Lecture Notes Chem 150 - K. Marr

Chapter 13Properties of Solutions

Silberberg 3 ed

Properties of SolutionsProperties of Solutions

13.113.1 Types of Solutions: IMF’s and Types of Solutions: IMF’s and Predicting SolubilityPredicting Solubility

13.213.2 Energy Changes in the Solution Energy Changes in the Solution ProcessProcess

13.313.3 Solubility as an Equilibrium Process Solubility as an Equilibrium Process

13.413.4 Quantitative Ways of Expressing Quantitative Ways of Expressing ConcentrationConcentration

13.513.5 Colligative Properties of Solutions Colligative Properties of Solutions

Will not coverWill not cover: :

13.613.6 The Structure and Properties of The Structure and Properties of ColloidsColloids

Definitions

1. Solution» homogeneous mixture with only one

phase present

2. Mixture » 2 or more substances physically mixed

together» Composition is variable» Properties of components are retained

Formation of Solutions

1. Driving Force: • Tendency toward Randomness (T1) • Nature favors processes that result in an

increase in entropy (more randomness or less order)– Explains why solutions of gases always form .

2. Why don’t solutions always form with solids and liquids?» Must consider IMF’s

Formation of Solutions

• Solute and solvent particles must be attracted to one another:

“Like Dissolves Like”• For a solution for form…..

Solute-solvent IMF’s > Solvent-Solvent & solute-solute IMFS

13-8

Figure B13.2

The mode of action of the

antibiotic, Gramicidin A

Destroys the Na+/K+ ionconcentration gradients

in the cell

13-9

Figure 13.4

The arrangement of atoms in two types of alloys

Solid-solid solutions: alloys (substitutional or interstitial)

Solutions involving Liquids

• Molecules of each liquid must be pushed apart for a solution to form

• Why doesn’t water form a solution with n-Hexane, C6H14?

» Water molecules too strongly attracted to each other to be pushed aside to make room for hexane molecules

Application Questions: Solutions involving Liquids

1. Why does a solution form between water and ethanol?• H-Bonding between water and ethanol

molecules is responsible for solution formation– Allows water molecules to be pushed aside to

make room for ethanol molecules

2. Why do all nonpolar liquids mix to form solutions? » e.g. Oil and n-Hexane

SAMPLE PROBLEM 13.1

Predicting Relative Solubilities of Substances

Predict which solvent will dissolve more of the given solute:

(a) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3) or in water.

(b)Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)

(c)Sodium chloride in methanol (CH3OH) or in propanol (CH3CH2CH2OH)

PLAN: Consider the intermolecular forces which can exist between solute molecules and consider whether the solvent can provide such interactions and thereby substitute.

Solids Dissolved in Liquids

1. Solid particles must separate for a solution to form (T2)

» Solute must be attracted to solvent– Ion-Dipole attractions– Solvation vs Hydration

2. What happens if the attractive forces within solute and solvent differ greatly?

» e.g. Hexane and NaCl

Figure 13.1

The major types of intermolecular forces in solutions

Figure 13.2

Hydration shells around an aqueous ion

What determines if the Hsoln for a solid with a liquid is exo- or

endothermic? (T3-5)

1. Hsoln = Lattice Energy + Solvation Energy

2. Lattice Energy » E needed to separate particles: Always

Endothermic

3. Solvation Energy» E released upon solvation» Always Exothermic

Figure 13.5

Solution cycles and the enthalpy components of the heat of solution

A. Exothermic Solution Process B. Endothermic Solution Process

Heats of solution and solution cycles

1. Solute particles separate from each other - endothermic

solute (aggregated) + heat solute (separated) Hsolute > 0

2. Solvent particles separate from each other - endothermic

3. Solute and solvent particles mix - exothermic

solvent (aggregated) + heat solvent (separated) Hsolvent > 0

solute (separated) + solvent (separated) solution + heat HSolv < 0

Hsoln = Hsolute + Hsolvent + Hsolvation

NaOH

Hsoln = -

NH4NO3

Hsoln = +

NaCl

Hsoln = +

Figure 13.6

Heats of Solution: Dissolving ionic compounds in water

Figure 13.7 Enthalpy diagrams for dissolving NaCl and octane in hexane

Table 13.3

Trends in Heats of Hydration for Various Ions

Heats of Solution: Application Questions

1. Why is the Hsoln always exothermic (negative) for solutions between gases and liquids?

2. Why is the Hsoln = 0 for solutions between gases?

3. Why is the dissolving of Calcium Chloride, CaCl2, in water exothermic?

» CaCl2 (along with NaCl) are used to salt roads4. Why is the dissolving of ammonium nitrate,

NH4NO3, in water endothermic?

» NH4NO3 is used in chemical cold packs

The Effect of Temperature on Solubility

Objectives:» Describe the effect of temperature

on solubility of gases, liquids, and solids in liquids

Solubility

1. Equation describing a saturated solution at equilibrium

Solute + Solvent Saturated Solution

2. Most Common Units:

» Mass solute/100 g solvent at a given temperature

Figure 13.8 Equilibrium in a saturated solution

Solute (undissolved) Solute (dissolved)

Figure 13.10

The relation between solubility and temperature for several ionic compounds

Effect of Temp. on the Solubility of a Gas in a Liquid

1. Solubility of a gas always decreases as Temp. increases. Why??

2. Le Chatelier’s Principle is used to predict how an increase in temp. affects the solubility of a gas in a liquid.

3. Recall Hsoln is exothermic for all gases in a liquid:

Gas + Liquid Gas dissolved + E

Solubility of a Gas in a Liquid: Applications

1. Thermal Pollution Decreases O2 Solubility

» Streams and Rivers– Salmon/Trout habitat restoration

» Deep lakes– Warm water at surface, cold water deep

2. Why are the richest fisheries in the coldest waters of the world?

Solubility of a Gas in a Liquid: Application Questions

1. Why do bubbles form on the side of a glass of water? What do these bubbles consist of?

2. Why do sodas get flat as they sit?3. Why are some ice cubes clear, others

cloudy?» How are clear ice cubes made?

PS

gasgas

:Law sHenry'

• gas volume decreases• gas pressure increases• more collisions with liquid surface• gas solubility increases

gas + solvent solution

13-32

Henry’s Law

The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.

kH = Henry’s law constant

for a gas; units of mol/L.atm Implications for scuba diving!

PS

PS

PkSPS

2

2

1

1

gasHgasgasgas

or

Application of Henry’s Law

1. Why does a soda start to bubble immediately after opening the bottle?

2. The solubility of methane , the chief component in natural gas, in water at 20.0 oC and 1.0 atm pressure is 0.025 g/L. What will the solubility be at 1.5 atm pressure and 20.0 oC ?

» Answer: 0.038 g/L

Section 13.4Concentrations of Solutions

1. Be able to convert from one concentration unit to another:

• Molarity: Review Section 3.5

• Molality• Mass Fraction and Mass Percent• Mole fraction and Mole Percent

2. Practice!! Practice!! Practice!!

Concentrations of Solultions: Molarity, M

Molarity = moles of solute divided by Liters of Solvent

M = mol solute / L of solution• Used in stoichiometric calculations

involving solutions since V x M = moles• Since Volume varies with temperature,

Molarity varies w/ temperature

Concentrations of Solutions: Molality, m

1. Molality =Moles of solute divided by kg of Solvent

m = mol solute / kg solvent2. Does not change with Temp.

– Used in BP elevation and FP depression calculations

Molality Practice #1Water freezes at a lower temperature when it contains solutes.

a. Calculate the number of grams of methanol, CH3OH, needed to prepare a 0.250 m solution, using 2000. grams of water. Methanol: 32.00 g/mol. Ans. = 16.0 g methanol

b. Calculate the Freezing point of 0.250 m Methanol.

Ans. = - 0.465 oC

FP Lowering

Tf = Kf mTf = amount FP is lowered

Kf = Freezing point depession constant

Kf is solvent Dependent1.86 oC/m for water5.07 oC/m for benzene20.0 oC/m for cyclohexane

Molality Practice #2a. If you prepare a solution by

dissolving 4.00 g of NaOH in 250. g of water, what is the molality of the solution? NaOH: 40.00 g/mol

Ans. = 0.400 m

b. At what temperature would the solution boil?

Ans. = 100.2 oC

BP Elevation

Tb = Kb m• Kb = Boiling point elevation

constant

• Kb is solvent Dependent

0.51 oC/m for water

2.53 oC/m for benzene

2.69 oC/m for cyclohexane

Parts by Mass (or Percent by Mass)

1. Mass of component divided by total mass of solution

Mass % = (msolute / m solution) x100

2. Also known as weight percent: w/w or % w/w

Mass Percent Practice #1

a. How many grams of NaOH are needed to prepare 250.0g of 1.00% NaOH in water? Ans. = 2.50 g NaOH

b. How many grams of water are needed? Ans. = 247.5 g

c. How many mL of water at 20.0 oC are needed? Ans.= 250.455 mL

• dwater at 20.0 oC = 0.9882 g/mL

d. What is the molality of the solution? Ans.= 0.2525 m NaOH

d. What is the approximate FP of the solution?Ans. = -0.939 oC

Mass Percent Practice #2

Concentrated hydrochloric acid can be purchased from chemical supply houses as a solution that is 37% HCl by mass. HCl = 36.46 g/mol

a. What mass of conc. HCl is needed to make 1.0 liter of 0.1 M HCl?

Ans. = 9.854 or 10 g conc HCl

b. How would you make the 0.1 M HCl solution?

Variations of % by Mass:ppm and ppb

1. Use ppm and ppb when concentrations of solute are very low

2. Parts per million» ppm = mass fraction x 106

3. Parts per billion» ppb = mass fraction x 109

Concentration Unit Conversion Problems

1. Strategies.........a. Determine the Units of Concentration

involved– What are the units you are starting with?– What are the units you are converting to?

b. Figure our what conversion factors are needed to go get you to the desired units of concentration

Concentration Unit Conversion Practice #1

a. Calculate the molality 2.00 % NaCl. NaCl = 58.4425 g/mol Ans. = 0.349 m NaCl

b. How would you prepare • 1.00 liter of 2.00 % NaCl (w/v)?• 500. mL of 2.00 % NaCl (w/v)?• 250. mL of 2.00 % NaCl (w/v)?

Concentration Unit Conversion Practice #2

Conc. hydrobromic acid can be purchased as 40.0% HBr. The density of the solution is 1.38 g/mL.

What is the molar concentration of 40.0% HBr? HBr = 80.912 g/mol

Ans. = 6.82 M HBr

Colligative Properties

Properties of a solution that depend on the number of solute particles, not on their identity:»Vapor Pressure Lowering»Freezing Point Lowering»Boiling Point Elevation»Osmotic Pressure (will not cover)

Vapor Pressures of Solutions

Which is higher, the vapor pressure of salt water or that of pure water?

Vapor Pressures of Solutions

1. A solution has a lower vapor pressure than that of the pure solvent (if the solute is nonvolatile). Why?

2. Solute particles impede evaporation, but do not affect condensation

Raoult’s Law:

Psoln = (Xsolvent)(Po

solvent)

Mole Fraction

Mole fraction • moles of component divided by total moles of

all components present in the mixture

Xa = na / (na + nb + nc +....)

• Used in Raoult’s Law Calculations

Calculating theVapor Pressure of a Solution

Raoult’s Law

Psoln = (Xsolvent)(Po

solvent)

Psoln = Vapor Pressure of Solution

Xsolvent = mole fraction of solvent

Po

solvent = Vapor pressure of the pure solvent

Raoult’s Law: Practice makes perfect?

Dibutyl phthalate, C16H22O4 (mw = 278 g/mol), is an oil sometimes worked into plastic articles to give them softness. It has a negligible vapor pressure (P = 1 torr @ 148 oC).

1. What is the vapor pressure at 20.0 oC of a solution of 20.0 g dibutyl phthalate in 50.0 g of octane, C8H18 (mw = 114 g/mol)?

» The vapor pressure of pure octane at 20.0 oC is 10.5 torr.

Ans. = 9.02 torr

BP Elevation and FP Depression

Objectives» Explain the effect of a solute on the

melting/freezing point and boiling point of a solution» Use F.P. depression and B.P elevation data to

calculate the molar mass of a compound.

BP Elevation and FP Depression

1. Nonvolatile Solutes Lower the Vapor Pressure of a Solvent (T11)

» Causes….

BP Elevation, Tb

FP Depression, Tf

» Colligative Properties: Depends on conc. of solute, not identity of solute

BP Elevation

1. Tb = Kb m

» Kb = Boiling point elevation constant

» Kb is solvent Dependent

0.51 oC/m for water

2.53 oC/m for benzene

2.69 oC/m for cyclohexane

FP Depression

Tf = Kf m

» Kf = Freezing point depession constant

» Kf is solvent Dependent

1.86 oC/m for water

5.07 oC/m for benzene

20.0 oC/m for cyclohexane

Practice makes perfect..........

1. A solution made by dissolving 3.46g of an unknown compound in 85.0 g of benzene (Kf = 5.07 oC/m, FP = 5.45 oC) froze at 4.13 oC. What is the molar mass of the compound? Answer: 157 g/mol

2. At what temperature will a 10.0 % aqueous sucrose, C12H22O11, solution boil?

Answer: 100.60 oC

Colligative Properties of Solutions of Electrolytes

Ionic Compounds (electrolytes) dissociate into ions when dissolved in water.

• NaCl(s) Na+ (aq) + Cl -

(aq)

1 mol 2 mol of Ions in solution

• CaCl2(s) Ca2+ (aq) + 2 Cl -

(aq)

1 mol 3 mol of Ions in solution

• (NH4) 2SO4 (s) 2 NH4 +

(aq) + SO42-

(aq)

1 mol 3 mol of Ions in solution

Colligative Properties of Solutions of Molecular Substances

1. Nonelectrolytes: Molecules separate when forming solutions

C12H22O11 (s) C12H22O11 (aq)

2. Weak Electrolytes: Incomplete ionization of

molecules in solution (e.g. acetic acid)

HC2H3O2 (aq) H +(aq) + C2H3O2

1- (aq)

Predicting Freezing Points

Estimate the FP of the following aqueous solutions Tf = Kf m

Kf for water = 1.86 oC/m)

1. 1.00 m sucrose, C12H22O11

2. 2.00 m sucrose, C12H22O11

3. 1.00 m NaCl

4. 1.00 m MgSO4

5. 1.00 m HCl

6. 1.00 m Acetic Acid, HC2H3O2

Why are the expected FP’s higher than the observed values for

some solutions?

Expected Observed

1. 1.00 m sucrose: -1.86 oC vs. -1.86 oC

2. 2.00 m sucrose: -3.72oC vs. -3.72 oC

3. 1.00 m NaCl: -3.72oC vs. -3.53 oC

4. 1.00 m MgSO4: -3.72 oC vs. -2.42 oC

5. 1.00 m HCl: -3.72 oC vs. -3.53 oC

6. 1.00 m HC2H3O2: -1.86 oC vs. -1.90 oC

Effects of Interionic Attractions

1. Dissociation into ions is not 100%2. Ion pairs exist in

solution....Thus....» Number of moles of ions in a 1.0

m NaCl solution is not double of the molality

» Causes Tb and Tf to be smaller than expected

van’t Hoff factor, iGives an indication of the % Dissociation of ions

in solution

tt

lyte)nonelectro a if as d(calculate f

(measured) f i

Use the FP’s to calculate the van’t Hoff factor

for each compound

1. 1.00 m NaCl: FP = -3.53 oC

2. 1.00 m MgSO4: FP = -2.42 oC

3. 1.00 m HC2H3O2: FP = -1.90 oC

tt

lyte)nonelectro a if as d(calculate f

(measured) f i

If 100 % dissociation, then for

a. NaCl : i = 2

b. KCl : i = 2

c. MgSO4 : i = 2

d. K2SO4 : i = 3

e. Na3PO4: i = 4

van’t Hoff factors for ideal solutions

Van’t Hoff factor:

Expected vs. Observed values

The End

Good luck on your final exams!!!