1. Solutions (Chapter 12) The Solution Process Why do things dissolve? -- driving force toward more random state (entropy) -- attractive forces between solute and solvent (enthalpy) “like dissolves like” (ionic and polar substances tend to be water soluble) Intermolecular Forces!!!

2. Intermolecular Forces in Solutions

3. Solubility the amount of a substance that will dissolve in a given amount of solvent How much will dissolve? –Units are often: g solute per 100 g of solution –“saturated” solution: maximum amount of solute is dissolved Supersaturated: more dissolved than saturated –miscible: soluble in all proportions Recall dissociation equations –e.g. NaCl (s)  NaCl (aq) = Na + (aq) + Cl – (aq) solute (undissolved) ⇆ solute (dissolved) molecular formionic form

4. Solubility the amount of a substance that will dissolve in a given amount of solvent How much will dissolve? –Units are often: g solute per 100 g of solution –“saturated” solution: maximum amount of solute is dissolved Supersaturated: more dissolved than saturated –miscible: soluble in all proportions Recall dissociation equations –e.g. NaCl (s) ⇆ NaCl (aq) = Na + (aq) + Cl – (aq) solute (undissolved) ⇆ solute (dissolved) molecular formionic form

5. Heats of Solution  H soln =  H solute +  H solvent +  H mix  H soln is a combination of two opposing effects: Lattice Energy (  H solute ) -- endothermic –energy required to separate solid particles Hydration (solvation) Energy (  H hydration ) -- exothermic –energy released as gaseous solute particles are surrounded by solvent molecules –e.g. K + (g) + Cl – (g)  K + (aq) + Cl – (aq)  H hydration = -819 kJ/mol ∴  H soln ≈  H solute +  H hydration (  H soln can be positive or negative) > 0 < 0  H hydration

6. Factors Affecting Solubility Effect of Temperature on Solubility –Most solids are more soluble at higher temp, most gases are less soluble at higher temp Effect of Pressure on Solubility –No significant effect for solid or liquid solutes, but major effect with gaseous solutes dissolved in liquid solvents Henry’s Law: gases are more soluble at higher pressure (e.g. carbonated beverage) S g ∝ P g or S g = k H P g or S 1 /S 2 = P 1 /P 2 where S = solubility, P = pressure, k H = Henry’s law constant (depends on gas)

7. Concentrations of Solutions I (See Table 12.5, p 529) Molarity (M) –M = moles of solute/liters of solution Mole Fraction (and mole percent) –X A = moles A/[moles A + moles B + …] –mole % = X A x 100% –Mixtures of gases: X A ∝ n A ∝ P A (at constant temp) where X = mol fraction, n = mol, P = pressure so, X A = P A /P total –(i.e. the mol fraction will equal the pressure fraction!) Mass %, parts per million, etc –ppm, ppb can be by mass or by volume, e.g. Multiplication factor = 100 mass % Multiplication factor = 10 6 ppm Multiplication factor = 10 9 ppb mass solute mass solution x multiplication factor

8. Concentrations of Solutions II Weight Fraction (and weight percent) –WF A = mass A/mass of solution –Wt % = WF A x 100% e.g. a 5.00% (by weight) solution of NaCl contains: 5 g NaCl in 100 g of solution (5.00 g NaCl and 95.00 g H 2 O) molality (m) -- don’t confuse it with Molarity (M)!!! –m = moles solute/kg of solvent –Independent of temperature e.g. molality of above 5.00% NaCl solution? m = [(5.00 g NaCl x (1 mole NaCl/58.4 g NaCl)]/(0.09500 kg H 2 O) = 0.90 mol NaCl/kg = 0.901 m NaCl

9. Conversions Between Concentration Methods Example: Commercial hydrobromic acid, HBr, is 40.0% by weight. The density of this solution is 1.38 g/mL. Calculate the molality, molarity, and mole fraction of this HBr solution. 40.0% HBr means that 100 g of solution contains: 40.0 g HBr and 60.0 g H 2 O Moles HBr = 40.0 g x (1 mole/80.9 g) = 0.494 mol Moles H 2 O = 60.0 g x (1 mole/18.0 g) = 3.333 mol X HBr = 0.494/(0.494 + 3.333) = 0.129 (or 12.9 mole %) m = moles HBr/kg H 2 O = 0.494 mole/0.0600 kg = 8.23 m To find molarity, need volume of solution (from density): Volume of 100 g of solution = 100 g x (1 mL/1.38 g) = 72.5 mL M = mole HBr/L soln = 0.494 mol/0.0725 L = 6.82 M

10. Sample Problem Commercial sulfuric acid is 96.0% H 2 SO 4 (formula mass = 98.07 g/mole) by weight and has a density of 1.85 g/mL. Calculate the molarity (M) and the molality (m) of the H 2 SO 4 solution.

11. Colligative Properties (depend on number of solute particles) Vapor Pressure –(related to mole fraction of solvent) –Vapor pressure of solution is always less than the pure solvent –For solutions of non-volatile solutes, Raoult’s Law applies: P solution = X solvent P° solvent where P solution = vapor pressure of soln, X solvent = mol fraction of solvent, P° solvent = vapor pressure of pure solvent OMIT -- mixtures of two or more volatile components (p 539-541)

12. Freezing and Boiling Points Freezing Point Depression and Boiling Point Elevation –(related to molality of the solution) –Change in freezing and boiling points:  T f = K f m  T b = K b m where K f and K b are properties of the solvent: K f = molal freezing point depression constant K b = molal boiling point elevation constant e.g. for water: K f = 1.86 °C/m K b = 0.51 °C/m

13. Example Problem A solution of 6.400 g of an unknown compound in 100.0 g of benzene (C 6 H 6 ) boils at 81.7 °C. Determine the molecular mass of the unknown. Data for benzene: K f = 5.07 °C/mT f = 5.07 °C K b = 2.53 °C/mT b = 80.2 °C  T b = K b m  T b = 81.7 - 80.2 = 1.5 °C m =  T b /K b = (1.5 °C)/(2.53 °C/m) = 0.593 m = = 0.0593 moles cmpd Molecular mass = g/mole = = 1.1 x 10 2 g/mol

14. Sample Problem Automobile antifreeze is a concentrated aqueous solution of ethylene glycol, C 2 H 6 O 2 (formula mass = 62.0 g/mol). Calculate the weight percent of an antifreeze solution that would have a freezing point of -25 °C (equivalent to -13 °F). The K f constant for water is 1.86 °C/m and the freezing point of water is 0.00 °C.

15. Osmotic Pressure (related to molarity) osmosis -- passage of solvent through a “semipermeable membrane” into a solution osmotic pressure (  ) -- back pressure required to stop osmosis  ∝ M (at constant temp) van’t Hoff equation:  V = nRT –since n/V = M, then  = MRT Used for determining MM of unknowns, especially large molecules, e.g. polymers, proteins, etc. Important in medical solutions; cell walls are semipermeable membranes! –hyperosmotic (  > body), hypoosmotic (  < body), isosmotic, or isotonic (  = body)

16. Osmotic Pressure Measurement

17. Real Solutions Strong electrolytes do not always dissociate 100%. van’t Hoff factors correct for ion pairing and other effects. “Corrected equations” –  T f = imK f –  T b = imK b –  = iMRT e.g., NaCl (s) --> Na + (aq) + Cl – (aq) # moles of ions = 2 x (moles of NaCl) so, colligative properties are about twice as large i = mol of particles in solution mol of formula units dissolved

18. Tyndall effect: scattering of light by a colloidal dispersion

19. Micelle Formation

20. How Soaps Work

21. Sample Problem A sample of a protein is dissolved in water to give a solution that contains 5.00 mg of protein per 1.00 mL. At 20.0 °C, this solution is found to have an osmotic pressure of 0.760 torr. Calculate the molecular mass of the protein.