13.1 - 13.7 Solutions

Notes on Solution Mixing, Enthalpy, and Intermolecular Forces

1. Why Do Two Nonpolar Liquids Mix?

• Entropy Increase: When two liquids mix, the number of possible molecular arrangements increases, leading to a higher entropy (ΔSmix). This is the main driving force behind mixing in cases where enthalpic changes are negligible.

• Like Dissolves Like: Since both substances are nonpolar, they only experience dispersion forces, which are similar in magnitude. This allows them to mix evenly without significant energy changes.

2. Predicting the Sign and Magnitude of ΔHsoln

ΔHsoln=ΔHsolute+ΔHsolvent+ΔHmix​

• For nonpolar liquids with similar dispersion forces, breaking solute-solute and solvent-solvent interactions requires energy (endothermic), while forming new solute-solvent interactions releases energy (exothermic).

• Since the forces are similar in magnitude, the energy needed to break the original bonds is almost equal to the energy released when new bonds form.

• Result: ΔHsoln\Delta H_{\text{soln}}ΔHsoln​ is close to zero or slightly negative, meaning there is no significant heat change when mixing.

3. Signs and Relative Magnitudes of Enthalpy Components

Since }ΔHmix​ compensates for ΔHsolute + ΔHsolvent​, the total ΔHsoln is approximately zero or slightly negative.

Example Problems and Solutions

Example 1: Mixing Two Nonpolar Liquids

Problem: Predict whether pentane (C5H12​) and hexane (C6H14) will mix, and determine ΔHsoln​.

Solution:

1. Both molecules are nonpolar and interact only through dispersion forces.

2. The solute-solute and solvent-solvent interactions are similar to the solute-solvent interactions.

3. Entropy drives the mixing process because breaking and forming forces require similar amounts of energy.

4. ΔHsoln is approximately zero or slightly negative.

5. Conclusion: Pentane and hexane will mix completely (miscible).

Example 2: Determining ΔHsoln\Delta H_{\text{soln}}ΔHsoln​ for Ionic Solutions

Problem: The lattice energy of potassium iodide (KIKIKI) is −632-632−632 kJ/mol, and its heat of hydration is −690-690−690 kJ/mol. Find ΔHsoln\Delta H_{\text{soln}}ΔHsoln​.

Solution:

ΔHsoln=ΔHsolute+ΔHhydration

ΔHsoln​=(+632 kJ/mol)+(−690 kJ/mol)

ΔHsoln=−58 kJ/mol

Since ΔHsoln​ is negative, dissolving KI in water is exothermic, meaning the solution warms up when KI dissolves.

Example 3: Predicting Solubility Based on Intermolecular Forces

Problem: Will water (H2O) dissolve hexane (C6H14​)?

Solution:

1. Water has strong hydrogen bonding.

2. Hexane has only dispersion forces.

3. Water prefers to interact with itself rather than with nonpolar molecules.

4. Since solvent-solute interactions are weaker than solvent-solvent interactions, ΔHmix​ is too small.

5. Conclusion: Water and hexane are immiscible (they do not mix).

Example 4: Calculating Freezing and Boiling Point Changes

Problem: A solution is made by dissolving 0.5 mol of NaCl in 1.0 kg of water. Calculate the freezing and boiling point of the solution.

Solution:

1. Determine the van 't Hoff factor (i):

   • NaCl dissociates into Na+ and Cl−, so i=2

2. Use the equations for freezing point depression and boiling point elevation:

   ΔTf = i⋅Kf⋅m

   ΔTb = i⋅Kb⋅m

   Kf for water is 1.86∘C⋅kg/mol

   Kb for water is 0.5∘C⋅kg/mol

3. Calculate molality (mmm):

   m = 0.5 mol/1.0 kg= 0.5m 

4. Calculate ΔTf\Delta T_fΔTf​ and ΔTb\Delta T_bΔTb​:

   ΔTf=(2)×(1.86)×(0.5)=1.86∘C

   ΔTb=(2)×(0.512)×(0.5)=0.512∘C

5. Final temperatures:

   Tf=0.00−1.86=−1.86∘C

   Tb=100.00+0.512=100.51∘C

Summary