Chapter 11. General Chemistry. Chapter 11/1

Chapter 11 Solutions and Their Properties Professor Sam Sawan General Chemistry 84.122 Chapter 11/1

Solutions Chapter 11/3

Energy Changes and the Solution Process

Solubility there is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible oil and water are immiscible the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility the solubility of one substance in another varies with temperature and pressure 5

Mixing and the Solution Process Entropy formation of a solution does not necessarily lower the potential energy of the system the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible yet the gases mix spontaneously the gases mix because the energy of the system is lowered through the release of entropy entropy is the measure of energy dispersal throughout the system energy has a spontaneous drive to spread out over as large a volume as it is allowed 6

Intermolecular Forces and the Solution Process Enthalpy of Solution energy changes in the formation of most solutions also involve differences in attractive forces between particles must overcome solute-solute l t attractive ti forces endothermic must overcome some of the solvent-solvent attractive forces endothermic at least some of the energy to do this comes from making new solute-solvent attractions exothermic 7

Intermolecular Attractions 8

Relative Interactions and Solution Formation Solute-to-Solvent > Solute-to-Solute + Solvent-to-Solvent to Solute-to-Solvent = Solute-to-Solute + Solvent-to-Solvent to Solute-to-Solvent < Solute-to-Solute + Solvent-to-Solvent Solution Forms Solution Forms when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent- to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy Solution May or May Not Form 9

Will It Dissolve? Chemist s Rule of Thumb Like Dissolves Like a chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, il the solvent molecules l will attract t the solute particles at least as well as the solute particles to each other 10

Classifying Solvents Structural Solvent Class Feature Water, H 2 O polar O-H Methyl Alcohol, CH 3 OH polar O-H Ethyl Alcohol, C 2 H 5 OH polar O-H Acetone, C 3 H 6 O polar C=O Toluene, C 7 H 8 nonpolar C-C & C-H Hexane, C 6 H 14 nonpolar C-C & C-H Diethyl Ether, C 4 H 10 O nonpolar C-C, CCH&CO C-H C-O, (nonpolar > polar) CarbonTetrachloride nonpolar C-Cl, Cl but symmetrical 11

Predict whether the following vitamin is soluble in fat or water The 4 OH groups make the molecule l highly polar and it will also H-bond to water. Vitamin C is water soluble H 2 C OH OH C H HO H C C O C OH Vitamin C C O 12

Predict whether the following vitamin is soluble in fat or water The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. HC HC H C C H C C O C CH 3 C CH C O Vitamin K 3 is fat soluble Vitamin K 3 13

Energy Changes and the Solution Process The sodium and chloride ions are hydrated.

Energy Changes and the Solution Process There is an entropy change for the solution process.

Energy Changes and the Solution Process There is an entropy change for the solution process. Solution #1 Solution #2 Chapter 11/16

Energetics of Solution Formation if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic if the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic 17

Energetics of Solution Formation overcome attractions between the solute particles endothermic overcome some attractions between solvent molecules endothermic for new attractions between solute particles and solvent molecules exothermic the overall ΔH depends on the relative sizes of the ΔH for these 3 processes ΔH sol n = ΔH solute + ΔH solvent + ΔH mix 18

Heat of Hydration 19

Energy Changes and the Solution Process

Units of Concentration Molarity (M) = Moles of solute Liters of solution Moles of component Mole Fraction (X) = Total moles making up solution Mass of component Mass Percent = Total mass of solution x 100% Moles of solute Molality y( (m) = Mass of solvent (kg) Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/22

Molarity and Dissociation the molarity of the ionic compound allows you to determine the molarity of the dissolved ions CaCl 2 (aq) = Ca +2 (aq) + 2 Cl -1 (aq) A 1.0 M CaCl 2 (aq) solution contains 1.0 moles of CaCl 2 in each liter of solution 1 L = 1.0 moles CaCl 2, 2 L = 2.0 moles CaCl 2 Because each CaCl 2 dissociates to give one Ca +2 = 1.0 M Ca +2 1L= 1.0 moles Ca +2,2L= 2.0 moles Ca +2 Because each CaCl 2 dissociates to give 2 Cl -1 = 2.0 M Cl -1 1 L = 2.0 moles Cl -1, 2 L = 4.0 moles Cl -1 23

Units of Concentration Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is 1.025 g/ml at 20 C, and the NaCl concentration is 3.50 mass %. Assuming 100.00 g of solution, calculate the volume: 100.00 g solution 1 ml 1 L x x 09756 1.025 g 1000 ml = 0.09756 L Convert the mass of NaCl to moles: 3.50 g NaCl 1 mol x = 0.0599 mol 58.4 g Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/24

Units of Concentration What is the molality of a solution prepared by dissolving 0.385 g of cholesterol, l C 27 H 46 O, in 40.00 g chloroform, CHCl 3? Convert the mass of cholesterol to moles: 0.385 g 1 mol x = 0.000 997 mol 386.0 g Calculate the mass of chloroform in kg: 40.0 g 1 kg x = 0.0400 kg 1000 g Calculate the molality of the solution: 0.000 997 mol = 0.0249 m 0.0400 kg Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/26

Mass of Percent Concentration Part (solute) Percent = 100% Whole (solution)( ) Mass of Solute, g Mass Percent = 100% Mass of Solution, g Solute + Mass of Solvent = Mass of Solution Mass of Solute, g Percent Mass/Volume = 100% Volume of Solution, ml Mass of Solute + Volume of Solvent Volume of Solution Volume of Solute, ml Volume Percent = 100% Volume of Solution, ml Volume of Solute + Volume of Solvent Volume of Solution 27

Using Concentrations as Conversion Factors concentrations show the relationship between the amount of solute and the amount of solvent 12%(m/m) sugar(aq) means 12 g sugar 100 g solution or 12 kg sugar 100 kg solution; or 12 lbs. 100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag 100 ml solution 22%(v/v) alcohol(aq) means 22 ml EtOH 100 ml solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice versa 28

How would you prepare 250.0 g of 5.00% by mass glucose solution (normal glucose)? Given: 250.00 g solution Find: g Glucose Equivalence: 5.00 g Glucose 100 g solution Solution Map: g solution g Glucose Apply Solution Map: 500 5.00 g Glucose 100 g solution 5.00 g glucose 250.0 g solution = 12.5 g glucose 100 g solution Answer: Dissolve 12.5 g of glucose in enough water to total 250.00 g 29

Solution Concentration PPM grams of solute per 1,000,000 g of solution mg of solute per 1 kg of solution 1 liter of water = 1 kg of water for water solutions we often approximate the kg of the solution as the kg or L of water mg solute kg solution grams solute grams solution x 106 mg solute L solution 30

What volume of 10.5% by mass soda contains 78.5 g of sugar? Given: Find: Concept Plan: Solve: : 78.5 g sugar volume, ml g solute g sol n ml sol n 100 g sol' n 1mL sol'n 10.5 g sugar 1.04 g sol' n 100 g sol n = 10.5 g sugar, 1 ml sol n = 1.04 g 100 g 1mL 78.5 g sugar = 78.5 g sugar 1.04 g 719 ml Check: the unit is correct, the magnitude seems reasonable as the mass of sugar 10% the volume of solution 31

Solution Concentrations Mole Fraction, X A the mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution total of all the mole fractions in a solution = 1 unitless the mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution = mole fraction x 100% mole fraction of A = X = moles of components A A total moles in the solution 32

What is the molarity of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 ml of solution? Given: Find: Concept Plan: Solve : 17.2 g C 2 H 6 O 2, 0.500 kg H 2 O, 515 ml sol n M g C 2 H 6 O 2 mol C 2 H 6 O 2 mol M = ml sol n L sol n M L M = mol/l, 1 mol C 2 H 6 O 2 = 62.07 g, 1 ml = 0.001 L 1mol C2H6O2 17.2 g C2 H6O2 = 62.07 g C H O 0.001L 515 ml = 0.515 L 1mL 0.2771 mol C 2H = 0.515 L M = 0.538 M 2 6 2 O M 6 2 0.2771 mol 33

What is the molality of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 ml of solution? Given: Find: Concept Plan: Solve : 17.2 g C 2 H 6 O 2, 0.500 kg H 2 O, 515 ml sol n m g C 2 H 6 O 2 mol C 2 H 6 O 2 mol m = kg H 2 O kg m m = mol/kg, 1 mol C 2 H 6 O 2 = 62.07 g 17.2 g C H O 1mol C2H6O2 62.07 g C H O 2 6 2 = 2 6 2 m = 0.2771 mol C2H6O 0.500 kg H O m = 0.554 M 2 0.2771 mol 2 34

What is the percent by mass of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 ml of solution? Given: Find: Concept Plan: 17.2 g C 2 H 6 O 2, 0.500 kg H 2 O, 515 ml sol n %(m/m) g solvent g solute g C 2 H 6 O 2 % = 100% g sol' n g sol n % Solve: 1 kg = 1000 g 1000 g H2O 2 0.500 kg H2 O = 5.00 10 g H2O 1kg H O 17.2 g C H6O2 + 500 g H2O 2 = 17.2 g C H6O % = 517.2 g sol' n 2 2 % % = 3.33% 2 100% 517.2 g sol'n 35

What is the mole fraction of a solution prepared by mixing 17.2 g of C 2 H 6 O 2 with 0.500 kg of H 2 O to make 515 ml of solution? Given: Find: Concept Plan: Solve: 17.2 g C 2 H 6 O 2, 0.500 kg H 2 O, 515 ml sol n Χ χ = mol g C 2 H 6 O 2 mol C 2 H 6 O 2 moltotal g H 2 O mol H 2 O Χ Χ = mol A /mol tot, 1 mol C 2 H 6 O 2 = 62.07 g, 1 mol H 2 O = 18.02 g 1mol C2H6O2 17.2 g C2 H6O2 = 62.07 g C H O 2 6 2 A 0.2771 mol 1000 g 1mol H2 O 0.500 kg H2 O = 27.75 mol H2O 1kg 18.02 g H O χ = 2 2H6O2 0.2771 mol C 0.2771 mol C H O + 27.75 mol H ( O) χ = 9.89 10-3 2 6 2 2 36

Units of Concentration Chapter 11/37

Some Factors Affecting Solubility Saturated Solution: A solution containing the maximum possible amount of dissolved d solute at equilibrium. Solute + Solvent dissolve crystallize Solution Supersaturated Solution: A solution containing a greater-than-equilibrium th i amount of solute. Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/38

Some Factors Affecting Solubility Chapter 11/40

Some Factors Affecting Solubility

Some Factors Affecting Solubility Henry s Law Solubility = k P Chapter 11/42

Henry s Law the solubility of a gas (S gas) ) is directly proportional to its partial pressure, (P gas) ) S gas = k H P gas k H is called Henry s Law Constant 43

What pressure of CO 2 is required to keep the [CO 2 ]=012Mat25 C? 0.12 25 Given: Find: Concept Plan: Solve: : P S = [CO 2 ] = 0.12 M, P of fco 2, atm [CO 2 ] P = S k H P S = k H P, k H = 3.4 x 10-2 M/atm S 0.12 M = = 2 k 3.14 10 M atm H -1 = 3.5 atm Check: the unit is correct, the pressure higher than 1 atm meets our expectation from general experience 44

Physical Behavior of Solutions: Colligative Properties Colligative Properties: Properties that depend on the amount of a dissolved d solute but not on its chemical identity. Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/45

Vapor-Pressure Lowering of Solutions: Raoult s Law Raoult s Law P soln = P solv X solv

Vapor-Pressure Lowering of Solutions: Raoult s Law The vapor pressure of pure water at 25 C is 23.76 mm Hg. What is the vapor pressure of a solution made from 1.00 mol glucose in 15.0 mol of water at 25 C? Glucose is a nonvolatile solute. P soln = P solv X solv = 23.76 mm Hg 15.0 mol x 1.00 mol + 15.0 mol = 22.3 mm Hg Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/47

Vapor-Pressure Lowering of Solutions: Raoult s Law Solutions of ionic substances often have a vapor pressure significantly ifi lower than predicted, d because the ion-dipole forces between the dissolved ions and polar water molecules are so strong. van t Hoff Factor: i = moles of particles in solution moles of solute dissolved NaCl(aq) Na 1+ (aq) + Cl 1- (aq) For sodium chloride, the predicted value of i is 2. For a 0.05 m solution of sodium chloride, the experimental value for i is 1.9. Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/48

Chapter 11/49

Vapor-Pressure Lowering of Solutions: Raoult s Law P total = P A + P B = (P A X A ) + (P B X B )

Boiling-Point Elevation and Freezing-Point Depression Chapter 11/51

Boiling-Point Elevation and Freezing-Point Depression Nonelectrolytes T b = K b m T f = K f m Electrolytes T b = K b m i T f = K f m i

Chapter 11/53

Chapter 11/54

Boiling-Point Elevation and Freezing-Point Depression What is the freezing point (in C) of a solution prepared by dissolving i 7.40 g of MgCl 2 in 110 g of water? The van t Hoff factor for MgCl 2 is i = 2.7. Calculate the moles of MgCl 2 : 7.40 g 1 mol x 95.3 g = 0.07760776 mol Calculate the molality of the solution: 0.0776 mol 1000 g mol x = 0.71 110 g 1k kg kg Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/55

Boiling-Point Elevation and Freezing-Point Depression Calculate the freezing point of the solution: Tf = K f mi= 1.86 C kg x mol 0.71 mol kg x 2.7 = 3.6 C T f = 0.0 C - 3.6 C = -3.6 C Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/56

Osmosis and Osmotic Pressure Osmosis: The passage of solvent through a semipermeable membrane from the less concentrated t side to the more concentrated side. Osmotic Pressure ( ): The amount of pressure necessary to cause osmosis to stop. Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/57

Chapter 11/58

= MRT Chapter 11/59

Osmosis and Osmotic Pressure Calculate the osmotic pressure of a 1.00 M glucose solution in water at 300 K. = MRT = 1.00 mol L x L atm 0.08206 K mol x 300 K = 24.6 atm Copyright 2010 Pearson Prentice Hall, Inc. Chapter 11/60

Some Uses of Colligative Properties Reverse Osmosis

Fractional Distillation of Liquid Mixtures

Chapter 11/63