Ch 13 The Properties of Mixtures: Solutions and Colloids
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1 Ch 13 The Properties of Mixtures: Solutions and Colloids Key equations: Concentration Unit - Quantitative Ways of Expressing Concentration Principles of Solubility Colligative Properties of Solutions nonelectrolyte and electrolyte The Structure and Properties of Colloids Henry s law: S gas = k H P gas Raoult s law: P solvent = χ solvent X P 0 solvent where P 0 solvent -the vapor pressure of the pure solvent Boiling Point Elevation and Freezing Point Depression T b = K b m T f = K f m Osmotic Pressure π = M R T where M is the molarity, R is the ideal gas law constant and T is the Kelvin temperature Table 1 Concentration Definitions Concentration Term Molarity (M) Molality (m) Parts by mass Parts by volume Mole fraction (χ) Ratio amount (mol) of solute volume (L) of solution amount (mol) of solute mass (kg) of solvent mass of solute mass of solution volume of solute volume of solution amount (mol) of solute amount (mol) of solute + amount (mol) of solvent 1
2 SAMPLE PROBLEM Calculating Molality PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl 2 in 271 g of water? PLAN: We have to convert the grams of CaCl 2 to moles and the grams of water to kg. Then substitute into the equation for molality. SOLUTION: mole CaCl g CaCl 2 x g CaCl 2 = mole CaCl 2 molality = mole CaCl g H 2 O x kg 10 3 g = 1.06 m CaCl 2 SAMPLE PROBLEM 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 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 (C 3 H 7 OH) and 58.0 g of water. What are the mole fractions of alcohol and water? PLAN: (a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply by (b) Knowing the % alcohol and total volume, we can find volume of alcohol. (c) Convert g of solute and solvent to moles; find the ratios of parts to the total. 2
3 SAMPLE PROBLEM continued Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction SOLUTION: (a) 40.5 mg Ca x 3.5 g g 10 3 mg x 10 6 = 1.16x10 4 ppm Ca (b) (c) L chianti x 11.5 L alcohol = L alcohol 100 L chianti moles ethylene glycol = 142 g moles water = 38.0g mole g = 2.36 mol C 2 H 6 mole g = 3.22 mol H 2 O 2.39 mol C 2 H mol C 2 H mol H 2 O 3.22 mol H 2 O 2.39 mol C 2 H mol H 2 O = c C 2 H 6 = c H 2 O Interconverting Concentration Terms To convert a term based on amount (mol) to one based on mass, you need the molar mass. These conversions are similar to mass-mole conversions. To convert a term based on mass to one based on volume, you need the solution density. Working with the mass of a solution and the density (mass/volume), you can obtain volume from mass and mass from volume. Molalityinvolves quantity of solvent, whereas the other concentration terms involve quantity of solution. 3
4 Take home message SAMPLE PROBLEM Converting Concentration Units PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution a a hair bleach. An aqueous solution H 2 is 30.0% by mass and has a density of 1.11 g/ml. Calculate its PLAN: (a) Molality (b) Mole fraction of H 2 (c) Molarity (a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide. (b) Convert g of solute and solvent to moles before finding mole fraction. (c) Use the density to find the volume of the solution. SOLUTION: (a) g of H 2 O = 100. g solution g H 2 = 70.0 g H 2 O molality= 30.0 g H 2 mol H g H2 O g H 2 O kg H 2 O 10 3 g mol H 2 = 12.6 m H 2 Take home message SAMPLE PROBLEM Converting Concentration Units continued (b) 70.0 g H 2 O mol H 2 O g H 2 O = 3.88 mol H 2 O mol H mol H mol H 2 O = χ of H 2 (c) g solution ml 1.11 g = 90.1 ml solution mol H ml solution L 10 3 ml = 9.79 M H 2 4
5 LIKE DISSOLVES LIKE Substances with similar types of intermolecular forces dissolve in each other. When a solute dissolves in a solvent, solute-solute interactions and solventsolvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur. Like dissolves like: solubility of methanol in water. water methanol A solution of methanol in water 5
6 SAMPLE PROBLEM PROBLEM: Predicting Relative Solubilities of Substances Predict which solvent will dissolve more of the given solute: (a) Sodium chloride (NaCl) in methanol (CH 3 OH) or in propanol(ch 3 CH 2 CH 2 OH) (b) Ethylene glycol (HOCH 2 CH 2 OH) in hexane (CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 ) or in water (H 2 O). (c) Diethyl ether (CH 3 CH 2 OCH 2 CH 3 ) in water (H 2 O) or in ethanol (CH 3 CH 2 OH) PLAN: Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute. SOLUTION: (a) Methanol -NaClis ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanolis subject to the dispersion forces to a greater extent. (b) Water -Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol. (c) Ethanol -Diethyl ether can interact through a dipole and dispersion forces. Ethanol can provide both while water would like to H bond. The effect of pressure on gas solubility. 6
7 Henry s Law S gas = k H X P gas The solubility of a gas (S gas ) is directly proportional to the partial pressure of the gas (P gas ) above the solution. SAMPLE PROBLEM Using Henry s Law to Calculate Gas Solubility PROBLEM: PLAN: The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25 0 C. What is the solubility of C? The Henry s law constant for C dissolved in water is 3.3 x10-2 mol/l*atmat 25 0 C. Knowing k H and P gas, we can substitute into the Henry s law equation. SOLUTION: S = (3.3 x10-2 mol/l*atm)(4 atm) = C 0.1 mol/l Colligative Properties : Property of solution is different from the those of the pure solvent. Property of solution depends on the concentration of solute rather than the nature of solute. Vapor pressure lowering, boiling point elevation, freezing pointdepression, osmotic pressure Raoult s Law (vapor pressure of a solvent above a solution, P solvent ) P solvent = χ solvent X P 0 solvent where P 0 solvent is the vapor pressure of the pure solvent P 0 solvent -P solvent = P = χ solute x P0 solvent Boiling Point Elevation and Freezing Point Depression T b = K b m T f = K f m K b is the modal boiling point constant, K f is the modal freezing point constant. Osmotic Pressure π = M R T where M is the molarity, R is the ideal gas law constant and T is the Kelvin temperature 7
8 SAMPLE PROBLEM Using Raoult s Law to Find the Vapor Pressure Lowering PROBLEM: PLAN: SOLUTION: Calculate the vapor pressure lowering, P, when 10.0 ml of glycerol (C 3 H 8 O 3 ) is added to 500. ml of water at C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is g/ml. The density of glycerol is 1.26 g/ml. Find the mol fraction, χ, of glycerol in solution and multiply by the vapor pressure of water g C 3 H 8 O 3 mol C 10.0 ml C 3 H 8 O 3 H 8 O 3 3 x = mol C ml C 3 H 8 O g C 3 H 8 O 3 H 8 O ml H 2 O g H 2 O ml H 2 O P = mol C 3 H 8 O mol C 3 H 8 O mol H 2 O x mol H 2 O g H 2 O x χ = torr = 27.4 mol H 2 O = torr SAMPLE PROBLEM Determining the Boiling Point Elevation and Freezing Point Depression of a Solution PROBLEM: You add 1.00 kg of ethylene glycol (C 2 H 6 ) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution? PLAN: Find the # mols of ethylene glycol; m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water. SOLUTION: 1.00x10 3 g C 2 H 6 mol C 2 H g C 2 H 6 = 16.1 mol C 2 H mol C 2 H kg H 2 O = 3.62 m C 2 H 6 T bp = C/m x 3.62m = C T fp = C/m x 3.62m BP = C FP = C 8
9 Osmotic Pressure The process, taking place thought a membrance permeable only to the solvent is called osmosis. The symbol for osmotic pressure is π. Pure solvent semipermeable membrane Solution π α n solute or π α M V solution n solute π = RT = MRT V solution Net movement of solvent causes the osmotic pressure. SAMPLE PROBLEM Determining Molar Mass from Osmotic Pressure PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studying a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at C to make 1.50 mlof solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin? PLAN: We know p as well as R and T. Convert pto atmand T to degrees K. Use the pequation to find M and then the amount and volume of the sampleto get to M. SOLUTION: M = π RT = 3.61 torr atm 760 torr = 2.08 x10-4 M ( L*atm/mol*K)(278.1 K) 2.08 x10-4 mol L (1.50 ml) L 10 3 ml = 3.12x10-8 mol 21.5 mg g 10 3 mg 1 = 6.89 x10 4 g/mol 3.12 x10-8 mol 9
10 The relation between solubility and temperature for several ionic compounds. Phase diagrams of solvent and solution. Vapor pressure lowering Boiling point elevation BP of a solution is the T at which the vapor pressure equals the external pressure. Higher pressure is required to raise the solution s vapor pressure. Freezing point depression FP of a solution is the T at which the vapor pressure equals that of pure solvent. P soln < P solv, the solution freezes at a lower T than that of solvent. 10
11 The effect of a solute on the vapor pressure of a solution. The vaporization occurs due to the tendency of a system to become more disordered. Equilibrium is established when the # of molecules vaporingand condensing are equal. The presence of dissolved solute already increases the disorder of the system, so fewer solvent molecules will vaporize. The equilibrium is attained at lower vapor pressure. Colligative Properties of Electrolyte Solutions For electrolyte solutions, the compound formula tells us how many particles are in the solution. The van t Hoft factor, i, tells us what the effective number of ions are in the solution. van t Hoff factor (i) i = measured value for electrolyte solution expected value for nonelectrolyte solution For vapor pressure lowering: P = i(χ solute xp 0 solvent ) For boiling point elevation: T b = i(κ b m) For freezing point depression: T f = i(κ f m) For osmotic pressure : π = i(mrt) 11
12 Osmotic Pressure The process, taking place thought a membrance permeable only to the solvent is called osmosis. The symbol for osmotic pressure is π. Pure solvent semipermeable membrane Solution π α n solute or π α M V solution n solute π = RT = MRT V solution Net movement of solvent causes the osmotic pressure. Colligative Properties of Electrolyte Solutions For electrolyte solutions, the compound formula tells us how many particles are in the solution. The van thoft factor, i, tells us what the effective number of ions are in the solution. Ideally, the van t Hoff van thoff factor (i) factor i = moles of particles / i = measured value for electrolyte solution moles of dissolved solute expected value for nonelectrolyte solution For vapor pressure lowering: P = i(χ solute xp 0 solvent ) For boiling point elevation: T b = i(κ b m) For freezing point depression: T f = i(κ f m) NaCl : 2 MgCl 2 : 3 Ba(NO 3 ) 2 : 3 For osmotic pressure : π = i(mrt) 12
13 The three types of electrolytes. STRONG NaCl Weak CH 3 COOH Nonelectrolyte Glucose SAMPLE PROBLEM Depicting a Solution to Find Its Colligative Properties PROBLEM: A g sample of magnesium chloride is dissolved in 100. g of water in a flask. (a) Which scene depicts the solution best? (b) What is the amount (mol) represented by cation ion and anion? (c) Assuming the solution is ideal, what is its freezing point (at 1 atm)? PLAN: (a) Consider the formula for magnesium chloride, an ionic compound. (b) Use the answer to part (a), the mass given, and the mol mass. (c) The total number of mols of cations and anions, mass of solvent, and equation for freezing point depression can be used to find the new freezing point of the solution. 13
14 SAMPLE PROBLEM continued Depicting a Solution to Find Its Colligative Properties (a) The formula for magnesium chloride is MgCl 2 ; therefore the correct depiction must be A with a ratio of 2 Cl - / 1 Mg 2+. (b) g MgCl 2 mols MgCl 2 = g MgCl 2 = mol MgCl 2 mol MgCl 2 mols Mg 2+ = mol MgCl 2 mols Cl - = mol MgCl 2 x 2 mols Cl - 1 mol MgCl 2 = mols Cl - SAMPLE PROBLEM continued Depicting a Solution to Find Its Colligative Properties (c) molality (m) = mol MgCl 2 1 kg 100. g x 10 3 g = m MgCl 2 Assuming this is an IDEAL solution, the van t Hoff factor, i, should be 3. T f = i (K f m) = 3( C/m x m) = C T f = C C = C 14
15 Colloid is the heterogeneous mixture containing particles larger enough to be seen by physical eyes. Colloidal particles diameter ranges from 10-9 to 10-6 m or nm nm = 1mm Types of Colloids Colloid Type Dispersed Substance Dispersing Medium Example Aerosol Aerosol Foam Solid foam Emulsion Solid emulsion Sol Solid sol Liquid Gas Fog Solid Gas Smoke Gas Liquid Whipped cream Gas Solid Marshmallow Liquid Liquid Milk Liquid Solid Butter Solid Liquid Paint; cell fluid Solid Solid Opal Light scattering and the Tyndall effect. Solution Colloid Tyndall effect : When a beam of light goes through the colloid, its pathway is broadened and is readily visible. This phenomenon results from the lightscattering, which is called Tyndall effect. The Sunlight is scattered as it shines through the misty air in a forest. Photo by C.A.Bailey, CalPoly SLO (Inlay Lake, Myanmar) Brownian motion : a characteristic movement in which particles change speed and direction erratically. 15
16 A Cottrell precipitator for removing particulates from industrial smokestack gases. Colloidal particles remain dispersed due to the interparticlesforces. Various methods can coagulate the particles and destroy the colloid. Heating a colloid can make the particles moving faster and collide more frequently, enhancing the heavier particles to be settled out. Adding electrolyte solution introduces the oppositely charged ions that neutralize the particle's surface charges, allowing the partilcesto agglomerate. 16
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