Solutions. Chapter 13. Properties of Solutions. Lecture Presentation

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1 Lecture Presentation Chapter 13 Properties of Yonsei University homogeneous mixtures of two or more pure substances: may be gases, liquids, or solids In a solution, the solute is dispersed uniformly throughout the solvent. aqueous solutions The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles. Solute-solute, Solvent-solvent, Solvent-solute interactions 2

2 Solution Process 3 How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them. Solvation, hydration 4

3 How Does a Solution Form? If an ionic salt is soluble in water, it is because the ion dipole interactions are strong enough to overcome the lattice energy of the salt crystal. 5 Energetics of Solution Formation Simply put, three processes affect the energetics of solution: Separation of solute particles, Separation of solvent particles, New interactions between solute and solvent. The enthalpy change of the overall process depends on H for each of these steps. 6

4 Why Do Endothermic Processes Occur? 7 Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered. Yet we know that in some processes, like the dissolution of NH 4 NO 3 in water, heat is absorbed, not released. MgSO 4 added to water has H soln = 91.2 kj/mol. NH 4 NO 3 added to water has H soln = kj/mol. entropy Solution Formation and Chemical Reactions Just because a substance disappears when it comes in contact with a solvent, it doesn t mean the substance dissolved. It may have reacted. Dissolution is a physical change you can get back the original solute by evaporating the solvent. If you can t get it back, the substance didn t dissolve, it reacted. Ni(s) + 2HCl(aq) NiCl 2 (aq) + H 2 (g) 8

5 Saturated Types of In a saturated solution, dissolved solute is in dynamic equilibrium with solid solute particles (T). Solubility Unsaturated solution 9 Types of 10 Supersaturated In supersaturated solutions, the solvent holds more solute than is normally possible at that temperature. These solutions are unstable; crystallization can usually be stimulated by adding a seed crystal or scratching the side of the flask.

6 Factors Affecting Solubility The tendency of a substance to dissolve in another depends on: the nature of the solute. the nature of the solvent. the temperature. the pressure (for gases). 11 Factors Affecting Solubility Nature of Solvent & Solute Chemists use the axiom like dissolves like. Solute-Solvent Interactions Polar substances tend to dissolve in polar solvents. Nonpolar substances tend to dissolve in nonpolar solvents. 12

7 Factors Affecting Solubility Nature of Solvent & Solute The more similar the intermolecular attractions, the more likely one substance is to be soluble in another. 13 Factors Affecting Solubility Glucose Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not. Miscible vs. immiscible 14

8 Factors Affecting Solubility Solubility of alcohols in water The number of carbon atoms in a chain affects solubility. The greater the number of carbons in the chain, the more the molecule behaves like a hydrocarbon. Thus, the more C atoms in the alcohol, the lower its solubility in water. Increasing the number of OH groups within a molecule increases its solubility in water. The greater the number of OH groups along the chain, the more solute-water H-bonding is possible. 15 Factors Affecting Solubility Vitamins Vitamin A is soluble in nonpolar compounds (like fats). Vitamin C is soluble in water. 16

9 Sample Exercise 13.1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4 ) or in water: C 7 H 16, Na 2 SO 4, HCl, and I 2. Solution C 7 H 16 and I 2 (the nonpolar solutes) would be more soluble in the nonpolar CCl 4 than in polar H 2 O, whereas water would be the better solvent for Na 2 SO 4 and HCl (the ionic and polar covalent solutes). 17 Sample Exercise 13.1 Predicting Solubility Patterns Continued Practice Exercise Arrange the following substances in order of increasing solubility in water: Answer: C 5 H 12 <C 5 H 11 Cl<C 5 H 11 OH<C 5 H 10 (OH) 2 (in order of increasing polarity and hydrogen-bonding ability) 18

10 Gases in Solution In general, the solubility of gases in water increases with increasing mass. Larger molecules have stronger dispersion forces. 19 Pressure Effects The solubility of liquids and solids does not change appreciably with pressure. But the solubility of a gas in a liquid is directly proportional to its pressure. 20

11 Henry s Law S g = kp g 21 where S g is the solubility of the gas, k is the Henry s Law constant for that gas in that solvent, and P g is the partial pressure of the gas above the liquid. Sample Exercise 13.2 A Henry s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25 C. The Henry s law constant for CO 2 in water at this temperature is mol/l-atm. Solution S CO2 = kp CO2 = ( mol/l-atm)(4.0 atm) = 0.14 mol/l = 0.14 M Practice Exercise Calculate the concentration of CO 2 in a soft drink after the bottle is opened and equilibrates at 25 C under a CO 2 partial pressure of atm. Answer: M 22

12 Temperature Effects Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature. The opposite for gases. Carbonated soft drinks are more bubbly if stored in the refrigerator. Warm lakes have less O 2 dissolved in them than cool lakes. Thermal pollution 23 Ways of Expressing Concentrations of Mass % of A = mass of A in solution total mass of solution Parts per million (ppm) ppm = Parts per billion (ppb) ppb = mass of A in solution total mass of solution mass of A in solution total mass of solution

13 ppm, ppb, ppt in dilute aqueous solution ppm ppb ppt gsolute 1mgsolute g of solution L solution 1g solute 1 g solute g of solution L solution 1g solute 1ng solute g of solution L solution 25 Mole Fraction (X) 26 X A = Molarity (M) Molality (m) moles of A total moles of all components M = moles of solute liters of solution Since volume is temperature-dependent, molarity can change with temperature. m = moles of solute kilograms of solvent Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent.

14 Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa. 27 Sample Problem -Conversions between unitsex) What is the molality of a M aluminum nitrate solution (d = 1.012g/mL)? Work with 1 liter of solution. mass = 1012 g mass Al(NO 3 ) 3 = mol g/mol = 42.6 g ; mass water = 1012 g -43 g = 969 g Molality 0.200mol mol / kg 0.969kg 28

15 Sample Exercise 13.4 Calculation of Molality A solution is made by dissolving 4.35 g glucose (C 6 H 12 O 6, MM=180.2g/mol) in 25.0 ml of water at 25 C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/ml. Solution (25.0 ml)(1.00 g/ml) = 25.0 g = kg Practice Exercise What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10 H 8 ) in 425 g of toluene (C 7 H 8 )? Answers: m 29 Sample Exercise 13.5 Calculation of Mole Fraction and Molarity An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution. Solution (a) (b) Practice Exercise A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution. Answer: (a) , (b) m 30

16 Sample Exercise 13.6 Calculation of Molarity Using the Density of the Solution A solution with a density of g/ml contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene. Calculate the molarity of the solution. Solution Practice Exercise A solution containing equal masses of glycerol (C 3 H 8 O 3 ) and water has a density of 1.10 g/ml. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution. Answer: (a) 10.9 m, (b) XC 3 H 8 O 3 = 0.163, (c) 5.97 M 31 Colligative Properties 32 Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. Among colligative properties are Vapor-pressure lowering Boiling-point elevation Melting-point depression Osmotic pressure

17 Vapor Pressure 33 Because of solute(non-volatile) solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Therefore, the vapor pressure of a solution is lower than that of the pure solvent. where Raoult s Law P P o solution solvent solvent X A is the mole fraction of compound A, and P A is the normal vapor pressure of A at that temperature. The vapor-pressure lowering, P, is P X P o solute solvent Note: This is one of those times when you want to make sure you have the vapor pressure of the solvent. 34

18 Sample Exercise 13.7 Calculation of Vapor Pressure of a Solution Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/ml at 25 C. Calculate the vapor pressure at 25 C of a solution made by adding 50.0 ml of glycerin to ml of water. The vapor pressure of pure water at 25 C is 23.8 torr (Appendix B), and its density is 1.00 g/ml. Solution Practice Exercise The vapor pressure of pure water at 110 C is 1070 torr. A solution of ethylene glycol and water has a vapor Pressure of 1.00 atm at 110 C. Assuming that Raoult s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: Boiling-Point Elevation and Freezing-Point Depression Nonvolatile solute solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent. 36

19 Boiling-Point Elevation The change in boiling point is proportional to the molality of the solution: T b = K b m where K b is the molal boiling-point elevation constant, a property of the solvent. T b is added to the normal boiling point of the solvent. 37 Freezing-Point Depression The change in freezing point can be found similarly: T f = K f m Here K f is the molal freezing-point depression constant of the solvent. T f is subtracted from the normal freezing point of the solvent. 38

20 Colligative Properties of Electrolytes of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes. However, a 1M solution of NaCl does not show twice the change in freezing point that a 1M solution of methanol does. 39 van t Hoff Factor One mole of NaCl in water does not really give rise to two moles of ions. Some Na + and Cl reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl. 40

21 Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentrationdependent. van t Hoff Factor We modify the previous equations by multiplying by the van t Hoff factor, i: T f = K f m i 41 Sample Exercise 13.8 Calculation Of Boiling-Point Elevation and Freezing-Point Depression Automotive antifreeze consists of ethylene glycol, CH 2 (OH)CH 2 (OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water. Solution T b = K b m = (0.51 C/m)(5.37 m) = 2.7 C T f = K f m = (1.86 C/m)(5.37 m) = 10.0 C Hence, the boiling and freezing points of the solution are Boiling point = (normal bp of solvent) + T b = C C = C Freezing point = (normal fp of solvent) Tf = 0.0 C 10.0 C = 10.0 C 42

22 Sample Exercise 13.9 Freezing-Point Depression in Aqueous List the following aqueous solutions in order of their expected freezing point: m CaCl 2, 0.15 m NaCl, 0.10 m HCl, m CH 3 COOH, 0.10 m C 12 H 22 O 11. Solution 0.15 m NaCl (lowest freezing point), < 0.10 m HCl < m CaCl 2 < 0.10 m C 12 H 22 O 11 < and m CH 3 COOH (highest freezing point). Practice Exercise Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(NO 3 ) 2, 2 mol of KCl, 3 mol of ethylene glycol (C 2 H 6 O 2 )? Answer: 2 mol of KCl because it contains the highest concentration of particles, 2 m K + and 2 m Cl, giving 4 m in all 43 Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so. 44

23 Osmosis 45 Osmosis 46 In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the area of lower solvent concentration (higher solute concentration).

24 Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure,, is = ( n V )RT = MRT where M is the molarity of the solution. 47 If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic. Sample Exercise Calculating Involving Osmotic Pressure The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose (C 6 H 12 O 6 ) will be isotonic with blood? Solution Practice Exercise What is the osmotic pressure at 20 C of a M sucrose (C 12 H 22 O 11 ) solution? Answer: atm, or 37 torr 48

25 Osmosis in Blood Cells If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic. 49 Water will flow out of the cell, and crenation results. Osmosis in Cells If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic. Water will flow into the cell, and hemolysis results. 50

26 Colloids Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity, are called colloids. Particle size: 5 to 1000 nm. A colloid particle may consist of a single giant molecule. Ex) hemoglobin has molecular dimensions of nm and a molar mass of 64,500 g/mol. 51 Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect. 52

27 Colloids in Biological Systems Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (waterhating) end. 53 Colloids in Biological Systems These molecules can aid in the emulsification of fats and oils in aqueous solutions. 54

28 Problems 10, 16, 38, 60, 80, 86, 108,

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