Power Point Lecture Presentation by J David Robertson
Power. Point Lecture Presentation by J. David Robertson University of Missouri Physical Properties of Solutions Chapter 12 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.
LEARNING OBJECTIVES • To understand the concept of solution & solution process; • To be able to describe the concentration of a solution in the most appropriate way;
A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount 12. 1
In the process of making solutions, intermolecular forces are rearranged. The H-bonds in water have to be interrupted. Ion-dipole forces form: Na+ … -OH 2 , Cl- … +H 2 O
A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12. 1
Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction H 1 = separation of solvent molecules H 2 = separation of solute molecules H 3 = formation of solute-solvent interaction DHsoln = DH 1 + DH 2 + DH 3 Hsoln can either be positive or negative depending on the intermolecular forces. 12. 2
Revision question: What is the strongest type of intermolecular forces between solute and solvent in each of the following solution? Answer: a) Cs. Cl (s) in H 2 O (ℓ) Ion-dipole forces b) CH 3 OH (ℓ) in CCl 4 (ℓ) Dipole-induced dipole forces c) CH 3 Cl (g) in CH 3 OCH 3 (g) Dipole-dipole forces d) CH 3 OCH 3 (g) in H 2 O (ℓ) Hydrogen bonding e) Br 2 (ℓ) in CCl 4 (ℓ) Dispersion forces
Energy Changes and Solution Formation Breaking intermolecular forces is always endothermic. Forming new intermolecular forces is always exothermic. To determine whether Hsoln is positive or negative, we consider the strengths of all solute-solute and solutesolvent interactions: 1. 2. 3. H 1 and H 2 are both positive. H 3 is always negative. It is possible to have either: H 3 > ( H 1 + H 2) or H 3 < ( H 1 + H 2).
Hsoln negative Ca. Cl 2 (s) : Hsoln = -81. 3 k. J/mol Mg. SO 4 (s) : Hsoln = -91. 2 k. J/mol Hsoln positive. NH 4 NO 3 (s) Hsoln = +25. 7 k. J/mol
Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass (w/w %) mass of solute x 100% % by mass = mass of solute + mass of solvent mass of solute x 100% = mass of solution Mole Fraction (X) moles of A XA = sum of moles of all components 12. 3
Concentration Units (cont. ) Molarity (M) M = moles of solute liters of solution Molality (m) m = moles of solute mass of solvent (kg) 12. 3
Concentration Units (cont. ) Parts per million (ppm): This unit of concentration may be expressed in a number of ways. It is often used to express the concentration of very dilute solutions. The "technical" definition of parts per million is: ppm = 1 g solute 1000 g solution Parts per million may also be expressed in the following two ways:
Concentration Units (cont. ) Parts per billion (ppb): This concentration unit is also used for very dilute solutions. The "technical" definition is as follows: 1 g solute ppb = 1000 g solution Owing to the dilute nature of the solution, once again, the density of the solution will be about the same as the density of the solvent. Thus, we may also express parts per billion as:
Question #1: Sterile saline solutions containing Na. Cl in water are often used in medicine. What is the percent by mass of Na. Cl in a solution made by dissolving 4. 6 g Na. Cl in 500 g of pure water? Answer: % by mass = = mass of solute + mass of solvent 4. 6 g + 500 g = 0. 91 % x 100%
Question #2: Calculate the molality of a solution containing 88. 4 g glycine (NH 2 COOH) dissolved in 1. 250 kg H 2 O Answer: Mole glycine = 88. 4 g 1 mol = 1. 178 mol 75. 07 g molality = Moles solute Mass solvent (kg) = 1. 178 mol 1. 250 kg = 0. 9424 molal
Question #3: A sample of water is found to contain 0. 010 ppm lead ions (Pb 2+). a) What is the mass of lead ions per liter of this solution? Assume density of the solution is 1. 0 g/m. L. b) What is the lead concentration in ppb? Answer: 0. 010 g Pb 2+ 1 g solution 1000 m. L a) x x 1000 g solution 1 m. L solution 1 L = 1 x 10– 5 g Pb 2+/L solution b) 1 x 10– 5 g Pb 2+ 1 L solution x 106 g 1 g = 10 g Pb 2+/L solution = 10 ppb Pb 2+
Question #4: Hydrochloric acid is sold as a concentrated aqueous solution of HCl with a density of 1. 18 g/m. L. The concentrated acid contains 38% HCl by mass. Calculate the molarity of HCl in this solution. Answer: 38% HCl by mass means that 100 g solution contains 38 g HCl Mole HCl = 38 g 1 mol HCl = 1. 041 mol HCl 36. 5 g HCl Volume of HCl = 100 g molarity = 1 m. L 1 L = 0. 0847 L 1. 18 g 1000 m. L Moles solute L solvent = 1. 041 mol 0. 0847 L = 12. 3 M
Question #5: What is the molality of a 5. 86 M ethanol (C 2 H 5 OH) solution whose density is 0. 927 g/m. L? moles of solute m = M = mass of solvent (kg) liters of solution Answer: Assume 1 L of solution Moles of ethanol = 5. 86 mol Mass of ethanol = 270 g Mass of solution = 0. 927 g/m. L 1000 m. L = 927 g Mass solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g molality = Moles solute Mass solvent (kg) = 5. 86 mol 0. 657 kg = 8. 92 molal 12. 3
Question #6: Vodka is essentially a solution of pure ethanol in water. A typical vodka is sold as “ 80 proof”, which means that it contains 40. 0% ethanol by volume. The density of pure ethanol is 0. 789 g/m. L at 20 o. C. If we assume that the volume of the solution is the sum of the volumes of the components, what are (a) the mass percent (b) the mole fraction (c) the molarity (d) the molality of ethanol in “ 80 proof” vodka?
Assume 1 L of solution: Volume of ethanol = 40. 0% x 1000 m. L = 400 m. L Volume of water = 1000 m. L – 400 m. L = 600 m. L Mass of ethanol = 40. 0% x 0. 789 g/m. L = 315. 6 g Mass of water = 600 m. L x 1. 0 g/m. L = 600 g (a) the mass percent of ethanol in vodka: 315. 6 g % ethanol = = 34. 5 % x 100% 315. 6 g + 600 g
(b) the mole fraction Mole ethanol = 315. 6 g 1 mol 46. 07 g = 6. 85 mol Mole water = 1 mol 18. 02 g = 33. 3 mol 600 g Mole fraction of ethanol in vodka XEt. OH = c) molarity = 6. 85 = 0. 171 6. 85 mol + 33. 3 mol 6. 85 mol = 6. 85 M 1 L d) molality = 6. 85 mol 0. 6 kg = 11. 4 m
LEARNING OBJECTIVES • To understand the relationship between solubility and molecular structure; • To understand the relationships between temperature, pressure and solubility;
Factors Affecting Solubility: (1) Molecular structure Like-Dissolves-Like Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar solutes are soluble in non-polar solvents CCl 4 in C 6 H 6 • polar solutes are soluble in polar solvents C 2 H 5 OH in H 2 O • ionic compounds are more soluble in polar solvents Na. Cl in H 2 O or NH 3 (l) If both solute and solvent are in liquid forms, solubility can be interpreted as ‘miscible’ or ‘immiscible’. 12. 2
e. g. of like-dissolves-like : soaps & detergents Polar head group (hydrophilic) Non-polar tail (hydrophobic) Head group dissolves in water and tail dissolves in grease. Tails sink into blob of grease up as far as head groups & head groups remain on surface. Head groups interact favourably with water and the whole blob can be dissolved in water and be washed away.
Action of soap on oil
QUESTION #7: Predict which of these substances will be most soluble and which will be the least soluble in water: Li. Cl ; Benzoic acid (C 6 H 5 COOH) ; Napthalene Answer: Li. Cl is an ionic compound, and it is highly soluble in water (polar). Benzoic acid has polar carboxylic group and a non-polar aromatic ring. The polar group is soluble in water (polar). Napthalene is a non-polar compound, it is not soluble in water (polar). Napthalene < benzoic acid < Li. Cl Increasing solubility 11. 2
QUESTION #8: Arrange the following compounds in order of their expected increasing solubility in water: Br 2 KBr Toluene (C 7 H 8, a constituent of gasoline). Answer: Toluene and Br 2 are both non-polar and toluene has a larger size than Br 2. KBr is ionic and easily dissolve in water. Toluene < Br 2 < KBr Increasing solubility
Factors Affecting Solubility: (2) Pressure effect (for gases in any solvent) Solubility of a gas in a solvent is a function of the pressure of the gas. The higher the pressure, the more molecules of gas becoming closer to the solvent and the greater the chance of a gas molecule to strike the surface and entering the solution. Therefore, the higher the pressure, the greater the solubility of gas. The lower the pressure, the fewer molecules of gas are close to the solvent and the lower the solubility.
The concentration of dissolved gas depends on the partial pressure of the gas. The partial pressure controls the number of gas molecule collisions with the surface of the solution. If the partial pressure is doubled the number of collisions with the surface will double. The increased number of collisions produce more dissolved gas. Low pressure Low concentration Double the pressure Double the concentration
Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas c = k. P P is the pressure of the gas over the solution k is a constant (mol/L • atm) that depends only on temperature low P high P low c high c 12. 5
Henry’s Law explains why soft drinks “fizz” and then go “flat” after being opened. Bubbles of CO 2 form as soon as a carbonated beverage is opened because the drink was bottled under CO 2 at a pressure higher than 1 atm. When the bottle is opened, the pressure of CO 2 above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles.
Factors Affecting Solubility: (3) Temperature effect (for gases in any solvent) Gas solubility and temperature solubility usually decreases with increasing temperature 12. 4
Experience tells us that carbonated beverages go flat as they get warm. Gases are less soluble at higher temperature. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution.
What is Thermal Pollution? The discharge of heated liquid or air into lakes, rivers, etc. , as by an industry or nuclear power plant, causing such a rise in the water temperature as to affect the life cycles within the water and disrupt the ecological balance. If lakes get too warm, CO 2 and O 2 become less soluble and are not available for plants or animals.
Factors Affecting Solubility: (3) Temperature effect (for solids in any solvent) Solid solubility and temperature: solubility increases with increasing temperature solubility decreases with increasing temperature 12. 4
Ionic compounds Experience tells us that sugar dissolves better in warm water than cold water. Generally, as temperature increases, solubility of solids increases. But sometimes, solubility decreases as temperature increases.
Although the solubility of a substance generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. In chemistry, fractional crystallization is a method of refining substances based on differences in solubility. If two or more substances are dissolved in a solvent, they will crystallize out of solution (precipitate) at different rates. Crystallization can be induced by changes in concentration, temperature or other means. This technique is often used in chemical engineering to obtain very pure substances, or to recover sellable products from waste solutions.
Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO 3 contaminated with 10 g Na. Cl. Fractional crystallization: 1. Dissolve sample in 100 m. L of water at 600 C 2. Cool solution to 00 C 3. All Na. Cl will stay in solution (s = 34. 2 g/100 g) 4. 78 g of PURE KNO 3 will precipitate (s = 12 g/100 g). 90 g – 12 g = 78 g 12. 4
LEARNING OBJECTIVES • To understand the colligative properties of a solution. • To understand the effect of intermolecular forces on the colligative properties of a solution.
Physical Behavior of Solutions: Colligative Properties When, comparing a pure solvent and a solution, the solution’s • vapour pressure is lower; • boiling point is elevated; • freezing point is lower; than the pure solvent. And, osmosis occurs from solvent to solution when separated by a membrane.
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 P 1 = X 1 P 0 1 Raoult’s law P 10 = vapor pressure of pure solvent X 1 = mole fraction of the solvent If the solution contains only one solute: X 1 = 1 – X 2 P 10 - P 1 = DP = X 2 P 10 X 2 = mole fraction of the solute 12. 6
Ideal Solution PA = XA P 0 A PB = XB P 0 B PT = PA + PB PT = XA P 0 A + XB P 0 B 12. 6
QUESTION #9: The vapor pressures of pure benzene and toluene at 25°C are 95. 1 and 28. 4 mm Hg, respectively. A solution is prepared in which the mole fractions of benzene and toluene are both 0. 500. (a) What are the partial pressures of the benzene and toluene above this solution? (b) What is the total vapor pressure? Answer: a) PBenzene = XBenzene x Po. Benzene = 0. 500 x 95. 1 mm. Hg = 47. 6 mm. Hg PToluene = XToluene x Po. Benzene = 0. 500 x 28. 4 mm. Hg = 14. 2 mm. Hg b) PTotal = PBenzene + PToluene = 61. 8 mm. Hg
Solutions that obey Raoult’s Law are called ‘ideal solution’. Most real solutions exhibit positive or negative deviation from Raoult’s Law, just like most real gases that do not obey the ‘Ideal gas Law’. Positive deviation or negative deviation can be distinguished depending on the intermolecular forces between solute-solvent, solvent-solvent and solute-solute.
Negative deviation: Methanol in acetone: Methanol contains Hbonding and acetone is a polar substance. PT is less than predicted by Raoults’s law Acetone can forms acetonemethanol (A-B) interactions that are stronger than acetone (A-A) or methanol (B-B) interactions. The A-B interactions effectively stabilize the solution thus, lowering the vapour pressure. Force > A-A & B-B A-B H soln = -ve 12. 6
PT is greater than predicted by Raoults’s law Force < A-A & B-B A-B H soln = +ve Positive deviation: Cyclohexane in ethanol: Cyclohexane is a non-polar substance and ethanol contains H-bonding. The non-polar cyclohexane cannot interact favourably with the polar ethanol, therefore the cyclohexane-ethanol interactions (A-B) is weaker than both ethanol-ethanol and cyclohexane-cyclohexane. Both cyclohexane & ethanol have the tendency to escape from the solution, thus, increasing the vapour pressure. 12. 6
Fractional Distillation Apparatus A procedure to separate liquid components based on their different boiling points. 12. 6
Boiling-Point Elevation DTb = Tb – T b 0 is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b 0 DTb > 0 Tb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0 C/m) 12. 6
QUESTION #10: Calculate the boiling point of a 30. 2% aqueous solution of ethylene glycol (EG). Given: Kb (water) = 0. 51 o. C/m. Answer: 30. 2% aqueous solution of EG means 30. 2 g of EG in 100 g solution mass water = 100 g – 30. 2 g = 69. 8 g 30. 2 g EG x Molality of EG = 1 mol EG 62. 01 g EG 0. 0698 kg = 6. 98 m ∆Tb = Kb m = (0. 51 o. C/m)(6. 98 m) = 3. 6 o. C Boiling point of pure water = 100 o. C Boiling point of EG solution = 100 o. C + 3. 6 o. C = 103. 6 o. C
Freezing-Point Depression DTf = T 0 f – Tf T 0 f is the freezing point of the pure solvent T f is the freezing point of the solution T 0 f > Tf DTf > 0 Tf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0 C/m) 12. 6
Question #11: What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62. 01 g. DTf = Kf m Kf water = 1. 86 0 C/m Answer: 478 g EG x Molality of EG = 1 mol EG 62. 01 g EG 3. 202 kg = 2. 41 m ∆Tf = Kf m = (1. 86 0 C/m)(2. 41 m) = 4. 48 o. C Freezing point of pure water = 0 o. C Freezing point of EG solution = 0 o. C – 4. 48 o. C = – 4. 48 o. C 12. 6
12. 6
Osmotic Pressure ( ) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semi-permeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. dilute more concentrated 12. 6
Osmotic Pressure ( ) High P Low P p = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 12. 6
Question #12: Assume that the fluids inside a sausage are approximately 0. 80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100 o. C to learn why experienced cooks pierced the semi-permeable skin of sausages before boiling them. Answer: = MRT = (0. 80 mol/L)(0. 0821 L. atm/mol. K)(373 K) = 24. 5 atm
Reverse Osmosis : Desalination • Application of a pressure to the solution (that is equal to or greater than the Osmotic Pressure) and the solvent flows from the more concentrated side to the other one. • This process is used to obtain pure water from salt water.
A cell in an: Isotonic solution Equal concentration, same osmotic pressure, the cell remains unchanged. Hypotonic solution Dilute solution outside the cell, the cell swells. Hypertonic solution Concentrated solution outside the cell, the cell shrinks. 12. 6
Example of Osmosis: Cucumber placed in Na. Cl solution loses water to shrivel up and become a pickle. Limp carrot placed in water becomes firm because water enters via osmosis. ‘Salt added to meat’ or ‘sugar added to fruit’ prevents bacterial infection (a bacterium placed on the salt will lose water through osmosis and die).
Colligative Properties of Non-electrolyte 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 P 1 = X 1 P 10 Boiling-Point Elevation Tb = Kb m Freezing-Point Depression Tf = Kf m Osmotic Pressure ( ) p = MRT 12. 6
Colligative Properties of Electrolyte Solutions 0. 1 m Na. Cl solution 0. 1 m Na+ ions & 0. 1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. 0. 1 m Na. Cl solution Van’t Hoff = factor, i 0. 2 m ions in solution actual number of particles in solution after dissociation number of formula units initially dissolved in solution i should be nonelectrolytes Na. Cl Ca. Cl 2 1 2 3 12. 7
Colligative Properties of Electrolyte Solutions Boiling-Point Elevation Tb = i Kb m Freezing-Point Depression Tf = i Kf m Osmotic Pressure ( ) p = i MRT 12. 7
Question #13: Calculate the van’t Hoff factor for a 0. 050 m aqueous solution of Mg. Cl 2 that has a measured freezing point of – 0. 25 o. C. Given Kf (H 2 O) = 1. 86 o. C/m. Answer: Tf = i Kf m – 0. 25 o. C = (i)(1. 86 o. C/m)(0. 050 m) i (measured) i = 2. 7 Mg. Cl 2 Mg 2+ + 2 Cl– i =3 i (calculated) i (measured) is lower than i (calculated) because of the formation of ion-pair : – electrostatic force between cation(s) and anion(s).
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