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Chapter 12: Solutions and Their Properties

Vanessa N. Prasad-Permaul CHM 1046 Valencia Community College. Chapter 12: Solutions and Their Properties. Introduction 1. A mixture is any intimate combination of two or more pure substances 2. Can be classified as heterogeneous or homogeneous Heterogeneous

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Chapter 12: Solutions and Their Properties

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  1. Vanessa N. Prasad-Permaul CHM 1046 Valencia Community College Chapter 12: Solutions and Their Properties

  2. Introduction 1. A mixture is any intimate combination of two or more pure substances 2. Can be classified as heterogeneous or homogeneous Heterogeneous -The mixing of components is visually nonuniform and have regions of different composition Homogenous -Mixing is uniform, same composition throughout -Can be classified according to the size of their particles as either solutions or colloids

  3. Solutions Solution 1. Homogeneous mixtures 2. Contain particles with diameters in the range of 0.1–2 nm 3. Transparent but may be colored 4. Do not separate on standing Colloids 1. Milk & fog 2. Diameters 2-500 nm 3. Do not separate on standing

  4. Types of Solutions

  5. Solution Formation Solute • Dissolved substance, or smaller quantity substance Solvent • Liquid dissolved in, larger quantity substance Saturated solution • Contains the maximum amount of solute that will dissolve in a given solvent.

  6. Solution Formation Unsaturated Contains less solute than a solvent has the capacity to dissolve. Supersaturated Contains more solute than would be present in a saturated solution. Crystallization The process in which dissolved solute comes out of the solution and forms crystals.

  7. SOLUTIONS EXAMPLE 12.1: A: GIVE AN EXAMPLE OF A SOLID SOLUTION PREPARED FROM A LIQUID AND A SOLID A dental filling made up of liquid mercury and solid silver is a solid solution B: GIVE AN EXAMPLE OF A LIQUID SOLUTION PREPARED BY DISSOLVING A GAS IN A LIQUID Aqueous ammonia

  8. SOLUTIONS • EXERCISE 12.1: IDENTIFY THE SOLUTE & SOLVENT IN THE FOLLOWING SOLUTIONS. • 80g of chromium & 5g of molybdenum • 5g of MgCl2 dissolved in 1000g of H2O • 39% N2, 41% Ar, and the rest O2

  9. Energy Changes and the Solution Process Three Types of interactions 1.      Solvent-solvent 2.      Solvent-solute 3.      Solute-solute “Like dissolves like” solutions will form when three types of interactions are similar in kind and magnitude

  10. Energy Changes and the Solution Process Example NaCl and water: Ionic solid NaCl dissolve in polar solvents like water because the strong ion-dipole attractions between Na+ and Cl- ions and polar water molecules are similar in magnitude to the strong dipole-dipole attractions between water molecules and to the strong ion-ion attractions between Na+ and Cl- ions Example Oil and water Oil does not dissolve in water because the two liquids have different kinds of intermolecular forces. Oil is not polar or an ionic solvent

  11. Energy Changes and the Solution Process NaCl in H2O 1. Ions that are less tightly held because of their position at a corner or an edge of the crystal are exposed to water molecules 2. Water molecules will collide with the NaCl until an ion breaks free 3. More water molecules then cluster around the ion, stabilizing it by ion-dipole attractions 4. The water molecules attack the weak part of the crystal until it is dissolved 5. Ions in solution are said to be solvated they are surrounded and stabilized by an ordered shell of solvent molecules

  12. SOLUBILITY EXAMPLE 12.2: WOULD NAPTHALENE (C10H8) BE MORE SOLUBLE IN ETHANOL OR IN BENZENE? EXPLAIN. Naphthalene is more soluble in benzene because nonpolar naphthalene must break the strong hydrogen bonds between ethanol molecules and replace them with weaker London forces.

  13. SOLUBILITY EXERCISE 12.2: WHICH OF THE FOLLOWING COMPOUNDS IS LIKELY TO BE MORE SOLUBLE IN WATER: C4H9OH or C2H9SH? EXPLAIN.

  14. Energy Changes and the Solution Process G, Free energy change 1. If G is negative the process is spontaneous, and the substance is dissolved 2. If G is positive the process is non-spontaneous, the substance is not dissolved 3. G = H -TS H, enthalpy, heat flow in or out of the system, Hsoln heat of solution S, entropy, disorder, Ssoln entropy of solution

  15. Energy Changes and the Solution Process

  16. Energy Changes and the Solution Process Ssoln Entropy of Solution Usually a positive number because when you dissolve something you are increasing disorder Hsoln Heat of Solution 1. Harder to predict because it could be exothermic (-Hsoln) or endothermic (+Hsoln) 2. The value of the heat of solution for a substance results from an interplay of the three kinds of interactions

  17. Energy Changes and the Solution Process • Solvent-solvent interactions: Energy is required (endothermic) to overcome intermolecular forces between solvent molecules because the molecules must be separated and pushed apart to make room for solute particles • Solute-solute interactions: Energy is required (endothermic) to overcome interactions holding solute particles together in a crystal. For an ionic solid, this is the lattice energy. Substances with higher lattice energies therefore tend to be less soluble than substances with lower lattice energies.

  18. Energy Changes and the Solution Process 3) Solvent-solute interactions: Energy is released (exothermic) when solvent molecules cluster around solute particles and solvate them. For ionic substances in water, the amount of hydration energy released is generally greater for smaller cations than for larger ones because water molecules can approach the positive nuclei of smaller ions more closely and thus bind more tightly. Hydration energy generally increases as the charge on the ion increases.

  19. Energy Changes and the Solution Process Exothermic -Hsoln The solute–solvent interactions are stronger than solute–solute or solvent–solvent. Favorable process Exothermic rxn.

  20. Energy Changes and the Solution Process Endothermic +Hsoln The solute–solvent interactions are weaker than solute–solute or solvent–solvent. Unfavorable process. Endothermic rxn

  21. Energy Changes and the Solution Process • Hydration • The attraction of ions for water molecules • Hydration Energy • The energy associated with the attraction between ions and water molecules • Lattice energy • The energy holding ions together in a crystal lattice

  22. HYDRATION ENERGY EXAMPLE 12.3: WHICH OF THE FOLLOWINGIONS WOULD BE EXPECTED TO HAVE THE GREATER ENERGY OF HYDRATION. Mg2+ OR Al3+? Al3+ because water molecules can approach the positive nuclei of smaller ions more closely and thus bind more tightly. Hydration energy generally increases as the charge on the ion increases.

  23. HYDRATION ENERGY EXERCISE 12.3: WHICH ION HAS THE LARGER HYDRATION ENERGY, Na+ or K+ EXPLAIN.

  24. Units of Concentration Molarity (M) M = mole of solute / Liter of solution Molality (m) m = moles of solute/mass of solvent (kg) Mole Fraction (x) X = mole of component / total moles

  25. CALCULATING THE MOLALITY OF SOLUTE EXAMPLE 12.6: GLUCOSE, C6H12O6, IS A SUGAR THAT IS IN FRUITS. IT IS ALSO FOUND IN BLOOD AND IS THE BODY’S MAIN SOURCE OF ENERGY. WHAT IS THE MOLALITY OF A SOLUTION CONTAINING 5.67g OF GLUCOSE DISSOLVED IN 25.2g OF WATER? 5.67g C6H12O6 x 1 mol C6H12O6 = 0.0315 mol C6H12O6 180.2g C6H12O6 0.0315 mol C6H12O6 =1.25 mC6H12O6 25.2g x 1kg 1000g

  26. CALCULATING THE MOLALITY OF SOLUTE EXERCISE 12.6: TOLUENE, C6H5CH3, IS A STARTING MATERIAL FOR TNT (TRINITROTLOUENE) . FIND THE MOLALITY OF TOLUENE IN A SOLUTION THAT CONTAINS 35.6g OF TOLUENE AND 125g OF BENZENE

  27. CALCULATING THE MOLE FRACTIONS OF COMPONENTS EXAMPLE 12.7: WHAT ARE THE MOLE FRACTIONS OF GLUCOSE AND WATER IN A SOLUTION CONTAINING 5.67g OF GLUCOSE DISSOLVED IN 25.2g OF WATER? 5.67g C6H12O6 x 1 mol C6H12O6 = 0.0315 mol C6H12O6 180.2g C6H12O6 25.2g H2O x 1 mol H2O = 1.40 mol H2O 18.0g H2O 1.40mol + 0.0315mol = 1.432mol MOLE FRACTION OF GLUCOSE 0.0315mol = 0.0220 x 100% = 2.20% 1.432mol MOLE FRACTION OF WATER 1.40mol = 0.978 x 100% = 97.8% 1.432mol

  28. CALCULATING THE MOLE FRACTIONS OF COMPONENTS EXERCISE 12.7: CALCULATE THE MOLE FRACTIONS OF TOLUENE AND BENZENE IN THE SOLUTION FROM EXERCISE 12.6

  29. Units of Concentration Mass Percent (mass %) Mass % = (mass of component / total mass of sol’n) x 100% Parts per million, ppm = (mass of component / total mass of solution) x 106 Parts per billion, ppb = (mass of component / total mass of solution) x 109

  30. CALCULATING WITH MASS PERCENTAGE OF SOLUTE EXAMPLE 12.5: HOW WOULD YOU PREPARE 425g OF AN AQUEOUS SOLUTION CONTAINING 2.40% BY MASS OF SODIUM ACETATE, NaC2H3O2? MASS OF SOLUTE: 0.0240 x 425g = 10.2g MASS OF WATER: 425g – 10.2g = 414.8 = 415g MASS OF SOLUTION: 425g YOU WOULD PREPARE THIS SOLUTION BY DISSOLVING 10.2g OF NaC2H3O2 IN 415g OF WATER FOR A TOTAL MASS OF 425g.

  31. CALCULATING WITH MASS PERCENTAGE OF SOLUTE EXERCISE 12.5: AN EXPERIMENT CALLS FOR 35.0g OF HYDROCHLORIC ACID THAT IS 20.2% HCl BY MASS. HOW MANY GRAMS OF HCl IS THIS? HOW MANY GRAMS OF WATER IS THIS?

  32. Some Factors Affecting Solubility Solubility The amount of solute per unit of solvent needed to form a saturated solution Miscible Mutually soluble in all proportions Effect of Temperature on Solubility 1. Most solid substances become more soluble as temperature rises 2. Most gases become less soluble as temperature rises

  33. Some Factors Affecting Solubility Effect of Pressure on Solubility 1. No effect on liquids or solids 2. The solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas over the solution, @ 25°C Henry’s Law solubility = k x P k = constant characteristic of specific gas, mol/Latm P = partial pressure of the gas over the sol’n

  34. Some Factors Affecting Solubility a) Equal numbers of gas molecules escaping liquid and returning to liquid b)   Increase pressure, increase # of gas molecules returning to liquid, solubility increases c) A new equilibrium is reached, where the #’s of escaping = # of returning

  35. APPLYING HENRY’S LAW EXAMPLE 12.4: 27g OF ACETYLENE (C2H2) DISSOLVES IN 1L OF ACETONE AT 1.0atm. IF THE PARTIAL PRESSURE OF ACETYLENE IS INCREASED TO 12atm, WHAT IS THE SOLUBILITY OF ACETONE? S2 = P2 S1 P1 S2 = 12atm 27g C2H2/L acetone 1.0atm S2 = 27g C2H2 x 12atm = 3.2x102g C2H2/L acetone L acetone 1.0atm 320g of acetylene will dissolve in 1L of acetone @ 12atm

  36. APPLYING HENRY’S LAW EXERCISE 12.4: A LITER OF WATER @ 250C DISSOLVES 0.0404g OF O2 WHEN THE PARTIAL PRESSURE OF THE OXYGEN IS 1.00atm. WHAT IS THE SOLUBILITY OF OXYGEN FROM AIR, IN WHICH THE PARTIAL PRESSURE OF O2 IS 159mmHg?

  37. Physical Behavior of Solutions: Colligative Properties Colligative properties Properties that depend on the amount of a dissolved solute but not its chemical identity There are four main colligative properties: 1. Vapor pressure lowering 2. Freezing point depression 3. Boiling point elevation 4. Osmotic pressure

  38. Physical Behavior of Solutions: Colligative Properties In comparing the properties of a pure solvent with those of a solution… 1. Vapor pressure of sol’n is lower 2. Boiling point of sol’n is higher 3. Freezing point of sol’n is lower 4. Osmosis, the migration of solvent molecules through a semipermeable membrane, occurs when solvent and solution are separated by the membrane

  39. Vapor-pressure Lowering of Solutions: Raoult’s Law 1.A liquid in a closed container is in equilibrium with its vapor and that the amount of pressure exerted by the vapor is called the vapor pressure. 2.  When you compare the vapor pressure of a pure solvent with that of a solution at the same temperature the two values are different. 3.  If the solute is nonvolatile and has no appreciable vapor pressure of its own (solid dissolved) the vapor pressure of the solution is always lower that that of the pure solvent. • If the solute is volatile and has a significant vapor pressure (2 liquids) the vapor pressure of the mixture is intermediate between the vapor pressures of the two pure liquids.

  40. Solutions with a Nonvolatile Solute When solute molecules displace solvent molecules at the surface, the vapor pressure drops since fewer gas molecules are needed to equalize the escape rate and capture rates at the liquid surface.

  41. Solutions with a Nonvolatile Solute!!! Raoult’s Law Psoln = Psolv· Xsolv Psoln = vapor pressure of the solution Psolv = vapor pressure of the pure solvent Xsolv = mole fraction of the solvent in the solution

  42. Raoult’s Law applies to only Ideal solutions 1.   Law works best when solute concentrations are low and when solute and solvent particles have similar intermolecular forces. 2.  If intermolecular forces between solute particles and solvent molecules are weaker than solvent molecules alone, solvent molecules are less tightly held, vapor pressure is higher than Raoult predicts 3.  If intermolecular forces between solute and solvent are stronger than solvent alone, solvent molecules are more tightly held and the vapor pressure is lower than predicted 4.      No Van’t Hoff factor!!! is a measure of the effect of a solute upon colligative properties

  43. Solutions with a Nonvolatile Solute Close-up view of part of the vapor pressure curve for a pure solvent and a solution of a nonvolatile solute. Which curve represents the pure solvent, and which the solution? Why?

  44. The lower vapor pressure of a sol’n relative to that of a pure solvent is due to the difference in their entropies of vaporization, Svap. Because the entropy of the solvent in a sol’n is higher to begin with, Svap is smaller for the sol’n than for the pure solvent. As a result vaporization of the solvent from the sol’n is less favored (less negative Gvap), and the vapor pressure of the solution is lower.

  45. Solutions with a Volatile Solute!! Ptotal = PA + PB Ptotal = (P°A · XA) + (P°B · XB) P°A = vapor pressure of pure A XA = mole fraction of A P°B = vapor pressure of pure B XB = mole fraction of B Ptotal should be intermediate to A & B

  46. Close-up view of part of the vapor pressure curves for two pure liquids and a mixture of the two. Which curves represent the mixture? • Red • Green • Blue

  47. CALCULATING VAPOR-PRESSURE LOWERING EXAMPLE 12.12: CALCULATE THE VAPOR-PRESSURE WHEN 5.67g OF GLUCOSE IS DISSOLVED 25.2g OF WATER @ 25oC. THE VAPOR PRESSURE OF WATER @ 25oC IS 23.8mmHg. WHAT IS THE VAPOR PRESSURE OF THE SOLUTION? P = PAOXB = 23.8 mmHg x 0.0220 = 0.524 mmHg THE VAPOR PRESSURE OF THE SOLUTION IS: PA= PAO- P = (23.8mmHg – 0.524mmHg) = 23.3mmHg

  48. CALCULATING VAPOR-PRESSURE LOWERING EXERCISE 12.12: NAPTHALENE IS USED TO MAKE MOTHBALLS. A SOLUTION IS MADE BY DISSOLVING 0.515g OF NAPTHALENE IN 60.8g OF CHLOROFORM. CALCULATE THE VAPOR PRESSURE LOWERING OF CHLOROFORM @ 20oC FROM NAPTHALENE. THE VAPOR PRESSURE OF CHLOROFORM @ 20oC IS 156mmHg. NAPTHALENE CAN BE ASSUMED TO BE NON-VOLATILE COMPARED WITH CHLOROFORM. WHAT IS THE VAPOR PRESSURE OF THE SOLUTION?

  49. Boiling Point Elevation and Freezing Point Depression of Solutions

  50. Boiling Point Elevation and Freezing Point Depression of Solutions 1. Red line is pure solvent 2. Green line solution of nonvolatile solute 3. Vapor pressure of sol’n is lower 4. Temp at which vapor pressure = 1 atm for sol’n is higher 5. Boiling point of sol’n is higher by Tb 6. Liquid/vapor phase transition line is lower for sol’n 7. Triple point temp is lower for sol’n 8. Solid/liquid phase transition has shifted to a lower temp. 9. The freezing point of the sol’n is lower by Tf

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