Transcript
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.
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Copyright © 2017 by Education, Inc.
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Copyright © 2017 by Education, Inc.