13.3 Intermolecular Forces and the Solution Process

1. June 16, 2009 – Class 43 and 44 Overview • 13.3 Intermolecular Forces and the Solution Process • Enthalpy of solution, ideal and nonideal solutions, heterogeneous mixture, formation of ionic solutions, hydration of ions and ion-dipole forces. • 13.4 Solution Formation and Equilibrium • Saturated, undersaturated, and supersaturated solutions, solubility as a function of temperature. • 13.5 Solubilities of Gases • Effect of temperature, effect of pressure and Henry's law • 13.6 Vapor Pressure of Solutions • Raoult's law, liquid-vapor equilibrium for ideal solutions

2. Enthalpy of solution • Pure solvent  separated solvent molecules DHa > 0 (endo) • Pure solute  separated solute molecules DHb > 0 (endo) • Separated solvent and solute molecules  solution DHc < 0 (exo) • Overall: Pure solvent + pure solute  solution DHsoln = DHa + DHb + DHc • (endo or exothermic)

3. DHb DHc DHa Enthalpy of solution A solution process may be endothermic, exothermic or have DHsoln = 0, depending on the magnitude of the enthalpy change in the mixing step.

4. Intermolecular Forces and the Solution Process All same • Ideal solution: has DHsoln = 0 and certain properties (notably vapor pressure) that are predictable from the properties of the solution components. Interactions between solvent molecules A-A; solute molecules B-B; solvent and solute molecules A-B. • If the intermolecular forces of attraction are of the same type and equal strength, the solvent and solute molecules randomly mix. A homogeneous solution results, and is an ideal solution. • “like dissolves like”

5. Intermolecular Forces and the Solution Process Force dominates • Nonideal solution: here DHsoln < 0 (exothermic) and the properties of the solution components cannot easily be predicted. Interactions between solvent molecules A-A; solute molecules B-B; solvent and solute molecules A-B. • If the intermolecular forces of attraction between unlike molecules exceed those of like molecules, a solution still forms. A nonideal solution results.

6. Intermolecular Forces and the Solution Process Intermolecular force between unlike molecules leading to a nonideal solution. Hydrogen bonding between CHCl3 (chloroform) and (CH3)2CO (acetone) molecules produces forces of attraction between unlike molecules that exceed those between like molecules. • If the intermolecular forces of attraction between unlike molecules exceed those of like molecules, a nonideal solution results.

7. Intermolecular Forces and the Solution Process Stronger • Nonideal solution: here DHsoln > 0 (endothermic) and the properties of the solution components cannot easily be predicted. Weaker Interactions between solvent molecules A-A; solute molecules B-B; solvent and solute molecules A-B. • If the intermolecular forces of attraction between unlike molecules is somewhat weaker than those of like molecules, a nonideal solution may still result.

8. Intermolecular Forces and the Solution Process • Dissolution does not occur to any significant extent. • Think: Water and gasoline. Interactions between solvent molecules A-A; solute molecules B-B; solvent and solute molecules A-B. • If the intermolecular forces of attraction between unlike molecules is significantly weaker than those of like molecules, the components remain segregated in a heterogeneous mixture.

9. Intermolecular Forces and the Solution Process Problem: Predict whether or not a solution will form in the following mixtures and whether the solution is likely to be ideal: • Ethanol and water (CH3CH2OH vs. H2O) • Hexane and octane (CH3(CH2)4CH3 vs. CH3(CH2)6CH3) • Octyl alcohol and water (CH3(CH2)4CH2OH vs. H2O). Problem: In which solvent is solid iodine more likely to be soluble, water (H2O) or carbon tetrachloride (CCl4)? Why?

10. Formation of ionic solutions

11. Intermolecular Forces and the Solution Process • Hydration of ions are ions surrounded by a cluster of water molecules. • Energy is released when ions become hydrated. • Ion-dipole forces: for hydrated molecules, the negative ends of water dipoles are pointed towards the cations and the positive ends of water dipoles are pointed towards the anions. • If these ion-dipole forces are sufficiently strong to overcome the interionic forces of attraction in the crystal, the crystal will dissolve.

12. Intermolecular Forces and the Solution Process An ionic crystal dissolving in water. Clustering of water dipoles around the surface of the ionic crystal and the formation of hydrated ions in solution are the key factors in the dissolving process.

13. Formation of ionic solutions NaCl(s)  Na+(g) + Cl-(g) DH1 = (-lattice energy NaCl) >0 H2O Na+(g)  Na+(aq) DH2 = (hydration energy Na+) < 0 H2O Cl-(g)  Cl-(aq) DH3 = (hydration energy Cl-) < 0 NaCl(s)  Na+(aq) + Cl-(aq) DHsoln = DH1 + DH2 + DH3

14. Solution Formation and Equilibrium (c) The solution becomes saturated when the rates of dissolving and crystallization become equal. (a) When solute is first placed in the solvent only dissolving occurs. (b) After a time the rate of crystallization becomes significant. • Saturated Solution: solution that contains the maximum quantity of solute that is normally possible at a given temperature.

15. Solution Formation and Equilibrium • Undersaturated Solution: solution that contains less than the maximum quantity of solute that the solvent is able to dissolve at a particular set of conditions. • Supersaturated Solution: contains more solute than normally expected for a saturated solution. • Supersaturated solutions are normally prepared by starting with a saturated solution, heating it to a temperature where more solute can be dissolved, and then slowly letting it cool to the original temperature.

16. Solution Formation and Equilibrium • For example, a solution of NaCl (the solute) in water (the solvent) has been prepared, but we are unsure if it is unsaturated, saturated, or supersaturated. We can distinguish between these three types of solutions by addition of a seed crystal (a single crystal of the solute, NaCl) • If the seed crystal dissolves, the solution is unsaturated. • If the seed crystal does not dissolve the solution is saturated. • If the seed crystal does not dissolve, and the excess solute recrystallizes (comes out of solution) the solution is supersaturated.

17. Solubility as a function of temperature For each solubility curve (as shown here for KClO4) points on the curve (S) represent saturated solutions. Regions above the curve (1) correspond to supersaturated solutions and below the curve (2), to unsaturated solutions.

18. Solubilities of Gases • Effect of temperature: difficult to generalize • For most gases, their solubility in water decreases when temperature increases. • You can see this when you heat a pot of water on the stove. Before water begins to boils, bubbles of dissolved gas escape. • For most gases, their solubility in organic solvents increases when temperature increases. • For noble gases, initially solubility decreases with increasing temperature. It reaches a minimum value at a particular temperature, and the solubility trend reverses.

19. Solubilities of Gases Effect of pressure: the solubility of a gas increases with increasing pressure. The concentration of dissolved gas (suggested by the depth of color) is proportional to the pressure of the gas above the solution (suggested by the density of the dots).

20. Solubilities of Gases Where C = solubility of a gas in a particular solvent at a fixed temperature Pgas = partial pressure of a gas above the solution k = proportionality constant • Henry's law: relates the solubility of a gas to the gas pressure maintained above a solution of the gaseous solute. The solubility is directly proportional to the pressure of the gas above the solution: C = k Pgas

21. Solubilities of Gases Problem: At 0 oC and an O2 pressure of 1.00 atm, the aqueous solubility of O2(g) is 48.9 mL O2 per liter. What is the molarity of O2 in a saturated water solution when the O2 is under its normal partial pressure in air, 0.2095 atm? Problem: A handbook lists the solubility of carbon monoxide in water at 0oC and 1 atm pressure as 0.0354 mL CO per millilitre of H2O. What pressure of CO(g) must be maintained above the solution to obtain 0.0100 mol/L CO?

22. Vapor Pressure of Solutions • For a solution composed of solvent A and solute B, it was observed that upon addition of B to A, the vapor pressure of A was lowered. • Raoult's law: states that the vapor pressure of a solution component is equal to the product of the vapor pressure of the pure liquid and its mole fraction in solution: PA = cA PoA

23. Vapor Pressure of Solutions Problem: The vapor pressures of pure benzene and pure toluene at 25 oC are 95.1 and 28.4 mmHg, respectively. A solution is prepared in which the mole fractions of benzene and toluene are both 0.500. • What are the partial pressures of the benzene and toluene above this solution. • What is the total vapor pressure. • What is the composition (mole fractions) of the vapor in equilibrium with the solution? Problem: Assume now that instead of identical mole fractions, that you have equal masses of the gases. Calculate the vapor pressures of pure benzene and toluene, the total vapor pressure and the equilibrium composition of the vapor at 25 oC.

24. Liquid-vapor equilibrium for ideal solutions • Liquid-vapor equilibrium for ideal solutions was explored in the previous example. • For ideal solutions of two components, the vapor phase is richer in the more volatile component than is the liquid phase. • Raoult’s Law applies only to ideal solutions and their volatile components, however, it also works reasonably well in dilute, nonideal solutions (that is, solutions where csolvent > 0.98)