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Jeffrey Mack California State University, Sacramento Chapter 14 Solutions and Their Behavior.

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1 Jeffrey Mack California State University, Sacramento Chapter 14 Solutions and Their Behavior

2 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

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

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

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

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

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

8 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. Concentration Units

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

10 m= mg solute kg solution Parts Per Million (ppm) Concentration Units Parts per million (ppm): grams of solute/grams of solution (then multiplied by 10 6 or 1 million)

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

12 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. Calculating Concentrations

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

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

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

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

17 17 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is g/mL? m =m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 g of solution (1000 mL x g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = kg m =m = moles of solute mass of solvent (kg) = 5.86 moles C 2 H 5 OH kg solvent = 8.92 m

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

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

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

22 Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)  lattice H Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.

23 Energetics of the Solution Process:  solution H Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)  lattice H 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 We can therefore represent the process of dissolving KF in terms of these chemical equations: Step 1: KF(s) → K + (g) + F  (g) =  lattice H Step 2: K + (g) + F  (g) → K + (aq) +F  (aq) =  hydration H Energetics of the Solution Process:  solution H

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

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

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

29 ENDOthermic Sodium acetate has an ENDOthermic heat of solution. heat NaCH 3 CO 2 (s) + heat  Na + (aq) + CH 3 CO 2 - (aq) EXOTHERMIC Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. heat Na + (aq) + CH 3 CO 2 - (aq)  NaCH 3 CO 2 (s) + heat Supersaturated Sodium Acetate

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

31 Gas solubility (mol/L) S g =K H  P g Factors Affecting: Solubility Pressure & Temperature— “Henry’s Law”

32 32 Temperature and Solubility O 2 gas solubility and temperature solubility usually decreases with increasing temperature Do you like your coke hot or cold? Why?

33 33 Chemistry In Action: The Killer Lake Lake Nyos, West Africa 8/21/86 CO 2 Cloud Released 1700 Casualties Trigger? earthquake landslide strong Winds

34 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 Colligative Properties

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

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

37 LIQUID-VAPOR EQUILIBRIUM Understanding Colligative Properties

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

39 P solution = X solvent  P o P solution = the vapor pressure of a mixture of solute and solvent P o = the vapor pressure of the pure solvent X solvent = 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. Raoult’s Law

40 40 P A = X A P A 0 P B = X B P B 0 P T = P A + P B P T = X A P A 0 + X B P B 0 Ideal Solution Special case of 2 Liquids

41 41 P T is greater than predicted by Raoults’s law P T is less than predicted by Raoults’s law Force A-B Force A-A Force B-B <& Force A-B Force A-A Force B-B >&

42 Vapor Pressure Lowering of Benzene + non volatile solute

43 Problem: Pure iodine (105 g) is dissolved in 325 g of CCl 4 at 65 °C. Given that the vapor pressure of CCl 4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl 4 –I 2 solution at 65 °C? (Assume that I 2 does not contribute to the vapor pressure.) Raoult’s Law

44 Problem: Pure iodine (105 g) is dissolved in 325 g of CCl 4 at 65 °C. Given that the vapor pressure of CCl 4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl 4 –I 2 solution at 65 °C? (Assume that I 2 does not contribute to the vapor pressure.)

45 Raoult’s Law Problem: Pure iodine (105 g) is dissolved in 325 g of CCl 4 at 65 °C. Given that the vapor pressure of CCl 4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl 4 –I 2 solution at 65 °C? (Assume that I 2 does not contribute to the vapor pressure.)

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

47 For boiling point elevation:K b > 0 (positive) For freezing point depression:K f < 0 (negative) Boiling Point Elevation & Freezing Point Depression Molecular Compounds

48 Boiling Point Elevation & Freezing Point Depression

49 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) compoundTypei CH 3 OHmolecular 1 NaClstrong electrolyte 2 Ba(NO 3 ) 2 strong electrolyte 3 HNO 2 weak electrolyte van’t Hoff Factor

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

51 LOWER The freezing point of a solution is LOWER than that of the pure solvent. Pure waterWater solution Change in Freezing Point

52 Water with and without antifreeze When a solution freezes, the solid phase is pure water. The solution becomes more concentrated. Lowering the Freezing Point

53 Problem: If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.) Freezing Point Depression

54 Problem: If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

55 Freezing Point Depression Problem: If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

56 Problem: Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl 3 ), the boiling point of the solution is °C. What is the molar mass of benzyl acetate? Molar Mass By Boiling Point Elevation

57 1.Solution: The molality of the solution can be found from the freezing point depression. 2.Knowing molality and mass of solvent, one can find the moles of solute. 3.Knowing mass and moles of solute, the molar mass of the compound can be determined. Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl 3 ), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

58 Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl 3 ), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

59 Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl 3 ), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

60 Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl 3 ), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

61 Dissolving the shell in vinegarEgg in corn syrupEgg in pure waterOsmosis

62 A semipermeable membrane allows only the movement of solvent molecules. Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute.Osmosis

63 Osmosis of solvent from one solution to another will occur as they try to equalize one another’s concentration. IsotonicAt the point where they have equal osmotic pressures, they are said to be Isotonic.Osmosis

64 Osmosis at the Particulate Level

65 Process of Osmosis

66 Equilibrium is reached when the internal pressure of the apparatus equals the external pressure of the open tube. Since the internal pressure, P int is greater then the atmospheric pressure, a column of liquid rises: P atm +P col = P int Recall that: P col = g  d  h Osmotic Pressure, 

67 Osmotic pressure ∏ P col is referred to as the Osmotic Pressure, ∏.   = cRT c = concentration of solute (mols/L) T = the absolute temperature (K)   is in units of atm. Osmotic Pressure, 

68 Osmosis & Living Cells

69 Water desalination plant Reverse Osmosis: Water Desalination

70 70 Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 0 Boiling-Point Elevation  T b = K b m Freezing-Point Depression  T f = K f m Osmotic Pressure (  )  = CRT Boiling-Point Elevation  T b = i K b m Freezing-Point Depression  T f = i K f m Osmotic Pressure (  )  = iCRT Electrolyte Solutions

71 Colloids In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution. In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension. In a colloid, the dispersed particles are on the order of 1– 1000 nm in size. Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely. A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

72 Colloidal dispersions scatter light, a phenomenon known as the “The Tyndall effect”Colloids

73 A hydrophobic colloid is stabilized by positive ions absorbed onto each particle and a secondary layer of negative ions. Because the particles bear similar charges, they repel one another, and precipitation is prevented. Hydrophobic Colloids

74 Soap molecules interact with water through the charged, hydrophilic end of the molecule. The long, hydrophobic end of the molecule binds through dispersion forces with nonpolar hydrocarbons and other non polar substances.Soap

75 An emulsifier (also known as an emulgent) is a substance which stabilizes an emulsion by increasing its kinetic stability. One class of emulsifiers is known as surface active substances, or surfactants. Examples of food emulsifiers are egg yolk (where the main emulsifying agent is lecithin) and honey. In some cases, particles can stabilize emulsions as well through a mechanism called Pickering stabilization. Both mayonnaise and Hollandaise sauce are oil-in- water emulsions that are stabilized with egg yolk lecithin or other types of food additives such as Sodium stearoyl lactylate. Emulsifiers & Surfactants


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