Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro
Chapter 12 Solutions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Thirsty Seawater Drinking seawater can cause you to dehydrate
Seawater is a homogeneous mixture of salts with water Seawater contains high concentrations of salts higher than the salt content of your cells As seawater passes through your body, it pulls water out of your cells; due mainly to nature’s tendency toward spontaneous mixing This reduces your cells’ water level and usually results in diarrhea as this extra liquid flows out with the seawater Tro: Chemistry: A Molecular Approach, 2/e

Seawater Tro: Chemistry: A Molecular Approach, 2/e

Seawater Drinking seawater will dehydrate you and give you diarrhea
The cell wall acts as a barrier to solute moving so the only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract Tro: Chemistry: A Molecular Approach, 2/e

Solutions Homogeneous mixtures
composition may vary from one sample to another appears to be one substance, though really contains multiple materials Most homogeneous materials we encounter are actually solutions e.g., air and seawater Nature has a tendency toward spontaneous mixing generally, uniform mixing is more energetically favorable Tro: Chemistry: A Molecular Approach, 2/e

Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can tell from the taste, or simply boiling away the water Homogeneous mixtures are called solutions The component of the solution that changes state is called the solute The component that keeps its state is called the solvent if both components start in the same state, the major component is the solvent Tro: Chemistry: A Molecular Approach, 2/e

Examples of Solutions Tro: Chemistry: A Molecular Approach, 2/e

Common Types of Solution
Solution Phase Solute Phase Solvent Phase Example Gaseous solutions Gas Air (mostly N2 & O2) Liquid solutions Liquid Solid Soda (CO2 in H2O) Vodka (C2H5OH in H2O) Seawater (NaCl in H2O) Solid solutions Brass (Zn in Cu) Solutions that contain Hg and some other metal are called amalgams Solutions that contain metal solutes and a metal solvent are called alloys Tro: Chemistry: A Molecular Approach, 2/e

Brass Tro: Chemistry: A Molecular Approach, 2/e

Solubility When one substance (solute) dissolves in another (solvent) it is said to be soluble salt is soluble in water bromine is soluble in methylene chloride When one substance does not dissolve in another it is said to be insoluble oil is insoluble in water The solubility of one substance in another depends on two factors – nature’s tendency toward mixing, and the types of intermolecular attractive forces Tro: Chemistry: A Molecular Approach, 2/e

Spontaneous Mixing When solutions with different solute concentrations come in contact, they spontaneously mix to result in a uniform distribution of solute throughout the solution Tro: Chemistry: A Molecular Approach, 2/e

Mixing and the Solution Process: Entropy
Most processes occur because the end result has less potential energy But formation of a solution does not necessarily lower the potential energy of the system When two ideal gases are put into the same container, they spontaneously mix even though the difference in attractive forces is negligible The gases mix because the energy of the system is lowered through the release of entropy Tro: Chemistry: A Molecular Approach, 2/e

Mixing and the Solution Process Entropy
Entropy is the measure of energy dispersal throughout the system Energy has a spontaneous drive to spread out over as large a volume as it is allowed By each gas expanding to fill the container, it spreads its energy out and lowers its entropy Tro: Chemistry: A Molecular Approach, 2/e

Intermolecular Forces and the Solution Process
Energy changes in the formation of most solutions also involve differences in attractive forces between the particles For the solvent and solute to mix you must overcome 1. all of the solute–solute attractive forces 2. some of the solvent–solvent attractive forces both processes are endothermic At least some of the energy to do this comes from making new solute–solvent attractions which is exothermic Tro: Chemistry: A Molecular Approach, 2/e

Intermolecular Attractions
Tro: Chemistry: A Molecular Approach, 2/e

Solution Interactions

Relative Interactions and Solution Formation
When the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the increase in entropy from mixing Tro: Chemistry: A Molecular Approach, 2/e

Solubility There is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible oil and water are immiscible The maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility The solubility of one substance in another varies with temperature and pressure Tro: Chemistry: A Molecular Approach, 2/e

Will It Dissolve? Like Dissolves Like Chemist’s Rule of Thumb –
A chemical will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles are attracted to each other Polar molecules and ionic compounds will be more soluble in polar solvents Nonpolar molecules will be more soluble in nonpolar solvents Tro: Chemistry: A Molecular Approach, 2/e

Classifying Solvents Tro: Chemistry: A Molecular Approach, 2/e

Example 12.1a: Predict whether the following vitamin is soluble in fat or water
Water is a polar solvent. Fat is mostly made of nonpolar molecules. The four OH groups make the molecule highly polar and it will also H-bond to water. Vitamin C is water soluble. Vitamin C Tro: Chemistry: A Molecular Approach, 2/e

Example 12.1b: Predict whether the following vitamin is soluble in fat or water
Water is a polar solvent. Fat is mostly made of nonpolar molecules. The two C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K3 is fat soluble. Vitamin K3 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Decide if the following are more soluble in hexane, C6H14, or water
nonpolar molecule more soluble in C6H14 polar molecule more soluble in H2O nonpolar part dominant more soluble in C6H14 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Explain the solubility trends seen in the table below

Practice – Explain the solubility trends seen in the table below
These alcohols all have a polar OH part and a nonpolar CHn part. As we go down the table the nonpolar part gets larger, but the amount of OH stays the same. We therefore expect that the solubility in water (polar solvent) should decrease and the solubility in hexane (nonpolar solvent) should increase, and it does. Tro: Chemistry: A Molecular Approach, 2/e

Heat of Solution When some compounds, such as NaOH, dissolve in water, a lot of heat is released the container gets hot When other compounds, such as NH4NO3, dissolve in water, heat is absorbed from the surroundings the container gets cold Why is this? Tro: Chemistry: A Molecular Approach, 2/e

Energetics of Solution Formation: the Enthalpy of Solution
To make a solution you must 1. overcome all attractions between the solute particles; therefore DHsolute is endothermic 2. overcome some attractions between solvent molecules; therefore DHsolvent is endothermic 3. form new attractions between solute particles and solvent molecules; therefore DHmix is exothermic The overall DH for making a solution depends on the relative sizes of the DH for these three processes DHsol’n = DHsolute + DHsolvent + DHmix Tro: Chemistry: A Molecular Approach, 2/e

Solution Process 2. Add energy in to overcome some solvent–solvent attractions 3. Form new solute–solvent attractions, releasing energy 1. Add energy in to overcome all solute–solute attractions Tro: Chemistry: A Molecular Approach, 2/e

Energetics of Solution Formation
If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is greater than the energy released in making the new attractions between the solute and solvent, the overall process will be endothermic If the total energy cost for breaking attractions between particles in the pure solute and pure solvent is less than the energy released in making the new attractions between the solute and solvent, the overall process will be exothermic Tro: Chemistry: A Molecular Approach, 2/e

Heats of Hydration For aqueous solutions of ionic compounds, the energy added to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration attractive forces between ions = lattice energy DHsolute = −DHlattice energy attractive forces in water = H-bonds attractive forces between ion and water = ion–dipole DHhydration = heat released when 1 mole of gaseous ions dissolves in water = DHsolvent + DHmix Tro: Chemistry: A Molecular Approach, 2/e

Ion-Dipole Interactions
When ions dissolve in water they become hydrated each ion is surrounded by water molecules The formation of these ion-dipole attractions causes the heat of hydration to be very exothermic Tro: Chemistry: A Molecular Approach, 2/e

Heats of Solution for Ionic Compounds
For an aqueous solution of an ionic compound, the DHsolution is the difference between the Heat of Hydration and the Lattice Energy DHsolution = DHsolute+ DHsolvent + DHmix DHsolution = −DHlattice energy+ DHsolvent + DHmix DHsolution = −DHlattice energy + DHhydration DHsolution = DHhydration− DHlattice energy Tro: Chemistry: A Molecular Approach, 2/e

Heat of Hydration DHsolution = DHhydration− DHlattice energy

Comparing Heat of Solution to Heat of Hydration
Because the lattice energy is always exothermic, the size and sign on the DHsol’n tells us something about DHhydration If the heat of solution is large and endothermic, then the amount of energy it costs to separate the ions is more than the energy released from hydrating the ions DHhydration < DHlattice when DHsol’n is (+) If the heat of solution is large and exothermic, then the amount of energy it costs to separate the ions is less than the energy released from hydrating the ions DHhydration > DHlattice when DHsol’n is (−) Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the lattice energy of KI if DHsol’n = +21
Practice – What is the lattice energy of KI if DHsol’n = kJ/mol and the DHhydration = −583 kJ/mol? Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the lattice energy of KI if DHsol’n = +21
Practice – What is the lattice energy of KI if DHsol’n = kJ/mol and the DHhydration = −583 kJ/mol? Given: Find: DHsol’n = kJ/mol, DHhydration = −583 kJ/mol DHlattice, kJ/mol Conceptual Plan: Relationships: DHsol’n = DHhydration− DHlattice DHsol’n, DHhydration DHlattice Solve: Check: the unit is correct, the lattice energy being exothermic is correct Tro: Chemistry: A Molecular Approach, 2/e

Solution Equilibrium The dissolution of a solute in a solvent is an equilibrium process Initially, when there is no dissolved solute, the only process possible is dissolution Shortly after some solute is dissolved, solute particles can start to recombine to reform solute molecules – but the rate of dissolution >> rate of deposition and the solute continues to dissolve Eventually, the rate of dissolution = the rate of deposition – the solution is saturated with solute and no more solute will dissolve Tro: Chemistry: A Molecular Approach, 2/e

Solution Equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Solubility Limit A solution that has the solute and solvent in dynamic equilibrium is said to be saturated if you add more solute it will not dissolve the saturation concentration depends on the temperature and pressure of gases A solution that has less solute than saturation is said to be unsaturated more solute will dissolve at this temperature A solution that has more solute than saturation is said to be supersaturated Tro: Chemistry: A Molecular Approach, 2/e

How Can You Make a Solvent Hold More Solute Than It Is Able To?
Solutions can be made saturated at non-room conditions – then allowed to come to room conditions slowly For some solutes, instead of coming out of solution when the conditions change, they get stuck in-between the solvent molecules and the solution becomes supersaturated Supersaturated solutions are unstable and lose all the solute above saturation when disturbed e.g. shaking a carbonated beverage Tro: Chemistry: A Molecular Approach, 2/e

Adding a Crystal of NaC2H3O2 to a Supersaturated Solution

Temperature Dependence of Solubility of Solids in Water
Solubility is generally given in grams of solute that will dissolve in 100 g of water For most solids, the solubility of the solid increases as the temperature increases when DHsolution is endothermic Solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line) Tro: Chemistry: A Molecular Approach, 2/e

Solubility Curves Tro: Chemistry: A Molecular Approach, 2/e

Temperature Dependence of Solid Solubility in Water (g/100 g H2O)

Purification by Recrystallization
One of the common operations performed by a chemist is removing impurities from a solid compound One method of purification involves dissolving a solid in a hot solvent until the solution is saturated As the solution slowly cools, the solid crystallizes out, leaving impurities behind Tro: Chemistry: A Molecular Approach, 2/e

Recrystallization of KNO3
KNO3 can be purified by dissolving a little less then 106 g in 100 g of water at 60 ºC then allowing it to cool slowly When it cools to 0 ºC only 13.9 g will remain in solution, the rest will precipitate out Tro: Chemistry: A Molecular Approach, 2/e

Practice – Decide if each of the following solutions is saturated, unsaturated, or supersaturated
50 g KNO3 in 100 g 34 ºC saturated 50 g KNO3 in 100 g 50 ºC unsaturated 50 g KNO3 in 50 g 50 ºC supersaturated 100 g NH4Cl in 200 g 70 ºC unsaturated 100 g NH4Cl in 150 g 50 ºC supersaturated Tro: Chemistry: A Molecular Approach, 2/e

Temperature Dependence of Solubility of Gases in Water
Gases generally have lower solubility in water than ionic or polar covalent solids because most are nonpolar molecules gases with high solubility usually are actually reacting with water For all gases, the solubility of the gas decreases as the temperature increases the DHsolution is exothermic because you do not need to overcome solute–solute attractions Tro: Chemistry: A Molecular Approach, 2/e

Temperature Dependence of Gas Solubility in Water (g/100 g H2O)

Tro: Chemistry: A Molecular Approach, 2/e

Pressure Dependence of Solubility of Gases in Water
The larger the partial pressure of a gas in contact with a liquid, the more soluble the gas is in the liquid Tro: Chemistry: A Molecular Approach, 2/e

Henry’s Law The solubility of a gas (Sgas) is directly proportional to its partial pressure, (Pgas) Sgas = kHPgas kH is called the Henry’s Law Constant Tro: Chemistry: A Molecular Approach, 2/e

Relationship between Partial Pressure and Solubility of a Gas

persrst Tro: Chemistry: A Molecular Approach, 2/e

Example 12. 2: What pressure of CO2 is required to keep the [CO2] = 0
Example 12.2: What pressure of CO2 is required to keep the [CO2] = 0.12 M in soda at 25 °C? Given: Find: S = [CO2] = 0.12 M, P of CO2, atm Conceptual Plan: Relationships: S = kHP, kH = 3.4 x 10−2 M/atm [CO2] P Solve: Check: the unit is correct, the pressure higher than 1 atm meets our expectation from general experience Tro: Chemistry: A Molecular Approach, 2/e

Practice – How many grams of NH3 will dissolve in 0
Practice – How many grams of NH3 will dissolve in 0.10 L of solution when its partial pressure is 7.6 torr? (kH = 58 M/atm) Tro: Chemistry: A Molecular Approach, 2/e

Practice – How many grams of NH3 will dissolve in 0
Practice – How many grams of NH3 will dissolve in 0.10 L of solution when its partial pressure is 7.6 torr? Given: Find: P of NH3 = 7.6 torr; 0.10 L mass of NH3, g Conceptual Plan: Relationships: S=kHP, kH= 58 M/atm, 1 atm = 760torr, 1 mol =17.04 g Solve: Tro: Chemistry: A Molecular Approach, 2/e

Concentrations Solutions have variable composition
To describe a solution, you need to describe the components and their relative amounts The terms dilute and concentrated can be used as qualitative descriptions of the amount of solute in solution Concentration = amount of solute in a given amount of solution occasionally amount of solvent Tro: Chemistry: A Molecular Approach, 2/e

Solution Concentration Molarity
Moles of solute per 1 liter of solution Used because it describes how many molecules of solute in each liter of solution If a sugar solution concentration is 2.0 M, 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar Tro: Chemistry: A Molecular Approach, 2/e

Molarity and Dissociation
The molarity of the ionic compound allows you to determine the molarity of the dissolved ions CaCl2(aq) = Ca2+(aq) + 2 Cl−(aq) A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2 Because each CaCl2 dissociates to give one Ca2+, a 1.0 M CaCl2 solution is 1.0 M Ca2+ 1 L = 1.0 moles Ca2+, 2 L = 2.0 moles Ca2+ Because each CaCl2 dissociates to give 2 Cl−, a 1.0 M CaCl2 solution is 2.0 M Cl− 1 L = 2.0 moles Cl−, 2 L = 4.0 moles Cl− Tro: Chemistry: A Molecular Approach, 2/e

Solution Concentration Molality, m
Moles of solute per 1 kilogram of solvent defined in terms of amount of solvent, not solution like the others Does not vary with temperature because based on masses, not volumes Tro: Chemistry: A Molecular Approach, 2/e

Parts Solute in Parts Solution
Parts can be measured by mass or volume Parts are generally measured in same units by mass in grams, kilogram, lbs, etc. by volume in mL, L, gallons, etc. mass and volume combined in grams and mL Percentage = parts of solute in every 100 parts solution if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution or 0.9 kg solute in every 100 kg solution Parts per million = parts of solute in every 1 million parts solution if a solution is 36 ppm by volume, then there are 36 mL of solute in 1 million mL of solution Tro: Chemistry: A Molecular Approach, 2/e

Percent Concentration

Parts Per Million Concentration
Tro: Chemistry: A Molecular Approach, 2/e 66

PPM grams of solute per 1,000,000 g of solution
mg of solute per 1 kg of solution 1 liter of water = 1 kg of water for aqueous solutions we often approximate the kg of the solution as the kg or L of water for dilute solutions, the difference in density between the solution and pure water is usually negligible Tro: Chemistry: A Molecular Approach, 2/e

Parts Per Billion Concentration
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Using Concentrations as Conversion Factors
Concentrations show the relationship between the amount of solute and the amount of solvent 12%(m/m) sugar(aq) means 12 g sugar  100 g solution or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution The concentration can then be used to convert the amount of solute into the amount of solution, or vice- versa Tro: Chemistry: A Molecular Approach, 2/e

Example 12. 3: What volume of 10. 5% by mass soda contains 78
Example 12.3: What volume of 10.5% by mass soda contains 78.5 g of sugar? Given: Find: 78.5 g sugar volume, mL Conceptual Plan: Relationships: g solute g sol’n mL sol’n 100 g sol’n = 10.5 g sugar, 1 mL sol’n = 1.04 g Solve: Check: the unit is correct, the magnitude seems reasonable as the mass of sugar  10% the volume of solution Tro: Chemistry: A Molecular Approach, 2/e

Preparing a Solution Need to know amount of solution and concentration of solution Calculate the mass of solute needed start with amount of solution use concentration as a conversion factor 5% by mass Þ 5 g solute  100 g solution “Dissolve the grams of solute in enough solvent to total the total amount of solution.” Tro: Chemistry: A Molecular Approach, 2/e

Practice – How would you prepare 250. 0 mL of 19. 5% by mass CaCl2
Practice – How would you prepare mL of 19.5% by mass CaCl2? (d = 1.18 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

Practice – How would you prepare 250. 0 mL of 19. 5% by mass CaCl2
Practice – How would you prepare mL of 19.5% by mass CaCl2? (d = 1.18 g/mL) Given: Find: 250.0 mL solution mass CaCl2, g Conceptual Plan: Relationships: mL sol’n g sol’n g solute 100 g sol’n = 19.5 g CaCl2, 1 mL sol’n = 1.18 g Solve: Answer: Dissolve 57.5 g of CaCl2 in enough water to total mL Tro: Chemistry: A Molecular Approach, 2/e

Solution Concentrations Mole Fraction, XA
The mole fraction is the fraction of the moles of one component in the total moles of all the components of the solution Total of all the mole fractions in a solution = 1 Unitless The mole percentage is the percentage of the moles of one component in the total moles of all the components of the solution = mole fraction x 100% Tro: Chemistry: A Molecular Approach, 2/e

Example 12.4a: What is the molarity of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n M mol C2H6O2, kg H2O, L M Conceptual Plan: Relationships: g C2H6O2 mol C2H6O2 mL sol’n L sol’n M M = mol/L, 1 mol C2H6O2 = g, 1 mL = L Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molarity of a solution made by dissolving 34
Practice – Calculate the molarity of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of solution (MMNH3 = g/mol) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molarity of a solution made by dissolving 34
Practice – Calculate the molarity of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of solution Given: Find: 2.00 mol NH3, 2.00 L sol’n M 34.0 g NH3, 2000 mL sol’n M Conceptual Plan: Relationships: g NH3 mol NH3 mL sol’n L sol’n M M = mol/L, 1 mol NH3 = g, 1 mL = L Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 77

Example 12.4b: What is the molality of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n m Conceptual Plan: Relationships: m = mol/kg, 1 mol C2H6O2 = g g C2H6O2 mol C2H6O2 kg H2O m Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molality of a solution made by dissolving 34
Practice – Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water (MMNH3 = g/mol, dH2O = 1.00 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water Given: Find: 34.0 g NH3, 2000 mL H2O m 2.00 mol NH3, 2.00 kg H2O m Conceptual Plan: Relationships: g NH3 mol NH3 mL H2O g H2O m kg H2O m=mol/kg, 1 molNH3=17.04 g, 1kg=1000 g, 1.00g=1 mL Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 80

Practice – Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 g of solution (MMNH3 = g/mol) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molality of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 g of solution Given: Find: 2.00 mol NH3, 1.97 kg H2O m 34.0 g NH3, 2000 g solution m Conceptual Plan: Relationships: g NH3 mol NH3 g sol’n g H2O m kg H2O m=mol/kg, 1 molNH3=17.04 g, 1kg=1000 g Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 82

Example 12.4c: What is the percent by mass of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n %(m/m) Conceptual Plan: Relationships: 1 kg = 1000 g g C2H6O2 g solvent g sol’n % Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the percent by mass of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water (MMNH3 = g/mol, dH2O = 1.00 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

the unit is correct, the magnitude is reasonable
Practice – Calculate the percent by mass of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water Given: Find: 34.0 g NH3, 2000 g H2O, 2034 g sol’n %(m/m) 34.0 g NH3, 2000 mL H2O %(m/m) Conceptual Plan: Relationships: g NH3 mL H2O % g sol’n g H2O % = g/g x 100%, 1.00 g=1 mL Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 85

Practice – Calculate the parts per million of a solution made by dissolving g of NH3 in 2.00 x 103 mL of water (MMNH3 = g/mol, dH2O = 1.00 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

the unit is correct, the magnitude is reasonable
Practice – Calculate the parts per million of a solution made by dissolving g of NH3 in 2.00 x 103 mL of water Given: Find: 0.340 g NH3, 2000 g H2O, 2000 g sol’n ppm 0.340 g NH3, 2000 mL H2O ppm Conceptual Plan: Relationships: g NH3 mL H2O ppm g sol’n g H2O ppm = g/g x 106, 1.00 g=1 mL Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 87

Example 12.4d: What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n c Conceptual Plan: Relationships: c = molA/moltot, 1 mol C2H6O2=62.07 g, 1 mol H2O=18.02 g g C2H6O2 mol C2H6O2 g H2O mol H2O c Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the mole fraction of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water (MMNH3 = g/mol, dH2O = 1.00 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the mole fraction of a solution made by dissolving 34.0 g of NH3 in 2.00 x 103 mL of water Given: Find: 2.00 mol NH3, mol H2O, tot mol c 34.0 g NH3, 2000 mL H2O c Conceptual Plan: Relationships: g NH3 mol NH3 mL H2O g H2O c mol H2O c=mol/mol, 1 mol NH3=17.04 g, 1mol H2O =18.02 g, 1.00 g =1 mL Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 90

Example 12.4d: What is the mole percent of a solution prepared by mixing 17.2 g of C2H6O2 with kg of H2O to make 515 mL of solution? Given: Find: 17.2 g C2H6O2, kg H2O, 515 mL sol’n c% Conceptual Plan: Relationships: c = molA/moltot, 1 mol C2H6O2 = 62.07g, 1 mol H2O=18.02 g g C2H6O2 mol C2H6O2 g H2O mol H2O c% Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e

Converting Concentration Units
1. Write the given concentration as a ratio 2. Separate the numerator and denominator separate into the solute part and solution part 3. Convert the solute part into the required unit 4. Convert the solution part into the required unit 5. Use the definitions to calculate the new concentration units Tro: Chemistry: A Molecular Approach, 2/e

Example 12. 5: What is the molarity of 6
Example 12.5: What is the molarity of 6.55% by mass glucose (C6H12O6) solution? Given: Find: 6.55 g C6H12O6, 100 g sol’n M mol C2H6O2, L M 6.55%(m/m) C6H12O6 M Conceptual Plan: Relationships: g C6H12O6 mol C6H12O6 mL L sol’n M g sol’n M =mol/L, 1mol C6H12O6=180.16g, 1mL=0.001L, 1mL=1.03g Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 94

Practice – Calculate the molality of 16. 2 M H2SO4(aq) (MMH2SO4 = 98
Practice – Calculate the molality of 16.2 M H2SO4(aq) (MMH2SO4 = g/mol, dsol’n = 1.80 g/mL) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molality of a 16.2 M H2SO4 solution
Given: Find: 16.2 M H2SO4 m 16.2 mol H2SO4, kg H2O m 16.2 mol H2SO4, 1.00 L sol’n m Conceptual Plan: Relationships: L mol H2SO4 g sol’n g H2O m kg H2O mL g H2SO4 m=mol/kg, 1molH2SO4=98.08g, 1kg=1000g, 1.80g=1mL Solve: Check: the unit is correct, the magnitude is reasonable Tro: Chemistry: A Molecular Approach, 2/e 96

Colligative Properties
Colligative properties are properties whose value depends only on the number of solute particles, and not on what they are value of the property depends on the concentration of the solution The difference in the value of the property between the solution and the pure substance is generally related to the different attractive forces and solute particles occupying solvent molecules positions Tro: Chemistry: A Molecular Approach, 2/e 97

Vapor Pressure of Solutions
The vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent the solute particles replace some of the solvent molecules at the surface Eventually, equilibrium is re-established, but with a smaller number of vapor molecules – therefore the vapor pressure will be lower Addition of a nonvolatile solute reduces the rate of vaporization, decreasing the amount of vapor The pure solvent establishes a liquid  vapor equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Thirsty Solutions Revisited
A concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix Similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix The result is reduction in vapor pressure Tro: Chemistry: A Molecular Approach, 2/e

Thirsty Solutions When equilibrium is established, the liquid level in the solution beaker is higher than the solution level in the pure solvent beaker – the thirsty solution grabs and holds solvent vapor more effectively Beakers with equal liquid levels of pure solvent and a solution are placed in a bell jar. Solvent molecules evaporate from each one and fill the bell jar, establishing an equilibrium with the liquids in the beakers. Tro: Chemistry: A Molecular Approach, 2/e

Psolvent in solution = csolvent∙P°
Raoult’s Law The vapor pressure of a volatile solvent above a solution is equal to its normal vapor pressure, P°, multiplied by its mole fraction in the solution Psolvent in solution = csolvent∙P° because the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent Tro: Chemistry: A Molecular Approach, 2/e

Example 12.6: Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C12H22O11 with mL of H2O Given: Find: 99.5 g C12H22O11, mL H2O PH2O Conceptual Plan: Relationships: P°H2O = 23.8 torr, 1mol C12H22O11 = g, 1mol H2O = 18.02g g C12H22O11 mol C12H22O11 g H2O mol H2O cH2O mL H2O PH2O Solve: Tro: Chemistry: A Molecular Approach, 2/e

Practice - Calculate the total vapor pressure of a solution made by dissolving 25.0 g of glucose (C6H12O6) in 215 g of water at 50 °C. (MMC6H12O6=180.2 g/mol, MMH2O = g/mol Vapor Pressure of 50 °C = 92.5 torr) Tro: Chemistry: A Molecular Approach, 2/e

Practice - Calculate the total vapor pressure of a solution made by dissolving 25.0 g of glucose (C6H12O6) in 215 g of water at 50 °C Given: Find: 25.0 g C6H12O6, 215 g H2O PH2O Conceptual Plan: Relationships: P°H2O = 92.5torr, 1mol C6H12O6 = 180.2g, 1mol H2O = 18.02g g C6H12O6 mol C6H12O6 g H2O mol H2O CH2O PH2O Solve: because glucose is nonvolatile, the total vapor pressure = vapor pressure H2O Tro: Chemistry: A Molecular Approach, 2/e 104

Vapor Pressure Lowering
The vapor pressure of a solvent in a solution is always lower than the vapor pressure of the pure solvent The vapor pressure of the solution is directly proportional to the amount of the solvent in the solution The difference between the vapor pressure of the pure solvent and the vapor pressure of the solvent in solution is called the vapor pressure lowering DP = Pºsolvent − Psolution = csolute  Pºsolvent Tro: Chemistry: A Molecular Approach, 2/e

Raoult’s Law for Volatile Solute
When both the solvent and the solute can evaporate, both molecules will be found in the vapor phase The total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent for an ideal solution Ptotal = Psolute + Psolvent The solvent decreases the solute vapor pressure in the same way the solute decreased the solvent’s Psolute = csoluteP°solute and Psolvent = csolventP°solvent Tro: Chemistry: A Molecular Approach, 2/e

Example 12.7: Calculate the component and total vapor pressure of a solution prepared by mixing 3.95 g of CS2 with 2.43 g of C3H6O Given: Find: molCS2, PCS2=285torr, molC3H6O, PC3H6O=148torr PCS2, PC3H6O, Ptotal molCS2, PºCS2=515torr, molC3H6O, PºC3H6O=332torr PCS2, PC3H6O, Ptotal 3.95 g CS2, PºCS2 = 515 torr, 2.43 g C3H6O, PºC3H6O = 332 torr PCS2, PC3H6O, Ptotal Conceptual Plan: Relationships: 1mol CS2 = g, 1mol C3H6O = 58.0 g g CS2 mol CS2 g C3H6O mol C3H6O c P Solve: Tro: Chemistry: A Molecular Approach, 2/e 107

Practice – Calculate the total vapor pressure of an ideal solution made by mixing mol of ether (C4H10O) with mol of ethanol (C2H6O) at 20 °C. (vapor pressure of 20 °C = 440 torr vapor pressure of 20 °C = 44.6 torr) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the total vapor pressure of an ideal solution made by mixing mol of ether (C4H10O) with mol of ethanol (C2H6O) at 20°C Given: Find: 0.500 mol C4H10O, Pºether=440 torr, mol C2H6O, Pºethanol = 44.6 torr Ptotal 0.500 mol C4H10O, PC4H10O=293 torr, mol C2H6O, PC2H6O=14.9 torr Ptotal Conceptual Plan: Relationships: mol C4H10O mol C2H6O c P Solve: Tro: Chemistry: A Molecular Approach, 2/e 109

Ideal vs. Nonideal Solution
In ideal solutions, the made solute–solvent interactions are equal to the sum of the broken solute–solute and solvent–solvent interactions ideal solutions follow Raoult’s Law Effectively, the solute is diluting the solvent If the solute–solvent interactions are stronger or weaker than the broken interactions the solution is nonideal Tro: Chemistry: A Molecular Approach, 2/e

Vapor Pressure of a Nonideal Solution
When the solute–solvent interactions are stronger than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be less than predicted by Raoult’s Law because the vapor pressures of the solute and solvent are lower than ideal When the solute–solvent interactions are weaker than the solute–solute  solvent–solvent, the total vapor pressure of the solution will be more than predicted by Raoult’s Law Tro: Chemistry: A Molecular Approach, 2/e

Deviations from Raoult’s Law

Example 12.7, cont’d: The experimentally measured total vapor pressure of the solution is 645 torr. Is the solution ideal? If not, what can you say about the relative strength of carbon disulfide–acetone interactions? Given: Find: Ptotal(expt) = 645 torr, Ptotal(ideal) = 443 torr is the solution ideal?, interaction strength if not ideal Solve: Ptotal(expt) = 645 torr > Ptotal(ideal) = 443 torr The solution is not ideal and shows positive deviations from Raoult’s law. Therefore, carbon disulfide–acetone interactions must be weaker than acetone–acetone and carbon disulfide–carbon disulfide interactions. Tro: Chemistry: A Molecular Approach, 2/e

Other Colligative Properties Related to Vapor Pressure Lowering
Vapor pressure lowering occurs at all temperatures This results in the temperature required to boil the solution being higher than the boiling point of the pure solvent This also results in the temperature required to freeze the solution being lower than the freezing point of the pure solvent Tro: Chemistry: A Molecular Approach, 2/e

Freezing Salt Water Pure water freezes at 0 ºC. At this temperature, ice and liquid water are in dynamic equilibrium. Adding salt disrupts the equilibrium. The salt particles dissolve in the water, but do not attach easily to the solid ice. When an aqueous solution containing a dissolved solid solute freezes slowly, the ice that forms does not normally contain much of the solute. To return the system to equilibrium, the temperature must be lowered sufficiently to make the water molecules slow down enough so that more can attach themselves to the ice. Tro: Chemistry: A Molecular Approach, 2/e

Freezing Point Depression
The freezing point of a solution is lower than the freezing point of the pure solvent therefore the melting point of the solid solution is lower The difference between the freezing point of the solution and freezing point of the pure solvent is directly proportional to the molal concentration of solute particles (FPsolvent – FPsolution) = DTf = mKf The proportionality constant is called the Freezing Point Depression Constant, Kf the value of Kf depends on the solvent the units of Kf are °C/m Tro: Chemistry: A Molecular Approach, 2/e

Kf Tro: Chemistry: A Molecular Approach, 2/e

Example 12. 8: What is the freezing point of a 1
Example 12.8: What is the freezing point of a 1.7 m aqueous ethylene glycol solution, C2H6O2? Given: Find: 1.7 m C2H6O2(aq) Tf, °C Conceptual Plan: Relationships: DTf = m Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C m DTf Solve: Check: the unit is correct, the freezing point lower than the normal freezing point makes sense Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molar mass of a compound if a solution of 12
Practice – Calculate the molar mass of a compound if a solution of 12.0 g dissolved in 80.0 g of water freezes at −1.94 °C (Kf water = 1.86 °C/m) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the molar mass of a compound if a solution of 12
Practice – Calculate the molar mass of a compound if a solution of 12.0 g dissolved in 80.0 g of water freezes at −1.94 °C Given: Find: masssolute = 12.0 g, massH2O= 80.0 g, FPsol’n = −1.94°C Tf, °C Conceptual Plan: Relationships: m DTf FPsol’n mol MM DTf =FPH2O−FPsol’n m  kgH2O= mol MM=g/mol DTf=mKf, m =mol/kg, MM=g/mol, Kf=1.86 °C/m, FPH2O=0.00 °C Solve: Tro: Chemistry: A Molecular Approach, 2/e 121

Boiling Point Elevation
The boiling point of a solution is higher than the boiling point of the pure solvent for a nonvolatile solute The difference between the boiling point of the solution and boiling point of the pure solvent is directly proportional to the molal concentration of solute particles (BPsolution – BPsolvent) = DTb = mKb The proportionality constant is called the Boiling Point Elevation Constant, Kb the value of Kb depends on the solvent the units of Kb are °C/m Tro: Chemistry: A Molecular Approach, 2/e

Example 12.9: How many grams of ethylene glycol, C2H6O2, must be added to 1.0 kg H2O to give a solution that boils at 105 °C? Given: Find: 1.0 kg H2O, Tb = 105 °C mass C2H6O2, g Conceptual Plan: Relationships: DTb m kg H2O mol C2H6O2 g C2H6O2 DTb = m Kb, Kb H2O = °C/m, BPH2O = °C MMC2H6O2 = g/mol, 1 kg = 1000 g Solve: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the boiling point of a solution made by dissolving 1.00 g of glycerin, C3H8O3, in 54.0 g of water Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the boiling point of a solution made by dissolving 1.00 g of glycerin, C3H8O3, in 54.0 g of water Given: Find: 1.00 g C3H8O3, 54.0 g H2O Tb, sol’n, °C Conceptual Plan: Relationships: m DTb g C3H8O3, kg H2O m = mol/kg, DTb = mKb, Kb for H2O = °C/m, BPH2O = °C, 1 mol C3H8O3 = g Solve: Tro: Chemistry: A Molecular Approach, 2/e 125

Osmosis Osmosis is the flow of solvent from a solution of low concentration into a solution of high concentration The solutions may be separated by a semi-permeable membrane A semi-permeable membrane allows solvent to flow through it, but not solute Tro: Chemistry: A Molecular Approach, 2/e

P = MRT Osmotic Pressure
The amount of pressure needed to keep osmotic flow from taking place is called the osmotic pressure The osmotic pressure, P, is directly proportional to the molarity of the solute particles R = (atm∙L)/(mol∙K) P = MRT Tro: Chemistry: A Molecular Approach, 2/e 127

Example 12. 10: What is the molar mass of a protein if 5
Example 12.10: What is the molar mass of a protein if 5.87 mg per 10 mL gives an osmotic pressure of 2.45 torr at 25 °C? Given: Find: 5.87 mg/10 mL, P = 2.45 torr, T = 25 °C molar mass, g/mol Conceptual Plan: Relationships: P,T M P=MRT, T(K)=T(°C) , R= atm∙L/mol∙K M = mol/L, 1 mL = L, MM=g/mol, 1atm =760 torr mL L mol protein Solve: Tro: Chemistry: A Molecular Approach, 2/e

Practice – Lysozyme is an enzyme used to cleave cell walls
Practice – Lysozyme is an enzyme used to cleave cell walls. A solution made by dissolving g of lysozyme in mL results in an osmotic pressure of 1.32 x 10−3 atm at 25 °C. Calculate the molar mass of lysozyme. Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the molar mass of lysozyme if 0. 0750 g per 100
Practice – What is the molar mass of lysozyme if g per mL gives an osmotic pressure of 1.32x 10−3 atm at 25 °C? Given: Find: g/100 mL, P = 1.32 x 10−3 atm, T = 25 °C molar mass, g/mol Conceptual Plan: Relationships: P,T M P=MRT, T(K)=T(°C) , R= atm∙L/mol∙K M = mol/L, 1 mL = L, MM=g/mol, 1atm =760 torr mL L mol protein Solve: Tro: Chemistry: A Molecular Approach, 2/e 131

van’t Hoff Factors Ionic compounds produce multiple solute particles for each formula unit The theoretical van’t Hoff factor, i, is the ratio of moles of solute particles to moles of formula units dissolved The measured van’t Hoff factors are generally less than the theoretical due to ion pairing in solution therefore the measured van’t Hoff factor often causes the DT to be lower than you might expect Tro: Chemistry: A Molecular Approach, 2/e

Example 12. 11: What is the measured van’t Hoff factor if 0
Example 12.11: What is the measured van’t Hoff factor if m CaCl2(aq) has freezing point of −0.27 ºC ? Given: Find: 0.050 m CaCl2(aq), Tf = −0.27°C i Conceptual Plan: Relationships: DTf = i∙m∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00 °C m, DTf i Solve: Tro: Chemistry: A Molecular Approach, 2/e 134

Example 12. 12: A solution contains 0. 102 mol Ca(NO3)2 and 0
Example 12.12: A solution contains mol Ca(NO3)2 and mol H2O. Calculate the vapor pressure of the solution at 55 ºC Given: Find: 0.102 mol Ca(NO3)2(aq), mol H2O, T = 55 °C, P°H2O = torr (at 55 °C) Psolution Psolution = i∙cH2O∙PºH2O Conceptual Plan: Relationships: c, PH2O Psolution Solve: Tro: Chemistry: A Molecular Approach, 2/e 135

Practice – Calculate the theoretical boiling point of a solution made by dissolving 10.0 g of NaCl (MM 58.44) in 54.0 g of water Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the theoretical boiling point of a solution made by dissolving 10.0 g of NaCl in 54.0 g of H2O Given: Find: 10.0 g NaCl, 54.0 g H2O Tb, sol’n, °C Conceptual Plan: Relationships: m DTb g NaCl, kg H2O m = mol/kg, DTb = imKb Kb for H2O = °C/m, BPH2O = 100.0°C, 1 mol C3H8O3 = g Solve: Tro: Chemistry: A Molecular Approach, 2/e 137

A hyposmotic solution has a lower osmotic pressure than the solution inside the cell – as a result there is a net flow of water into the cell, causing it to swell A hyperosmotic solution has a higher osmotic pressure than the solution inside the cell – as a result there is a net flow of water out of the cell, causing it to shrivel An isosmotic solution has the same osmotic pressure as the solution inside the cell – as a result there is no net flow of water into or out of the cell Tro: Chemistry: A Molecular Approach, 2/e

Mixtures Solutions = homogeneous
Suspensions = heterogeneous, separate on standing Colloids = heterogeneous, do not separate on standing particles can coagulate cannot pass through semi-permeable membrane hydrophilic stabilized by attraction for solvent (water) hydrophobic stabilized by charged surface repulsions Show the Tyndall Effect and Brownian motion Tro: Chemistry: A Molecular Approach, 2/e

Mixtures Suspenion Colloid Solution none visible none visible
Brownian movement scatter all light Tyndall Effect Light scattering invisible Visibility of particle w/ elec. mcrscp. barely visible Visibility of particle w/ lite microscope small large Colligative properties retained pass through Ultrafilters pass through Ordinary filters settle out Centrifuge effect Gravity effect > 1000 nm nm < 1 nm Particle size negligible many maybe Electrical charge Tro: Chemistry: A Molecular Approach, 2/e 140

Brownian Motion Tro: Chemistry: A Molecular Approach, 2/e

Types of Colloidal Suspensions

Soaps Triglycerides can be broken down into fatty acid salts and glycerol by treatment with a strong hydroxide solution Fatty acid salts have a very polar “head” because it is ionic and a very nonpolar “tail” because it is all C and H hydrophilic head and hydrophobic tail This unique structure allows the fatty acid salts, called soaps, to help oily substances be attracted to water micelle formation emulsification Tro: Chemistry: A Molecular Approach, 2/e

Soap Tro: Chemistry: A Molecular Approach, 2/e

Soap Tro: Chemistry: A Molecular Approach, 2/e 145

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Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro

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