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Lecture Notes Chem 150 - K. Marr

Lecture Notes Chem 150 - K. Marr. Chapter 13 Properties of Solutions Silberberg 3 ed. Properties of Solutions. 13.1 Types of Solutions: IMF’s and Predicting Solubility 13.2 Energy Changes in the Solution Process 13.3 Solubility as an Equilibrium Process

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Lecture Notes Chem 150 - K. Marr

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  1. Lecture Notes Chem 150 - K. Marr Chapter 13 Properties of Solutions Silberberg 3 ed

  2. Properties of Solutions 13.1 Types of Solutions: IMF’s 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 Will not cover: 13.6 The Structure and Properties of Colloids

  3. Definitions • Solution • homogeneous mixture with only one phase present • Mixture • 2 or more substances physically mixed together • Composition is variable • Properties of components are retained

  4. Formation of Solutions • Driving Force: • Tendency toward Randomness(T1) • Nature favors processes that result in an increase in entropy (more randomness or less order) • Explains why solutions of gases always form . • Why don’t solutions always form with solids and liquids? • Must consider IMF’s

  5. Formation of Solutions • Solute and solvent particles must be attracted to one another: “Like Dissolves Like” • For a solution for form….. Solute-solvent IMF’s > Solvent-Solvent & solute-solute IMFS

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

  7. The arrangement of atoms in two types of alloys Solid-solid solutions: alloys (substitutional or interstitial) Figure 13.4

  8. Solutions involving Liquids • Molecules of each liquid must be pushed apart for a solution to form • Why doesn’t water form a solution with n-Hexane, C6H14? • Water molecules too strongly attracted to each other to be pushed aside to make room for hexane molecules

  9. Application Questions:Solutions involving Liquids • Why does a solution form between water and ethanol? • H-Bonding between water and ethanol molecules is responsible for solution formation • Allows water molecules to be pushed aside to make room for ethanol molecules • Why do all nonpolar liquids mix to form solutions? • e.g. Oil and n-Hexane

  10. SAMPLE PROBLEM 13.1 • Predicting Relative Solubilities of Substances • Predict which solvent will dissolve more of the given solute: (a) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3) or in water. • Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH) • Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH) PLAN: Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute.

  11. Solids Dissolved in Liquids • Solid particles must separate for a solution to form(T2) • Solute must be attracted to solvent • Ion-Dipole attractions • Solvation vs Hydration • What happens if the attractive forces within solute and solvent differ greatly? • e.g. Hexane and NaCl

  12. Figure 13.1 The major types of intermolecular forces in solutions

  13. Figure 13.2 Hydration shells around an aqueous ion

  14. What determines if the DHsoln for a solid with a liquid is exo- or endothermic?(T3-5) • DHsoln = Lattice Energy + Solvation Energy • Lattice Energy • E needed to separate particles: Always Endothermic • Solvation Energy • E released upon solvation • Always Exothermic

  15. Figure 13.5 Solution cycles and the enthalpy components of the heat of solution B. Endothermic Solution Process A. Exothermic Solution Process

  16. solute (aggregated) + heat solute (separated) DHsolute > 0 solvent (aggregated) + heat solvent (separated) DHsolvent > 0 solute (separated) + solvent (separated) solution + heat DHSolv < 0 Heats of solution and solution cycles 1. Solute particles separate from each other - endothermic 2. Solvent particles separate from each other - endothermic 3. Solute and solvent particles mix - exothermic DHsoln = DHsolute + DHsolvent + DHsolvation

  17. Figure 13.6 Heats of Solution: Dissolving ionic compounds in water NaCl DHsoln = + NH4NO3 DHsoln = + NaOH DHsoln = -

  18. Figure 13.7 Enthalpy diagrams for dissolving NaCl and octane in hexane

  19. Table 13.3 Trends in Heats of Hydration for Various Ions

  20. Heats of Solution: Application Questions • Why is the DHsoln always exothermic (negative) for solutions between gases and liquids? • Why is the DHsoln = 0 for solutions between gases? • Why is the dissolving of Calcium Chloride, CaCl2, in water exothermic? • CaCl2 (along with NaCl) are used to salt roads • Why is the dissolving of ammonium nitrate, NH4NO3, in water endothermic? • NH4NO3 is used in chemical cold packs

  21. The Effect of Temperature on Solubility Objectives: • Describe the effect of temperature on solubility of gases, liquids, and solids in liquids

  22. Solubility • Equation describing a saturated solution at equilibrium Solute + Solvent Saturated Solution • Most Common Units: • Mass solute/100 g solvent at a given temperature

  23. Solute (undissolved)Solute (dissolved) Figure 13.8 Equilibrium in a saturated solution

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

  25. Effect of Temp. on the Solubility of a Gas in a Liquid • Solubility of a gas always decreases as Temp. increases. Why?? • Le Chatelier’s Principle is used to predict how an increase in temp. affects the solubility of a gas in a liquid. • Recall DHsoln is exothermic for all gases in a liquid: Gas + Liquid Gas dissolved + E

  26. Solubility of a Gas in a Liquid: Applications • Thermal Pollution  Decreases O2 Solubility • Streams and Rivers • Salmon/Trout habitat restoration • Deep lakes • Warm water at surface, cold water deep • Why are the richest fisheries in the coldest waters of the world?

  27. Solubility of a Gas in a Liquid: Application Questions • Why do bubbles form on the side of a glass of water? What do these bubbles consist of? • Why do sodas get flat as they sit? • Why are some ice cubes clear, others cloudy? • How are clear ice cubes made?

  28. gas volume decreases • gas pressure increases • more collisions with liquid surface • gas solubility increases gas + solvent solution

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

  30. Application of Henry’s Law • Why does a soda start to bubble immediately after opening the bottle? • The solubility of methane , the chief component in natural gas, in water at 20.0 oC and 1.0 atm pressure is 0.025 g/L. What will the solubility be at 1.5 atm pressure and 20.0 oC ? • Answer: 0.038 g/L

  31. Section 13.4Concentrations of Solutions • Be able to convert from one concentration unit to another: • Molarity: Review Section 3.5 • Molality • Mass Fraction and Mass Percent • Mole fraction and Mole Percent • Practice!! Practice!! Practice!!

  32. Concentrations of Solultions: Molarity, M Molarity = moles of solute divided by Liters of Solvent M = mol solute / L of solution • Used in stoichiometric calculations involving solutions since V x M = moles • Since Volume varies with temperature, Molarity varies w/ temperature

  33. Concentrations of Solutions: Molality, m • Molality =Moles of solute divided by kg of Solvent m = mol solute / kg solvent • Does not change with Temp. • Used in BP elevation and FP depression calculations

  34. Molality Practice #1 Water freezes at a lower temperature when it contains solutes. • Calculate the number of grams of methanol, CH3OH, needed to prepare a 0.250 msolution, using 2000. grams of water. Methanol: 32.00 g/mol. Ans. = 16.0 g methanol • Calculate the Freezing point of 0.250 m Methanol. Ans. = - 0.465 oC

  35. FP Lowering DTf = Kf m DTf = amount FP is lowered Kf = Freezing point depession constant Kf is solvent Dependent 1.86 oC/m for water 5.07 oC/m for benzene 20.0 oC/m for cyclohexane

  36. Molality Practice #2 • If you prepare a solution by dissolving 4.00 g of NaOH in 250. g of water, what is the molality of the solution? NaOH: 40.00 g/mol Ans. = 0.400 m • At what temperature would the solution boil? Ans. = 100.2 oC

  37. BP Elevation DTb = Kb m • Kb = Boiling point elevation constant • Kb is solvent Dependent 0.51 oC/m for water 2.53 oC/m for benzene 2.69 oC/m for cyclohexane

  38. Parts by Mass (or Percent by Mass) • Mass of component divided by total mass of solution Mass % = (msolute / m solution) x100 • Also known as weight percent: w/w or % w/w

  39. Mass Percent Practice #1 • How many grams of NaOH are needed to prepare 250.0g of 1.00% NaOH in water? Ans. = 2.50 g NaOH • How many grams of water are needed?Ans. = 247.5 g • How many mL of water at 20.0 oC are needed?Ans.= 250.455 mL • dwater at 20.0 oC = 0.9882 g/mL • What is the molality of the solution? Ans.= 0.2525 m NaOH • What is the approximate FP of the solution? Ans. = -0.939 oC

  40. Mass Percent Practice #2 Concentrated hydrochloric acid can be purchased from chemical supply houses as a solution that is 37% HCl by mass. HCl = 36.46 g/mol • What mass of conc. HCl is needed to make 1.0 liter of 0.1 M HCl? Ans. = 9.854 or 10 g conc HCl • How would you make the 0.1 M HCl solution?

  41. Variations of % by Mass:ppm and ppb • Use ppm and ppb when concentrations of solute are very low • Parts per million • ppm = mass fraction x 106 • Parts per billion • ppb = mass fraction x 109

  42. Concentration Unit Conversion Problems • Strategies......... • Determine the Units of Concentration involved • What are the units you are starting with? • What are the units you are converting to? • Figure our what conversion factors are needed to go get you to the desired units of concentration

  43. Concentration Unit Conversion Practice #1 • Calculate the molality 2.00 % NaCl. NaCl = 58.4425 g/mol Ans. = 0.349 m NaCl • How would you prepare • 1.00 liter of 2.00 % NaCl (w/v)? • 500. mL of 2.00 % NaCl (w/v)? • 250. mL of 2.00 % NaCl (w/v)?

  44. Concentration Unit Conversion Practice #2 Conc. hydrobromic acid can be purchased as 40.0% HBr. The density of the solution is 1.38 g/mL. • What is the molar concentration of 40.0% HBr? HBr = 80.912 g/mol Ans. = 6.82 M HBr

  45. Colligative Properties Properties of a solution that depend on the number of solute particles, not on their identity: • Vapor Pressure Lowering • Freezing Point Lowering • Boiling Point Elevation • Osmotic Pressure (will not cover)

  46. Vapor Pressures of Solutions Which is higher, the vapor pressure of salt water or that of pure water?

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