11.3 Factors Affecting Solubility 11.4 The Vapor Pressures of Solutions 11.5 Boiling-Point Elevation and Freezing-Point Depression 11.6 Osmotic Pressure 11.7 Colligative Properties of Electrolyte Solutions Copyright © Cengage Learning. All rights reserved

Solution Composition Copyright © Cengage Learning. All rights reserved

Solution = Solvent + Solute So, Mole Fraction Solution = Solvent + Solute So, Mole solution = mole solvent + mole solute Copyright © Cengage Learning. All rights reserved

Exercise A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 (molar mass 98.0 g/mol) 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 (molar mass 98.0 g/mol) 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

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

Pressure Effects 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 vs. Temperature Copyright © Cengage Learning. All rights reserved

Vapor pressure results from dynamic equilibrium between liquid state and gas state An Aqueous Solution and Pure Water in a Closed Environment: Eventually, pure solvent be transferred into solution state via evaporation/condensation equilibrium Copyright © Cengage Learning. All rights reserved

Vapor Pressures of Solutions Mixing of nonvolatile solute reduces the chance of solvent molecule entering gas state, thus reducing vapor pressure. Nonvolatile solute lowers the vapor pressure of a solvent. Raoult’s Law: Psoln = observed vapor pressure of solution solv = mole fraction of solvent Psolv° = vapor pressure of pure solvent Copyright © Cengage Learning. All rights reserved

Practice: Vapor pressure for solution The vapor pressure of a solution containing 107. g glycerin (molar mass 92.1 g/mol) in 266. g ethanol (molar mass 46.1 g/mol) is 104. torr at 35°C. Calculate the vapor pressure of pure ethanol at 35°C, assuming glycerin is nonvolatile nonelectrolyte in ethanol. Mole fraction of solvent = 0.832 Vapor pressure of pure ethanol = 104/0.832 = 125 torr Copyright © Cengage Learning. All rights reserved

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

Liquid-liquid solutions where both components are volatile. 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, but deviate between 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 Hsoln T 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

Colligative Properties Boiling-point elevation: Because of depressed vapor pressure of solvent in a solution, the boiling point of solution is higher than pure solvent. 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

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

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

100.35 °C ΔT = m Kb = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C Exercise A solution was prepared by dissolving 25.00 g 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. Kb = 0.51 °C·kg/mol. mol glucose = 0.1388 mol 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. Molality m = 0.1388 mol/0.2000 kg = 0.694 m ΔT = m Kb = (0.51 °C·kg/mol)(0.6938 mol/kg) = 0.35 °C 100.35 °C Copyright © Cengage Learning. All rights reserved

Osmotic pressure (): The pressure needed to stop the flow of solvent Osmosis – flow of solvent into the solution through a semipermeable membrane. Osmotic pressure (): The pressure needed to stop the flow of solvent  = MRT (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

Salt causes minced water-rich plants/fruits to dehydrate Osmosis in life Salt causes minced water-rich plants/fruits to dehydrate What happens when salt is applied to live tissues/organs/animals with thin membrane? High concentration of salt from exterior of membrane causes water (solvent) leaving the interior of membrane. The following videos are for scientific observation only, DO NOT imitate!!! https://www.youtube.com/watch?v=XXVc7HXjRp8 https://www.youtube.com/watch?v=2YZJt_Bw3eo https://www.youtube.com/watch?v=OpBuR0ERLfc Copyright © Cengage Learning. All rights reserved

Exercise: Determine molar mass using Colligative Properties 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  = MRT Find molarity of solution: M= /(RT) The molar mass is 111 g/mol. Note: Use the red box animation to assist in explaining how to solve the problem. 0.734/(0.08206*298) = 0.300 M Mol unknown = 0.300 M x 0.0100 L = 3.00E-4 mol Molar mass = mass/mol =3.34E-2 g/3.00E-4 mol = 111 g/mol Copyright © Cengage Learning. All rights reserved

Ionic compound dissociates in solution, resulting more particles. van’t Hoff Factor, i Ionic compound dissociates in solution, resulting more particles. Example: 1 mole NaCl  1 mol Na+ + 1 mol Cl- 1 mole Fe(NO3)3  1 mol Fe3+ + 3 mol NO3- These “extra” ions enhance the colligative properties, thus a calibration factor i was introduced. 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 = Na+ + Cl- i = 2 NH4NO3 = NH4+ + NO3- i = 2 K3PO4 = 3 K+ + PO43- i = 4 Copyright © Cengage Learning. All rights reserved

Considering the dissociation of ionic solute, freezing-point depression, boiling-point elevation, and osmotic pressure equations become Because of the dissociation, ionic compounds are more effective than nonelectrolytes with similar concentration in causing freezing-point depression, boiling-point elevation, and enhancing osmotic pressure. Homemade icecream so easy  https://www.youtube.com/watch?v=N4ztYjFxwmI Salt helps melt snow/ice on the road Salt causes freshly cut vegetables to “liquefy” (osmosis through cell membrane) Copyright © Cengage Learning. All rights reserved

Practice: Predict the freezing point depression The old fashion to make homemade ice-cream is to churn the ingredient (milk, sugar, vanilla, etc) in a container placed in a mixture of ice and salt. Which solution, 500. mL 1.0 M NaCl solution or 250. mL 0.80 M CaCl2 solution, will give the lower temperature? Freezing point constant Kf for water is 1.86 °C/kg. The molarity of solution is treated approximately same as the molality of solution. 500. mL NaCl: T  2x1.0x1.86 = 3.72°C 250. mL CaCl2 : T  3x0.80x1.86 = 4.46°C Copyright © Cengage Learning. All rights reserved

Practice: Predict the osmotic pressure Seawater contains 3.5% salt, roughly 0.60 M NaCl. Calculate the minimal pressure needed at 298 K to purify the water by osmosis. Assume NaCl is completely dissociated.  = 2 x 0.60 x 0.08206 x 298 = 30 atm. Copyright © Cengage Learning. All rights reserved

Incomplete separation of Ions: Ion Pairing In diluted solutions, the ions are farther apart (free ions). Ion Pairing: As concentration increases or charges are high, counter ions are more likely to attract each other, leading to “pairing”. Ion pairing occurs to some extent in all electrolyte solutions, but is most important in concentrated solutions, and/or for highly charged ions. The result of ion pairing is van’t Hoff facotr i < ideal value. For example, i = 1.3 for 0.05 m MgSO4 solution (ideal value = 2.0) 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

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

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

Processes that require large amounts of energy tend not to occur. 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

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

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