Chemistry Chapter 11 Properties of Solutions PDF

Chapter 11 Properties of Solutions Copyright ©2018 Cengage Learning. All Rights Reserved. Chapter 11 Table of Contents Table of Contents  (11.1) Solution composition  (11.2) The energies of solution formation  (11.3) Factors affecting solubi...

Chapter 11 Properties of Solutions Copyright ©2018 Cengage Learning. All Rights Reserved. Chapter 11 Table of Contents Table of Contents  (11.1) Solution composition  (11.2) The energies of solution formation  (11.3) Factors affecting solubility  (11.4) The vapor pressures of solutions Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Solution Composition  As mixtures have variable compositions, relative amounts of substances in a solution must be specified.  Qualitative terms - Dilute and concentrated  The qualitative term dilute (relatively little solute present) and concentrated (relatively large amount of solute present) are often used to describe the solution content  Molarity (M): Number of moles of solute per litre of solution. CopyrightIt isCengage ©2018 useful to Reserved. Learning. All Rights describe solution Copyright © Cengage Learning. All rights reserved 3 Section 11.1 Solution Composition Solution Composition (continued) moles of solute Molarity (M ) = liters of solution mass of solute Mass (weight) percent = 100% mass of solution moles A Mole fraction (  A ) = total moles of solution moles of solute Molality ( m ) = kilogram of solvent Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Various Methods for Describing Solution Composition  A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 g water to give a final volume of 101 mL  Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution  Molarity  The moles of ethanol can be obtained from its molar mass (46.07 g/mol): 1 mol C 2 H 5OH 1.00 g C 2 H 5OH  2.17  10  2 mol C 2H 5OH 46.07 g C 2 H 5OH 1L Volume 101 mL  0.101 L 1000 mL Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 1) moles of C 2 H 5OH 2.17  10  2 mol Molarity of C 2 H 5OH =  liters of solution 0.101 L Molarity of C 2 H 5OH = 0.215 M  Mass percent  mass of C2 H 5OH  Mass percent C 2 H 5OH =    100%  mass of solution   1.00 g C 2 H 5OH     100%  0.990% C 2 H 5OH  100.0 g H 2O + 1.00 g C 2 H 5OH  Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 2)  Mole fraction nC2H5OH Mole fraction of C 2 H 5OH = nC2H5OH  nH 2O 1 mol H 2 O nH2O 100.0 g H 2 O  5.56 mol 18.0 g H 2 O 2.17  10 2 mol 2.17  10 2  C H OH   0.00389 2 5 2 2.17  10 mol  5.56 mol 5.58 Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 3)  Molality moles of C 2 H 5OH 2.17 10 2 mol Molality of C 2 H 5OH =  kilogram of H 2O 100.0 g  1 kg 1000 g 2.17 10 2 mol  0.217 m 0.1000 kg Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Normality (N)  Measure of concentration  Another concentration measure sometimes encountered is normality (symbolized by N)  Number of equivalents per liter of solution where:  Definition of an equivalent depends on the reaction that takes place in a solution  For acid–base reactions, the equivalent is the mass of acid or base that can accept or provide exactly 1 mole of protons  For oxidation–reduction reactions, the equivalent is the quantity of oxidizing or reducing agent that can accept or Copyright ©2018 provide Cengage 1 mole Learning. All Rights Reserved. of electrons Section 11.1 Solution Composition Interactive Example - Calculating Various Methods of Solution Composition from the Molarity  The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL  Calculate the mass percent, molality, and normality of the sulfuric acid Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution  What is the density of the solution in grams per liter? g 1000 mL 1.230  1.230  103 g/L mL 1L  What mass of H2SO4 is present in 1.00 L of solution?  We know 1 liter of this solution contains 1230 g of the mixture of sulfuric acid and water Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 1)  Since the solution is 3.75 M, we know that 3.75 moles of H2SO4 is present per liter of solution  The number of grams of H2SO4 present is 98.0 g H 2SO 4 3.75 mol  368 g H 2SO 4 1 mol Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 2)  How much water is present in 1.00 L of solution?  The amount of water present in 1 liter of solution 1230 g is obtained solution  368from g H SOthe difference = 862 gH O 2 4 2  What is the mass percent?  Since we now know the masses of the solute and solvent, we can calculate the mass percent Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 3) mass of H 2SO 4 Mass percent H 2SO 4   100% mass of solution 368 g   100% = 29.9% H 2SO 4 1230 g  What is the molality?  From the moles of solute and the mass of solvent, we can calculate the molality Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 4) moles H 2SO 4 Molality of H 2SO 4  kilogram of H 2 O 3.75 mol H 2SO 4  4.35 m 1 kg H 2O 862 g H 2O  1000 g H 2O Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.1 Solution Composition Interactive Example - Solution (Continued 5)  What is the normality?  normality is described as the number of gram or mole equivalents of solute present in one litre of a solution.  Since each sulfuric acid molecule can furnish two protons, 1 mole of H2SO4 represents 2 equivalents  Thus, a solution with 3.75 moles of H2SO4 per liter contains 2×3.75 = 7.50 equivalents per Copyright ©2018 liter Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Formation of a Liquid Solution 1. Dissolving solutes in liquids is very common such as we dissolve salt in water, sugar in water for tea and coffee etc., 2. Separating the solute into its individual components (expanding the solute). 3. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 4. Allowing the solute and solvent to interact Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Steps in the Dissolving Process Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Steps Involved in the Formation of a Liquid Solution (continued)  Steps 1 and 2 are endothermic  Forces must be overcome to expand the solute and solvent  Step 3 is often exothermic  What factors affect solubility? The cardinal rule of solubility is “like dissolve like”  Like polar solvent to dissolve a polar or ionic solute  a non-polar solvent to dissolve non-polar solute. Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 20 Section 11.2 The Energies of Solution Formation Steps Involved in the Formation of a Liquid Solution (continued)  Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.  Step 3 usually releases energy.  Step 1 and 2 are endothermic, and step 3 is often exothermic. Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 21 Section 11.2 The Energies of Solution Formation Enthalpy (Heat) of Solution (ΔHsoln)  Enthalpy change associated with the formation of the solution called the Enthalpy (heat) of solution ΔHsoln which is the sum of the ΔH values for the steps: H soln H1  H 2  H 3  ΔHsoln can have a positive sign when energy is absorbed or a negative sign when energy is releasedCopyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 22 Section 11.2 The Energies of Solution Formation Table - The Energy Terms for Various Types of Solutes and Solvents Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Concept Check Explain why water and oil (a long chain hydrocarbon) do not mix Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Chromatography is the technique to separate mixture Thin Layer Chromatography (TLC)  Uses a TLC plate as the stationary phase  TLC plate consists of a plastic sheet covered with a thin layer of silica gel  Silica gel is very polar and is capable of hydrogen bonding  Mixture to be analyzed is placed on the plate, and the plate is dipped into a solvent (the mobile phase) Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Figure - Thin Layer Chromatography (a) The plate is spotted and (b) After some time, the solvent (mobile placed into the solvent phase) will travel up the plate and the less polar component travels farther than the more polar component Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Interactive Example - Differentiating Solvent Properties  Decide whether liquid hexane (C6H14) or liquid methanol (CH3OH) is the more appropriate solvent for the substances grease (C20H42) and potassium iodide (KI) Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Interactive Example - Solution  Hexane is a nonpolar solvent because it contains C—H bonds  Hexane will work best for the nonpolar solute grease  Methanol has an O—H group that makes it significantly polar  Will serve as the better solvent for the ionic solid KI Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.2 The Energies of Solution Formation Example  Which of the following chemical or physical changes is an endothermic process? a. Combustion of gasoline b. Evaporation of water c. Freezing of water d. Mixing of sulfuric acid and water Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.3 Factors Affecting Solubility Factors Affecting Solubility  Structural Effects:  Polarity  Pressure Effects:  Henry’s law  Temperature Effects:  Affecting aqueous solutions Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.3 Factors Affecting Solubility Structure Effects  Vitamins can be used to study the relationship among molecular structure, polarity, and solubility  Fat-soluble vitamins (A, D, E, and K) are nonpolar  Considered to be hydrophobic (water-fearing)  Can build up in the fatty tissues of the body  Water-soluble vitamins (B and C) are polar  Considered to be hydrophilic (water-loving)  Must be consumed Copyright ©2018 Cengageregularly Copyright © Cengage Learning. All rights reserved as they31 are Learning. All Rights Reserved. Section 11.3 Factors Affecting Solubility Pressure Effects  Pressure increases the solubility of a gas  Henry’s law: Amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution C kP  C - Concentration of the dissolved gas  k - Constant  P - Partial pressure of the gaseous solute above the solution Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 32 Section 11.3 Factors Affecting Solubility Pressure Effects  While pressure has little effect on the solubility of solids and liquids, it does significantly increase the solubility of gas.  Carbonated beverages, for example, are always bottled at high pressure of carbon dioxide CO2 to ensure a high concentration of CO2 in the liquid. Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.3 Factors Affecting Solubility Figure - Schematic Diagram That Depicts the Increase in Gas Solubility with Pressure (a) A gaseous solute in equilibrium with a solution. (b) The piston is pushed in, which increases pressure of the gas and the number of gas molecules per unit volume. This causes an increase in the rate at which the gas enters the solution so the concentration of dissolved gas increases. (c) The greater gas concentration in the solution causes an increase in the rate of escape. A new equilibrium is reached. Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 34 Section 11.3 Factors Affecting Solubility Temperature Effects (for Aqueous Solutions)  Every day experiences of dissolving substances such as sugar may lead to think that solubility always increases with temperature. This is not the case:  Solids dissolve rapidly at higher temperatures  Amount of solid that can be dissolved may increase or decrease with increasing temperature  Solubilities of some substances decrease with increasing temperature  Predicting temperature dependence of solubility is very difficult. The only sure way to determine the temperature Copyright dependence © Cengage Learning. All rights of a solid’s solubility Copyright ©2018 Cengage Learning. All Rights Reserved. reserved 35 is by Section 11.3 Factors Affecting Solubility Temperature Effects (for Aqueous Solutions) (continued)  Solubility of a gas in water decreases with increasing temperature. This temperature affect has importance environmental implications because of the widespread use of water from lakes and rivers for industrial cooling.  Water used for industrial cooling is returned to its natural source at higher than ambient temperatures  Causes thermal pollution occurs.  Warm Copyright © Cengage Learning. All water rights reservedtends to float over the 36 Copyright ©2018 Cengage Learning. All Rights Reserved. colder water, Section 11.4 The Vapor Pressures of Solutions Vapor Pressure  Liquid solutions have physical properties different from those of the pure solvent.  A fact that has great practical importance.  For an example, we add antifreeze to the water in a car ‘s cooling system to prevent freezing in winter and boiling in summer.  To explore non-volatile solute affects a Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Figure - An Aqueous Solution and Pure Water in a Closed Environment An aqueous solution and pure water in a closed environment (a) Initial stage (b) After a period of time, the water is transferred to the solution Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Vapor Pressures of Solutions  Presence of a nonvolatile solute lowers the vapor pressure of a solvent  Inhibits the escape of solvent molecules from the liquid Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 39 Section 11.4 The Vapor Pressures of Solutions Raoult's Law 0 Psoln  P solvent solvent  Psoln - Observed vapor pressure of the solution  χsolvent - Mole fraction of the solvent  P0solvent - Vapor pressure of the pure solvent  Nonvolatile solute simply dilutes the solvent Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 40 Section 11.4 The Vapor Pressures of Solutions Graphical Representation of Raoult's Law  Can be represented as a linear equation of the form y = mx + b  y = Psoln  x = χsolvent  m = P0solvent b=0  Slope of the graph is a straight line with a slope equal to P0solvent Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 41 Section 11.4 The Vapor Pressures of Solutions Figure - Plot of Raoult's Law Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Interactive Example - Calculating the Vapor Pressure of a Solution  Calculate the expected vapor pressure at 25°C for a solution prepared by dissolving 158.0 g common table sugar (sucrose,(C12H22O11) molar mass = 342.3 g/mol) in 643.5 cm3 of water  At 25°C, the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Interactive Example - Solution  What is Raoult’s law for this case? Psoln  H2O PH02 O  To calculate the mole fraction of water in the solution, we must first determine the number of moles of sucrose and the moles of water present Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Interactive Example - Solution (Continued 1)  What are the moles of sucrose? 1 mol sucrose Moles of sucrose 158.0 g sucrose  0.4616 mol sucrose 342.3 g sucrose  What are the moles of water?  To determine the moles of water present, we first convert volume to mass using the density: 0.9971 g H 2 O 643.5 cm3 H 2 O  3 641.6 g H 2 O cm H 2 O Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Interactive Example - Solution (Continued 2)  The number of moles of water 1 mol H 2 O 641.6 g H 2 O  35.60 mol H2 O 18.02 g H 2 O  What is the mole fraction of water in the solution? mol H 2O 35.60 mol H O   2 mol H 2O + mol sucrose 35.60 mol + 0.4616 mol 35.60 mol  0.9873 36.06 mol Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Interactive Example - Solution (Continued 3)  The vapor pressure of the solution is: Psoln  H 2O PH02O 0.9872 23.76 torr  23.46 torr  The vapor pressure of water has been lowered from 23.76 torr in the pure state to 23.46 torr in the solution  The vapor pressure has been lowered by 0.30 torr Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Nonideal Solutions  So far we have assumed that the solute is non-volatile and so does not contribute to the vapor pressure over the solution. However if:  Both components are volatile in liquid–liquid solutions  It does Contribute to the total vapor pressure PTOTAL PA  PB   A PA0   B PB0  Hence the Modified Raoult’s law is applied here Copyright ©2018 Cengage Learning. All Rights Reserved. Section 11.4 The Vapor Pressures of Solutions Nonideal Solutions (continued)  P0A and P0B - Vapor pressures of pure A and pure B  PA and PB - Partial pressures resulting from molecules of A and of B in the vapor above the solution  Liquid–liquid solution that obeys Raoult’s law is called Ideal solution.  Raoult’s law is to solutions, what the ideal gas law is to gases. As with gases, ideal behavior for solutions is never perfectly achieved but is sometimes closely approached. Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 49