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 Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. See Chapter 14 Video Presentation Slide 5

Solutions See Chapter 14 Video Presentation Slide 6 An unsaturated solution can still take on more solute at a given temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

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

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

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

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.

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

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

Solution Concentration Amount of solute Most concentration units are expressed as: Amount of solvent or solution 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%) 11

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.

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

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

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

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

5.86 moles ethanol = 270 g ethanol What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? 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 0.927 g/mL) mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m 17

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 be miscible. In contrast, immiscible liquids do not mix to form a solution; they exist in contact with each other as separate layers.

The Solution Process

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.

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

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.

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.

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

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

Energetics of the Solution Process: solutionH

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! See Chapter 14 Video Presentation Slide 7

Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.” The enthalpy of solution for sodium acetate is ENDOthermic. See Chapter 14 Video Presentation Slide 8

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

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

Factors Affecting: Solubility Pressure & Temperature— “Henry’s Law” See Chapter 14 Video Presentation Slide 9 Gas solubility (mol/L) Sg =KHPg

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

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

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

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.

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

Understanding Colligative Properties See Chapter 14 Video Presentation Slides 10 & 11 LIQUID-VAPOR EQUILIBRIUM

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 decreased by the presence of a solute. More solute particles, lower vapor pressure of solution.

Psolution = Xsolvent  Po 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.

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

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

Vapor Pressure Lowering of Benzene + non volatile solute

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.)

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.)

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.)

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 decreased by: Where the K’s are the respective boiling and freezing point constants and msolute is the molality of the solution.

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

Boiling Point Elevation & Freezing Point Depression

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

The Boiling Point of a Solution is Higher Than That of a Pure Solvent See Chapter 14 Video Presentation Slide 13

Change in Freezing Point Pure water Water solution See Chapter 14 Video Presentation Slides 13 & 14 The freezing point of a solution is LOWER than that of the pure solvent.

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

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.)

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.)

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.)

Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate? Solution: The molality of the solution can be found from the freezing point depression. Knowing molality and mass of solvent, one can find the moles of solute. 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 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Molar Mass By Boiling Point Elevation Problem: Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is 61.82 °C. What is the molar mass of benzyl acetate?

Osmosis Dissolving the shell in vinegar Egg in pure water Egg in corn syrup See Chapter 14 Video Presentation Slide 4

Osmosis 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. See Chapter 14 Video Presentation Slides 15 & 16

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

Osmosis at the Particulate Level

Process of Osmosis

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

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

Osmosis & Living Cells

Reverse Osmosis: Water Desalination Water desalination plant

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 P1 = X1 P 1 Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = CRT Electrolyte Solutions Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iCRT 70

Colloids A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. 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.

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

Hydrophobic Colloids 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.

Soap 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.

Emulsifiers & Surfactants 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.