The Properties of Mixtures: Solutions and Colloids Dr.ssa Rossana Galassi

Transcription

1 The Properties of Mixtures: Solutions and Colloids Dr.ssa Rossana Galassi

2 The Properties of Mixtures: Solutions and Colloids 1 Types of Solutions: Intermolecular Forces and Predicting Solubility 2 Intermolecular Forces and Biological Macromolecules 3 Why Substances Dissolve: Understanding the Solution Process 4 Solubility as an Equilibrium Process 5 Quantitative Ways of Expressing Concentration 6 Colligative Properties of Solutions 7 The Structure and Properties of Colloids

3 Solutions Solutions are homogeneous mixtures obtained by mixing two or more substances in a single phase. By convention the component in the largest amount is identified as the solvent and the other component as the solute.

4 Types of solutions Solid solid Gas solid Gas gas Gas liquid Liquid liquid Solid liquid alloy clathrate air river/sea water alcoholic drinks salty water

5 The arrangement of atoms in two types of alloys. brass - a substitutional alloy carbon steel -an interstitial alloy

6 Clathrate hydrates constitute a class of solids in which the guest molecules occupy, fully or partially, cages in host structures made up of H-bonded water molecules. The usually unstable empty clathrate is stabilised by inclusion of the guest species. In case of guest molecules which are gaseous at ambient conditions the resulting clathrate hydrate is often called a gas hydrate

7 Preparing a solution : dissolving a solid in a liquid Solvent = water Solute = CuCl 2 Interactions between water molecules and Cu 2+ and Cl - ions allow the solid to dissolve

8 Preparing a solution : diluting a more concentrated solution

9 A solution does not scatter light A colloidal dispersion does.

10 Units of concentration The concentration of a solution represents the amount of solute dissolved in a certain amount of solvent or solution at a fixed temperature. Molarity (M, mol/l) Molality (m, mol/kg) Percent (w/w; w/v; V/V) Normality (N, eq/l) Mole fraction (, n/n+n) Parts per milion (ppm, mg/l)

11 Molarity Normality molality Mole fraction Weight percent M = number of mole of solute / 1 liter of solution N = number of equivalent / 1 liter of solution m = number of mole of solute / 1Kg of solvent = n / n + N %W/W = mass of A / mass of A + mass of B + etc.. Volume percent %V/V = volume of A / volume of A + volume of B + Weight/volume % W/V = weight of A / volume of A + volume of B

12 Normality number of equivalent /1 liter of solutions This unit of concentration is particularly usefull in titrations. The number of equivalent takes into account the valence of the chemical system in the chemical reaction occurring in the titration. In example : Zn 0 + 2H + => H 2 + Zn 2+ In this redox reaction occurs an exchange of 2 electrons, so the equivalent mass is given by MM/2 and the number of equivalent are equal to eq = grams /EM 2NaOH + H 2 SO 4 => Na 2 SO 4 + H 2 O The equivalent number of H 2 SO 4 is equal to eq = grams / MM/2 taking into account the two H + exchanged during the acid base reaction.

13 The solution process LIKE DISSOLVES LIKE Substances giving similar types of intermolecular forces dissolve in each other. When a solute dissolves in a solvent, solutesolute interactions and solvent-solvent interactions are being replaced with solutesolvent interactions. The forces must be comparable in strength in order to have a solution.

14 The major types of intermolecular forces in solutions.

15 Hydration shells around an aqueous ion. Salt in water..

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17 Like dissolves like: solubility of methanol in water. water methanol A solution of methanol in water

18 Liquids dissolving in liquids If two liquids mix to an appreciable extent to form a solution, they are said to be miscibile. In constrast, immiscibile liquids do not mix to form a solution, they exist in contact with each other as separated layers.

19 I 2 is a non polar molecule. In the left tube it dissolves in water (superior layer) after mixing it prefers the CCl 4 (it is non polar and non miscibile with water) ess dense octane uso4 in water ore dense CCl4 CuSO4 in water CCl4 and octane

20 The solution process When the solid is added to the water it starts to dissolve till a point when no additional solid seems to dissolve. The solution is saturated and the concentration is the maximum reachable in those conditions (Solubility of CuCl 2, 70.6 g in 100 ml of H 2 O at 20 C)

21 Equilibrium in a saturated solution. solute (undissolved) solute (dissolved)

22 Heats of solution and solution cycles 1. Solute particles separate from each other - endothermic solute (aggregated) + heat solute (separated) H solute > 0 2. Solvent particles separate from each other - endothermic solvent (aggregated) + heat solvent (separated) H solvent > 0 3. Solute and solvent particles mix - exothermic solute (separated) + solvent (separated) solution + heat H mix < 0 H soln = H solute + H solvent + H mix

23 Solution cycles and the enthalpy components of the heat of solution. Exothermic solution process Endothermic solution process

24 Heats of solution so The enthalpy of solution is the algebric sum of all the enthalpies involved in the solution process. When a solid dissolves in a liquid the solid separation is actually: H solute = H lattice. Then, high H lattice generally corresponds to low solubility high H lattice are for highly charged and small ions H latticemgs > H latticekcl

25 Heats of Hydration The solvation of ions by water is always exothermic H 2 O M + (g) [or X - (g)] M + (aq) [or X - (aq)] H hydr of the ion < 0 H hydr is related to the charge density of the ion, that is, coulombic charge and size matter (ion radius) Lattice energy is the H involved in the formation of an ionic solid from its gaseous ions. M + (g) + X - (g) MX(s) H lattce is always (-)

26 Trends in Ionic Heats of Hydration Ion Ionic Radius (pm) H hydr (kj/mol) Group 1A(1) Li + Na + K + Rb + Cs + Group 2A(2) Mg 2+ Ca 2+ Sr 2+ Ba 2+ Group 7A(17) F - Cl - Br - I

27 Dissolving ionic compounds in water. watch?v=gyy9b6vncti NaCl NH 4 NO 3 NaOH e.com/watch?v=g3s 4YvhhAic om/watch?v=qa_odwi t_tw

28 Enthalpy diagrams for dissolving NaCl and octane in hexane.

29 The relation between solubility and temperature for several ionic compounds. For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions. The increased vibration (kinetic energy) of the solute molecules causes them to dissolve more readily because they are less able to hold together Na2SO4 and Ce2(SO4)3 solubilities decreases with the increase of temperature

30 Saturated solutions.

31 Supersaturated solution. Supersaturation is a unstable solution that contains more solute than could be dissolved by the solvent under normal circumstances. It can also refer to a vapor of a compound that has a higher (partial) pressure than the vapor pressure of that compound.

32 Saturated versus Supersaturated solution.

33 The structure and function of a soap.

34 The structure and function of a soap.

35 Factors affecting the solubility of gases in liquid: pressure and temperature 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. Some values for k H for gases dissolved in water at 298 K include: oxygen (O 2 ) : L atm/mol carbon dioxide (CO 2 ) : L atm/mol hydrogen (H 2 ) : L atm/mol

36 The effect of pressure on gas solubility.

37 Why when you open a can of coke, bubbles come out?

38 May CO 2, a linear apolar molecule, dissolve in water to form a solution? p=kh c where: p is the partial pressure of the solute above the solution c is the concentration of the solute in the solution (in one of its many units) kh is the Henry's Law constant, which has units such as L atm/mol or atm/(mole fraction) or Pa m3/mol

39 Does the temperature affect the solubility of a gas in water?

40 SAMPLE PROBLEM Using Henry s Law to Calculate Gas Solubility PROBLEM: PLAN: The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25 0 C. What is the solubility of CO 2? The Henry s law constant for CO 2 dissolved in water is 3.3 x10-2 mol/l*atm at 25 0 C. Knowing k H and P gas, we can substitute into the Henry s law equation. SOLUTION: S = (3.3 x10-2 mol/l*atm)(4 atm) = CO mol/l

41 Table Concentration Definitions Concentration Term Molarity (M) Molality (m) Parts by mass Parts by volume Mole fraction Ratio amount (mol) of solute volume (L) of solution amount (mol) of solute mass (kg) of solvent mass of solute mass of solution volume of solute volume of solution amount (mol) of solute amount (mol) of solute + amount (mol) of solvent

42 SAMPLE PROBLEM Calculating Molality PROBLEM: PLAN: What is the molality of a solution prepared by dissolving 32.0 g of CaCl 2 in 271 g of water? 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. SOLUTION: mole CaCl g CaCl 2 x g CaCl 2 = mole CaCl 2 molality = mole CaCl 2 kg 271 g H 2 O x 10 3 g = 1.06 m CaCl 2

43 SAMPLE PROBLEM Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction 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 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? PLAN: (a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply by (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.

44 SAMPLE PROBLEM continued Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction SOLUTION: (a) 40.5 mg Ca x 3.5 g g 10 3 mg x 10 6 = 1.16x10 4 ppm Ca (b) (c) 11.5 L alcohol L chianti x = L alcohol 100 L chianti moles ethylene glycol = 142 g moles water = 38.0g mole g = 2.36 mol C 2H 6 O 2 mole g = 3.22 mol H 2O 2.39 mol C 2 H 8 O mol C 2 H 8 O mol H 2 O 3.22 mol H 2 O 2.39 mol C 2 H 8 O mol H 2 O = C 2 H 6 O 2 = H 2 O

45 Interconverting Concentration Terms 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.

46 SAMPLE PROBLEM Converting Concentration Units 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 PLAN: (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. SOLUTION: (a) g of H 2 O = 100. g solution g H 2 O 2 = 70.0 g H 2 O 30.0 g H 2 O 2 mol H 2 O g H2O mol H 2 O 2 molality = 70.0 g H 2 O kg H 2 O 10 3 g = 12.6 m H 2 O 2

47 SAMPLE PROBLEM Converting Concentration Units continued (b) 70.0 g H 2 O mol H 2 O g H 2 O = 3.88 mol H 2O mol H 2 O mol H 2 O mol H 2 O = of H 2 O 2 (c) g solution ml 1.11 g = 90.1 ml solution mol H 2 O ml solution L 10 3 ml = 9.79 M H 2 O 2

48 Dilution law When a solution is diluted (or concentrated) the amount of solute does not change (just the solvent is added or removed). If C is the molar concentration term of a solution and V its volume, the product of CV is the number of mole CV = n (number of mole of solute). Upon dilution or concentration C and V change their values but n is equal. Accordingly to that : C 1 V 1 = C 2 V 2 = n Es: If 200 ml of a 0.1 M solution, 200 ml of water are added, how much will be the final concentration? V1 = 200 ml, V2 = 400 ml; C1 = 0.1 M, C2 =? C2 = C1V1/C2 = 200 x 0.1 /400 = 0.05 M

49 Colligative Properties By considering the properties of liquids, such as the vapour pressure, the boiling and the freezing temperatures the colligative properties are involved in the study of the change of these properties upon addition of a solute to a solvent (pure liquid) to form a solution.

50 Colligative Properties Raoult s Law (vapor pressure of a solvent above a solution, P solvent ) P solvent = solvent X P 0 solvent where P 0 solvent is the vapor pressure of the pure solvent P 0 solvent - P solvent = 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

51 The three types of electrolytes. STRONG nonelectrolyte weak

52 The effect of a solute on the vapor pressure of a solution.

53 SAMPLE PROBLEM Using Raoult s Law to Find the Vapor Pressure Lowering PROBLEM: PLAN: SOLUTION: Calculate the vapor pressure lowering, P, when 10.0 ml of glycerol (C 3 H 8 O 3 ) is added to 500. ml of water at C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is g/ml. The density of glycerol is 1.26 g/ml. Find the mol fraction,, of glycerol in solution and multiply by the vapor pressure of water g C 3 H 8 O 3 mol C 3 H 8 O ml C 3 H 8 O 3 x = mol C ml C 3 H 8 O g C 3 H 8 O 3 H 8 O ml H 2 O g H 2 O ml H 2 O P = mol C 3 H 8 O mol C 3 H 8 O mol H 2 O x mol H 2 O g H 2 O x = torr = 27.4 mol H 2 O = torr

54 Phase diagrams of solvent and solution.

55 Table Molal Boiling Point Elevation and Freezing Point Depresssion Constants of Several Solvents Solvent Boiling Melting Point ( 0 C)* K b ( 0 C/m) Point ( 0 C) K b ( 0 C/m) Acetic acid Benzene Carbon disulfide Carbon tetrachloride Chloroform Diethyl ether Ethanol Water *at 1 atm.

56 SAMPLE PROBLEM Determining the Boiling Point Elevation and Freezing Point Depression of a Solution 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. SOLUTION: 1.00x10 3 g C 2 H 6 O 2 mol C 2 H 6 O g C 2 H 6 O 2 = 16.1 mol C 2 H 6 O mol C 2 H 6 O kg H 2 O = 3.62 m C 2 H 6 O 2 T bp = C/m x 3.62m = C T fp = C/m x 3.62m BP = C FP = C

57 Osmotic Pressure The symbol for osmotic pressure is. n solute or M V solution n solute = RT = MRT V solution

58 The development of osmotic pressure. osmotic pressure Applied pressure needed to prevent volume increase pure solvent solution semipermeable membrane net movement of solvent solute molecules solvent molecules

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61 SAMPLE PROBLEM Determining Molar Mass from Osmotic Pressure 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 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. SOLUTION: 3.61 torr atm M = = 760 torr = 2.08 x10-4 M RT ( L*atm/mol*K)(278.1 K) 2.08 x10-4 mol L (1.50 ml) L 10 3 ml = 3.12x10-8 mol 21.5 mg g 10 3 mg x10-8 mol = 6.89 x10 4 g/mol

62 Colligative Properties of Electrolyte Solutions For electrolyte solutions, the compound formula tells us how many particles are in the solution. The van t Hoft factor, i, tells us what the effective number of ions are in the solution. van t Hoff factor (i) i = measured value for electrolyte solution expected value for nonelectrolyte 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)

63 Figure Nonideal behavior of electrolyte solutions.

64 An ionic atmosphere model for nonideal behavior of electrolyte solutions.

65 SAMPLE PROBLEM Depicting a Solution to Find Its Colligative Properties PROBLEM: A g sample of magnesium chloride is dissolved in 100. g of water in a flask. (a) Which scene depicts the solution best? (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)? 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.

66 SAMPLE PROBLEM 13.9 continued Depicting a Solution to Find Its Colligative Properties (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) g MgCl 2 mols MgCl 2 = g MgCl 2 = mol MgCl 2 mol MgCl 2 mols Cl - = mol MgCl 2 x 2 mols Cl - 1 mol MgCl 2 = mols Cl - mols/sphere = mols Cl - 8 spheres = 2.50 x 10-3 mols/sphere

67 SAMPLE PROBLEM continued Depicting a Solution to Find Its Colligative Properties mol MgCl 2 (c) = m MgCl molality (m) = 2 1 kg 100. g x 10 3 g Assuming this is an IDEAL solution, the van t Hoff factor, i, should be 3. T f = i (K f m) = 3( C/m x m) = C T f = C C = C

68 Exercise: A L solution is made by dissolving a sample of CaCl2(s) in water. (a) Is CaCl2 an electrolyte or a nonelectrolyte? (b) The solution has an osmotic pressure of 3.55 atm at 27 C. What is the approximate molarity of CaCl2 in the solution? (c) The van't Hoff factor for CaCl2 is 2.6 in the concentration range M. Using this value, calculate the molarity of the NON Ideal CaCl2 solution. (d) The enthalpy of solution for CaCl2 is H = kj/mol. If the final temperature of the solution was 27.0 C, what was its initial temperature? (Assume that the density of the solution is 1.0 g/ml, that its specific heat is 4.18 J/g-K, and that the solution loses no heat to its surroundings.) SOLUTION (a) Soluble ionic compounds are strong electrolytes. CaCl2 consists of metal cations (Ca2+) and nonmetal anions (Cl ) and is hence a strong electrolyte. (b) we have : This concentration is the total effective concentration of particles in the solution. Because each CaCl2 unit ionizes to form three ions (one Ca2+ and two Cl ), the concentration of CaCl2 is approximately M/3 = M.

69 (c) In the concentration range of M CaCl2, the ion pairing reduces the effective number of free ions per CaCl2 unit from the ideal or limiting value of 3 to an actual value of 2.6. This number is the van't Hoff factor, i. Using this value, we calculate that the concentration of CaCl2 equals M/2.6 = M, just a bit higher than what we estimated in part (b). (d) If the solution is M in CaCl2 and has a total volume of L, the number of moles of solute is (0.100 L)(0.055 mol/l) = mol. Hence the quantity of heat generated in forming the solution is ( mol)(-81.3 kj/mol) = kj. The solution absorbs this heat, causing its temperature to increase. The relationship between temperature change and heat is given by The heat absorbed by the solution is q = kj = 450 J. The mass of the L of solution is (100 ml)(1.0 g/ml) = 100 g (to 2 significant figures). Thus the temperature change is: A kelvin has the same size as a degree Celsius. Because the solution temperature increases by 1.1 C, the initial temperature was 27.0 C C = 25.9 C.

70 Types of Colloids Colloid Type Dispersed Substance Dispersing Medium Example Aerosol Aerosol Foam Solid foam Emulsion Solid emulsion Sol Solid sol Liquid Gas Fog Solid Gas Smoke Gas Liquid Whipped cream Gas Solid Marshmallow Liquid Liquid Milk Liquid Solid Butter Solid Liquid Paint; cell fluid Solid Solid Opal

71 Light scattering and the Tyndall effect. Photo by C.A.Bailey, CalPoly SLO (Inlay Lake, Myanmar)

72 Colloids. A colloid is one of the three primary types of mixtures, with the other two being a solution (homogeneous) and suspension (heterogeneous). A colloid is a mixture that has particles ranging between 1 and 1000 nanometers in diameter, yet are still able to remain evenly distributed throughout the mixture. These are also known as colloidal dispersions because the substances remain dispersed and do not settle to the bottom of the container. In colloids, one substance is evenly dispersed in another. The types of colloids includes sol, emulsion, foam, and aerosol. Sol is a colloidal suspension with solid particles in a liquid. Emulsion is between two liquids. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas.

73 Solutions toward suspensions.

74 Comparison among the properties of solutions toward suspensions

75 The most important colloids are those in which the dispersing medium is water. Such colloids are frequently referred to as hydrophilic (water loving) or hydrophobic (water fearing). In the human body the extremely large molecules that make up such important substances as enzymes and antibodies are kept in suspension by interaction with surrounding water molecules. The molecules fold in such a way that the hydrophobic groups are away from the water molecules, on the "inside" of the folded molecule, while the hydrophilic, polar groups are found on the surface, interacting with the water molecules. These hydrophilic groups generally contain oxygen or nitrogen

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78 A Cottrell precipitator for removing particulates from industrial smokestack gases.

79 The steps in a typical municipal water treatment plant.

80 Ion exchange for removal of hard-water cations.

81 Reverse osmosis for the removal of ions.