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The Properties of Mixtures: Solutions and Colloids Chapter 13
The Properties of Mixtures: Solutions and Colloids 13.1 Types of Solutions: Intermolecular Forces and Predicting Solubility 13.2 Intermolecular Forces and Biological Macromolecules 13.3 Why Substances Dissolve: Understanding the Solution Process 13.5 Quantitative Ways of Expressing Concentration 13.6 Colligative Properties of Solutions 13.7 The Structure and Properties of Colloids 13.4 Solubility as an Equilibrium Process
Solution : homogenous mixtures of two or more components that can be varied in composition Solvent : component present in greatest amount; substance in which a solute dissolves Solute : other components; substance that is dissolved Aqueous solutions : solution in which water is the solvent Solvation : interaction between solute and solvent molecules; due to IMF; e.g. Na+ and Cl- ion surrounded byH2O molecules Hydration : solvation when solvent is water
Types of Solution dilute – solution with low solute concentration concentrated – one with high solute concentration solubility : maximum amount of solute that can be dissolved in a given amount of solvent saturated solution – a solution that contains the maximum amount of solute the solvent can dissolves; no more solute can dissolve in it unsaturated solution – solution containing amount of solute less than its solubility; more solute can dissolve in it supersaturated solution – solution containing an amount of solute greater than the solubility; unstable solution
Nature of Solute and Solvent (Solute-Solvent Interactions)
miscible liquids – pair of liquids that mix in all proportion, e.g. ethanol-H 2 O
immiscible liquids – liquids that do not mix, e.g. oil-H 2 O
“ Like dissolves like” Substances with similar intermolecular attractive forces tend to be soluble in one another: polar ↔ polar nonpolar ↔ nonpolar
Pressure (for Gases)
↑ pressure, ↑ solubility
Henry’s Law: C g = k P g C g ≡ solubility of gas in solution
K ≡ Henry’s law constant
P g ≡ partial pressure of the gas over solution
Correlation Between Boiling Point and Solubility in Water Table 13.3 Gas Solubility (M)* bp (K) He Ne N 2 CO O 2 NO 4.2 x 10 -4 4.2 6.6 x 10 -4 27.1 10.4 x 10 -4 77.4 15.6 x 10 -4 81.6 21.8 x 10 -4 90.2 32.7 x 10 -4 121.4 * At 273K and 1 atm
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.
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 solvent-solvent interactions are being replaced with solute-solvent interactions. The forces must be comparable in strength in order to have a solution occur.
Figure 13.3 Like dissolves like: solubility of methanol in water. water methanol A solution of methanol in water
Figure 13.8 The structure and function of a soap.
Figure 13.9 The structure of lecithin. lecithin phospholipid found in all cell membranes
SAMPLE PROBLEM 13.1 Predicting Relative Solubilities of Substances SOLUTION: (a) Sodium chloride 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. (c) Diethyl ether (CH 3 CH 2 OCH 2 CH 3 ) in water or in ethanol (CH 3 CH 2 OH) (c) Ethanol - Diethyl ether can interact through a dipole and dispersion forces. Ethanol can provide both while water would like to H bond. (b) Water - Hexane has no dipoles to interact with the -OH groups in ethylene glycol. Water can H bond to the ethylene glycol. PROBLEM: Predict which solvent will dissolve more of the given solute: PLAN: Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute. (a) Methanol - NaCl is ionic and will form ion-dipoles with the -OH groups of both methanol and propanol. However, propanol is subject to the dispersion forces to a greater extent.
Classification of Solutions A. Based on elemental composition Organic – compounds containing carbon (except CO2, CO, carbonates and cyanides) Inorganic – compounds of the other elements including acids, bases, and salts B. Based on Ionization/Electrolytic Property of Solute Electrolytic property – the ability of the solution to conduct electricity Electrolytes – substances whose aqueous solutions contain ions and thus conduct electricity strong electrolytes – substances which completely dissociates into ions e.g. salts, strong acids, strong bases NaCl -> Na+ (aq) + Cl- (aq)
* weak electrolytes – substances which produce small amounts of ions; partially dissociated into ions
e.g. weak acids, weak bases
CH3COOH ↔ CH3COO- (aq) + H+ (aq)
CH3COOH ionizes to form CH3COO- abd H+. While this happens, some ions also combine to form back CH3COOH. This results to partial dissociation
Nonelectrolytes – substances that does not dissociate into ions; form nonconducting solutions
e.g. most molecular compounds are nonelectrolytes
Figure 13.25 The three types of electrolytes. STRONG weak nonelectrolyte
Table 13.5 Concentration Definitions Concentration Term Ratio Molarity (M) amount (mol) of solute volume (L) of solution Molality ( m ) amount (mol) of solute mass (kg) of solvent percent by mass mass of solute mass of solution Percent by volume volume of solute volume of solution Mole fraction amount (mol) of solute amount (mol) of solute + amount (mol) of solvent
SAMPLE PROBLEM 13.3 Calculating Molality PLAN: SOLUTION: 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. molality = = 0.288 mole CaCl 2 = 1.06 m CaCl 2 PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl 2 in 271 g of water? 271 g H 2 O 0.288 mole CaCl 2 kg 10 3 g x 32.0 g CaCl 2 mole CaCl 2 110.98 g CaCl 2 x
SAMPLE PROBLEM 13.4 Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction PLAN: 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 0.750-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? (a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply by 10 6 . (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.
SAMPLE PROBLEM 13.4 Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction SOLUTION: continued (a) 3.5 g = 1.16x10 4 ppm Ca (b) = 0.0862 L alcohol (c) moles ethylene glycol = 142 g = 2.36 mol C 3 H 8 O moles water = 38.0g = 3.22 mol H 2 O 2.39 mol C 3 H 8 O 2.39 mol C 3 H 8 O 2 + 3.22 mol H 2 O 3.22 mol H 2 O 2.39 mol C 3 H 8 O 2 + 3.22 mol H 2 O 10 3 mg g 40.5 mg Ca x 10 6 x 11.5 L alcohol 100 L chianti 0.750 L chianti x mole 60.09 g mole 18.02 g = 0.423 C 2 H 6 O 2 = 0.577 H 2 O
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.
Molality involves quantity of solvent , whereas the other concentration terms involve quantity of solution.
SAMPLE PROBLEM 13.5 Converting Concentration Units PLAN: SOLUTION: (a) Molality (b) Mole fraction of H 2 O 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 . (c) Use the density to find the volume of the solution. (a) g of H 2 O = 100. g solution - 30.0 g H 2 O 2 = 70.0 g H 2 O molality = 30.0 g H 2 O 2 34.02 g H 2 O 2 mol H 2 O 2 70.0 g H 2 O kg H 2 O 10 3 g = 12.6 m H 2 O 2 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 O 2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its 0.882 mol H 2 O 2
SAMPLE PROBLEM 13.5 Converting Concentration Units continued (b) 0.882 mol H 2 O 2 70.0 g H 2 O = 3.88 mol H 2 O 0.882 mol H 2 O 2 + 3.88 mol H 2 O = 0.185 of H 2 O 2 (c) 100.0 g solution = 90.1 mL solution 0.882 mol H 2 O 2 90.1 mL solution = 9.79 M H 2 O 2 mol H 2 O 18.02 g H 2 O mL 1.11 g L 10 3 mL
Dilution of Solution When a solution is diluted: the volume is increased by adding more solvent the concentration is decreased, and the total amount of solute remains constant Dilution Formula: M1V1 = M2V2 where M1 ≡ initial concentration M2 ≡ final concentration V1 ≡ initial volume V2 ≡ final volume
Polyprotic Acids – conatins more than one 2 replaceable H+ i. H3PO4 + 3 NaOH -> Na3PO4 + 3 H2O a = 3 eq/mol ii. H3PO4 + 2 NaOH -> Na2HPO4 + 2 H2O a = 2 eq/mol iii. H3PO4 + NaOH -> NaH2PO4 + H2O a = 1 eq/mol
COLLIGATIVE PROPERTIES properties that are dependent only on the concentrations of solute, not on the nature applicable only to non-volatile, non-electrolytic solutions Vapor Pressure Lowering/Depression Boiling Point Elevation Freezing Point Depression Osmotic Pressure, π
Colligative Properties Raoult’s Law (vapor pressure of a solvent above a solution, P solvent ) P solution = solvent X P 0 solvent where P 0 solvent is the vapor pressure of the pure solvent P 0 solvent - P solution = P = solute x P 0 solvent Boiling Point Elevation and Freezing Point Depression T b = K b m T f = K f m Osmotic Pressure M R T where M is the molarity, R is the ideal gas law constant and T is the Kelvin temperature
When a liquid is placed in a sealed container, a certain amount will evaporate as vapor to completely occupy the container.
The vapors exerts a pressure (vapor pressure) over the liquid
The VP of a liquid depends on temperature:
hi T -> hi VP
Addition of nonvolatile solute to a liquid reduces the capacity of the solvent molecules to evaporate (lower VP)
Vapor pressure lowering α concentration of solute
RAOULT’S LAW – predicts the VP of solutions containing nonvolatile solutes
P A = χ A P A o P A ≡ VP of solution
χ A ≡ mole fraction of solvent
P A o ≡ VP of pure solvent
Figure 13.23 The effect of pressure on gas solubility. Pure solvent Solution
Ideal Solution – a solution that obeys Raoult’s Law; ideal behavior can at low solute concentration, solute and solvent Idea Solutions with two or more volatile components (A and B) -> partial pressures of A and B P A = χ A P A o P B = χ B P B o P TOTAL = P A + P B P TOTAL = χ A P A o + χ B P B o
SAMPLE PROBLEM 13.6 Using Raoult’s Law to Find the Vapor Pressure Lowering SOLUTION: 10.0 mL C 3 H 8 O 3 = 0.137 mol C 3 H 8 O 3 500.0 mL H 2 O = 27.4 mol H 2 O P = 27.4 mol H2O 0.137 mol C 3 H 8 O 3 + 27.4 mol H 2 O 92.5 torr x x x = 92.0 torr PROBLEM: Calculate the vapor pressure of the solution when 10.0 mL of glycerol (C 3 H 8 O 3 ) is added to 500. mL of water at 50. 0 C. 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. PLAN: Find the mol fraction, , of glycerol in solution and multiply by the vapor pressure of water. 1.26 g C 3 H 8 O 3 mL C 3 H 8 O 3 mol C 3 H 8 O 3 92.09 g C 3 H 8 O 3 0.988 g H 2 O mL H 2 O mol H 2 O 18.02 g H 2 O = 0.00498
SAMPLE PROBLEM 13.7 Determining the Boiling Point Elevation and Freezing Point Depression of a Solution SOLUTION: 1.00x10 3 g C 2 H 6 O 2 = 16.1 mol C 2 H 6 O 2 T bp = 0.512 0 C/m 16.1 mol C 2 H 6 O 2 4.450 kg H 2 O = 3.62 m C 2 H 6 O 2 3.62 m x = 1.85 0 C BP = 101.85 0 C T fp = 1.86 0 C/m 3.62 m x FP = -6.73 0 C PROBLEM: You add 1.00 kg of ethylene glycol (C 2 H 6 O 2 ) 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. mol C 2 H 6 O 2 62.07 g C 2 H 6 O 2
Freezing Point, o C K f , o C/m Boiling Point, o C K b , o C/m Acetic Acid 16.6 3.90 118.1 3.07 Benzene 5.51 4.90 80.1 2.53 Water 0.00 1.86 100.0 0.512 CCl 4 -22.8 31.8 76.8 5.03 Ethanol -117.3 1.99 78.5 1.22
Table 13.6 Molal Boiling Point Elevation and Freezing Point Depresssion Constants of Several Solvents Solvent Boiling Point ( 0 C)* K b ( 0 C/ m ) K b ( 0 C/ m ) Melting Point ( 0 C) Acetic acid Benzene Carbon disulfide Carbon tetrachloride Chloroform Diethyl ether Ethanol Water 117.9 80.1 46.2 76.5 61.7 34.5 78.5 100.0 3.07 16.6 3.90 2.53 5.5 4.90 2.34 -111.5 3.83 5.03 -23 30. 3.63 -63.5 4.70 2.02 -116.2 1.79 1.22 -117.3 1.99 0.512 0.0 1.86 *at 1 atm.
4 . Osmotic Pressure, π - pressure required to prevent osmosis from occurring Osmosis – net movement of solvent molecules from a region of low solute concentration to a region of high solute concentration through a semi-permeable membrane (allows selective passage of water molecules) π = MRT = Where M = molarity R = gas constant = 0.0821 L·atm/K·mol T = absolute temperature
When Papplied < π : osmosis takes place in the normal way and water moves through the membrane in the solution When Papplied > π : reverse osmosis occurs wherein water molecules move through the membrane from the solution to pure water. *Note: Reverse osmosis is employed in making freshwater from seawater, where the applied pressure is large enough to reverse the normal precess.
Figure 13.28 The development of osmotic pressure. pure solvent solution net movement of solvent solvent molecules solute molecules osmotic pressure Applied pressure needed to prevent volume increase semipermeable membrane
Figure B13.3 Reverse osmosis for the removal of ions.
SAMPLE PROBLEM 13.8 Determining Molar Mass from Osmotic Pressure SOLUTION: M = = 3.61 torr (0.0821 L*atm/mol*K)(278.1 K) = 2.08 x10 -4 M (1.50 mL) = 3.12x10 -8 mol 21.5 mg = 6.89 x10 4 g/mol PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studing a variety associated with a fatal disease first finds its molar mass ( M ). She dissolves 21.5 mg of the protein in water at 5.0 0 C 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? PLAN: We know as well as R and T. Convert to atm and T to degrees K. Use the equation to find M and then the amount and volume of the sample to get to M . RT atm 760 torr 2.08 x10 -4 mol L 10 3 mL L g 10 3 mg 1 3.12 x10 -8 mol
DEFINITIONS: Isotonic Solutions – two solutions having the same osmotic pressure; equal concentrations Hypotonic Solutions – less concentrated solutions Hypertonic Solutions – more concentrated solutions Crenation – caused by movement of water from a (hypotonic) cell Hemolysis – caused by movement of water into a (hypertonic) cell Active transport – opposite of osmosis; movement of substances from a region of low concentration to a region of high concentration
Colligative Properties of Electrolyte Solutions For electrolyte solutions, the compound formula tells us how many particles are in the solution. For vapor pressure lowering: P = i ( solute x P 0 solvent ) For boiling point elevation: T b = i ( b m ) For freezing point depression: T f = i ( f m ) For osmotic pressure : = i (MRT) The van’t Hoft factor, i , tells us what the “effective” number of ions are in the solution. The more ions, the larger the value of vant hoff facor. van’t Hoff factor ( i ) i = measured value for electrolyte solution expected value for nonelectrolyte solution
Colligative Properties of Electrolytic Solutions different from nonelectrolytes since ions (of opposing charges) have the tendency to stick together and from ION PAIRS formation of ion pairs causes slight changes in π, ΔTf, ΔTb, and VP van’t Hoff factor, i – measure of the extent of dissociation of electrolytes
Types of Homogeneous Mixture Kind of Mixture Particle Size, nm Examples Characteristics Solution 0.2 – 2.0 Air, seawater, gasoline, wine Transparent to light; does not separate on standing; nonfilterable Colloid 2.0 – 1000 Butter, milk, fog, pearl Often murky or opaque to light (Tyndall Effect); does not separate on standing; nonfilterable Suspension > 1000 Blood, paint Murky or opaque to light; separate on standing; filterable
Colloid – a dispersion of particles of one substance (the dispersed phase) throughout another substance (the continuous phase) Tyndal Effect – the scattering of light of by colloidal-sized particles Coagulation – process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase, e.g. curdling of milk when in sours (lactose, milk sugar, ferments to lactic acid)
Figure 13.32 Light scattering and the Tyndall effect. Photo by C.A.Bailey, CalPoly SLO (Inlay Lake, Myanmar)
SAMPLE PROBLEM 13.9 Depicting a Solution to Find Its Colligative Properties (b) What is the amount (mol) represented by each green sphere? (c) Assuming the solution is ideal, what is its freezing point (at 1 atm)? PROBLEM: 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? 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.
SAMPLE PROBLEM 13.9 Depicting a Solution to Find Its Colligative Properties continued (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) mols MgCl 2 = = 0.0100 mol MgCl 2 0.952 g MgCl 2 95.21 g MgCl 2 mol MgCl 2 mols Cl - = 0.0100 mol MgCl 2 x 2 mols Cl - 1 mol MgCl 2 = 0.0200 mols Cl - mols/sphere = 0.0200 mols Cl - 8 spheres = 2.50 x 10 -3 mols/sphere
SAMPLE PROBLEM 13.9 Depicting a Solution to Find Its Colligative Properties continued (c) molality (m) = = 0.100 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(1.86 0 C/m x 0.100 m) = 0.558 0 C T f = 0.000 0 C - 0.558 0 C = - 0.558 0 C 0.0100 mol MgCl 2 100. g x 10 3 g 1 kg