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The Properties of Solutions

Chapter 13. The Properties of Solutions. If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time. TYPES OF SOLUTIONS.

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The Properties of Solutions

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  1. Chapter 13 The Properties of Solutions If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

  2. TYPES OF SOLUTIONS Most of us think of a solid dissolved in a liquid as a solution, but any type of solute dissolved in any type of solvent is a solution S/L = L phase (sugar water) G/L = L phase (O2 in water for fish) L/L = L phase (beer) G/G = G phase (air) G/S = S phase (H2 in Pt) S/S = S alloy (like stainless steel) If 1-phase = homogeneous mixture If 2-phase = heterogeneous (like cold tea with ice cubes)

  3. Figure 13.1 The major types of intermolecular forces in solutions. Forces are listed in decreasing order of strength (with values in kJ/mol), and an example of each is shown with space-filling models

  4. DEFINITIONS : Two fluids that mix in all proportions are miscible. Solubility = maximum amount of solute that can be dissolved in a given amount of solvent at a given fixed T, at equilibrium. Saturated = solution containing maximum amount of solute. Any additional solute appears as a precipitate or a gas, or a separate liquid phase. Unsaturated = more solute can be added. Supersaturated = temporary condition where more solute has dissolved, but add just 1 crystal to this, many will precipitate.

  5. More types of solutions and definitions Types of Solutions: Solubility and Saturation Part 2 - YouTube

  6. solute (undissolved) solute (dissolved) Figure 13.19 from 4th ed. Equilibrium in a saturated solution.

  7. Figure 13.8 Sodium acetate crystallizing from a supersaturated solution. Supersaturated solution: add just one little “seed” crystal. From one “seed” excess solute will crystallize from the solution. Now it’s just a saturated solution with some solute present.

  8. Factors affecting solubility: Natural inclination of the universe towards disorder so substances do mix Strength of Force of Attraction between molecules and ions and the solvent IP Force of attraction affects & limits solubility 1. Gases - small forces, mix freely, completely miscible

  9. Factors affecting solubility: 2. Liquids: Similar liquid molecules - heptane in octane - both have only London forces involved, about the same strength, both nonpolar Different IP forces - octane and water – will NOT mix The less dense liquid will rise and stay in separate phase on top of water. LEADS TO GENERAL RULE: LIKE DISSOLVES LIKE. Polar to polar, nonpolar to nonpolar.

  10. Factors affecting solubility: Using a drawing, explain which solvent is better for dissolving ethanol, water or octane?

  11. water methanol A solution of methanol in water Figure 13.3 Like dissolves like: solubility of methanol in water. The H-bonding force of attraction in water and in methanol are similar in type and strength, so they can substitute for one another. Thus, methanol is soluble in water; in fact, the two substances are 100% miscible in each other.

  12. Factors affecting solubility: 3. Solids - must also be "like" the solvent Glucose dissolves in water because of extensive Hydrogen-bonding Energy from attraction approx = energy to break water to water H-bonding I2 is not as soluble - London forces involved CCl4 also London forces. I2 dissolves in carbon tetrachloride, but glucose does not

  13. PROBLEM: Predict which solvent will dissolve more of the given solute: (a) Methanol. NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to London dispersion forces to a greater extent. (b) Water. Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can experience Hydrogen bonding attraction with ethylene glycol. (c) Ethanol. Diethyl ether can interact through a dipole and London dispersion forces. Ethanol can provide both while water would like to Hydrogen bond. SAMPLE PROBLEM 13.1 Predicting Relative Solubilities of Substances (a) Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH) (b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3) or in water. (c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH) SOLUTION:

  14. Practice • Work on problems 7 and 11 in chapter 13

  15. The cyclic structure of b-D-glucopyranose in aqueous solution. Figure 13.14 from 4th ed. Solid glucose is named just D-glucose and is a linear arrangement. In water it reacts with itself, C1 and C5, to form a hemiacetal.

  16. There are three processes involved in dissolving a solute in a solvent 1. Have to separate solvent molecules: an endothermic process 2. Have to separate solute molecules or ions: another endothermic process 3. Surround solute particles with solvent particles, an exothermic process. If solvent is water, this is called Hydration. This step may give back a lot of the energy that was taken in by the first two processes.

  17. Figure 13.2 Hydration shells around an aqueous ion. When an ionic compound dissolves in water, ion-dipole forces orient water molecules around the separated ions to form hydration shells. The cation shown here is octahedrally surrounded by six water molecules, which form H-bonding attractions with water molecules in the next hydration shell, and those form H-bonding attractions with others farther away.

  18. Enthalpy of solution For NaCl: NaCl(s) Na+(aq) + Cl-(aq) lattice DHsoln enthalpy of energy hydration (DHhyd) Na+(g) + Cl-(g) Hess’s Law tells us that DHsoln = sum of LE and DHhyd Given that Lattice Energy is 786 kJ/mol and DHhyd is –783 kJ/mol, then DHsoln will be +3 kJ/mol or slightly endothermic. If heat of hydration is about equal to Lattice Energy, the ionic compound will dissolve. Look at AgCl: LE = 916 kJ/mol, DHhyd = -851 kJ/mol, sum = DHsoln = +65 kJ/mol to dissolve it, so AgCl is not soluble

  19. NH4NO3 NaOH Figure 13.5 Dissolving ionic compounds in water. Endothermic for NH4NO3 NaCl –slightly endothermic The enthalpy diagram for an ionic compound in water includes DHlattice (DHsolute; always positive) and the combined ionic heats of hydration (DHhydr; always negative). EXOTHERMIC!

  20. Figure 13.6 Enthalpy diagrams for dissolving NaCl and octane in hexane. A, Because attractions between Na (or Cl) ions and hexane molecules are weak, DHmix is much smaller than DHsolute. Thus, DHsoln is so positive that NaCl does not dissolve in hexane. B, Intermolecular forces in octane and in hexane are so similar that DHsoln is very small. Octane dissolves in hexane because the solution has greater entropy than the pure components. Very endothermic – NaCl will not dissolve in hexane.

  21. Temperature does cause solubility to vary for all substances Temperature increase ALWAYS causes gas solubility to decrease Sometimes increase in T can increase solubility of a solid or liquid in liquid Dissolving solids can be endothermic, so adding heat causes solubility to increase - If it was exothermic, adding heat may cause solubility to decrease

  22. The relation between solubility and temperature for several ionic compounds. Figure 13.9 Most ionic compounds have higher solubilities at higher temperatures. Cerium sulfate is one of several exceptions.

  23. EFFECT OF P & T ON SOLUBILITY: Le Chatelier's Principle - a change in any of the factors determining an Equilibrium will cause the system to adjust in order to counteract the effect of the change as much as possible. (You will see this again soon!) Pressure has little effect of liquid or solid in water, but gases are different. Picture two cylinders with pistons, same volume of water, same moles of CO2 gas total. ALL (nonreactive nondissociative) gases become more soluble at higher Pressure.

  24. The effect of pressure on gas solubility. Figure 13.10 A, A saturated solution of a gas is in equilibrium at pressure P1. B, If the pressure is increased to P2, the volume of the gas decreases. Therefore, the frequency of collisions with the surface increases. C, As a result, more gas is in solution when equilibrium is re-established.

  25. Henry’s Law Sgas = kH X Pgas The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.

  26. PROBLEM: The partial pressure of carbon dioxide gas inside a bottle of cola is 4.0 atm at 250C. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3 x10-2 mol/L*atm at 250C. S = (3.3 x10-2 mol/L*atm)(4 atm) = CO2 SAMPLE PROBLEM 13.2 Using Henry’s Law to Calculate Gas Solubility SOLUTION: 0.13 mol/L (Yes, it’s really this easy!)

  27. Practice • Work on problems 15, 16, 25, and 34 in chapter 13.

  28. CONCENTRATION: AMOUNT OF SOLUTE IN GIVEN QUANTITY OF SOLVENT OR SOLUTION Units of concentration: M = Molarity = moles of solute per liter of solution moles/L mass-percent = (mass of solute per mass of solution)x100 %-wt(or mass) vol-percent = (vol of solute per volume of sol'n)x100 %-vol m = molality = moles of solute per kilogram of solvent moles/kg mole fraction = moles of solute per total moles fraction,c ***Solubility = grams of solute per 100 grams of solvent g/100g ***Normality: old concept, skip text, use N = M * (#H) if acid or N = M * (#OH) if base Units: N (normal) ***LEARN SOLUBILITY IN g/100g and NORMALITY: NOT IN TEXTBOOK

  29. Examples: Mass percent: 3.5% NaCl solution means 3.5 g NaCl in 100 g solution Preparation: put in 3.5 g, add water to make 100 grams. (96.5g, NOT 100 g of water!) How would you prepare 425 g of a 2.40%-wt aqueous solution of sodium acetate? Molar mass NOT needed!!!

  30. EXAMPLES Molality: molal = mol solute/kg solvent What is molality of a solution where 0.20 mol ethylene glycol is dissolved in 2.0 x 103 g of H2O? molality = 0.20 mol/2.0 kg = 0.10 molal (little m can mean meter or milli, so write out molal) TRY TO KEEP M AND m STRAIGHT! Calculate the molality of a solution which has 4.57 g glucose in 25.2 g of H2O. (Turn the answer in to the instructor.)

  31. EXAMPLES Mole fraction: ci = molessubstance A/total moles What is the mole fraction of ethylene glycol if a solution has 1 mole ethylene glycol in 9 mol H2O? Total moles = 10, ci = 1/10 = 0.10 What is the mole fraction of glucose if 4.57 g glucose is dissolved in 25.2 g H2O? (Turn the answer in to the instructor.)

  32. SAMPLE PROBLEM 13.4 Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca. (b) The label on a 0.750-L bottle of Italian Chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain? (c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water?

  33. 40.5 mg Ca x g x 106 103 mg x 0.750 L Chianti 11.5 L alcohol 100 L Chianti mole mole 18.02 g 60.09 g = 0.423 = 0.577 C2H6O2 H2O SAMPLE PROBLEM 13.4 Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction continued SOLUTION: (a) = 1.16x104 ppm Ca 3.5 g (b) = 0.0862 L alcohol (c) moles ethylene glycol = 142 g = 2.36 mol C2H6O2 moles water = 38.0g = 3.22 mol H2O 2.39 mol C2H8O2 3.22 mol H2O 2.39 mol C2H8O2 + 3.22 mol H2O 2.39 mol C2H8O2 + 3.22 mol H2O

  34. CONVERSIONS BETWEEN CONCENTRATION UNITS: Given that you have 0.100 L of ethanol/H2O solution made with 10.00 mL of ethanol and the Deth = 0.789g/mL, Dsoln= 0.982g/mL. Find a) vol %, b) mass %, c) molarity, d) molality, and e) mole fraction.

  35. Other conversions: 1. Converting molality to mole fraction: given a 0.120 molal glucose solution (0.120 mol glucose/1.00 kg H2O), find ci for glucose and water. mol H2O = 1000 g/18.015 g/mol = 55.51 mol Total moles = 0.120 mol + 55.51 mol = 55.63 cglucose = 0.120 mol glucose/55.63 mol = 0.00216 cwater = 55.51 mol/55.63 mol = 0.998 2. Converting mole fraction to molal: Given a solution that is 0.150 mole fraction glucose and 0.850 mole fraction water, find molality. (So assume you have a total of 1.000 moles.) Convert water to kg: 0.850 mol x 18.015 g/mol = 15.3 g molality = 0.150 mol/0.0153 kg = 9.80 molal

  36. Other conversions: 3. Convert molality to Molarity: need D of solution, Given a 0.273 molal KCl sol'n, D = 1.011 g/mL, find Molarity. molality = 0.273 mol KCl/1.00 kg water Find mass KCl: 0.273 mol x 74.553 g/mol = 20.35 g Total mass = 1000.0 g + 20.35 g = 1020.35 g Volume = 1020.35g/1.011 g/mL = 1009.25 mL Molarity = 0.273 mol/1.00925 L = 0.2705 M

  37. Other conversions: 4. Convert Molarity to molality: Given a 0.907 M Pb(NO3)2 solution with D = 1.252 g/mL, find molality. Mass of solution: 1000 mL x 1.252 g/mL = 1252 g 0.907 mol x 331.2 g/mol = 300.4 g Mass water = 1252 – 300.4 = 951.6 g molality = .907 mol/0.9516 kg = 0.953 molal

  38. Practice • Do problem 56 in chapter 13.

  39. COLLIGATIVE PROPERTIES COLLIGATIVE PROPERTIES: the effects of solutes on four physical properties of the solvents: VP, BP, FP (MP), and osmotic pressure VP, vapor pressure: rate of evaporation decreases and VP at a given T decreases ***Raoult's Law: VP lowering is proportional to mole fraction of solvent, which is always <1.0 Psoln = csolv * Posolv for a solid nonvolatile solute (Or you can use two steps: DP = csolute*Posolv and then Psoln = Posolv - DP

  40. STRONG nonelectrolyte weak Figure 13.25 from 4th ed. The three types of electrolytes. Review: what makes a strong vs. a weak electrolyte?

  41. Figure 13.11 The effect of a solute on the vapor pressure of a solution. Solute particles are interfering with solvent particles trying to go into vapor phase: affects VP, BP, etc. A, Equilibrium is established between a pure liquid and its vapor when the numbers of molecules vaporizing and condensing in a given time are equal. B, The presence of a dissolved solute decreases the number of solvent molecules at the surface so fewer solvent molecules vaporize in a given time. Therefore, fewer molecules need to condense to balance them, and equilibrium is established at a lower vapor pressure.

  42. PROBLEM: Calculate the vapor pressure lowering, DP, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL. 0.988 g H2O 1.26 g C3H8O3 mol H2O mol C3H8O3 mL H2O mL C3H8O3 18.02 g H2O 92.09 g C3H8O3 SAMPLE PROBLEM 13.6 Using Raoult’s Law to Find the Vapor Pressure Lowering SOLUTION: DP = csolute*Posolv 10.0 mL C3H8O3 = 0.137 mol C3H8O3 x = 27.4 mol H2O 500.0 mL H2O x 0.137 mol C3H8O3 DP = x 92.5 torr = 0.461 torr 0.137 mol C3H8O3 + 27.4 mol H2O cglycerol = 0.00498 Psoln = 92.5 - 0.461 = 92.039 torr

  43. Raoult’s Law for two liquids If 2 or more liquids are involved, say A & B, then PA=cA*PoA and PB=cB*PoB THIS HOLDS TRUE FOR IDEAL SOLUTIONS WHICH OBEY THESE LAWS: - Intermolecular forces between solute particles and solvent are same as between two solvent particles, leading to NO volume change and no DHsoln HOWEVER, NOT ALL LIQ/LIQ SOLNS ARE IDEAL! If there's an enhanced attraction like Hydrogen-bonding between solute and solvent, the observed vapor pressure is even lower than expected.

  44. Raoult’s Law for two liquids EXAMPLE: A solution of 0.500 mol benzene and 0.500 mol toluene is in a distillation apparatus. At 25°C (before heating), what will be the equilibrium VP's and concentrations of each? Given PoB = 95.1 torr, PoToluene = 28.4 torr; PB+PToluene=Ptot 0.500 mol * 95.1 torr + 0.500 mol * 28.4 torr = 61.8 torr Mole fraction in vapor using partial pressure of gas: cToluene = PTolene/Ptot = 14.2/61.8 = 0.230 toluene cB = PB/Ptot = 47.6/61.8 = 0.770 benzene If we heated the solution, the mole fraction of benzene would be even higher! Then if we cooled the vapor through a condenser, collected it and repeated the process, we could approach making pure benzene. This is called fractional distillation.

  45. Practice Raoult’s Law A solution contains 20.00 grams of methanol in 100.0 g of ethanol. The vapor pressures are: MeOH = 94 torr, EtOH = 44 torr. Find the vapor pressure of each alcohol and the total pressure over the solution. Also find the mole fraction of each alcohol in the vapor phase. (PMeOH = 21 torr, PEtOH = 34 torr, Pt = 55 torr XMeOH = 21/55 = 0.38, XEtOH = 34/55 = 0.62

  46. Phase diagrams of solvent and solution. Figure 13.12 Phase diagrams of an aqueous solution (dashed lines) and of pure water (solid lines) show that, by lowering the vapor pressure (P), a dissolved solute elevates the boiling point (Tb) and depresses the freezing point (Tf).

  47. More Colligative Properties:FP & BP The VP effect causes the FP and BP of solutions to change also. FP depression: solvent begins to crystallize before solute does, solute interferes and gets in the way of the crystal lattice. Two step calculation: Step 1. DTf = Kf*molality and Step 2. new FP = FPo-DTf for nonvolatile covalent solutes BP elevation: because VP is lower at any given T, must have higher T at 1 atm to boil. Similar two-step calculation: Step 1.DTb = Kb*molality and Step 2. new BP = BPo+DTb

  48. Table 13.5 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents Typo in text Melting Point (0C) Boiling Point (0C)* Solvent Kb (0C/m) Kf (0C/m) Acetic acid 117.9 3.07 16.6 3.90 Benzene 80.1 2.53 5.5 4.90 Carbon disulfide 46.2 2.34 -111.5 3.83 Carbon tetrachloride 76.5 5.03 -23 30. Chloroform 61.7 3.63 -63.5 4.70 Diethyl ether 34.5 2.02 -116.2 1.79 Ethanol 78.5 1.22 -117.3 1.99 Water 100.0 0.512 0.0 1.86 *at 1 atm.

  49. Practice with FP & BP Calculate both the new FP & BP for a solution that has 100.0 g of glucose in 500.0 mL of water, given the density of water is 0.998 g/mL at room temperature kg water = 500.0 mL (0.998 g/mL)(1 kg/1000g) = 0.499 kg molality = 100.0 g (1mol/180.16 g)/0.499 kg = 1.112 molal DTf = 1.86oC/molal (1.112 molal) = 2.068oC new FP = 0.00 – 2.068oC = -2.068oC DTb = 0.512oC/molal (1.112 molal) = 0.569oC new BP = 100.000 + 0.569 = 100.569oC

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