Chapter 14 - Colligative Properties

Physical Propertiesof SolutionsChapter 11

A saturated solution contains the maximum amount of asolute that will dissolve in a given solvent at a specifictemperature.An unsaturated solution contains less solute than the solventhas the capacity to dissolve at a specific temperature.A supersaturated solution contains more solute than is presentin a saturated solution at a specific temperature.Sodium acetate crystals rapidly form when a seed crystal isadded to a supersaturated solution of sodium acetate.11.7

Three types of interactions in the solution process:• solvent‐solvent interaction (breaking IMF)• solute‐solute interaction (breaking IMF)• solvent‐solute interaction (called Solvalysis, making IMF)ΔH soln = ΔH 1 + ΔH 2 + ΔH 311.7

“like dissolves like”Two substances with similar intermolecular forces are likely to besoluble in each other.• non‐polar molecules are soluble in non‐polar solventsCCl 4 in C 6 H 6• polar molecules are soluble in polar solventsC 2 H 5 OH in H 2 O• ionic compounds are more soluble in polar solventsNaCl in H 2 O or NH 3 (l)11.7

Concentration UnitsThe concentration of a solution is the amount of solutepresent in a given quantity of solvent or solution.Percent by Massmass of solute% by mass = x 100%mass of solute + mass of solventMole Fraction (X)mass of solute= x 100%mass of solutionX A =moles of Asum of moles of all components11.7

Molarity (M)Concentration Units ContinuedM =moles of soluteliters of solutionMolality (m)m =moles of solutemass of solvent (kg)11.7

Molality ProblemCalculate the molality of a solution of 13.5g ofKF dissolved in 250. g of water.m=mol of solutekg solvent=1 mol KF⎜ ⎟⎝ 58.1 ⎠0.250 kg( 13.5g) ⎛ ⎞0.929 m11.7

What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solutionwhose density is 0.927 g/mL?m =moles of solutemass of solvent (kg)M =moles of soluteliters of solutionAssume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)mass of solvent = mass of solution –mass of solute= 927 g – 270 g = 657 g = 0.657 kgm =moles of solutemass of solvent (kg)= 5.86 moles C 2H 5 OH0.657 kg solvent= 8.92 m11.7

Mole Fraction• Mole fraction (χ) is the ratio of the numberof moles of a substance over the totalnumber of moles of substances in solution.χ=inumber of moles of itotal number of moles11.7

Mole Fraction ProblemCalculate the mole fraction of 10.0g of NaCldissolved in 100. g of water.χ =NaClmol of NaClmol of NaCl + mol H O2=⎛1 mol NaCl ⎞( 10.0g)⎜⎟⎝ 58.5g NaCl ⎠⎛1 mol NaCl ⎞ ⎛1 mol H ⎞⎜⎟ +2O2 ⎜ ⎟⎝58.5g NaCl ⎠ ⎝ 18.0g H2O⎠( 10.0g) ( 100.g H O)= 0.029911.7

Temperature and SolubilitySolid solubility and temperaturesolubility decreases increases withincreasing temperature11.8

Fractional crystallization is the separation of a mixture of substancesinto pure components on the basis of their differing solubilities.Suppose you have 90 g KNO 3contaminated with 10 g NaCl.Fractional crystallization:1. Dissolve sample in 100 mL ofwater at 60 0 C2. Cool solution to 0 0 C3. All NaCl will stay in solution (s =34.2g/100g)4. 78 g of PURE KNO 3 willprecipitate (s = 12 g/100g). 90g –12 g = 78 g11.8

Temperature and SolubilityGas solubility and temperaturesolubility usually decreaseswith increasingtemperature11.8

Pressure and Solubility of Gases• Henry’s LawP gas = k gas C gasP gas = pressure of the gas above the solutionC gas = concentration of the gask gas = Henry’s law constantlow Phigh Plow chigh c11.8

Henry’s Law ProblemA liter of water dissolves 0.0404 g of oxygen at 25 o Cat a pressure of 760. torr. What would be theconcentration of oxygen (in g/L) if the pressurewere increased to 1880 torr at the sametemperature?P gas = k gas C gas or P gas C gas = k gasSince the ratio is equal to a constant, 2 conditionscan be set equal.P1 P2=C C1 2C 20.0404 g/L =760. torr 1880 torrC2= 0.0999 g/L11.8

Colligative PropertiesColligative properties are properties that depend only on thenumber of solute particles in solution and not on the nature ofthe solute particles.11.9

Colligative Properties of Nonelectrolyte Solutions• Vapor pressure lowering –the vaporpressure of a solvent is lowered by theaddition of a nonvolatile solute.P 1 = X 1 P 10Raoult’s lawP 1 0 = vapor pressure of pure solventX 1 = mole fraction of the solventIf the solution contains only one solute:X 1 = 1 – X 2P 1 0 - P 1 = ΔP = X 2 P 10X 2 = mole fraction of the solute11.9

V.P. Lowering ProblemWhat will be the vapor pressure of a solution made bydissolving 6.25g of glucose, C 6 H 12 O 6 , in 50.0g of water at25 o C? How much was the vapor pressure of the purewater lowered? The vapor pressure of water at 25 o C is23.8 torrmol glucose = 6.25 g( )( )P = χ P o = 0.988 23.8 torr = 23.5 torrsoln water water⎛ 1 mol ⎞⎜ ⎟ = 0.0347 mol glucose⎝180. g ⎠⎛1 mol ⎞mol water = 50.0 g ⎜ ⎟ = 2.78 mol water⎝18.0g⎠2.78χwater= = 0.9880.0347+2.78vapor pressure lowering = 23.8 torr - 23.5 torr = 0.2 torr11.9

Ideal SolutionP A = X A P A0P B = X B P B0P T = P A + P BP T = X A P A0+X B P B011.9

P T is greater thanpredicted by Raoults’s lawP T is less thanpredicted by Raoults’s lawForceA-BForceA-A< &ForceB-BForceA-BForceA-A> &ForceB-B11.9

Boiling-Point ElevationΔT b = T b – T b00T b is the boiling point ofthe pure solventT b is the boiling point ofthe solutionT b > T b0ΔT b > 0ΔT b = K b mm is the molality of the solutionK b is the molal boiling-pointelevation constant ( 0 C/m)11.9

Freezing-Point Depression0ΔT f = T f0T fT f0T f > T f– T fis the freezing point ofthe pure solventis the freezing point ofthe solutionΔT f > 0ΔT f = K f mm is the molality of the solutionK f is the molal freezing-pointdepression constant ( 0 C/m)11.9

11.9

What is the freezing point of a solution containing 478 gof ethylene glycol (antifreeze) in 3202 g of water? Themolar mass of ethylene glycol is 62.01 g.m =ΔT f = K f mmoles of solute=mass of solvent (kg)K f water = 1.86 0 C/m478 g x1 mol62.01 g3.202 kg solvent= 2.41 mΔT f = K f m = 1.86 0 C/m x 2.41 m = 4.48 0 C0ΔT f = T f– T f0T f = T f– ΔT f= 0.00 0 C – 4.48 0 C = -4.48 0 C11.9

Osmotic Pressure (π)Osmosis is the selective passage of solvent molecules through a porous membranefrom a dilute solution to a more concentrated one.A semipermeable membrane allows the passage of solvent molecules but blocks thepassage of solute molecules.Osmotic pressure (π) is the pressure required to stop osmosis.П = iMRTi= sum of the coefficients of the ions (i= 1 for molecular compounds)M = molarityR = gas constant (0.0821 L • atm/mol • K)T = temperature in Kelvin11.9

A cell in an:isotonicsolutionhypotonicsolutionhypertonicsolution11.9

Osmotic Pressure ProblemWhat is the osmotic pressure of a 100.0 mLsolution containing 9.50 g of glucose, C 6 H 12 O 6 , at20.0 o C?⎛ 1 mol ⎞mol glucose = 9.50 g ⎜ ⎟=0.0528 mol⎝180. g ⎠0.0528 molMglucose= = 0.528 mol/L0.100 L⎛ L⋅atm ⎞π = (1)( 0.528 mol/L) ⎜0.08206 ⎟( [20.0+273] K ) = 12.7 atm⎝ K⋅mol⎠11.9

Colligative Properties of Nonelectrolyte SolutionsColligative properties are properties that depend only on the numberof solute particles in solution and not on the nature of the soluteparticles.Vapor-Pressure Lowering P 1 = X 1 P 10Boiling-Point ElevationFreezing-Point DepressionOsmotic Pressure (π)ΔT b = K b mΔT f = K f mπ = MRT11.9

Colligative Properties of Electrolyte Solutions0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl - ionsColligative properties are properties that depend only on thenumber of solute particles in solution and not on the nature ofthe solute particles.0.1 m NaCl solution 0.2 m ions in solutionvan’t Hoff factor (i) =actual number of particles in soln after dissociationnumber of formula units initially dissolved in solnnonelectrolytesNaClCaCl 2i should be12311.9

Colligative Properties of Electrolyte SolutionsBoiling-Point ElevationFreezing-Point DepressionOsmotic Pressure (π)ΔT b = i K b mΔT f = i K f mπ = iMRT11.9