A Molecular Approach, 4e - Notes for Chapter (13).doc

Chapter 13. Solutions Chapter 13. Solutions Chapter 13. Solutions Student Objectives 13.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater Define solution, solute, and solvent. Understand that water in tissue will combine with saltwater in order to dilute it, causing dehydration. 13.2 Types of Solutions and Solubility Know and understand that common solutions can have a gas, liquid, or solid as either the solvent or solute. Define entropy and know that it causes the mixing that results in solution formation. Know that the common types of intermolecular forces determine whether solutions will form when components are mixed. Identify some common laboratory solvents. Identify organic substances as water soluble or fat-soluble. 13.3 Energetics of Solution Formation Know and understand the components of solution formation and the energy changes associated with them. Define and understand heat of hydration and heat of solution. 13.4 Solution Equilibrium and Factors Affecting Solubility Know and understand the differences among the states that describe solution equilibrium and dissolution: unsaturated, saturated and supersaturated. Know the temperature dependence of the solubility of solids. Know that the solubility of gases is a function of both temperature and pressure. Use Henry’s law to calculate molar concentrations of gases in solution. 13.5 Expressing Solution Concentration Define the different expressions of solution concentration: molarity, molality, parts by mass, parts by volume, mole fraction, and mole percent. Know how to prepare a solution of known concentration. Convert between the different units of concentration. 13.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure Define colligative property. Understand dynamic equilibrium with respect to vapor pressure in solutions and the effect of a solute on the rate of vaporization. Use Raoult’s law to calculate the vapor pressure of a solution. Understand vapor pressure for solutions containing two volatile components, and understand deviations from Raoult’s law for non-ideal solutions. Calculate the vapor pressure of a solution containing two volatile components. Understand the basis for freezing point depression. Calculate the freezing point depression of a solution from its molality and vice versa. Understand the basis for boiling point elevation. Calculate the boiling point elevation of a solution from its molality and vice versa. Understand osmosis and osmotic pressure. Calculate the osmotic presure of a solution. Use colligative properties to calculate the molar mass of an unidentified solute. 13.7 Colligative Properties of Strong Electrolyte Solutions Understand the difference between colligative properties of nonelectrolytes and electrolytes. Calculate the van’t Hoff factor from deviations in freezing point depression, boiling point elevation, and osmotic pressure. Define hyperosmotic, hyposmotic, and isosmotic in relation to biological cells. 13.8 Colloids Define colloidal dispersion or colloid. Know the differences between the different kinds of colloid: aerosol, solid aerosol, foam, emulsion, and solid emulsion. Understand Brownian motion in solutions. Know the structure of a soap molecule and a micelle. Know and understand the Tyndall effect. Section Summaries Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Teaching Tips Suggestions and Examples Misconceptions and Pitfalls Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 13.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater Life of Pi by Yann Martel solution solvent solute spontaneous mixing two liquids solutions of different concentration saltwater in biological tissues Intro figure: rowboat on an ocean, osmotic pressure, Life of Pi (book) Figure 13.1 A Typical Solution Figure 13.2 The Tendency to Mix unnumbered figure: illustration of water flow across a cell membrane 13.2 Types of Solutions and Solubility Common types of solutions gas–gas liquid gas–liquid liquid–liquid solid–liquid solid–solid Solubility entropy intermolecular forces solvents unnumbered figure: photo and illustration of dissolved CO2 in soda water Table 13.1 Common Types of Solutions Figure 13.3 Spontaneous Mixing of Two Ideal Gases Figure 13.4 Intermolecular Forces Involved in Solutions Table 13.2 Relative Interactions and Solution Formation Figure 13.5 Forces in a Solution Table 13.3 Common Laboratory Solvents Example 13.1 Solubility 13.3 Energetics of Solution Formation Energetics Hsolute Hsolvent Hsoln = Hsolute + Hsolvent + Hmix Aqueous solutions heat of solution heat of hydration Hhydration = Hsolvent + Hmix Hsoln = Hsolute + Hhydration unnumbered figure: separation of solute particles unnumbered figure: separation of solvent particles unnumbered figure: mixing of solute and solvent particles Figure 13.6 Energetics of the Solution Process Figure 13.7 Heat of Hydration and Heat of Solution Figure 13.8 Ion–Dipole Interactions Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 13.1 Thirsty Solutions: Why You Shouldn’t Drink Seawater The Life of Pi mention explains why drinking seawater is detrimental despite the fact that the mixture is mostly water. Spontaneous mixing is closely related to effusion and diffusion in gases. The flow of water out of tissue or cells may not appear to be the same principle as spontaneous mixing. 13.2 Types of Solutions and Solubility Solutions involve the intermolecular forces between solute particles, between solvent particles, and between the two kinds. This is a good opportunity to have the class predict the magnitude of these interactions. Predicting the solubility of vitamins (Example 13.1) will require a consideration of which bonds and (functional) groups are polar. Conceptual Connection 13.1 Solubility Solutions can involve solute and solvent from each of the three phases. Many students may only have thought about liquids as solvents. Solubility is a continuum. Many “insoluble” compounds are soluble to a small extent. 13.3 Energetics of Solution Formation Have the students predict which energies are endothermic and exothermic. Solubility can be predicted from knowledge of the heat of hydration and the lattice energy. Conceptual Connection 13.2 Energetics of Aqueous Solution Formation Heats of solution can be exothermic or endothermic, though students may presume only the former from practical experience. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 13.4 Solution Equilibrium and Factors Affecting Solubility Equilibrium and solubility dissolution recrystallization solution states unsaturated saturated supersaturated Factors affecting solubility temperature pressure Henry’s law Figure 13.9 Dissolution of NaCl Figure 13.10 Precipitation from a Supersaturated Solution Figure 13.11 Solubility and Temperature unnumbered figure: photo of rock candy unnumbered figure: photos of cold and warm soda pop Figure 13.12 Soda Fizz unnumbered figures: illustration of gas solubility as a function of pressure Table 13.4 Henry’s Law Constants for Several Gases in Water at 25 oC Example 13.2 Henry’s Law 13.5 Expressing Solution Concentration Solution concentration units molarity (M) molality (m) mole fraction () mole percent (mol %) parts by mass percent (%) parts per million (ppm) parts per billion (ppb) parts by volume percent (%) parts per million (ppm) parts per billion (ppb) Making a solution of known concentration Interconverting units Chemistry in the Environment: Lake Nyos Table 13.5 Solution Concentration Terms Figure 13.13 Preparing a Solution of Known Concentration Example 13.3 Using Parts by Mass in Calculations Chemistry in the Environment: The Dirty Dozen Table13.6 The Dirty Dozen Table 13.7 EPA Maximum Contaminant Level (MCL) for Several “Dirty Dozen” Chemicals Example 13.4 Calculating Concentrations Example 13.5 Converting between Concentration Units Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 13.4 Solution Equilibrium and Factors Affecting Solubility The representation of solubility as a dynamic equilibrium helps explain saturation and the process of recrystallization. This requires a balance between dissolving and precipitating. Solubility can be manifested in two distinct properties: the amount of solute that dissolves and the rate at which the solute dissolves. The phenomenon of supersaturation is related to other ‘super’ phenomena: heating, cooling, etc. Conceptual Connection 13.3 Solubility and Temperature Conceptual Connection 13.4 Henry’s Law The solubility of gases is anomalous relative to that of most substances but logically so when considering Hsolute. Saturation especially is a dynamic process in which solute can reversibly enter and leave the solution. 13.5 Expressing Solution Concentration The most common solution concentration terms are presented. It should be pointed out that molarity is most widely used by chemists, but the others are common in other contexts and disciplines. Medicine uses a variety of units including ones not included in the text. Parts per million and parts per billion don’t make much sense for large concentrations. Give the students practical examples of these, such as in the expression of ground-water contaminants. Environmental pollutants are often evaluated at low or very low concentrations. Every water plant in the U.S. must report certain solutes and pollutants; the data for locality makes for an engaging discussion, as does detection limits and the meaning of ‘zero’ concentration. Students sometimes question the need for molarity and molality instead of one or the other. The issue there is conservation of mass and thus concentration for molality but the opposite for molarity. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 13.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure Solute effect on vapor pressure dynamic equilibrium solute effect on solvent–solvent interactions Raoult’s law Vapor pressure of solution with two volatile components ideal solutions non-ideal solutions Freezing point depression Tf = m × Kf Boiling point elevation Tb = m × Kb Osmosis = MRT unnumbered figure: photo of road salting Figure 13.14 unnumbered figure: illustration of dynamic equilibrium between liquid and vapor unnumbered figures: illustration of effect of solute on vapor pressure of a liquid unnumbered figures: dilution in a closed system Example 13.6 Calculating the Vapor Pressure of a Solution Containing a Nonelectrolyte and a Nonvolatile Solute Figure 13.15 Behavior of Ideal and Nonideal Solutions Example 13.7 Calculating the Vapor Pressure of a Two-Component Solution unnumbered figure: phase diagrams of a pure liquid and a solution unnumbered figure: photo of antifreeze/coolant Table 13.8 Freezing Point Depression and Boiling Point Elevation Constants for Several Liquid Solvents Example 13.8 Freezing Point Depression Example 13.9 Boiling Point Elevation Chemistry in Your Day: Antifreeze in Frogs Figure 13.16 An Osmosis Cell Example 13.10 Osmotic Pressure 13.7 Colligative Properties of Strong Electrolyte Solutions Van’t Hoff factor, i medical solutions isosmotic hyperosmotic hyposmotic Table 13.9 Van’t Hoff Factors at 0.05 m Concentration in Aqueous Solution Figure 13.17 Ion Pairing Example 13.11 Van’t Hoff Factor and Freezing Point Depression Example 13.12 Calculating the Vapor Pressure of a Solution Containing an Ionic Solute Figure 13.18 Red Blood Cells and Osmosis unnumbered figure: photos of a nurse and 0.9% sodium chloride solution Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 13.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure Models show the effect of a solute on solvent–solvent interactions. Solutes reduce the vapor pressure of a solution relative to the pure solvent. Vapor pressure lowering is the basis for the other colligative properties. Conceptual Connection 13.6 Raoult’s Law Freezing point depression and boiling point elevation are easy to demonstrate and explain with examples from the students’ experience. The equations provide a quantitative measure of the effects. Conceptual Connection 13.7 Boiling Point Elevation Plants and animals use several mechanisms to protect themselves from freezing. In addition to producing higher solute concentrations that take advantage of freezing point depression, some also have proteins that inhibit ice crystal formation, a different kinetic mechanism. Osmosis is an important concept in biology and medicine. The last section shows the form of cells upon being subjected to different concentrations. The vapor pressure of a solution is lower than the vapor pressure of the pure solvent. The actual pressure is given by Raoult’s law. Solutions can be ideal or non-ideal depending on the strength of solute–solvent interactions. The equations for freezing point depression and boiling point elevation give the change in temperature but not the final temperature. 13.7 Colligative Properties of Strong Electrolyte Solutions Freezing point depression, boiling point elevation, and osmotic pressure depend on the number of particles produced by the solute upon dissociation. Conceptual Connection 13.8 Colligative Properties For strong electrolytes, the equations for boiling point elevation, freezing point depression, and osmotic pressure are modified only by including the multiplicative van’t Hoff factor, i. Strong electrolytes are considered to dissociate completely only in very dilute solutions, hence the variation in values of i. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 13.8 Colloids Types of colloidal dispersions aerosol solid aerosol foam emulsion solid emulsion Properties Brownian motion specific structure soap structure micelle structure Tyndall effect Figure 13.19 A Colloid Table 13.10 Types of Colloidal Dispersions Figure 13.20 Brownian Motion Figure 13.21 Structure of a Soap Figure 13.22 Micelle Structure Figure 13.23 The Tyndall Effect unnumbered figure: photo of the Tyndall effect Figure 13.24 Micelle Repulsions Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 13.8 Colloids Colloids can be complex, but there are a number of common examples. It is not possible to show the molecular-level representations since colloidal materials involve macromolecular-sized particles. The soap example is one that everyone has experienced. The Tyndall effect can be demonstrated in several contexts. A colloid is a substance that is not a solution but has a solute-like substance evenly distributed or dispersed through the solvent. Additional Problem for Henry’s Law (Example 13.2) What pressure of nitrogen is required to keep the nitrogen concentration in a bottle of water at 0.12 M at 25 °C? Is this a reasonable value? Sort You are given the solubility of nitrogen and asked to find the pressure required to achieve this solubility. Given Snitrogen = 0.012 M Find Pnitrogen Strategize Use Henry’s law to find the required pressure from the solubility. You will need the Henry’s law constant for nitrogen. Conceptual Plan Snitrogen Pnitrogen Snitrogen = kH Pnitrogen Relationships Used Sgas = kH Pgas (Henry’s law) kH (nitrogen) = 6.1 104 M/atm (from Table 12.4) Solve Solve the Henry’s law equation for Pnitrogen and substitute the other quantities to compute it. Solution Check The units (atm) are correct. The magnitude of the answer (197) seems correct, because the concentration requested is the same as the one in a previous example for carbon dioxide, but CO2 is far more soluble than nitrogen. The outcome is not a reasonable pressure of nitrogen to use because it is dangerously high. Additional Problem for Calculating Concentrations (Example 13.4) A solution is prepared by dissolving 114 g of glucose (C6H12O6) in 0.500 kg of water. The final volume of the solution is 590 mL. Calculate each value for this solution: a) molarity b) molality c) percent by mass d) mole fraction e) mole percent. Molarity To calculate molarity, first find the amount of glucose in moles from the mass and molar mass. Then divide the amount in moles by the volume of the solution in liters. Molality To calculate molality, use the amount of glucose in moles and divide by the mass of water in kilograms. Percent by Mass To calculate the percent by mass, divide the mass of the solute glucose by the sum of the masses of the solute and the solvent (water) and multiply by 100%. Mole Fraction To calculate the mole fraction, first determine the amount of water in moles from the mass of water and its molar mass. Then divide the amount of glucose in moles by the total number of moles. Mole Percent To calculate the mole percent, multiply the mole fraction by 100%. mol % = nsolute 100% = 0.0223 100% = 2.23% Additional Problem for Boiling Point Elevation (Example 13.9) A solution is prepared from 445 g of ethylene glycol (C2H6O2) and 500 g of water. This solution represents one that is 50% by volume ethylene glycol. At what temperature will the water in this solution boil? Sort You are given the mass of the solute ethylene glycol and the mass of the water solvent. You are asked to find the new boiling point. Given 445 g of ethylene glycol 500 g of water Find boiling point of water in solution Strategize You need the molality of the solution to calculate the boiling point elevation. First, calculate the number of moles of ethylene glycol using its mass and molar mass. Next, calculate the molality of the solution using the moles of solute and mass of solvent water. Calculate the boiling point elevation and add it to the normal boiling point of water. Conceptual Plan g C2H6O2 mol C2H6O2 molality molality Tb boiling point Tb = m Kb bp = 100 °C + Tb Relationships Used 1 mol C2H6O2 = 62.07 g C2H6O2 Tb = m Kb Kb (water) = 0.512 °C/m bp = 100 °C + Tb Solve Follow the conceptual plan. Calculate the number of moles of ethylene glycol and then the molality. Determine the boiling point elevation and add it to the normal boiling point of water. Solution bp = 100 °C + 8.86 °C = 109 °C Check The units are correct. The magnitude of the answer (109) seems to make physical sense. Additional Problem for Calculating the Vapor Pressure of a Solution Containing an Ionic Solute (Example 13.12) A solution contains 0.481 mol of Na2SO4 and 10.0 mol water. Calculate the vapor pressure of the solution at 25 °C. The vapor pressure of pure water at 25 °C is 23.8 torr. Sort You are given the number of moles of each component of a solution and asked to find the vapor pressure of the solution. You are also given the vapor pressure of pure water at the desired pressure. Given 0.481 mol Na2SO4 10.0 mol water Powater = 23.8 torr at 25 °C Find Psolution Strategize Use Raoult’s law to solve this problem. Calculate solvent from the given amounts of solute and solvent. The key to the problem is to understand the dissociation of sodium sulfate. Conceptual Plan water & Powater Psolution Psolution = water Powater Relationships Used Psolution = water Powater Solve Write an equation for the dissociation of sodium sulfate. Since one mole of sodium sulfate dissociates into three moles of ions, the total number of moles of solute must be multiplied by 3 when computing the mole fraction. Use the mole fraction of water and vapor pressure of pure water to compute the vapor pressure of the solution. Solution Na2SO4(s) 2 Na+(aq) + SO42(aq) Psolution = water Powater = 0.874 23.8 torr = 20.8 torr Check The units of the answer (torr) are correct. The magnitude of the answer (20.8) makes physical sense since water dominates the mole fraction. The vapor pressure of the solution is very similar to pure water. 178 Copyright © 2017 by Education, Inc. 179 Copyright © 2017 by Education, Inc.