1 / 74

Chapter 13

Chapter 13. Properties of Mixtures: Solutions. Properties of Mixtures: Solutions. 13.1 Types of solutions: Intermolecular forces and predicting solubility. 13.2 Energy changes in the solution process. 13.3 Solubility as an equilibrium process.

fred
Download Presentation

Chapter 13

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 13 Properties of Mixtures: Solutions

  2. Properties of Mixtures: Solutions 13.1 Types of solutions: Intermolecular forces and predicting solubility 13.2 Energy changes in the solution process 13.3 Solubility as an equilibrium process 13.4 Quantitative ways of expressing concentration 13.5 Colligative properties of solutions

  3. The major types of intermolecular forces in solutions (from Chapter 12) Figure 13.1 (energies in parenthesis)

  4. “LIKE DISSOLVES LIKE” Substances with similar types of intermolecular forces dissolve in each other. When a solute dissolves in a solvent, solute-solute interactions and solvent-solvent interactions are partly replaced with solute-solvent interactions. The new forces created between solute and solvent must be comparable in strength to the forces destroyed within the solute and the solvent.

  5. A major factor that determines whether a solution forms: The relative strengths of the intermolecular forces within and between solute and solvent molecules

  6. Some Definitions Solvent: the most abundant component of a given solution Solute: component dissolved in the solvent Solubility (S): the maximum amount of solute that dissolves in a fixed quantity of solvent at a given temperature (in the presence of excess solute) Dilute and concentrated solutions: qualitative terms

  7. Hydration shells around an aqueous ion Formation of ion-dipole forces when a salt dissolves in water Figure 13.2

  8. Liquid Solutions • Liquid-Liquid • Gas-Liquid Gas and Solid Solutions • Gas-Gas • Gas-Solid • Solid-Solid

  9. hexane = CH3(CH2)4CH3 Competition between H-bonding and dispersion forces

  10. Molecular Basis for the Solubility of CH3OH in H2O H-bonding: CH3OH can serve as a donor and acceptor (maximum number of three H-bonds / molecule) Figure 13.3

  11. PROBLEM: Predict which solvent will dissolve more of the given solute: PLAN: Consider the intermolecular forces that exist between solute molecules and consider whether the new solvent-solute interactions can substitute for them. (a) NaCl is ionic and forms ion-dipoles with the OH groups of both methanol and propanol. However, propanol is subject to greater dispersion forces (more CH bonds than methanol). (b) Hexane has no dipoles to interact with the OH groups of ethylene glycol. Water can H-bond to ethylene glycol. (c) Diethyl ether can interact through dipole and dispersion forces. Ethanol can provide both while water can only H-bond. SAMPLE PROBLEM 13.1 Predicting relative solubilities of substances (a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH) (b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3) or in water. (c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH) SOLUTION:

  12. Structure-Function Correlations: A Soap Soap: the salt form of a long-chain fatty acid; is amphipathic in character (has polar and non-polar components) Figure B13.1

  13. The mode of action of the antibiotic, Gramicidin A Destroys the Na+/K+ ion concentration gradients in the cell Figure B13.2

  14. Gas-Liquid Solutions Non-polar gas solubility in water is directly related to the boiling point of the gas. important to aquatic life

  15. Gas-gas solutions: All gases are infinitely soluble in one another. Gas-solid solutions: The gas molecules occupy the spaces between the closely packed particles of the solid. Solid-solid solutions: alloys (substitutional or interstitial)

  16. The arrangement of atoms in two types of alloys Figure 13.4

  17. solute (aggregated) + heat solute (separated) DHsolute > 0 solvent (aggregated) + heat solvent (separated) DHsolvent > 0 solute (separated) + solvent (separated) solution + heatDHmix < 0 Heats of solution and solution cycles Dissolution of a solid: breaking down the process into three steps 1. Solute particles separate from each other - endothermic 2. Solvent particles separate from each other - endothermic 3. Separate solute and solvent particles mix - exothermic

  18. Calculating the heat of solution, DHsoln The total enthalpy change that occurs when a solution forms by dissolving a solute into a solvent. DHsoln=DHsolute+DHsolvent+DHmix A thermochemical solution cycle

  19. Solution cycles and the enthalpy components of the heat of solution Figure 13.5

  20. H2O M+ (g) [or X- (g)] M+ (aq) [or X- (aq)] DHhydr of the ion < 0 M+ (g) + X- (g) MX(s) DHlattice is always (-) Heats of Hydration The solvation of ions by water is always exothermic. (for 1 mole of gaseous ions) DHhydr is related to the charge density of the ion, that is, both coulombic charge and ion size are important. Lattice energy is the DH involved in the formation of an ionic solid from its gaseous ions. Thus, DHsoln = -DHlattice + DHhydr

  21. Heats of Hydration and Ionic Character • For a given size, greater charge leads to a more (-) DHhydr • For a given charge, smaller size leads to a more (-) DHhydr

  22. Table 13.4Trends in Ionic Heats of Hydration ion ionic radius (pm) DHhydr (kJ/mol) Group 1A Li+ 76 -510 Na+ 102 -410 K+ 138 -336 Rb+ 152 -315 Cs+ 167 -282 Group 2A Mg2+ 72 -1903 Ca2+ 100 -1591 Sr2+ 118 -1424 Ba2+ 135 -1317 Group 7A F- 133 -431 Cl- 181 -313 Br- 196 -284 I- 220 -247

  23. Enthalpy Diagrams for Dissolving Three Different Ionic Compounds in Water NaCl NH4NO3 Figure 13.6 NaOH

  24. Entropy Considerations The natural tendency of most systems is to become more disordered; entropy increases. Entropy always favors the formation of solutions. Dissolution: involves a change in enthalpy and a change in entropy.

  25. Enthalpy diagrams for dissolving NaCl and octane in hexane NaCl in insoluble in hexane! In this case, dissolution is entropy-driven! Figure 13.7

  26. More Definitions When excess undissolved solute is in equilibrium with the dissolved solute: a saturated solution An unsaturated solution: more solute can be dissolved, ultimately producing a saturated solution A supersaturated solution: a solution that contains more than the equilibrium amount of dissolved solute

  27. solute (undissolved) solute (dissolved) Equilibrium in a saturated solution Figure 13.8

  28. Sodium acetate crystallizing from a supersaturated solution nucleation a saturated solution results Figure 13.9

  29. Solubility and Temperature Most solids are more soluble at higher temperatures. The sign of the heat of solution, however, does not predict reliably the effect of temperature on solubility; e.g., NaOH and NH4NO3 have DHsoln of opposite signs, yet their solubility in H2O increases with temperature.

  30. The relation between solubility and temperature for several ionic compounds Figure 13.10

  31. Gas Solubility in Water: Temperature Effects For all gases, DHsolute = 0, DHhydr < 0; thus, DHsoln < 0 solute(g) + water(l) saturated solution(aq) + heat Implications: gas solubility in water decreases with increasing temperature

  32. Thermal Pollution Leads to O2 deprivation in aquatic systems Figure 13.11

  33. Pressure Effects on Solubility Essentially zero for solids and liquids, but substantial for gases! gas + solvent saturated solution

  34. The effect of pressure on gas solubility gas volume is reduced; pressure (concentration!) increases; more collisions occur with liquid surface Figure 13.12

  35. Henry’s Law A quantitative relationship between gas solubility and pressure Sgas = kHx Pgas The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution. Implications for scuba diving! kH = Henry’s law constant for a gas; units of mol/L.atm

  36. PROBLEM: The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25 oC. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3 x 10-2 mol/L.atm at 25 oC. S = (3.3 x 10-2 mol/L.atm)(4 atm) = CO2 SAMPLE PROBLEM 13.2 Using Henry’s Law to calculate gas solubility PLAN: Knowing kH and Pgas, we can substitute into the Henry’s Law equation. SOLUTION: 0.1 mol / L

  37. amount (mol) of solute molarity (M) volume (L) of solution amount (mol) of solute molality (m) mass (kg) of solvent mass of solute parts by mass mass of solution volume of solute parts by volume volume of solution amount (mol) of solute mole fraction  amount (mol) of solute + amount (mol) of solvent Table 13.5Concentration Definitions concentration term ratio

  38. PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water? mole CaCl2 32.0 g CaCl2 x 110.98 g CaCl2 0.288 mole CaCl2 kg x 271 g H2O 103 g SAMPLE PROBLEM 13.3 Calculating molality PLAN: Convert grams of CaCl2 into moles and grams of water to kg. Then substitute into the equation for molality. SOLUTION: = 0.288 mole CaCl2 = 1.06 m CaCl2 molality =

  39. The Sex Attractant of the Gypsy Moth Potent at Extremely Low Concentrations! 100-300 molecules/mL air 100 parts per quadrillion by volume! Practical Implications: a strategy used to target and trap specific insects (Japanese beetles) Figure 13.13

  40. Other Expressions of Concentration mass percent (% w/w) = mass solute / mass of solution x 100 (related to parts per million (ppm) or parts per billion (ppb)) volume percent (% (v/v) = volume solute / volume of solution x 100 % (w/v) = solute mass / solution volume x 100 mole percent (mol%) = mole fraction x 100

  41. SAMPLE PROBLEM 13.4 Expressing concentration in parts by mass, parts by volume, and mole fraction PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50 g pill that contains 40.5 mg of Ca. (b) The label on a 0.750 liter bottle of Italian chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain? (c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? PLAN: (a) Convert mg to g of Ca, find the ratio of g Ca to g pill, and multiply by 106. (b) Knowing the % alcohol and the total volume, the volume of alcohol can be calculated. (c) Convert g of solute and solvent to moles, and find the ratios of each part to the total.

  42. 40.5 mg Ca x g x 106 103 mg x 0.750 L chianti 11.5 L alcohol 100. L chianti mole mole 18.02 g 60.09 g = 0.577 = 0.423 H2O C3H7OH SAMPLE PROBLEM 13.4 (continued) SOLUTION: (a) = 1.16 x 104 ppm Ca 3.5 g (b) = 0.0862 L alcohol (c) moles isopropyl alcohol = 142 g x = 2.36 mol C3H7OH x moles water = 58.0 g = 3.22 mol H2O 2.36 mol C3H7OH 3.22 mol H2O 2.36 mol C3H7OH + 3.22 mol H2O 2.36 mol C3H7OH + 3.22 mol H2O

  43. PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its: SAMPLE PROBLEM 13.5 Converting concentration units (a)molality (b)mole fraction (c)molarity PLAN: (a) To find the mass of solvent, assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution to determine the mass of solvent. (b) Convert g of solute and solvent to moles before finding c. (c) Use the density to find the volume of the solution. SOLUTION: (a) g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O mol H2O2 30.0 g H2O2 x 34.02 g H2O2 molality = = 12.6 m H2O2 kg H2O x 70.0 g H2O 103 g

  44. mol H2O 18.02 g H2O mL 1.11 g L 103 mL SAMPLE PROBLEM 13.5 (continued) (b) 70.0 g H2O x = 3.88 mol H2O 0.882 mol H2O2 = 0.185 = c of H2O2 0.882 mol H2O2 + 3.88 mol H2O (c) = 90.1 mL solution 100.0 g solution x 0.882 mol H2O2 = 9.79 M H2O2 90.1 mL solution x

  45. Colligative Properties Physical properties of solutions dictated by the number of solute particles present. Their chemical structures are not factors in determining these properties! • vapor pressure lowering • boiling point elevation • freezing point depression • osmotic pressure

  46. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. strong non-electrolyte weak Three types of electrolytes Figure 13.14

  47. Vapor Pressure Lowering The vapor pressure of a solution of a nonvolatile nonelectrolyte is always lower than the vapor pressure of the pure solvent. An entropy argument! Figure 13.15

  48. Quantitative Treatment of VP Lowering Raoult’s Law (vapor pressure of a solvent above a solution, Psolvent) Psolvent = csolvent x Posolvent where Posolvent = vapor pressure of the pure solvent How does the amount of solute affect the magnitude of the VP lowering? ( substitute 1- csolute forcsolventin the above equation and rearrange) Posolvent - Psolvent = DP = csolutex Posolvent (change in VP is proportional to the mole fraction of solute)

  49. PROBLEM: Calculate the vapor pressure lowering, DP, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50. oC. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL. PLAN: Find the mol fraction, c, of glycerol in solution and multiply by the vapor pressure of water. 0.988 g H2O 1.26 g C3H8O3 mol H2O mol C3H8O3 mL C3H8O3 mL H2O 18.02 g H2O 92.09 g C3H8O3 c = 0.00498 SAMPLE PROBLEM 13.6 Using Raoult’s Law to find the vapor pressure lowering SOLUTION: 10.0 mL C3H8O3 = 0.137 mol C3H8O3 x x x = 27.4 mol H2O 500.0 mL H2O x 0.137 mol C3H8O3 DP = x 92.5 torr = 0.461 torr 0.137 mol C3H8O3 + 27.4 mol H2O

More Related