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Chapter 17

Chapter 17. Properties of Solutions. Chapter 17: Properties of Solutions. 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 The Vapor Pressure of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression

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Chapter 17

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  1. Chapter 17 Properties of Solutions

  2. Chapter 17: Properties of Solutions 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation 17.3 Factors Affecting Solubility 17.4 The Vapor Pressure of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolytic Solutions 17.8 Colloids

  3. The left beaker contains copper sulfate solution. In the right beaker, ammonia is being added to create a precipitate of copper(II) hydroxide.

  4. Definitions for Solutions Solute- The smaller (in mass) of the components in a solution, the material dispersed into the solvent. Solvent - The major component of the solution, the material that the solute is dissolved into. Solubility - The maximum amount that can be dissolved into a particular solvent to form a stable solution at a specified temperature. Miscible - Substances that can dissolve in any proportion, so that it is difficult to tell which is the solvent or solute!

  5. Hydration shells around an aqueous ion

  6. Solution Composition Mass percent: weight percent Mass percent = x 100% Mole fraction: symbolized by the Greek letter, chi = X mole fraction of component A = XA = Molarity: M ; (chapter 4) M = Molality: m m = grams of solute grams of solution nA nA + nB moles of solute liter of solution moles of Solute Kg of solvent

  7. Like Dissolves Like • Polar molecules - dissolve best in Polar solvents. • Polar molecules can hydrogen bond with polar solvents, such as water, hence increasing their solubility. • Non-polar molecules - dissolve best in non - polar solvents. • Hydrocarbons, non - polar molecules, do not dissolve, or mix with water!

  8. Like dissolves Like: Solubility of methanol in water

  9. Fig 13.4 (P 486) The structure and function of a soap

  10. Predicting Relative Solubility's of Substances -I Problem: Predict which solvent will dissolve more of the given solute. (a) Sodium Chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH). (b) Ethylene glycol (HOCH2CH2OH) in water or in hexane (CH3CH2CH2CH2CH2CH3). (c) Diethyl ether (CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in water. Plan: Examine the formulas of each solute and solvent to determine which forces will occur. A solute tends to be more soluble in a solvent which has the same type of forces binding its molecules.

  11. Predicting Relative Solubility's of Substances - II Solution: (a)Methanol - NaCl is an ionic compound that dissolves through ion- dipole forces. Both methanol and propanol contain a polar hydroxyl group, and propanol’s longer hydrocarbon chain would form only weak forces with the ions, so it would be less effective at replacing the ionic attractions of the solvent. (b) Water Ethylene glycol molecules have two -OH groups, and the molecules interact with each other through H bonding. They would be more soluble in water, whose H bonds can replace solute H bonds better than can the dispersion forces in hexane. (c) Ethanol Diethyl ether molecules interact with each other through dipole and dispersion forces and could form H bonds to both water and ethanol. the ether would be more soluble in ethanol because the solvent can form H bonds and replace the dispersion forces in the solute, whereas the H bonds in water must be partly replaced with much weaker dispersion forces.

  12. An Energy Solution? – Solar Ponds Solar ponds are shallow bodies of salt water ( a very high salt content) designed to collect solar energy as it water the water in the ponds, and then it can be used for heating, or converted into other forms of energy. Sufficient salt must be added to establish a salt gradient in a pool 2-3 meters deep, with a dark bottom. A salt gradient will be established in which the upper layer, called the conductive layer, has a salt content of about 2% by mass. The bottom layer, called the heat storage layer has a salt content of about 27%. The middle layer called the nonconvective layer has an intermediate salt content, and acts as an insulator between the two layers. The water in the deepest layer can reach temperatures of between 90 and 100oC, temperatures as high as 107oC have been reported. A 52 acre pond near the Dead sea in Israel can produce up to 5 mega watts of power.

  13. Three Steps in making a Solution Step #1 : Breaking up the solute into individual components: (expanding the Solute) Step #2 : Overcoming intermolecular forces in the solvent to make room for the solute: (expanding the solvent) Step #3 : Allowing the solvent and solute to interact and form the solution.

  14. Figure 17.1: The formation of a liquid solution can be divided into three steps

  15. Figure 17.2: (a) Enthalpy of solution Hsoln has a negative sign (the process is exothermic) if Step 3 releases more energy than is required by Steps 1 and 2. (b) Hsoln has a positive sign (the process is endothermic) if Steps 1 and 2 require more energy than is released in Step 3.

  16. H of solution for Sodium Chloride NaCl(s) Na+(g) + Cl –(g) Ho1 = 786 kJ/mol H2O(l) + Na+(g) + Cl –(g) Na+(aq) + Cl –(aq) Hohyd = Ho2 + Ho3 = _____________ kJ/mol Hhyd = enthalpy (heat) of hydration Hosoln = 786 kJ/mol – _______ kJ/mol = _______ kJ/mol The dissolving process is positive, requiring energy. Then why is NaCl so soluble? The answer is in the Gibbs free energy equation from chapter 10, G = H – T S The entropy term – T S is Negative, and the result is that G becomes negative, and as a Result, NaCl dissolves very well in the polar solvent water.

  17. French Navy ship L'Ailette equipped with vacuum pumps approaches an oil slick. Source: AP/Wide World Photos

  18. Solution Cycle Step 1: Solute separates into Particles - overcoming attractions Therefore --Endothermic Step 2: Solvent separates into Particles - overcoming intermolecular attractions Therefore -- Endothermic solvent (aggregated) + heat solvent (separated) Hsolvent> 0 Step 3: Solute and Solvent Particles mix - Particles attract each other Therefore -- Exothermic solute (separated) + solvent (separated) solution + heat Hmix< 0 The Thermochemical Cycle Hsolution = Hsolute + Hsolvent + Hmix If HEndothermic Rxn< HExothermic Rxn solution becomes warmer If HEndothermic Rxn> HExothermic Rxn solution becomes colder

  19. Solution Cycles and the Enthalpy Components ofthe Heat of Solution

  20. Multistage supercritical fluid extraction apparatus Source: USDA

  21. Chicken fat obtained by using supercritical carbon dioxide as the extracting agent. Source: USDA

  22. Figure 17.3: Vitamin A, C A Fat – Soluble Vitamin A Water – Soluble Vitamin A Hydrophobic Vitamin A Hydrophilic Vitamin

  23. Carbonation in a bottle of soda

  24. Figure 17.4: (a) a gaseous solute in equilibrium with a solution. (b) the piston is pushed in, which increases the pressure of the gas and the number of gas molecules per unit volume. (c) greater gas

  25. Henry’s Law of Gas solubilities in Liquids P = kHX P = Partial pressure of dissolved gas X = mole fraction of dissolved gas kH = Henry’s Law Constant

  26. Henry’s Law of Gas Solubility Problem: The lowest level of oxygen gas dissolved in water that will support life is ~ 1.3 x 10 - 4 mol/L. At the normal atmospheric pressure of oxygen is there adaquate oxygen to support life? Plan: We will use Henry’s law and the Henry’s law constant for oxygen in water with the partial pressure of O2 in the air to calculate the amount. Solution: The Henry’s law constant for oxygen in water is 1.3 x 10 -3 mol liter atm and the partial pressure of oxygen gas in the atmosphere is 21%, or 0.21 atm. . Soxygen = kH x PO2 = 1.3 x 10 -3 mol x ( 0.21 atm) liter atm SOxygen = mol O2 / liter . This is adaquate to sustain life in water!

  27. Figure 17.5: The solubilities of several solids as a function of temperature.

  28. Predicting the Effect of Temperature on Solubility - I Problem: From the following information, predict whether the solubility of each compound increases or decreases with an increase in temperature. (a) CsOH Hsoln = -72 kJ/mol (b) When CsI dissolves in water the water becomes cold (c) KF(s) K+(aq) + F -(aq) + 17.7 kJ Plan: We use the information to write a chemical reaction that includes heat being absorbed (left) or released (right). If heat is on the left, a temperature shifts to the right, so more solute dissolves. If heat is on the right, a temperature increase shifts the system to the left, so less solute dissolves. Solution: (a) The negative H indicates that the reaction is exothermic, so when one mole of Cesium Hydroxide dissolves 72 kJ of heat is released. H2O

  29. Predicting the Effect of Temperature on Solubility - II (a) continued H2O CsOH(s) Cs+(aq) + OH -(aq) + Heat A higher temperature (more heat) decreases the solubility of CsOH. (b) When CsI dissolves, the solution becomes cold, so heat is absorbed. H2O CsI(s) + Heat Cs+(aq) + I -(aq) A higher temperature increases the solubility of CsI. (c) When KF dissolves, heat is on the product side, and is given off so the reaction is exothermic. H2O KF(s) K+(aq) + F -(aq) + 17.7 kJ A higher temperature decreases the solubility of KF

  30. Figure 17.6: The solubilities of several gases in water as a function of temperature at a constant pressure of 1 atm of gas above the solution.

  31. Figure 17.7: Pipe with accumulated mineral deposits (left) lengthwise section (right) Source: Visuals Unlimited

  32. Lake Nyos in Cameroon Source: Corbis

  33. Figure 17.8: An aqueous solution and pure water in a closed environment

  34. Figure 17.9: The presence of a nonvolatile solute inhibits the escape of solvent molecules from the liquid

  35. Figure 17.10: For a solution that obeys Raoult’s law, a plot of Psoln versus xsolvent yields a straight line.

  36. Fig. 13.15

  37. Vapor Pressure Lowering -I Problem: Calculate the vapor pressure lowering when 175g of sucrose is dissolved into 350.00 ml of water at 750C. The vapor pressure of pure water at 750C is 289.1 mm Hg, and it’s density is 0.97489 g/ml. Plan: Calculate the change in pressure from Raoult’s law using the vapor pressure of pure water at 750C. We calculate the mole fraction of sugar in solution using the molecular formula of sucrose and density of water at 750C. Solution: molar mass of sucrose ( C12H22O11) = 342.30 g/mol 175g sucrose 342.30g sucrose/mol = 0.51125 mol sucrose 350.00 ml H2O x 0.97489g H2O = 341.21g H2O ml H2O 341.21 g H2O 18.02g H2O/mol = ______ molH2O

  38. Vapor Pressure Lowering - II mole sucrose moles of water + moles of sucrose Xsucrose = 0.51125 mole sucrose 18.935 mol H2O + 0.51125 mol sucrose Xsurose = = 0.2629 P = Xsucrose x P 0H2O = 0.2629 x 289.1 mm Hg = ________ mm Hg

  39. Like Example 17.1 (P 841-2) A solution was prepared by adding 40.0g of glycerol to 125.0g of water at 25.0oC, a temperature at which pure water has a vapor pressure of 23.76 torr. The observed vapor pressure of the solution was found to be 22.36 torr. Calculate the molar mass of glycerol! Solution: Roults Law can be rearranged to give: XH2O = = = 0.9411 = mol H2O = = 6.94 mol H2O 0.9411 = mol gly = = 0.4357 mol Psoln PoH2O 22.36 torr 23.76 torr mol H2O mol gly + mol H2O 125.0 g 18.0 g/mol 6.94 mol mol gly + 6.96 mol 6.94 mol – (6.94 mol)(0.9411) 0.9411 40.0 g 0.4357 mol = g/mol (MMglycerol = 92.09 g/mol)

  40. Figure 17.11: Vapor pressure for a solution of two volatile liquids.

  41. - - - - - - - - - - - H O H H-C-C-C-H H H - - - - - - - - - - -H H H H H H H H-C-C-C-C-C-C-H H H H H H H Hexane H H + H-C-C-O-H Ethanol H H H-O Acetone + Water

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