# chapter 10 solution lecture notes

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Chapter 10 Properties of Solutions 17.1 Solution Composition 17.2 The Thermodynamics of Solution Formation (skip) 17.3 Factors Affecting Solubility 17.4 The Vapor Pressures of Solutions 17.5 Boiling-Point Elevation and Freezing-Point Depression 17.6 Osmotic Pressure 17.7 Colligative Properties of Electrolyte Solutions 17.8 Colloids os it ion tio n De p Su b li ma ap Ev SOLID n tio sa en on nd ati Co or GAS Melting Freezing LIQUID Colligative Properties of Solutions • • For Colligative properties, the difference between a pure solvent and dilute solution depends only on the number of solute particles present and not on their chemical identity Examples – – – – Vapor Pressure Depression Boiling Point Elevation Melting Point Depression Osmotic Pressure Lowering of Vapor Pressure – Vapor Pressure of a solvent above a dilute solution is always less than the vapor pressure above the pure solvent Elevation of Boiling Point – The boiling point of a solution of a non-volatile solute in a volatile solvent always exceeds the boiling point of a pure solvent Boiling • liquid in equilibrium with its vapor at the external pressure Boiling Point • Vapor press = external pressure Normal boiling point • Vap press = atm Elevation of Boiling Point & Vapor Pressure Depression ∆T = K m b solute ΔT is the boiling point elevation Kb is molal boiling - point elevation constant m is the molality of the solute in solution solute Phase diagrams for pure water (red lines) and for an aqueous solution containing a nonvolatile solution (blue lines) Solution Composition The solute and solvent can be any combination of solid (s), liquid (l), and gaseous (g) phases Dissolution: Two (or more) substances mix at the level of individual atoms, molecules, or ions Solution: A homogeneous mixture (mixed at level of atoms molecules or ions Solvent: The major component Solute: The minor component Solution Composition Mass Fraction, Mole Fraction, Molality and Molarity Mass percentage (weight percentage): mass percentage of the component = mass of component total mass of mixture X 100% Mole fraction: The amount of a given component (in moles) divided by the total amount (in moles) X1 = n1/(n1 + n2) for a two component system X2 = n2/(n1 + n2) = – X1 or X1+X2=1 Molality msolute = moles solute per kilogram solvent = moles per kg or (mol kg-1) Molarity (biochemists pay attention) csolute = moles solute per volume solution = moles per liter of solution (mol L-1) Factors Affecting Solubility Molecular Interactions – Review chapter – Polar molecules, water soluble, hydrophilic (water loving) • E.g., Vitamins B and C; water-soluble – Non-polar molecules, soluble in non-polar molecules, hydrophobic (water fearing) • E.g., Vitamins A, D, K and E; fat-soluble Factors Affecting Solubility of Gases Structure Effects Pressure Effects Henry’s Law (for dilute solutions) The mole fraction of volatile solute is proportional to the vapor pressure of the solute P = kH X kH = Henry’s Law constant, X = mole fraction Increasing the partial pressure of a gas over a liquid increases the amount of gas disolved in the liquid kH depends on temperature Vapor pressure for a solution of two volatile liquids Positive deviation = solute-solvent attractions < solvent-solvent attractions For non-ideal Solutions Negative deviation = solute-solvent attractions > solvent-solvent attractions boiling point : ΔT = K m b solute freezing point : ΔT = K m f solute Osmotic Pressure Fourth Colligative Property • Important for transport of molecules across cell membranes, called semipermeable membranes • Osmotic Pressure = Π Π = M RT ΠV = n RT PV = nRT Molarity (M) = moles/L or n/V Osmotic Pressure The normal flow of solvent into the solution (osmosis) can be prevented by applying an external pressure to the solution Osmotic Pressure useful for  Determining the Molar Mass of protein and other macromolecules  small concentrations cause large osmotic pressures  Can prevent transfer of all solute particles  Dialysis at the wall of most plant and animal cells Dialysis: Representation of the functioning of an artificial kidney A cellophane (polymeric) tube acts as the semipermeable membrane  Purifies blood by washing impurities (solutes) into the dialyzing solution A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C Compute the molar mass of the compound Strategy 1) use Π = MRT to find M (mol/L) mass 2) Recall that # of moles = mwt g g L 3) Rearrange mwt = = mole mole L 1.19 = M A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C Compute the molar mass of the compound Solution Π RT Π 0.0288 atm c= = RT (0.0820 L atm mol −1K −1 )(37 + 273.15K) 1) use Π = MRT or M = M = 1.132x10 −3 mol/L g 2) Rearrange M = M = 1.19 g L g L = mole mole 1.132x10 −3 mol/L L = 1.05x10 g/mol The Person Behind the Science J.H van’t Hoff (1852-1901) Highlights – – – – Discovery of the laws of chemical dynamics and osmotic pressure in solutions Mathematical laws that closely resemble the laws describing the behavior of gases his work led to Arrhenius's theory of electrolytic dissociation or ionization Studies in molecular structure laid the foundation of stereochemistry Moments in a Life – 1901 awarded first Noble Prize in Chemistry van’t Hoff Factor (i) moles of particles in solution ι= moles of solute dissolved ΔT = − i m K Colligative Properties of Electrolyte Solutions Elevation of Boiling Point ΔTb = m Kb Where m = molality (Molality is moles of solute per kilogram of solvent) The Effect of Dissociation ΔTb = i m Kb i = the number of particles released into the solution per formula unit of solute e.g., NaCl dissociates into i = e.g., Na2SO4 dissociates into i = (2 Na+ + SO4-2) e.g., acetic acid (a weak acid and weak electrolyte) does not dissociate i = also Depression of Freezing Point ΔTf = − m Kf ΔTf = − i m Kf Which aqueous solution would be expected to have the highest boiling point? 1) 0.100 m NaCl 2) 0.100 m CaCl2 3) 0.080 m Fe(NO3)3 4) 0.080 m Fe(NO3)2 5) 0.080 m Co(SO4) Which aqueous solution would be expected to have the highest boiling point? 1) 0.100 m NaCl 2) 0.100 m CaCl2 3) 0.080 m Fe(NO3)3 ΔTb = (4)(0.080) Kb = 0.320 Kb 4) 0.080 m Fe(NO3)2 ΔTb = (3)(0.080) Kb = 0.240 Kb 5) 0.080 m Co(SO4) ΔTb = (2)(0.100) Kb = 0.200 Kb ΔTb = (3)(0.100) Kb = 0.300 Kb ΔTb = (2)(0.080) Kb = 0.160 Kb Elevation of Boiling Point The Effect of Dissociation ΔTb = i m Kb Colloids: Colloidal Dispersions • Colloids are large particles dispersed in solution • Examples • Characteristics – 1nm to 1000 nm in size – E.g., Globular proteins 500 nm – – – – – – – Opal (water in solid SiO2) Aerosols (liquids in Gas) Smoke (solids in Air) Milk (fat droplets & solids in water) Mayonnaise (water droplets in oil) Paint (solid pigments in liquid) Biological fluids (proteins & fats in water) – Large particle size colloids: translucent, cloudy, milky) – Small particle size colloids: can be clear Colloidal Dispersions – Tyndall Effect • Light Scattering [...]... aqueous solution would be expected to have the highest boiling point? 1) 0 .100 m NaCl 2) 0 .100 m CaCl2 3) 0.080 m Fe(NO3)3 4) 0.080 m Fe(NO3)2 5) 0.080 m Co(SO4) Which aqueous solution would be expected to have the highest boiling point? 1) 0 .100 m NaCl 2) 0 .100 m CaCl2 3) 0.080 m Fe(NO3)3 ΔTb = (4)(0.080) Kb = 0.320 Kb 4) 0.080 m Fe(NO3)2 ΔTb = (3)(0.080) Kb = 0.240 Kb 5) 0.080 m Co(SO4) ΔTb = (2)(0 .100 )... in water at this temperature Given 2 kN XN kN 2 2 ≈ nN nH 2 2O 2 = XN = Find 2 5.76x10 −3 mol/l =  100 0g/l    18g/mol     = 1.0378 x 10 − 4 PN Given 9.20 atm 2 = = = XN Find 1.0378x10 − 4 2 = 8.86 x 10 4 atm Factors Affecting Solubility 1 2 3 Structure Effects Pressure Effects Temperature Effects for Aqueous Solutions The solubility of some solids as a function of temperature The aqueous solubilities... contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C Compute the molar mass of the compound Solution Π RT Π 0.0288 atm c= = RT (0.0820 L atm mol −1K −1 )(37 + 273.15K) 1) use Π = MRT or M = M = 1.132x10 −3 mol/L g 2) Rearrange M = M = 1.19 g L g L = mole mole 1.132x10 −3 mol/L L = 1.05x10 3 g/mol The Person Behind the Science J.H van’t... pressure above the solution increases 5) decreases as the gas pressure above the solution increases According to Henry's Law, the solubility of a gas in a liquid 1) 2) 3) 4) depends on the polarity of the liquid depends on the liquid's density remains the same at all temperatures increases as the gas pressure above the solution increases 5) decreases as the gas pressure above the solution increases... into the dialyzing solution A dilute aqueous solution of a non-dissociating compound contains 1.19 g of the compound per liter of solution and has an osmotic pressure of 0.0288 atm at a temperature of 37°C Compute the molar mass of the compound Strategy 1) use Π = MRT to find M (mol/L) mass 2) Recall that # of moles = mwt g g L 3) Rearrange mwt = = mole mole L 1.19 = M A dilute aqueous solution of a non-dissociating... nitrogen in the water is 5.76 x 10- 3 mol L-1 Compute Henry’s law constant for nitrogen in water at this temperature Given PN = 9.20 atm 2 c N = [N 2 ] = 5.76x10 −3 mol/L 2 Henry's Law PN = k N X N 2 XN = 2 2 2 nN nN + nH 2 ≈ 2 2O nN nH 2 2O When the partial pressure of nitrogen over a sample of water at 19.4°C is 9.20 atm, then the concentration of nitrogen in the water is 5.76 x 10- 3 mol L-1 Compute Henry’s... Effects for Aqueous Solutions The solubility of some gases in water as a function of temperature at a constant pressure of 1 atm The greatest gas solubility for a gas in solution is predicted under what conditions? 1) 2) 3) 4) 5) low T, low P low T, high P high T, low P high T, high P solubility of gases does not depend upon temperature The greatest gas solubility for a gas in solution is predicted... (3)(0.080) Kb = 0.240 Kb 5) 0.080 m Co(SO4) ΔTb = (2)(0 .100 ) Kb = 0.200 Kb ΔTb = (3)(0 .100 ) Kb = 0.300 Kb ΔTb = (2)(0.080) Kb = 0.160 Kb Elevation of Boiling Point The Effect of Dissociation ΔTb = i m Kb Colloids: Colloidal Dispersions • Colloids are large particles dispersed in solution • Examples • Characteristics – 1nm to 100 0 nm in size – E.g., Globular proteins 500 nm – – – – – – – Opal (water in solid... = k N X N Given PN = 9.20 atm 2 2 c N = [N 2 ] = 5.76x10− 3 mol/l 2 Henry' s Law PN = k N X N 2 XN = 2 2 nN 2 nN + nH 2 ≈ 2 O 2 nN nH 2 2 rearrange PN Given 2 kN = = XN Find 2 2 2 O 2 Next assume 1 Liter When the partial pressure of nitrogen over a sample of water at 19.4°C is 9.20 atm, then the concentration of nitrogen in the water is 5.76 x 10- 3 mol L-1 Compute Henry’s law constant for nitrogen... with a solution is proportional to the ratio of the number of solvent molecules to non-volatile solute molecules allows molecular weights to be determined, and provides the explanation for freezing point depression and boiling point elevation Moments in a Life – Raoult was a prominent member of the group which created physical chemistry, including Arrhenius, Nernst, van t'Hoff, Planck For ideal solutions
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