Physical Properties of Solutions Chapter 13 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Homework - pages 426 - 429 1-11, 14, 16, 18, 20, 22, 27, 31, 32, 34, 37, 38, 40 - 44, 56, 58, 62, 64, 67, 71 - 74, 80, 85, 87

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

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

Temperature and Solubility Solid solubility and temperature 12.4

PREVIEW There are two factors that govern whether or not a process will occur spontaneously. 1. There is a tendency to decrease energy. 2. There is a tendency to increase disorder (entropy).

“like dissolves like” Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl4 in C6H6 polar molecules are soluble in polar solvents C2H5OH in H2O ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l) 12.2

Like Dissolves Like: Solubility of Methanol in Water

Equilibrium in a Saturated Solution Fig. 13.9

Predicting Relative Solubilities of Substances Problem: Predict which solvent will dissolve more of the given solute. (a) Sodium Chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH). (b) Ethylene glycol (HOCH2CH2OH) in water or in hexane (CH3CH2CH2CH2CH2CH3). (c) Diethyl ether (CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in water.

The Cleansing Action of Soap 12.8

Test on chapters 12 & 13 – EH Assignment Due Unit 15 Properties of Solutions Mininum score Sec 1 Solvents and Solutions 90 Sec 2 Concentration Calculations Sec 3 Pressure and Temp Effects Sec 4 Vapor Pressure of Solutions 60 Sec 5 FP Lowering and BP Rise Unit 8 Section 2 only (Phase Changes and Heat) 80 Test on chapters 12 & 13 –

Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass x 100% mass of solute mass of solute + mass of solvent % by mass = = x 100% mass of solute mass of solution Mole Fraction (X) XA = moles of A sum of moles of all components 12.3

Concentration Units Continued Molarity (M) M = moles of solute liters of solution Molality (m) m = moles of solute mass of solvent (kg) 12.3

Calculating Molality Problem: What is the molality of a solution prepared by dissolving 75.0 g Ba(NO3)2 (s) in to 374.00 g of water at 250C. Solution: molar mass of Ba(NO3)2 = 261.32 g/mol 75.0 g Ba(NO3)2 x 1 mole = 0.28700 mole 261.32 g 0.28700 mole molality = = 0.76739 m = 0.767 m 0.37400 kg

Expressing Concentration by Parts by Mass, Parts by Volume, and Mole Fraction Problem: a) Calculate the PPB by mass of Iron in a 1.85 g Iron supplement pill that contains 0.0543 g of Iron. b) The label on a can of beer (340 mL) indicates “4.5% alcohol by volume”. What is the volume in liters of alcohol it contains? c) A sample of alcohol contains 118g of ethanol (C2H5OH), and 375.0 g of water. What are the mole fractions?

Converting Concentration Units Problem: Commercial concentrated hydrochloric acid is 11.8 M HCl and has a density of 1.190g/ml. Calculate the (a) mass% HCl, (b) molality and (c) mole fraction of HCl. Calculate the molarity of a 1.74 m sucrose (C12H22O11) solution whose density is 1.12 g/mL

5.86 moles ethanol = 270 g ethanol What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? m = moles of solute mass of solvent (kg) M = moles of solute liters of solution Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol 927 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 kg m = moles of solute mass of solvent (kg) = 5.86 moles C2H5OH 0.657 kg solvent = 8.92 m 12.3

Temperature and Solubility Solid solubility and temperature solubility increases with increasing temperature solubility decreases with increasing temperature 12.4

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 KNO3 contaminated with 10 g NaCl. Fractional crystallization: Dissolve sample in 100 mL of water at 60oC Cool solution to 0oC All NaCl will stay in solution (s = 34.2g/100g) 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g 12.4

Temperature and Solubility – O2 Gas solubility and temperature solubility usually decreases with increasing temperature 12.4

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 = kP 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

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Colligative Properties Properties of solutions that depend on the number of solute particles (atoms, molecules, ions) and not on the nature of the solute I ) Vapor Pressure Lowering - Raoult’s Law II ) Boiling Point Elevation III ) Freezing Point Depression IV ) Osmotic Pressure

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 P1 = X1 P 1 P 1 = vapor pressure of pure solvent X1 = mole fraction of the solvent Raoult’s law If the solution contains only one solute: X1 = 1 – X2 P 1 - P1 = DP = X2 X2 = mole fraction of the solute 12.6

PA = XA P A Ideal Solution PB = XB P B PT = PA + PB PT = XA P A Ideal Solution PB = XB P B PT = PA + PB PT = XA P A + XB P B 12.6

< & > & PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law Force A-B A-A B-B < & Force A-B A-A B-B > & 12.6

Fractional Distillation Apparatus 12.6

Next Test From 1999 – test #1 Spring 2001 – test #2

Boiling-Point Elevation DTb = Tb – T b T b is the boiling point of the pure solvent T b is the boiling point of the solution Tb > T b DTb > 0 DTb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6

Freezing-Point Depression DTf = T f – Tf T f is the freezing point of the pure solvent T f is the freezing point of the solution T f > Tf DTf > 0 DTf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6

Problem: Calculate the the boiling point and freezing point of a benzene solution if 257g of napthalene (C10H8, mothballs) is dissolved in 500.00g of benzene (C6H6). napthalene = 128.16g/mol

DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m 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 0C/m = 3.202 kg solvent 478 g x 1 mol 62.01 g m = moles of solute mass of solvent (kg) = 2.41 m DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C Tf = -4.48 0C 12.6

Osmotic Pressure (p) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (p) is the pressure required to stop osmosis. more concentrated dilute 12.6

A cell in an: isotonic hypotonic hypertonic solution hypotonic solution (less conc.) hypertonic Solution (more conc.) 12.6

Determining Molar Mass from Osmotic Pressure Problem: A physician studying a type of hemoglobin formed during a fatal disease dissolves 21.5 mg of the protein in water at 5.00C to make 1.5 ml of solution in order to measure its osmotic pressure. At equilibrium, the solution has an osmotic pressure of 3.61 torr. What is the molar mass of the hemoglobin? Plan: We know the osmotic pressure (),R, and T. We convert  from torr to atm and T from 0C to K and use the osmotic pressure equation to solve for molarity (M). Then we calculate the moles of hemoglobin from the known volume and use the known mass to find M. Solution: 1 atm 760 torr P = 3.61 torr x = 0.00475 atm Temp = 5.00C + 273.15 = 278.15 K

Determining Molar Mass from Freezing Point Depression Problem: A 7.85 g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of pure benzene. The freezing point is 4.50 °C. What are the molar mass and molecular formula of this compound?

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 P1 = X1 P 1 o Boiling-Point Elevation DTb = Kb m Freezing-Point Depression DTf = Kf m Osmotic Pressure (p) p = MRT 12.6

Colligative Properties of Ionic Solutions For ionic solutions we must take into account the number of ions present i = van’t Hoff factor or the number of ions present For vapor pressure lowering: P = i XsoluteP 0solvent For boiling point elevation: Tb = i Kb m For freezing point depression: Tf = i Kf m For osmotic pressure:  = i MRT im = conc. of particles

Colligative Properties of Electrolyte Solutions 0.1 m NaCl 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 NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln i should be nonelectrolytes 1 NaCl 2 CaCl2 3 12.6

Colligative Properties of Electrolyte Solutions Boiling-Point Elevation DTb = i Kb m Freezing-Point Depression DTf = i Kf m Osmotic Pressure (p) p = iMRT 12.7

Arrange the solutions in order of increasing FP a. 0.1 m CaCl2, 0.1 m C12H22O11, 0.1m NaCl b. 0.05 m HCl, 0.1m HCl, 0.1m HC2H3O2 What is the FP of 0.010 m K2SO4 ? The osmotic pressure of 0.010 M KI is 0.456 atm at 25 °C. What is the van’t Hoff factor at this concentration?

Colloid versus solution A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance. Colloid versus solution collodial particles are larger than solute molecules (1-1000 nm) collodial suspension is not as homogeneous as a solution 12.8

Fig. 13.22A