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Physical Properties of Solutions

Physical Properties of Solutions. Chapter 13. Outline of the Chapter. The fundamental properties of Solutions Types Energetics Working with solutions Concentration Units Saturated Solutions and Equilibrium Colligative Properties Freezing point depression Boiling point elevation.

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Physical Properties of Solutions

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  1. Physical Properties of Solutions Chapter 13

  2. Outline of the Chapter • The fundamental properties of Solutions • Types • Energetics • Working with solutions • Concentration Units • Saturated Solutions and Equilibrium • Colligative Properties • Freezing point depression • Boiling point elevation

  3. Definitions to Know Solution Solute Solvent Saturated, unsaturated, supersaturated solutions Solubility Dynamic Equilibrium Types of solutions

  4. Solutions • Solutions are homogeneous mixtures of two or more pure substances. • Solute is dispersed uniformly throughout the solvent ( we will deal with water solutions). • Liquids that mix in all proportions are called miscible

  5. 12.1 Some Types Of Solutions 12.1

  6. 12.1 Types of Solutions • Saturated • Solvent holds as much solute as is possible (maximum) at that temperature. • Dissolved solute is in dynamic equilibrium with solid solute particles. Dynamic equilibrium: rate of crystallization = rate of dissolving Solubility: concentration of saturated solution.

  7. 12.1 Types of Solutions • Unsaturated • Less than the maximum amount of solute for that temperature is dissolved in the solvent.

  8. 12.1 Types of Solutions • Supersaturated • Solvent holds more solute than is normally possible at that temperature. • Solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask. • Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

  9. A Molecular View of the Solution Process

  10. Intermolecular Forces In Solution Formation 1. In the following processes differentiate between physical process that leads to solution formation and chemical reaction a) Ni(s) + 2HCl → NiCl2(aq) + H2(g) b) NaCl (s) → Na+(aq) + Cl-(aq) c) CH3OH(l) → CH3OH(aq)

  11. Intermolecular Forces In Solution Solute-solute Solute-solvent Solvent-solvent

  12. Formation Of Solution: Energy Considerations Energy needed to break the: solute-soluteinteraction: ΔH1 > 0 Solvent-solvent interactions: ΔH2 > 0 Energy released when: Solvent-solute interaction ΔH3 < 0 DHsoln = DH1 + DH2 + DH3

  13. Energy Changes in Solution The enthalpy change of the overall process depends on H for each of these steps.

  14. Why Do Endothermic Processes Occur? Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

  15. Why Do Endothermic Processes Occur? Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

  16. Enthalpy Is Only Part of the Picture The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

  17. Enthalpy Is Only Part of the Picture So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

  18. Intermolecular Forces In Solution Formation • Ideal Solution: All intermolecular forces are of comparable strengthΔHsoln = 0. (N2 and O2; gasoline; benzene + toluene) Enthalpy driven dissolution: • Intermolecular forces between solute and solvent molecules are stronger than other intermolecular forces, |ΔH3|> ΔH1 +ΔH2, ΔHsoln < 0(exothermic) MgSO4, ΔHsolution = -92. kJ/mol Sometimes, volume of solution is smaller than volume of individual components (C2H5OH + H2O)

  19. Intermolecular Forces In Solution Formation Entropy driven dissolution: • Intermolecular forces between solute and solvent molecules are weaker than other intermolecular forces, • |ΔH3|< ΔH1 +ΔH2, • ΔHsoln > 0Endothermic. Entropy driven. • NH4NO3,ΔH = 26.4 kJ/mol • Intermolecular forces between solute and solvent are much weaker (ΔH1 + ΔH2 << |ΔH3| ) than other intermolecular forces, the solute does not dissolve in the solvent. The compound is relatively insoluble in the solvent.

  20. Example • Predict the relative solubilities in the following case: • Bromine in benzene (μ = 0 D) • Bromine in water (μ = 1.87 D) • KCl in carbon tetrachloride (μ = 0 D) • KCl in liquid ammonia (μ = 1.46 D) • Formaldehyde (CH2O) in carbon disulfide CS2, μ = 0 D) • Formaldehyde (CH2O) in water (μ = 1.87 D) • Urea, (NH2)2CO in carbon disulfide • Urea in water

  21. Factors Affecting Solubility • Chemists use the axiom “like dissolves like”: • Polar substances tend to dissolve in polar solvents. • Nonpolar substances tend to dissolve in nonpolar solvents: CCl4 in C6H6

  22. Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

  23. Liquid-Liquid Solutions • Miscible: completely soluble • Immiscible: do not dissolve in each other • Solvent: component with greater concentration • Like dissolves like • Acetic acid dimers

  24. Solutions of Solids in Liquids Things to know: • aqueous solutions • hydration • polar-polar solutions • Solvation • Crystallization • Solubility of metals

  25. Aqueous Solutions • Two forces - Break ionic attractions(positive and negative ions) in the solid - Form ion-dipole forces between solute particles and solvent particles (form hydration shell). Solubility determined by the strength of the solid-solid IMfs and solute-solvent interactions Network and metallic + water: does not happen

  26. Ion-Dipole Forces in Dissolution Hydration shell 12.4

  27. Example 3. Predict whether each of the following is likely to be a solution or a heterogeneous mixture. a) Ethanol, CH3OH, and water, HOH b) Pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3 c) Sodium chloride, NaCl, and carbon tetrachloride, CCl4 d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH 12.4

  28. Examples 4. Some solution processes are endothermic and some are endothermic. Provide a molecular interpretation for these differences. 5. Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible? Give examples. Answers: • (IMFs: solid-solid; liqud-liquid, solid-liquid; Energy diagrams) • (miscible: completely soluble in all proportions; IMFs comparable in strength, butane in octane; methanol in water, describe the IMFs))

  29. Ways of Expressing Concentrations of Solutions

  30. Concentration Units • Things to know: • Percent by mass • Mole fraction • Molarity • Molality • Comparison of Concentration Units • Parts per million (ppm) • Parts per billion (ppb)

  31. Mass Percentage mass of A in solution total mass of solution Mass % of A =  100

  32. Parts per Million andParts per Billion mass of A in solution total mass of solution mass of A in solution total mass of solution ppm = Parts per Million (ppm), 1 ppm = 1 mg/L  106 Parts per Billion (ppb), 1 ppb = 1 µg/L  109 ppb =

  33. Mole Fraction (X) moles of A total moles in solution XA = • In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need! • xi < 1; x1 + x2 + x3 + … = 1

  34. Molarity (M) mol of solute L of solution M = • Volume is temperature dependent. Therefore • molarity can change with temperature.

  35. Molality (m) mol of solute kg of solvent m = • Moles and mass do not change with temperature. Therefore, • Molality (unlike molarity) is not temperature dependent.

  36. Concentration Units: Examples 6. A sample of 0.892 g KCl dissolved in 54.6 g H2O. Calculate % by mass. 7. A solution is prepared by mixing 200.4 g C2H5OH to 143.9 g of H2O. Calculate mole fractions of both ingredients of the solution. (Xalcohol 0.3539; xH2O = 0.647 8. Calculate molality of H2SO4 solutions containing 24.4 g H2SO4 in 198 g of H2O. (1.26 m)

  37. Example 9. How many milliliters of water (d = 0.998 g/mL) are required to dissolve 25.0 g of urea and thereby produce a 1.65 m solution of urea, CO(NH2)2? (253 mL of H2O) 10. The density of a 2.45 M aqueous methanol solution is 0.976 g/mL. What is the molality of the solution? (2.73 m) 11. Calculate the molality of a 34.5 % (by mass) aqueous solution of phosphoric acid, H3PO4. The molar mass of phosphoric acid is 98.00 g/mol. (5.37 m)

  38. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent 12. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? (8.92 m) Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m

  39. Example (H@P, 516) 13. An aqueous solution of ethylene glycol used as an automobile engine coolant is 40.0% HOCH2CH2OH by mass and has a density of 1.05 g/mL. What are the (a) molarity, (b) molality, and (c) mole fraction of HOCH2CH2OH. (a. 6.77 M b. 10.7 m c. 0.162)

  40. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

  41. Effect of Temperature on Solubility • Solubility: the amount of solute that will dissolve in a given amount of solvent at a given temperature • Solubilities of ionic compounds increase significantly with increasing temperature(About 95% of compounds). Rest do not change or very slightly (NaCl). • A very few have solubilities that decrease with increasing temperature. • A supersaturatedsolution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute. 12.6

  42. Temperature and Solubility Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature. Exceptions?

  43. Fractional Crystallization When KNO3(s) is crystallized from an aqueous solution of KNO3 containing CuSO4 as an impurity, CuSO4 remains in the solution. 12.6

  44. Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: • Dissolve sample in 100 mL of water at 600C • Cool solution to 00C • All NaCl will stay in solution (s = 34.2g/100g) • 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 12.6

  45. Solubility: Examples 15. The solubility of ammonium formate, NH4HCO2, in 100.0 g of water is 102.0 g at 0ºC and 546 g at 80ºC. A solution is prepared by dissolving NH4HCO2 in 200.0 g of water until no more can dissolve at 80ºC. The solution is then cooled to 0ºC. What mass of ammonium formate precipitates? (Assume that no water evaporates and the solution is not supersaturated.

  46. Gases in Solution • In general, the solubility of gases in water increases with increasing mass. • Larger molecules have stronger dispersion forces.

  47. Gas Solubility and Temperature 12.6

  48. Effect of Pressure on Solubility of Gases Things to know: Henry’s law Molecular Interpretation of Henry’s Law Check online tutorial

  49. Gases in Solution • The solubility of liquids and solids does not change appreciably with pressure. • The solubility of a gas in a liquid is directly proportional to its pressure.

  50. Pressure and Solubility of Gases Explain the pictures.

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