Physical Properties of Solutions Chemistry Second Edition Julia Burdge Lecture Powerpoints Jason A. Kautz University of Nebraska-Lincoln 13 Physical Properties of Solutions Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Physical Properties of Solutions 13 13.1 Types of Solutions 13.2 A Molecular View of the Solution Process The Importance of Intermolecular Forces Energy and Entropy in Solution Formation 13.3 Concentration Units Molality Percent by Mass Comparison of Concentration Units 13.4 Factors that Affect Solubility Temperature Pressure 13.5 Colligative Properties Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure Electrolyte Solutions 13.6 Calculations Using Colligative Properties 13.7 Colloids

Types of Solutions 13.1 A solution is a homogeneous mixture of two or more substances. A solution consists of a solvent and one or more solutes.

Types of Solutions Solutions can be classified by the amount of solute dissolved. An unsaturated solution is one that contains less solute than the solvent has the capacity to dissolve at a specific temperature.

Types of Solutions Solutions can be classified by the amount of solute dissolved. A saturated solution is one that contains the maximum amount of solute that will dissolve in a solvent at a specific temperature.

Types of Solutions Supersaturated solutions are generally unstable.

The Solution Process 13.2 Solvation occurs when solute molecules are separated from one another and surrounded by solvent molecules. Solvation depends on three types of interactions: 1) Solute-solute interactions 2) Solvent-solvent interactions 3) Solute-solvent interactions

The Solution Process ΔHsoln = ΔH1 + ΔH2 + ΔH3 Separated solute Separated solvent Step 3 ΔH3 < 0 Energy Step 2 ΔH2 > 0 Separated solute Solvent Solution Step 1 ΔH1 > 0 ΔHsoln > 0 Solute Solvent

Both non-polar liquids, solution forms when mixed The Solution Process “Like dissolves like” Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other. Toluene, C7H8 Octane, C8H18 Both non-polar liquids, solution forms when mixed Two liquids are said to be miscible if they are completely soluble in each other in all proportions.

Polar and non-polar liquids, solution does not form when mixed The Solution Process “Like dissolves like” Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other. Water, H2O Octane, C8H18 Polar and non-polar liquids, solution does not form when mixed

solution forms when mixed The Solution Process “Like dissolves like” Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other. Water, H2O Ethanol, C2H6O Both polar liquids, solution forms when mixed

Concentration Units 13.3 The amount of solute relative to the volume of a solution or to the amount of solvent in a solution is called concentration. Molarity: Mole fraction:

Concentration Units Molality (m) is the number of moles of solute dissolved in 1 kg (1000 g) solvent: Percent by Mass:

Concentration Units An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of 1.109 g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C. Solution: Step 1: Assume 100 grams of solution, and use mass percent to determine the grams and moles of the solute.

Concentration Units An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of 1.109 g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C. Solution: Step 2: Assume 100 g solution and determine the volume of the solution using the density of the solution.

Concentration Units An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of 1.109 g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C. Solution: Step 3: Use the equation below to determine the molarity.

Concentration Units An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of 1.109 g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C. Solution: Step 4: Determine the kg of water (solvent). grams solution = grams solute + grams solvent 100 g solution  16 g solute = 84 g solvent = 0.084 kg solvent

Concentration Units An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of 1.109 g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C. Solution: Step 5: Use the equation below to determine the molality.

Factors That Affect Solubility 13.4 Temperature affects the solubility of most substances.

Factors That Affect Solubility Pressure greatly influences the solubility of a gas. Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution. c molar concentration (mol/L) P pressure (atm) k proportionality constant called Henry’s law constant. c = kP

Factors That Affect Solubility Calculate the concentration of CO2 in water at 25°C when the pressure of CO2 over the solution is 4.0 atm. Solution: Step 1: Use the equation below to calculate concentration. The Henry’s law constant for CO2 at 25°C is 3.1 x 10‒3 mol/L atm. c = (3.1 x 10‒3 mol/L atm)(4.0 atm) c = 0.012 mol/L c = kP

Colligative Properties 13.5 Colligative properties are properties that depend on the number of solute particles in solution. Colligative properties do not depend on the nature of the solute particles. The colligative properties are: vapor-pressure lowering boiling-point elevation freezing-point depression osmotic pressure

Colligative Properties Raoult’s law states that the partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. P1 partial pressure of solvent over solution P° vapor pressure of pure solvent χ1 mole fraction of solvent ΔP vapor pressure lowering χ2 mole fraction of solute

Colligative Properties Calculate the vapor pressure of a solution made by dissolving 114 g of urea (molar mass = 60.06 g/mol) in 485 g of water at 25°C. (At 25°C the vapor pressure of water is 23.8 mmHg) Solution: Step 1: Determine the mole fraction of water in the solution.

Colligative Properties Calculate the vapor pressure of a solution made by dissolving 114 g of urea (molar mass = 60.06 g/mol) in 485 g of water at 25°C. (At 25°C the vapor pressure of water is 23.8 mmHg) Solution: Step 2: Determine the vapor pressure of the solution using the equation below. Pwater = 0.9336(23.8 mmHg) Pwater = 22.2 mmHg

Colligative Properties If both components of a solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures. Benzene Toluene

Colligative Properties Benzene Toluene An ideal solution obeys Raoult’s law.

Colligative Properties Solutions boil at a higher temperature than the pure solvent. ΔTb boiling point elevation Kb boiling point elevation constant (°C/m) m molality

Colligative Properties Solutions freeze at a lower temperature than the pure solvent. ΔTf freezing point depression Kf freezing point depression constant (°C/m) m molality

Colligative Properties

Colligative Properties Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m. Solution: Step 1: Determine the moles of ethylene glycol.

Colligative Properties Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m. Solution: Step 2: Use the equation below to determine the molality of the solution.

Colligative Properties Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m. Solution: Step 3: Use the equation below to determine Tf for the solution and the freezing point of the solution.

Colligative Properties Osmosis is the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated one.

Colligative Properties Osmotic pressure () of a solution is the pressure required to stop osmosis.  Osmotic pressure (atm) M molarity (moles/L) R gas constant (0.08206 L atm/mol K) T absolute temperature (Kelvin)

Colligative Properties Electrolytes undergo dissociation when dissolved in water. The van’t Hoff factor (i) accounts for this effect.

Colligative Properties The van’t Hoff factor (i) is 1 for all nonelectrolytes: For strong electrolytes i should be equal to the number of ions: C12H22O11(s) C12H22O11(aq) H2O 1 particle dissolved, i = 1 NaCl(s) Na+(aq) + Cl–(aq) H2O 2 particles dissolved, i = 2 Na2SO4(s) 2Na+(aq) + SO42–(aq) H2O 3 particle dissolved, i = 3

Colligative Properties The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs. An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces. ion pair

Colligative Properties The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs. An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces. Table 13.3

Colligative Properties Concentration has an effect on experimentally measured van’t Hoff factors (i). Table 13.4

Colligative Properties The osmotic pressure of a 0.0150 M NaCl solution at 25C is 0.693 atm. Determine the experimental van’t Hoff factor for NaCl at this concentration. Solution: Step 1: Solve for the van’t Hoff factor (i) using the equation below.

Calculations Using Colligative Properties 13.6 Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m. Solution: Step 1: Use the freezing point depression equation below to calculate the molality of the solution.

Calculations Using Colligative Properties Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m. Solution: Step 2: Using the equation for molality, determine the moles of naphthalene.

Calculations Using Colligative Properties Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m. Solution: Step 3: Calculate the molar mass of naphthalene.

Calculations Using Colligative Properties Percent dissociation is the percentage of dissolved molecules (or formula units, in the case of an ionic compound) that separate into ions in a solution. Strong electrolytes should have complete, or 100%, dissociation, however, experimentally determined van’t Hoff factors indicate that this is not the case. Percent dissociation of a strong electrolyte is more complete at lower concentration. Percent ionization of weak electrolytes is also dependent on concentration.

Colloids 13.7 A colloid is a dispersion of particles of one substance throughout another substance. Colloid particles are much larger than the normal solute molecules. Categories of colloids: aerosols foams emulsions sols gels

Colloids Examples of colloids

Colloids Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing). Hydrophilic groups on the surface of a large molecule stabilize the molecule in water.

Colloids Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing). Negative ions are adsorbed onto the surface of hydrophobic colloids. The repulsion between like charges prevents aggregation of the articles.

Colloids Hydrophobic colloids can be stabilized by the presence of hydrophilic groups on their surface.

Colloids Emulsification is the process of stabilizing a colloid that would otherwise not stay dispersed.

13 Key Concepts Types of Solutions A Molecular View of the Solution Process The Importance of Intermolecular Forces Energy and Entropy in Solution Formation Concentration Units Molality Percent by Mass Comparison of Concentration Units Factors that Affect Solubility Temperature Pressure Colligative Properties Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure Electrolyte Solutions Calculations Using Colligative Properties Colloids