Chap 13B - Chem 210 / WANG / Chapter 13B Colligative...

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Unformatted text preview: Chem 210 / WANG / Chapter 13B Colligative Properties of Solution Chapter 13 Properties of Solutions 13.1 The Solution Process 13.2 Saturated Solutions and Solubility 13.3 Factors Affecting Solubility 13.4 Ways of Expressing Concentration 13.5 Colligative Properties 13.6 Colloids Group of properties caused by the same reason. the • vapor pressure lowering • freezing point depression • boiling point elevation • osmotic pressure “The same reason = property based on reason” the number the number of solute particles but not on the type (cation, anion, molecule, etc.) of solute. type cation, NOTE: electrolytes dissociate into ions and produce larger number of particles per mole larger than non-electrolytes. non- Colligative Property: Vapor Vapor Pressure Lowering The vapor pressure of a solution is lower lower than that of the pure solvent. There are two general classes of solutes. Electrolytic • ionic compounds in polar solvents • dissociate in solution to make ions • 100% dissociation = strong electrolyte • less than 100% dissociation = weak electrolyte • conduct electricity Nonelectrolytic • do not conduct electricity • solute is dispersed but does not dissociate Raoult’s Raoult’s law The effect of a solute on the vapor pressure of a solution. Based on Raoult’s law, the lowering of the vapor pressure of the solvent is directly proportional to the mole fraction (XB) of the (nonvolatile) solute. solute. Given Raoult’s law: PA = XA PAº law: Lowering = PAº - PA = PAº - XA PAº = PAº (1- XA) = PAº XB PAº - PA = PAº XB mole fraction of B Fall 2009 UM-SJTU JI (A = solvent) The vapor pressure of the vapor solvent solvent (PA) in a solution is equal to the mole fraction mole of the solvent (XA) solvent vapor multiplied by the vapor pressure of the pure pure solvent (PAº) at the same temperature. In a sealed chamber, water vapor from the clear cylinder condenses condenses and adds to the blue solution, because the adds blue blue solution has lower vapor pressure. lower Raoult’s law and vapor vapor pressure lowering Figure 13.20 vapor vapor pressure ∝ mole fraction of the solvent. PA = XAPAº Fewer water molecules are at the surface to escape as vapor, because solute particles are taking up space at the surface. lowering of PA = Properties of aqueous solutions Colligative properties of A Solution SAMPLE PROBLEM 1 Calculate the vapor pressure of water from a solution which consists of 3.00×102 g of urea (NH2CONH2) dissolved in 5.00×102 g of water. The vapor pressure of water is 23.8 mmHg at 25°C. Roult’s law: PH2O = XH2O PH2Oº moleH2O = 5.00×102 g H2O / 18.0 g = 27.8 mole moleurea = 3.00×102 g urea / 60.0g = 5.00 mole Total moles = 32.8 mole XH2O = moleH2O ÷(total moles) = 27.8/32.8 = 0.8476 (keep more sig fig ‘till the end) PH2O = (0.8476) x 23.8 mmHg = 20.2 mmHg 20.2 SAMPLE PROBLEM 2 Calculate the vapor pressure lowering, ∆P, 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. 10.0 mL 1.26 g C3H8O3 mol C3H8O3 C3H8O3 mL C H O x 92.09 g C3H8O3 383 = 0.137 mol C3H8O3 500.0 mL 0.988 g H2O x mol H2O H 2O 18.02 g H2O mL H2O = 27.4 mol H2O ∆P = 0.137 mol C3H8O3 0.137 mol C3H8O3 + 27.4 mol H2O mf = 0.00498 x92.5 torr = 0.461 torr Page 1 Chem 210 / WANG / Chapter 13B Colligative Properties of Solution Ref: Phase diagrams of solvent and solution. Ref: Phase diagrams of solvent and solution. Ref: Phase diagrams of solvent and solution. Figure Figure Figure 13.22 13.22 13.22 FreezingFreezing-Point Depression VaporVapor-Pressure Lowering ∆PA = XB P Ao BoilingBoiling-Point Elevation When you add a solute (nonvolatile) to a solvent, the boiling point will go up. • Solute mixed in with solvent molecules • Binds solvent molecules and escaping prevent them from escaping into gaseous state, thus lowering lowering vapor pressure • Requires higher temp to reach atm. pressure • Boiling point elevates to elevates higher temps Colligative Property: Freezing Freezing point depression Adding salt to ice water lowers the temperature to below freezing point of pure water. Example “Salting” roads in winter prevents freezing Fall 2009 UM-SJTU JI Colligative Property: Freezing Freezing point depression ∆Tb = BP of solution – BP of pure solvent ∆Tb = Kbm Kb = molal boiling point elevation constant For water, Kb = 0.52ºC / m 0.52ºC When you add a solute to a solvent, the melting point goes down. “m” = molality molality Freezing Point of 1 kg water is lowered 1.86 ºC for every mole of solute particles. (BP increases 0.52 ºC for 1 mole solute particles per 1 kg water). • • Molality inovles only mass measurements, which will not change with temperature. It is more precise than Molarity (M, moles per Molarity Liter Liter) because when temperature changes, volume may contract or expand, leading to changes in “L” “M” also changes! Freezing Point (FP) Depression ∆Tf = FP of solution – FP of pure solvent (∆Tf is a negative number!) ∆Tf = - Kf m Salty solution draws out H2O molecules in ice to make the solution more dilute. This induces melting which requires a lot of energy, resulting in lowering the temperature of the solution. ∆ Tb = K b m Boiling point (BP) elevation Colligative Property: Boiling Boiling Point Elevation Example Antifreeze mixed with water in radiator prevents boiling as well as freezing! ∆Tf = -Kf m Example Presence of solute makes it “Salting” roads “hard” for water molecules in winter prevents freezing to line up and settle into ice structure. Ref. Ref. Table 13.4 Molal Boiling Point Elevation and Freezing Point Depresssion Constants of Several Solvents Solvent Boiling Point (0C)* Kb (0C/m) Melting Point (0C) Kf (0C/m) (“-” because ∆Tf is negative) (“Kf = molal freezing point depression constant. For water, Kf = 1.86 ºC/m 1.86 Freezing Point of water is lowered 1.86 ºC for lowered every mole of solute particles per 1 kg of water. Acetic acid 117.9 3.07 16.6 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 Chloroform 61.7 3.63 -23 -63.5 30. 4.70 m = molality NOTE: Both ∆Tb and ∆Tf are directly proportional directly to the number of solute particles per kg of number per solvent. Diethyl ether 34.5 2.02 -116.2 1.79 78.5 1.22 -117.3 1.99 100.0 0.512 0.0 1.86 Ethanol Water *at 1 atm. Page 2 3.90 Chem 210 / WANG / Chapter 13B Colligative Properties of Solution Answer to SAMPLE PROBLEM 3 Page 557 textbook Compare moles solute particles per kg solvent 0.050 m MgCl2 MgCl MgCl2 makes 3 particles: (Mg+2 and 2 Cl-1) particles: 0.050 mole x 3 = 0.15 mole 0.10 m sucrose molecular molecular compound 0.10 mole 0.12 m ethylene glycol molecular molecular compound 0.12 mole 0.070 m NaCl Ionic Ionic 0.070 mole x 2 = 0.14 mole 0.050 m MgCl2 has most particles /kg solvent. solvent. ∆Tf = - kfm = -1.86 x 0.15 = - 0.28 oC oC ∆Tb = kbm = 0.52 x 0.15 = 0.078 F.P. = - 0.28 oC & B.P. = 100.078 oC Modify answers to SAMPLE PROBLEM 3 Lowest freezing point: In 1.0 kg solvent 0.050 m MgCl2 2.7 2.7 instead MgCl MgCl2 makes 3 particles: particles: 2.7 = = 1.5 0.050 mole x 3 0.135 mole 0.10 m sucrose molecular molecular compound 0.10 mole No change NonNon-ideal Solution After dissociation, ions may not be completely “free” from each other due to “ion pairing.” What effect does “ion pairing” have on the number of solute particles present in an electrolyte solution? A 0.952-g sample of magnesium chloride is dissolved in 100. g of water in a flask. (a) formula = MgCl2 correct correct depiction must be “A” with ratio = 2 Cl-/ 1 Mg2+ (b) moles represented by each large sphere of Cl0.952 g MgCl2 mols MgCl2 = 95.21 g/mol MgCl = 0.0100 mol MgCl2 2 mols Cl- =0.0100 mol MgCl2 x 2 mols Cl = 0.0200 mols Cl1 mol MgCl2 mols/sphere = 0.0200 mols Cl= 2.50 x 10-3 moles/sphere 2.50 8 spheres Fall 2009 UM-SJTU JI (ideal electrolyte) i = number of solute particles per mole of solute SAMPLE PROBLEM 4 SAMPLE PROBLEM 5 You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution? A 0.952-g sample of magnesium chloride is dissolved in 100. g of water in a flask. (a) Which scene depicts the solution best? 1.00x103 g C2H6O2 mol C2H6O2 = 16.1 mol C2H6O2 62.07 g C2H6O2 (b) What is the amount (mol) represented by each large sphere of Cl-? (c) Assuming the solution is ideal, what is its freezing point (at 1 atm)? 16.1 mol C2H6O2 = 3.62 m C2H6O2 4.450 kg H2O ∆Tbp = 0.512 0C/m -∆Tfp = 1.86 0C/m x 3.62m x 3.62m = 1.850C BP = 101.85 0C SAMPLE PROBLEM 5 Ion-pairs effectively decrease the number Ioneffectively of particles in electrolyte solutions decrease decrease Colligative Properties expressed expressed by “i = Van’t Hoff Factor” SAMPLE PROBLEM 5 FP = -6.73 0C Colligative Property: Osmotic Osmotic pressure (c) freezing point? molality (m) = 0.0100 mol MgCl2 100. g x 1 kg 103 g = 0.100 m MgCl2 Assuming this is an IDEAL solution, the van’t Hoff factor, i, should be 3. • ∆Tf = i (Kfm) = 3(1.86 0C/m x 0.100 m) = 0.558 0C Tf = 0.000 0C - 0.558 0C = - 0.558 0C • • Pure solvent and a solution are divided by solution a semi permeable membrane; only solvent semi solvent can pass through this membrane Solute particles draws solvent molecules solvent across the membrane osmosis osmosis Increasing volume at the solution side volume solution creates pressure on the membrane pressure osmotic osmotic pressure! Page 3 Chem 210 / WANG / Chapter 13B Colligative Properties of Solution Osmotic Pressure in the body Isotonic - Red blood cell and plasma have the same number of solute particles (no osmotic pressure). Hypertonic plasma – higher number of solute particles outside cell than inside cell w ater moves out & cell collapses. Hypotonic – Lower number of solute particles outside cell cell swells and ruptures. = hemolysis hemolysis - water drawn into the cell and the cell bursts. Osmosis and Osmotic Pressure Osmosis and I.V. (intravenous) solutions Normal Saline (NS) used to replace or replenish body fluids Hypertonic More concentrated than plasma Used to draw water from tissue (decrease fluid build-up). Hypotonic More on Osmotic Pressure equation πV= nRT V = volume of solution N = moles of solute solute. Since “n/V = Molarity (M)” n/V Molarity Re-write the osmotic pressure equation: π= molar mass = 4÷2.87 x10-4 Solve for molar mass = 13900 g/mole Fall 2009 UM-SJTU JI nsolute Vsolution RT = MRT SAMPLE PROBLEM 6 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 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this variety of hemoglobin? 7.04 x 10-3 = MRT = (mole/L) x RT 7.04 x 10-3 = [4.÷ molar mass÷ 1] x 0.0821x(298) [ 7.04 x 10-3 = [4.÷ molar mass÷ 1] x 24.47 [ 7.04 x 10-3 / 24.47 = 4÷ molar mass 24.47 2.87 x10-4 pure solvent SemiSemipermeable membrane net movement of solvent solute molecules Less concentrated than plasma Used to rehydrate tissue πV = nRT Question: Osmotic pressure for a solution of 4.00 g ribonuclease in 1.00 L water is 7.04× 10-3 atm at 25°C. What is the molar mass of the enzyme? solution solvent molecules • V is the volume of the solution. solution. • n is the number of moles of the solute. Osmotic Pressure: used to find the molar used mass of biological molecules Applied pressure needed to prevent volume increase osmotic pressure Isotonic The osmotic pressure of a solution is the pressure required to prevent osmosis. The osmotic pressure (π) of a solution idealobeys a law similar to the ideal-gas law (PV = nRT) as shown in: nRT) Ref: Figure The development of osmotic pressure. 13.23 M= Osmotic Pressure: used to find the molar used mass of biological molecules Question: Osmotic pressure for a solution of 4.00 g ribonuclease in 1.00 L water is 7.04× 10-3 atm at 25°C. What is the molar mass of the enzyme? = MRT = (mole/L) x RT = [g÷ molar mass÷ L] x RT π = 7.04 x 10-3 atm T = 273+25 = 298 K R = 0.0821 atm-L/mole-K g = 4.00g L = 1.00 L Solve for molar mass = 13900 g/mole π Reverse osmosis for the removal of ions. External pressure >> osmotic pressure Reverse Reverse the direction of pure solvent flow! Use: “desalination”, “domestic water treatment” atm 3.61 torr π 760 torr = RT = 2.08 x10-4 M (0.0821 L*atm/mol*K)(278.1 K) L 2.08 x10-4 mol (1.50 mL) = 3.12x10-7 mol L 103 mL g 1 = 6.89 x104 g/mol 21.5 mg 3 10 mg 3.12 x10-7 mol Figure B13.3 Page 4 Chem 210 / WANG / Chapter 13B Colligative Properties of Solution Colligative properties Aqueous Aqueous Solutions Summary vapor pressure: lowering Po– P = XB Po (note: (note: use mole fraction of solute) freezing point: depression ∆Tf = -kf m kf for water = 1.86oC/m (note: use molality. Negative sign due to molality. solution FP is lower than pure solvent.) boiling point: elavation ∆Tb = kb m, kb for water = 0.52oC/m osmotic pressure: π = MRT Figure 13.37 Solutions are only one of two categories of homogeneous homogeneous mixtures Homogeneous mixtures mixtures: Uniform mixture that has the same composition same throughout. • Solution – solute particle sizes very small very example: saline, air, alloy • Colloid – solute particle sizes bigger (2-500 nm) bigger example: soapy water, milk, fog, pearl Heterogeneous mixtures mixtures: NonNon-uniform mixture that has variable variable compositions in different parts. particle sizes = large enough to filter out large Example: blood Light scattering and the Tyndall effect. Difference between Solution and Colloid: Tyndal effect Unlike solutions, colloids exhibit light scattering. Which is colloidal? 1. purple gold sol 2. copper sulfate solution 3. iron(III) hydroxide colloid 1 2 3 Colloids exist in various states Dispersing Dispersed medium phase Name Example Gas Gas Liquid Solid Aerosol Fog Smoke Liquid Liquid Liquid Gas Liquid Solid Foam Emulsion Sol Whipped cream Milk, mayo Paint, ink Solid Solid Solid Gas Liquid Solid Solid foam Marshmallow Butter Pearls, opals Photo by C.A.Bailey, CalPoly SLO (Inlay Lake, Myanmar) Fall 2009 UM-SJTU JI Page 5 ...
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