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Solutions (Chapter 12)

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)

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Solutions (Chapter 12)

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  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 form ionic 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 form ionic form

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

  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) Sg∝Pg or Sg = kHPg or S1/S2 = P1/P2 where S = solubility, P = pressure, kH = 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) • XA = moles A/[moles A + moles B + …] • mole % = XA x 100% • Mixtures of gases: XA∝nA∝ PA (at constant temp) where X = mol fraction, n = mol, P = pressure so, XA = PA/Ptotal • (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 = 106ppm • Multiplication factor = 109ppb mass solute x multiplication factor mass solution

  8. Concentrations of Solutions II • Weight Fraction (and weight percent) • WFA = mass A/mass of solution • Wt % = WFA 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 H2O) • 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 H2O) = 0.90 molNaCl/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 H2O Moles HBr = 40.0 g x (1 mole/80.9 g) = 0.494 mol Moles H2O = 60.0 g x (1 mole/18.0 g) = 3.333 mol XHBr = 0.494/(0.494 + 3.333) = 0.129 (or 12.9 mole %) m = moles HBr/kg H2O = 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% H2SO4 (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 H2SO4 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: Psolution = Xsolvent• P°solvent • where Psolution = vapor pressure of soln, Xsolvent = 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: DTf = Kfm DTb = Kbm where Kf and Kb are properties of the solvent: Kf = molal freezing point depression constant Kb = molal boiling point elevation constant e.g. for water: Kf = 1.86 °C/m Kb = 0.51 °C/m

  13. Example Problem • A solution of 6.400 g of an unknown compound in 100.0 g of benzene (C6H6) boils at 81.7 °C. Determine the molecular mass of the unknown. Data for benzene: Kf = 5.07 °C/m Tf = 5.07 °C Kb = 2.53 °C/m Tb = 80.2 °C DTb = Kbm DTb = 81.7 - 80.2 = 1.5 °C m = DTb/Kb = (1.5 °C)/(2.53 °C/m) = 0.593 m = = 0.0593 moles cmpd Molecular mass = g/mole = = 1.1 x 102 g/mol

  14. Sample Problem Automobile antifreeze is a concentrated aqueous solution of ethylene glycol, C2H6O2 (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 Kf 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(P) -- back pressure required to stop osmosis P∝ M (at constant temp) • van’t Hoff equation: PV = nRT • since n/V = M, then P = 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 (P > body), hypoosmotic (P < body), isosmotic, or isotonic (P = 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” • DTf = imKf • DTb = imKb • P = 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 mol of particles in solution i = 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.

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