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Chapter 14 Solutions and Their Behavior

Chapter 14 Solutions and Their Behavior. Solutions. A solution is a homogeneous mixture of two or more substances in a single phase. By convention, the component present in largest amount is identified as the solvent and the other component(s) as the solute (s). Solutions.

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Chapter 14 Solutions and Their Behavior

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  1. Chapter 14Solutions and Their Behavior

  2. Solutions • A solution is a homogeneous mixture of two or more substances in a single phase. • By convention, the component present in largest amount is identified as the solvent and the other component(s) as the solute(s)

  3. Solutions • Solutions can be classified as saturatedor unsaturated. • A saturated solution contains the maximum quantity of solute that dissolves at that temperature.

  4. Solutions • An unsaturated solution can still take on more solute at a given temperature. • SUPERSATURATED SOLUTIONScontain more than is possible and are unstable.

  5. Concentration Units Molarity (M) moles solute M = L of solution

  6. Concentration Units x 100% mass of solute mass of solute + mass of solvent Weight (mass) % mass solute  100 % Total mass of solution = % by mass =

  7. Concentration Units Mole Fraction (X) moles solute (i) Xi = Total moles in solution

  8. Concentration Units Mole Fraction (X) • Where “i” is the moles of one component of the solute. • Total moles are all species: • mols solute (i) + mols solvent. • The sum of all of the mole fractions for each component are 1 exactly.

  9. Concentration Units moles of solute liters of solution moles of solute m = mass of solvent (kg) M = Molarity(M) Molality(m)

  10. Concentration Units Parts Per Million (ppm) mg solute m = kg solution • Parts per million (ppm): grams of solute/gramsof solution (then multiplied by 106 or 1 million)

  11. Solution Concentration Amount of solute Most concentration units are expressed as: Amount of solvent or solution • Molarity: molesof solute/literof solution • Percent by mass: grams of solute/gramsof solution (then multiplied by 100%) • Percent by volume: milliliters of solute/millilitersof solution (then multiplied by 100%) • Mass/volume percent: gramsof solute/milliliters of solution (then multiplied by 100%) 11

  12. Calculating Concentrations Problem: 62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water. Calculate mol fraction, molality, and weight % of the solution.

  13. Calculating Concentrations Problem: 62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water. Mole Fraction:

  14. Calculating Concentrations Problem: 62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water. Mole Fraction:

  15. Calculating Concentrations Problem: 62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water. Molality:

  16. Calculating Concentrations Problem: 62.1 g (1.00 mol) of ethylene glycol is dissolved in 250. g of water. Wt. %:

  17. 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 What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? 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 17

  18. The Solution Process • Solutes dissolve in solvents by a process called solvation. • Polar solvent dissolve polar solutes, non-polar solvents dissolve non-polar solutes. (aka: “like dissolves like”. • If two liquids mix to an appreciable extent to form a solution, they are said to bemiscible. • In contrast, immiscible liquids do not mix to form a solution; they exist in contact with each other as separate layers.

  19. The Solution Process

  20. Solvation of Ions + When a cation exists in solution, it is surrounded by the negative dipole ends of water molecules. When as anion exists in solution, it is surrounded by the positive dipole ends of water molecules.

  21. Energetics of the Solution Process: solutionH Energy must be supplied to separate the ions from their attractive forces. (an endothermic process) latticeH

  22. Energetics of the Solution Process: solutionH Energy must be supplied to separate the ions from their attractive forces. (an endothermic process) latticeH Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.

  23. Energetics of the Solution Process: solutionH Energy must be supplied to separate the ions from their attractive forces. (an endothermic process) latticeH Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation. This process, referred to as the Energy of Hydration when water is the solvent, is strongly exothermic.

  24. Energetics of the Solution Process: solutionH We can therefore represent the process of dissolving KF in terms of these chemical equations: Step 1: KF(s) → K+(g) + F(g) = latticeH Step 2: K+(g) + F(g) → K+(aq) +F(aq) =hydrationH

  25. Energetics of the Solution Process: solutionH The overall reaction is the sum of these two steps. The enthalpy of the overall reaction, called the enthalpy of solution (solnH), is the sum of the two enthalpies. Overall: KF(s)→ K+(aq) + F(aq) solnH =latticeH+ hydrationH

  26. Energetics of the Solution Process: solutionH

  27. Energetics of the Solution Process If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic!

  28. Supersaturated Sodium Acetate • One application of a supersaturated solution is the sodium acetate “heat pack.” • The enthalpy of solution for sodium acetate is ENDOthermic.

  29. Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH3CO2 (s) + heatf Na+(aq) + CH3CO2-(aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na+(aq) + CH3CO2-(aq) f NaCH3CO2 (s) + heat

  30. solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility Solid solubility and temperature 30

  31. Factors Affecting: Solubility Pressure & Temperature— “Henry’s Law” Gas solubility (mol/L) Sg =KHPg

  32. Temperature and Solubility O2 gas solubility and temperature solubility usually decreases with increasing temperature Do you like your coke hot or cold? Why? 32

  33. Chemistry In Action: The Killer Lake 8/21/86 CO2 Cloud Released 1700 Casualties • Trigger? • earthquake • landslide • strong Winds Lake Nyos, West Africa 33

  34. Colligative Properties • Relative to a pure solvent, a solution has: • Lower vapor pressure • Higher boiling point • Lower freezing point • A higher osmotic pressure • Example: Pure water: b.p.=100°C f.p.= 0°C 1.00 m NaCl (aq) b.p.= 101°C f.p. = -3.7 °C

  35. Colligative properties Upon adding a solute to a solvent, the properties of the solvent are affected: • Vapor pressure decreases • Melting point decreases • Boiling point increases • Osmosis is possible (osmotic pressure) Collectively these changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

  36. Understanding Colligative Properties To understand colligative properties, one must consider the LIQUID-VAPOR EQUILIBRIUM for a solution.

  37. Understanding Colligative Properties LIQUID-VAPOR EQUILIBRIUM

  38. Changes in Vapor Pressure: Raoult’s Law • Vapor pressure: pressure exerted by the vapor when a liquid in a closed container is at equilibrium with the vapor. • Vapor pressure of solution is decreasedby the presence of a solute. • More solute particles, lower vapor pressure of solution.

  39. Raoult’s Law Psolution = Xsolvent Po Psolution = the vapor pressure of a mixture of solute and solvent Po = the vapor pressure of the pure solvent Xsolvent = the mole fraction of the solvent. The expression can also be written in the form: P is the change to the vapor pressure of the pure solvent.

  40. 0 0 0 0 PA = XA P A PB = XB P B PT = XA P A +XB P B Ideal Solution PT = PA + PB Special case of 2 Liquids 40

  41. Force A-A Force A-A Force B-B Force B-B Force A-B Force A-B < > & & PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law 41

  42. Vapor Pressure Lowering of Benzene + non volatile solute

  43. Raoult’s Law Problem: Pure iodine (105 g) is dissolved in 325 g of CCl4 at 65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

  44. Raoult’s Law Problem: Pure iodine (105 g) is dissolved in 325 g of CCl4 at 65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

  45. Raoult’s Law Problem: Pure iodine (105 g) is dissolved in 325 g of CCl4 at 65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

  46. Boiling Point Elevation & Freezing Point Depression Molecular Compounds The temperature of the normal boiling point of a solution is increased by: The temperature of the normal freezing point of a solution is decreasedby: Where the K’s are the respective boiling and freezing point constants and msolute is the molality of the solution.

  47. Boiling Point Elevation & Freezing Point Depression Molecular Compounds For boiling point elevation: Kb > 0 (positive) For freezing point depression: Kf < 0 (negative)

  48. Boiling Point Elevation & Freezing Point Depression

  49. van’t Hoff Factor Boiling and Freezing point effects involving ions: When solutions containing ions are involved, the total concentration of solute particles must be considered. The change in b.p. or f.p. is given by the equation: Where m is the calculated molaity based on formula wt. “i” = the number of ions (van’t Hoff factor) compound Type i CH3OH molecular 1 NaCl strong electrolyte 2 Ba(NO3)2 strong electrolyte 3 HNO2 weak electrolyte 1 - 2

  50. The Boiling Point of a Solution is Higher Than That of a Pure Solvent

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