.

This formula directly tells us the solubility of a compound under standard conditions: if $\Delta G_{soln}^{\circ }<0$, dissolution is spontaneous, so $K_{sp}>1$, and the compound is soluble. If $\Delta G_{soln}^{\circ }>0$, dissolution is non-spontaneous under standard conditions, so $K_{sp}<1$, and the compound is classified as sparingly soluble or insoluble. At $\Delta G_{soln}^{\circ }=0$, $K_{sp}=1$, which is the boundary between soluble and insoluble behavior.

Worked Example

The $K_{sp}$ of lead(II) chloride ($PbC{l}_2$) at 298 K is $1.6\times 10^{-5}$. Calculate $\Delta G_{soln}^{\circ }$ for $PbC{l}_2$ at this temperature, and classify it as soluble or insoluble under standard conditions.

1. List given values: $K_{sp}=1.6\times 10^{-5}$, $T=298\text{ K}$, $R=0.008314\text{ kJ/(molcdotpK)}$.
2. Calculate $\ln K_{sp}$: $\ln (1.6\times 10^{-5})=\ln (1.6)+\ln (10^{-5})\approx 0.470-11.513=-11.043$.
3. Substitute into the formula: $\Delta G_{soln}^{\circ }=-(0.008314)(298)(-11.043)\approx +27.4\text{ kJ/mol}$.
4. Interpret the result: $\Delta G_{soln}^{\circ }$ is positive, so dissolution is non-spontaneous under standard conditions, meaning $PbC{l}_2$ is sparingly soluble.

Exam tip: When calculating $\Delta G_{soln}^{\circ }$ from $K_{sp}$, always confirm your units for $R$ match your desired final answer: use $R=0.008314\text{ kJ/(molcdotpK)}$ for $\Delta G$ in kJ/mol, which is what nearly all AP exam questions request.

3. Calculating $\Delta G_{soln}^{\circ }$ from $\Delta H_{soln}^{\circ }$ and $\Delta S_{soln}^{\circ }$

When $K_{sp}$ is not provided, you can calculate $\Delta G_{soln}^{\circ }$ using the fundamental Gibbs free energy relation: $$\Delta G_{soln}^{\circ }=\Delta H_{soln}^{\circ }-T\Delta S_{soln}^{\circ }$$ Here, $\Delta H_{soln}^{\circ }$ is the standard enthalpy change when 1 mole of solute dissolves, and $\Delta S_{soln}^{\circ }$ is the standard entropy change of dissolution.

This formula lets you quantitatively predict how solubility changes with temperature, a common AP exam question. If $\Delta H_{soln}^{\circ }$ is positive (endothermic dissolution), increasing $T$ makes $-T\Delta S^{\circ }$ more negative, so $\Delta G_{soln}^{\circ }$ becomes more negative, $K_{sp}$ increases, and solubility increases. If $\Delta H_{soln}^{\circ }$ is negative (exothermic dissolution), increasing $T$ makes $\Delta G_{soln}^{\circ }$ more positive, $K_{sp}$ decreases, and solubility decreases. This matches Le Chatelier’s principle but gives a quantitative framework for calculations.

Worked Example

The dissolution of ammonium nitrate is $N{H}_{4}N{O}_3(s)\rightleftharpoons N{H}_4^{+}(aq)+N{O}_3^{-}(aq)$. At 298 K, $\Delta H_{soln}^{\circ }=+25.7\text{ kJ/mol}$ and $\Delta S_{soln}^{\circ }=+108.7\text{ J/(molcdotpK)}$. Calculate $\Delta G_{soln}^{\circ }$ at 298 K and 350 K, then predict how solubility changes with increasing temperature.

1. Convert $\Delta S_{soln}^{\circ }$ to kJ to match the units of $\Delta H^{\circ }$: $\Delta S_{soln}^{\circ }=108.7/1000=0.1087\text{ kJ/(molcdotpK)}$.
2. Calculate $\Delta G^{\circ }$ at 298 K: $\Delta G^{\circ }=25.7-(298)(0.1087)=25.7-32.4=-6.7\text{ kJ/mol}$.
3. Calculate $\Delta G^{\circ }$ at 350 K: $\Delta G^{\circ }=25.7-(350)(0.1087)=25.7-38.0=-12.3\text{ kJ/mol}$.
4. Interpret the result: $\Delta G^{\circ }$ becomes more negative as temperature increases, so $K_{sp}$ increases, meaning solubility increases with increasing temperature.

Exam tip: Always convert $\Delta S^{\circ }$ from J/(mol·K) to kJ/(mol·K) before substituting into $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$; a unit mismatch is the most common calculation error on this type of problem.

4. Non-Standard Free Energy and Classifying Solution Saturation

When a solution is not at equilibrium (not saturated), we use the non-standard Gibbs free energy change of dissolution, which follows the general relation: $$\Delta G_{soln}=\Delta G_{soln}^{\circ }+RT\ln Q$$ Where $Q$ is the reaction quotient, calculated the same way as $K_{sp}$ but using current ion concentrations instead of equilibrium concentrations.

The sign of $\Delta G_{soln}$ tells us which direction the reaction proceeds to reach equilibrium:

• $\Delta G_{soln}<0$: Dissolution is spontaneous, so more solid will dissolve → solution is unsaturated ($Q<K_{sp}$)
• $\Delta G_{soln}=0$: System is at equilibrium, no net change → solution is saturated ($Q=K_{sp}$)
• $\Delta G_{soln}>0$: Precipitation (reverse reaction) is spontaneous, so ions will precipitate → solution is supersaturated ($Q>K_{sp}$)

Worked Example

For $PbC{l}_2$ dissolution at 298 K, $\Delta G_{soln}^{\circ }=+27.4\text{ kJ/mol}$. A solution has $[P{b}^{2+}]=0.001\text{ M}$ and $[C{l}^{-}]=0.001\text{ M}$. Calculate $\Delta G_{soln}$ and classify the solution.

1. Write the expression for $Q$: $Q=[P{b}^{2+}][C{l}^{-}{]}^2$.
2. Calculate $Q$: $Q=(0.001)(0.001{)}^2=1\times 10^{-9}$.
3. Substitute into the $\Delta G$ formula: $\Delta G_{soln}=27.4+(0.008314)(298)(\ln 1\times 10^{-9})\approx 27.4+(2.478)(-20.72)\approx -23.9\text{ kJ/mol}$.
4. Interpret: $\Delta G_{soln}$ is negative, so dissolution is spontaneous, meaning the solution is unsaturated.

Exam tip: If you forget what the sign of $\Delta G$ means for saturation, cross-check with $Q$: $Q<K_{sp}$ always means unsaturated, which always matches $\Delta G<0$.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $R=0.0821\text{ Lcdotpatm/(molcdotpK)}$ instead of $8.314\text{ J/(molcdotpK)}$ when calculating $\Delta G$ from $K_{sp}$. Why: Students remember $R$ from gas law problems and use it by mistake, leading to a $\Delta G$ value that is 3 orders of magnitude incorrect. Correct move: Always confirm the value of $R$ before starting: use $R=8.314\text{ J/(molcdotpK)}$ (or $0.008314\text{ kJ/(molcdotpK)}$) for all Gibbs free energy calculations.
• Wrong move: Forgetting to convert $\Delta S^{\circ }$ from J/(mol·K) to kJ/(mol·K) when calculating $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$. Why: $\Delta H^{\circ }$ is almost always reported in kJ/mol, while $\Delta S^{\circ }$ is reported in J/(mol·K), so leaving $\Delta S$ in J gives a $\Delta G$ value 1000x too large. Correct move: As a first step, divide $\Delta S^{\circ }$ by 1000 to convert to kJ/(mol·K) before substituting into the formula.
• Wrong move: Claiming a positive $\Delta G_{soln}^{\circ }$ means the compound never dissolves in water. Why: Students confuse standard $\Delta G^{\circ }$ with non-standard $\Delta G$; many sparingly soluble compounds dissolve to a small extent even if $\Delta G^{\circ }$ is positive. Correct move: Interpret $\Delta G_{soln}^{\circ }$ as telling you the equilibrium extent of dissolution: positive $\Delta G^{\circ }$ means $K_{sp}<1$, so only small amounts dissolve at equilibrium, not that no dissolution occurs.
• Wrong move: Getting the sign of $\Delta G_{soln}^{\circ }$ wrong when starting from $K_{sp}$ (e.g., getting a negative $\Delta G^{\circ }$ for $K_{sp}=1.6\times 10^{-5}$). Why: Students forget that $\ln$ of a number less than 1 is negative, so the two negative signs multiply to a positive $\Delta G^{\circ }$. Correct move: After calculating, always check your sign: $K_{sp}<1$ always gives positive $\Delta G_{soln}^{\circ }$, $K_{sp}>1$ always gives negative $\Delta G_{soln}^{\circ }$.
• Wrong move: Claiming a positive $\Delta G_{soln}$ means the solution is unsaturated. Why: Students mix up the direction of spontaneity: positive $\Delta G$ means the reverse reaction (precipitation) is spontaneous. Correct move: Always pair $Q$ and $\Delta G$: $Q>K_{sp}\to \Delta G>0\to$ precipitation spontaneous $\to$ supersaturated.
• Wrong move: Using Celsius temperature instead of Kelvin in $\Delta G$ calculations. Why: Students are used to Celsius for enthalpy problems where temperature differences are identical, but $\Delta G$ calculations require absolute temperature. Correct move: Always convert any given Celsius temperature to Kelvin by adding 273 (acceptable for AP exams) before substituting into the formula.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

A student measures the $K_{sp}$ of silver sulfate ($A{g}_{2}S{O}_4$) at 298 K and finds it is $1.2\times 10^{-5}$. Which of the following correctly gives $\Delta G_{soln}^{\circ }$ and identifies the solubility of $A{g}_{2}S{O}_4$ at 298 K under standard conditions? A) $\Delta G_{soln}^{\circ }=-28\text{ kJ/mol}$, soluble B) $\Delta G_{soln}^{\circ }=+28\text{ kJ/mol}$, sparingly soluble C) $\Delta G_{soln}^{\circ }=+2.8\text{ kJ/mol}$, sparingly soluble D) $\Delta G_{soln}^{\circ }=+28\text{ kJ/mol}$, soluble

Worked Solution: Use the relation $\Delta G_{soln}^{\circ }=-RT\ln K_{sp}$. Since $K_{sp}=1.2\times 10^{-5}<1$, $\ln K_{sp}$ is negative, so $\Delta G_{soln}^{\circ }$ must be positive, eliminating option A. Calculate $\ln (1.2\times 10^{-5})\approx -11.3$, then substitute to get $\Delta G_{soln}^{\circ }=-(0.008314)(298)(-11.3)\approx +28\text{ kJ/mol}$, eliminating options A and C. A positive $\Delta G_{soln}^{\circ }$ means $K_{sp}<1$, so the compound is sparingly soluble, eliminating D. Correct answer: B

Question 2 (Free Response)

Potassium hydroxide (KOH) dissolves in water via the reaction: $KOH(s)\rightleftharpoons K^{+}(aq)+O{H}^{-}(aq)$. At 298 K, $\Delta H_{soln}^{\circ }=-43\text{ kJ/mol}$ and $\Delta S_{soln}^{\circ }=-25\text{ J/(molcdotpK)}$. (a) Calculate $\Delta G_{soln}^{\circ }$ for KOH at 298 K. (b) Using your value from (a), calculate $K_{sp}$ for KOH at 298 K. (c) A student claims that the solubility of KOH will increase as temperature increases. Is the student correct? Justify your answer with reference to $\Delta G^{\circ }$ and temperature.

Worked Solution: (a) Convert $\Delta S^{\circ }$ to kJ/(mol·K): $\Delta S^{\circ }=-25/1000=-0.025\text{ kJ/(molcdotpK)}$. Substitute into $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$: $$\Delta G^{\circ }=-43-(298)(-0.025)=-43+7.45=-35.55\approx -36\text{ kJ/mol}$$

(b) Rearrange $\Delta G^{\circ }=-RT\ln K_{sp}$ to solve for $\ln K_{sp}$: $$\ln K_{sp}=-\frac{\Delta G^{\circ }}{RT}=-\frac{(-35.55)}{(0.008314)(298)}\approx 14.35$$ $$K_{sp}=e^{14.35}\approx 1.7\times 10^6$$

(c) The student is incorrect. $\Delta H_{soln}^{\circ }$ for KOH is negative (exothermic dissolution). As temperature increases, the term $-T\Delta S^{\circ }$ becomes more positive (since $\Delta S^{\circ }$ is negative: $-T(-0.025)=+0.025T$, which increases as $T$ increases). This makes $\Delta G_{soln}^{\circ }$ less negative (more positive) as temperature increases, so $K_{sp}$ decreases, meaning solubility decreases.

Question 3 (Application / Real-World Style)

Disposable hand warmers rely on the crystallization of sodium acetate from a supersaturated solution. The dissolution reaction of sodium acetate is: $C{H}_{3}COONa(s)\rightleftharpoons C{H}_{3}CO{O}^{-}(aq)+N{a}^{+}(aq)$. At 298 K, $\Delta G_{soln}^{\circ }=-10.0\text{ kJ/mol}$, and a supersaturated hand warmer solution has an ion product $Q=100M^2$. Calculate $\Delta G_{soln}$ for the dissolution reaction at 298 K, and use your answer to explain why crystallization occurs spontaneously.

Worked Solution: Use the non-standard $\Delta G$ formula: $$\Delta G_{soln}=\Delta G_{soln}^{\circ }+RT\ln Q$$ Substitute values with $R=0.008314\text{ kJ/(molcdotpK)}$ and $T=298\text{ K}$: $$\Delta G_{soln}=-10.0+(0.008314)(298)(\ln 100)\approx -10.0+(2.478)(4.605)\approx +1.4\text{ kJ/mol}$$ $\Delta G$ for the forward dissolution reaction is positive, meaning dissolution is non-spontaneous. The reverse reaction (crystallization of solid sodium acetate from solution ions) has $\Delta G=-1.4\text{ kJ/mol}$, which is spontaneous. This explains why crystallization occurs spontaneously when the supersaturated solution is triggered, releasing heat to warm the hand warmer.

7. Quick Reference Cheatsheet

8. What's Next

Free energy of dissolution bridges core thermodynamics and solubility equilibrium, and it is a prerequisite for multiple key topics that follow in AP Chemistry. Immediately after mastering this topic, you will move on to predicting precipitation reactions, where you compare $Q$ and $K_{sp}$ to determine if a precipitate forms when two solutions are mixed. Without understanding how $\Delta G$ relates to $Q$ and $K_{sp}$ for dissolution, you cannot correctly predict precipitation behavior, a common AP FRQ topic. This topic also feeds into the broader study of colligative properties, where free energy changes of dissolution drive boiling point elevation and freezing point depression, and into acid-base equilibrium, where the dissociation of weak acids and bases follows the same $\Delta G^{\circ }$-$K$ relationship derived here.

• Solubility product equilibrium
• Predicting precipitation reactions
• Gibbs free energy and equilibrium
• Colligative properties