Various Types of Solutions Copyright © Cengage Learning. All rights reserved

Solution Composition Copyright © Cengage Learning. All rights reserved

Molarity Copyright © Cengage Learning. All rights reserved

EXERCISE! You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity. 8.00 M 1.00 mol / (125.0 / 1000) = 8.00 M Copyright © Cengage Learning. All rights reserved

EXERCISE! You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? 0.200 L 2.00 mol / 10.0 M = 0.200 L Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity. 10.0 M NaOH 5.37 M KCl [100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH [100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl Copyright © Cengage Learning. All rights reserved

Mass Percent Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6% [5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% Copyright © Cengage Learning. All rights reserved

Mole Fraction Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.) 0.0145 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 mol / 18.016 g H2O) = 5.55 mol H2O Mole Fraction (H3PO4) = 0.0816 mol H3PO4 / [0.0816 mol H3PO4 + 5.55 mol H2O] = 0.0145 Copyright © Cengage Learning. All rights reserved

Molality Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) 0.816 m 8.00 g H3PO4 × (1 mol / 97.994 g H3PO4) = 0.0816 mol H3PO4 100.0 mL H2O × (1.00 g H2O / mL) × (1 kg / 1000 g) = 0.1000 kg H2O Molality = 0.0816 mol H3PO4 / 0.1000 kg H2O] = 0.816 m Copyright © Cengage Learning. All rights reserved

Formation of a Liquid Solution Separating the solute into its individual components (expanding the solute). Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). Allowing the solute and solvent to interact to form the solution. Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process Copyright © Cengage Learning. All rights reserved

Steps in the Dissolving Process Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔHsoln = ΔH1 + ΔH2 + ΔH3 ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released). Copyright © Cengage Learning. All rights reserved

Enthalpy (Heat) of Solution Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. Oil is a mixture of nonpolar molecules that interact through London dispersion forces, which depend on molecule size. ΔH1 will be relatively large for the large oil molecules. The term ΔH3 will be small, since interactions between the nonpolar solute molecules and the polar water molecules will be negligible. However, ΔH2 will be large and positive because it takes considerable energy to overcome the hydrogen bonding forces among the water molecules to expand the solvent. Thus ΔHsoln will be large and positive because of the ΔH1 and ΔH2 terms. Since a large amount of energy would have to be expended to form an oil-water solution, this process does not occur to any appreciable extent. Copyright © Cengage Learning. All rights reserved

The Energy Terms for Various Types of Solutes and Solvents ΔHsoln Outcome Polar solute, polar solvent Large Large, negative Small Solution forms Nonpolar solute, polar solvent Large, positive No solution forms Nonpolar solute, nonpolar solvent Polar solute, nonpolar solvent Copyright © Cengage Learning. All rights reserved

In General One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed. Processes that require large amounts of energy tend not to occur. Overall, remember that “like dissolves like”. Copyright © Cengage Learning. All rights reserved

Affecting aqueous solutions Structure Effects: Polarity Pressure Effects: Henry’s law Temperature Effects: Affecting aqueous solutions Copyright © Cengage Learning. All rights reserved

Structure Effects Hydrophobic (water fearing) Non-polar substances Hydrophilic (water loving) Polar substances Copyright © Cengage Learning. All rights reserved

Pressure Effects C = concentration of dissolved gas k = constant Little effect on solubility of solids or liquids Henry’s law: C = kP C = concentration of dissolved gas k = constant P = partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. Copyright © Cengage Learning. All rights reserved

A Gaseous Solute Copyright © Cengage Learning. All rights reserved

Temperature Effects (for Aqueous Solutions) Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature. Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Solids as a Function of Temperature Copyright © Cengage Learning. All rights reserved

The Solubilities of Several Gases in Water Copyright © Cengage Learning. All rights reserved

An Aqueous Solution and Pure Water in a Closed Environment Copyright © Cengage Learning. All rights reserved

Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Vapor Pressure Lowering: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Vapor Pressures of Solutions Nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved

A Solution Obeying Raoult’s Law Copyright © Cengage Learning. All rights reserved

Nonideal Solutions Liquid-liquid solutions where both components are volatile. Modified Raoult’s Law: Nonideal solutions behave ideally as the mole fractions approach 0 and 1. Copyright © Cengage Learning. All rights reserved

Vapor Pressure for a Solution of Two Volatile Liquids Copyright © Cengage Learning. All rights reserved

Summary of the Behavior of Various Types of Solutions Interactive Forces Between Solute (A) and Solvent (B) Particles ΔHsoln ΔT for Solution Formation Deviation from Raoult’s Law Example A  A, B  B  A  B Zero None (ideal solution) Benzene-toluene A  A, B  B < A  B Negative (exothermic) Positive Negative Acetone-water A  A, B  B > A  B Positive (endothermic) Ethanol-hexane Copyright © Cengage Learning. All rights reserved

Hexane (C6H14) and chloroform (CHCl3) Ethyl alcohol (C2H5OH) and water CONCEPT CHECK! For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation? Hexane (C6H14) and chloroform (CHCl3) Ethyl alcohol (C2H5OH) and water Hexane (C6H14) and octane (C8H18) a) Positive deviation; Hexane is non-polar, chloroform is polar. b) Negative deviation; Both are polar, and the ethyl alcohol molecules can form stronger hydrogen bonding with the water molecules than it can with other alcohol molecules. c) Ideal; Both are non-polar with similar molar masses. Copyright © Cengage Learning. All rights reserved

Colligative Properties Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure Copyright © Cengage Learning. All rights reserved

Boiling-Point Elevation Nonvolatile solute elevates the boiling point of the solvent. ΔT = Kbmsolute ΔT = boiling-point elevation Kb = molal boiling-point elevation constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Liquid/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Boiling Point Elevation: Solution/Vapor Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing-Point Depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = Kfmsolute ΔT = freezing-point depression Kf = molal freezing-point depression constant msolute = molality of solute Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Liquid Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Addition of a Solute To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Freezing Point Depression: Solid/Solution Equilibrium To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Changes in Boiling Point and Freezing Point of Water Copyright © Cengage Learning. All rights reserved

EXERCISE! A solution was prepared by dissolving 25.00 g of glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. 100.35 °C The change in temperature is ΔT = Kbmsolute. Kb is 0.51 °C·kg/mol. To solve formsolute, use the equation m = moles of solute/kg of solvent. Moles of solute = (25.00 g glucose)(1 mol / 180.16 g glucose) = 0.1388 mol glucose Kg of solvent = (200.0 g)(1 kg / 1000 g) = 0.2000 kg water msolute = (0.1388 mol glucose) / (0.2000 kg water) = 0.6938 mol/kg ΔT = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C. The boiling point of the resulting solution is 100.00 °C + 0.35 °C = 100.35 °C. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313 The solution is 72.8% sucrose and 27.2% sodium chloride. The mole fraction of the sucrose is 0.313. To solve this problem, the students must assume that i = 2 for NaCl. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

EXERCISE! A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing? The cell will explode (or at least expand). The concentration of the solution is 0.186 m. Thus, the cell has a higher concentration, and water will enter the cell. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

M = molarity of the solution R = gas law constant Osmosis – flow of solvent into the solution through a semipermeable membrane. = MRT = osmotic pressure (atm) M = molarity of the solution R = gas law constant T = temperature (Kelvin) Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

Osmosis To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

EXERCISE! When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. 111 g/mol The molar mass is 111 g/mol. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

van’t Hoff Factor, i The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as: Copyright © Cengage Learning. All rights reserved

Ion Pairing At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle. Copyright © Cengage Learning. All rights reserved

Examples The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2 Na3PO4 i = 4 Copyright © Cengage Learning. All rights reserved

Ion Pairing Ion pairing is most important in concentrated solutions. As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs. Ion pairing occurs to some extent in all electrolyte solutions. Ion pairing is most important for highly charged ions. Copyright © Cengage Learning. All rights reserved

Modified Equations Copyright © Cengage Learning. All rights reserved

A suspension of tiny particles in some medium. Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm. Copyright © Cengage Learning. All rights reserved

Types of Colloids Copyright © Cengage Learning. All rights reserved

Coagulation Destruction of a colloid. Usually accomplished either by heating or by adding an electrolyte. Copyright © Cengage Learning. All rights reserved