Overview. Types of Solutions. Intermolecular forces in solution. Concentration terms. Colligative properties. Osmotic Pressure 2 / 46

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2 2 / 46 Overview Types of Solutions. Intermolecular forces in solution Concentration terms Colligative properties Osmotic Pressure

3 3 / 46 Solutions and Colloids A solution is a homogeneous mixture and exists as a single phase. The particles in a solution are individual atoms, ions, or small molecules. A colloid is a heterogeneous mixture and exists as two or more phases, which may be visibly distinct. The particles in a colloid are typically macromolecules or aggregations of small molecules, their diametrer varies between nm.

4 4 / 46 Solutions and Solubility A solute dissolves in a solvent to form a solution. Usually, the solvent is the most abundant component. The solubility (S) of a solute is the maximum amount that dissolves in a fixed quantity of solvent at a given temperature. Substances that exhibit similar types of intermolecular force dissolve in each other. This is often expressed by saying like dissolves in like.

5 Types of intermolecular forces in solutions 5 / 46

6 6 / 46 Solutions and Intermolecular Forces When a solution forms, solute-solute attractions and solvent-solvent attractions are replaced by solute-solvent attractions. This can only occur if the forces within the solute and solvent are similar to the forces that replace them.

7 Hydration shells around a Na + ion Ion-dipole forces orient water molecules around an ion. In the innermost shell here, six water molecules surround the cation octahedrally. 7 / 46

8 Dual Polarity and Effects on Solubility Solubility* of a Series of Alcohols in Water and in Hexane. 8 / 46

9 9 / 46 Like dissolves like Both water and methanol have H-bonding between their molecules.

10 10 / 46 Problem 1: Predicting relative solubilities Which solute is more soluble in the given solvent? 1. 1-butanol (CH 3 CH 2 CH 2 CH 2 OH) or 1,4-butanediol (HOCH 2 CH 2 CH 2 CH 2 OH in water 2. chloroform (CHCl 3 ) or carbon tetrachloride (CCl 4 ) in water

11 11 / 46 Gas-liquid solutions Correlation between boiling point and solubility in water Gas Solubility ( mol L 1 ) Boiling point (K) He Ne N CO O NO at 273 K and bar

12 Intermolecular forces in biological molecules Structure of proteins 12 / 46

13 Intermolecular forces in biological molecules Dual polarity and soap action 13 / 46

14 Intermolecular forces in biological molecules Phospholipids and the cell membrane 14 / 46

15 15 / 46 Intermolecular forces in biological molecules DNA double helix

16 16 / 46 Thermochemistry of solution formation H solution = H solute + H solvent + H mix H solution = H lattice + H hydration of the ions

17 17 / 46 Energy changes in Solution Formation Step 1: Solute particles separate from each other. This process is endothermic. Solute (aggregated) + heat solute (separated) H solute > 0 Step 2: Solvent particles separate from each other. This process is endothermic. Solvent (aggregated) + heat solvent (separated) H solvent > 0 Step 3: Solute and solvent particles mix and form a solution. This step is exothermic.

18 18 / 46 Heat of Solution H soln = H solute + H solvent + H mix The overall solution process may be either exothermic or endothermic. Exothermic process: H soln < 0 because the sum of the endothermic processes ( H solute + H solvent ) is smaller than the exothermic term ( H mix ). Endothermic process: H soln > 0 because the sum of the endothermic processes ( H solute + H solvent ) is larger than the exothermic term ( H mix.

19 19 / 46 Thermochemistry of solution formation Enthalpy changes in the dissolution of ionic compounds

20 Effect of temperature on solubility 20 / 46

21 21 / 46 Effect of pressure on the solubility of gases: Henry s Law S gas = k H P gas

22 22 / 46 Problem If air contains 78% N 2 by volume, what is the solubility of N 2 in water at 25 C and bar? (k H for N 2 is mol L 1 bar 1 ) P N2 = bar = bar S N2 = k H P gas = ( mol L 1 bar 1 ) (0.790 bar) = mol L 1

23 23 / 46 Problem If air contains 78% N 2 by volume, what is the solubility of N 2 in water at 25 C and bar? (k H for N 2 is mol L 1 bar 1 ) P N2 = bar = bar S N2 = k H P gas = ( mol L 1 bar 1 ) (0.790 bar) = mol L 1

24 24 / 46 Amount-of-substance Concentration Amount concentration = amount of solute mol) volume of solution (L) c B n B (mol) V (L)

25 Mole Fraction For a substance B, the mole fraction of B in a solution is χ B = n B Σ i n i From the definition, it follows that: χ is a pure number Σ i χ i = 1 25 / 46

26 26 / 46 Molality Molality, m, is defined as the amount of solute (mol) per kilogram of solvent m = n solute m solvent,kg

27 27 / 46 Colligative Properties of Solutions Colligative properties are properties that depend on the number of solute particles, not their chemical identity. The number of particles in solution can be predicted from the formula and type of the solute. An electrolyte separates into ions when it dissolves in water. Strong electrolytes dissociate completely while weak electrolytes dissociate very little. A nonelectrolyte does not dissociate to form ions.

28 Colligative properties: Vapor pressure decrease 28 / 46

29 29 / 46 Raoult s Law P solventabove solution = χ solvent P solvent The vapor pressure lowering can be calculated as P = χ solute P solvent

30 Colligative properties: Boiling point elevation 30 / 46

31 31 / 46 Colligative properties: Boiling point elevation T b = T boiling,solution T boiling,solvent T b = K B m

32 32 / 46 Colligative properties: Freezing point depression A solution always freezes at a lower temperature than the pure solvent. The freezing point depression is proportional to the molality of the solution. T f = K f m K f is the molal freezing point depression constant for the solvent.

33 33 / 46 Colligative properties: K b and K f Molal boiling point elevation and freezing point depression constants of several solvents B.P. K b M.P. K f Solvent ( C) (kg mol 1 K) ( C) (kg mol 1 K) acetic acid benzene carbon disulfide carbon tetrachloride trichloromethane ethoxyethane ethanol water at bar

34 34 / 46 Problem What is the lowest molality of ethylene glycol that will protect your car s coolant from freezing at 0.00 F? (Assume the solution is ideal) F = 17.8 C T = ( ) C = 17.8 K m = T K f 17. K m = 1.86 kg mol 1 K m = 9.56 mol kg 1

35 35 / 46 Problem What is the lowest molality of ethylene glycol that will protect your car s coolant from freezing at 0.00 F? (Assume the solution is ideal) F = 17.8 C T = ( ) C = 17.8 K m = T K f 17. K m = 1.86 kg mol 1 K m = 9.56 mol kg 1

36 36 / 46 Problem What is the lowest molality of ethylene glycol that will protect your car s coolant from freezing at 0.00 F? (Assume the solution is ideal) F = 17.8 C T = ( ) C = 17.8 K m = T K f 17. K m = 1.86 kg mol 1 K m = 9.56 mol kg 1

37 Semipermeable membrane 37 / 46

38 38 / 46 Osmotic Pressure Osmotic Pressure, Π: Excess pressure required to maintain osmotic equilibrium between a solution and the pure solvent separated by a membrane permeable only to the solvent; Π = c B RT where R is the gas constant, T the thermodynamic temperature and c B the amount concentration of individually moving solute molecules, different ions etc. regardless of their nature.

39 Osmotic Pressure 39 / 46

40 Reverse Osmosis 40 / 46

41 41 / 46 Determination of Molar Mass of Protein To determine the molar mass of a protein, g of the protein was dissolved in enough water to make 1.00 ml of solution. The osmotic pressure of this solution was found to be kpa at 25.0 C. Calculate the molar mass of the protein. c = Π RT = c = mol L bar bar L mol 1 K 1 298K n = mol M = g mol 1

42 42 / 46 Determination of Molar Mass of Protein To determine the molar mass of a protein, g of the protein was dissolved in enough water to make 1.00 ml of solution. The osmotic pressure of this solution was found to be kpa at 25.0 C. Calculate the molar mass of the protein. c = Π RT = c = mol L bar bar L mol 1 K 1 298K n = mol M = g mol 1

43 43 / 46 Determination of Molar Mass of Protein To determine the molar mass of a protein, g of the protein was dissolved in enough water to make 1.00 ml of solution. The osmotic pressure of this solution was found to be kpa at 25.0 C. Calculate the molar mass of the protein. c = Π RT = c = mol L bar bar L mol 1 K 1 298K n = mol M = g mol 1

44 44 / 46 Determination of Molar Mass of Protein To determine the molar mass of a protein, g of the protein was dissolved in enough water to make 1.00 ml of solution. The osmotic pressure of this solution was found to be kpa at 25.0 C. Calculate the molar mass of the protein. c = Π RT = c = mol L bar bar L mol 1 K 1 298K n = mol M = g mol 1

45 45 / 46 Determination of Molar Mass of Protein To determine the molar mass of a protein, g of the protein was dissolved in enough water to make 1.00 ml of solution. The osmotic pressure of this solution was found to be kpa at 25.0 C. Calculate the molar mass of the protein. c = Π RT = c = mol L bar bar L mol 1 K 1 298K n = mol M = g mol 1

46 46 / 46 Colloids and Tyndall Effect The narrow, barely visible light beam that passes through a solution (left), is scattered and broadened by passing through a colloid (right). Sunlight is scattered by dust in air.

Properties of Solutions. Chapter 13

Properties of Solutions. Chapter 13 Properties of Solutions Chapter 13 Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. Saturated solution: contains the maximum amount of a