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Daniel L. Reger Scott R. Goode David W. Ball http://academic.cengage.com/chemistry/reger Chapter 12 Solutions
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Chapter 12 Solutions

Jan 03, 2016

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Chapter 12 Solutions. Solution Concentration. There are a number of ways to express concentration. You have seen: molarity; mole fraction; mass percentage: for expressing the composition of a compound; and can be used for solutions, as well. normality. Solution Concentration. - PowerPoint PPT Presentation
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Page 1: Chapter 12 Solutions

Daniel L. RegerScott R. GoodeDavid W. Ball

http://academic.cengage.com/chemistry/reger

Chapter 12Solutions

Page 2: Chapter 12 Solutions

• There are a number of ways to express concentration. You have seen:• molarity;• mole fraction;• mass percentage:• for expressing the composition of a compound;• and can be used for solutions, as well.

• normality

Solution Concentration

Page 3: Chapter 12 Solutions

• All concentration units are fractions.• The numerator contains the quantity of

solute.• The denominator is the quantity of either

solution or solvent.

• They differ in the units used to express these two quantities.

Solution Concentration

Page 4: Chapter 12 Solutions

Units of Concentration Used Earlier

solution of liters

solute of moles

4) (ChapterMolarity

M

C... mol B mol A mol

Amol

6) (Chapter fraction Mole

Page 5: Chapter 12 Solutions

Mass Percent Composition

%100solution grams

solute gramspercent mass

Page 6: Chapter 12 Solutions

• A solution is prepared by dissolving 3.00 g of NaCl (molar mass = 58.44 g/mol) in 150 g of water. Express its concentration as mass percent.

Example: Percent Composition

Page 7: Chapter 12 Solutions

Molality• Molality (m or molal) is defined as

solvent of kilograms

solute of moles molality

Page 8: Chapter 12 Solutions

Example: Calculate Molality

• What is the molality of a solution prepared by dissolving 3.00 g NaCl (molar mass = 58.44 g/mol) in 150 g of water?

Page 9: Chapter 12 Solutions

• Express the concentration of a 3.00% H2O2 solution as

(a) molality;

(b) mole fraction.

Example: Concentration Conversion

Page 10: Chapter 12 Solutions

Test Your Skill

• Calculate (a) the molality, and (b) the mole fraction of alcohol (C2H5OH; molar mass = 46.07 g/mol) in a wine that has an alcohol concentration of 7.50 mass percent.

Page 11: Chapter 12 Solutions

• Conversion of most concentration units to molarity usually involve using the density of the solution to convert units of mass to units of volume.• The density of a 12.0% sulfuric acid

(H2SO4; molar mass = 98.08 g/mol) is 1.080 g/mL. What is the molarity of this solution?

Example: Conversion to Molarity

Page 12: Chapter 12 Solutions

• For most substances, there is a limit to the quantity of solute that dissolves in a specific quantity of solvent.• A dynamic equilibrium exists between the

solute particles in solution and the undissolved solute.

Principles of Solubility

Page 13: Chapter 12 Solutions

Normality• Normality, (N) is defined as the number of equivalents per

liter of solution.• Problem – what are equivalents – depends on reaction• Not used very often

• One place that normality is sometimes encountered is with acid-base chemistry.

• How many protons are release by an acid defines the equivalents• EX:• HCl – releases 1 H+ so… 1M = 1N

• H2SO4 – releases 2H+ so… 1M = 2N

• We will not be using normality, but you might encounter it in other chemistry classes!!

Page 14: Chapter 12 Solutions

• Solubility is the concentration of solute that exists in equilibrium with an excess of that substance.• A saturated solution has a

concentration of solute equal to its solubility.

Definitions

Page 15: Chapter 12 Solutions

Definitions (continued)• An unsaturated solution is one that has a solute concentration less than

the solubility.• A supersaturated solution is one with a solute concentration that is greater

than the solubility.• Supersaturation is an unstable condition.

Page 16: Chapter 12 Solutions

• Many spontaneous processes are exothermic.• Enthalpy of solution: the H that

accompanies the dissolution of one mole of solute.• When a solid and a liquid form a

solution, the enthalpy change arises mainly from changes in the intermolecular attractions.

Solute-Solvent Interactions

Page 17: Chapter 12 Solutions

The Solution Process

• Steps 1 and 2 are endothermic; step 3 is exothermic.

Page 18: Chapter 12 Solutions

• A decrease in enthalpy is an important factor in causing spontaneous change.• However, many endothermic processes

are spontaneous, suggesting another contribution to spontaneity.

• In increase in disorder also favors spontaneous change.

Spontaneity

Page 19: Chapter 12 Solutions

• An example of increasing disorder as a driving force is illustrated by the mixing of gases.

Spontaneous Mixing of Gases

(a) separated gases (b) spontaneously mixed

Page 20: Chapter 12 Solutions

• An increase in disorder generally accompanies the mixing of molecules in the formation of a solution.• Ammonium nitrate is very soluble in water

because of the increase in disorder upon mixing, even though the process is quite exothermic (Hsoln = +26.4 kJ/mol).

Disorder and Spontaneity

Page 21: Chapter 12 Solutions

• Relative solubilities can often be predicted by comparing the relative strengths of the intermolecular attractions of solute-solute, solvent-solvent, and solute-solvent interactions.

Solubility of Molecular Compounds

Page 22: Chapter 12 Solutions

Like Dissolves Like

• In general, substances that have similar intermolecular forces have strong solute-solvent interactions and tend to form solution.

Page 23: Chapter 12 Solutions

• (a) Is iodine (I2) more soluble in water or in hexane (C6H14)?

• (b) Is methanol (CH3OH) more soluble in water or in octane (C8H18)?

Example: Relative Solubility

Page 24: Chapter 12 Solutions

• Hydration is the interaction of water molecules with ions, and is very exothermic.

Interaction of Ions with Water

Page 25: Chapter 12 Solutions

• When an ionic compound dissolves in water, disorder changes because:• separating the ions increases disorder;• separating the water molecules increases disorder;• hydrating the ions, which restricts some water molecules, decreases disorder.

• A few examples are known where disorder decreases on dissolving ionic compounds.

Ionic Compounds in Water

Page 26: Chapter 12 Solutions

Pressure and Solubility• Pressure has very little effect on the

solubilities of liquids and solids.• The solubility of gases in a liquid

depends on the pressure of the gas.• Henry’s Law: The solubility of a gas

is directly proportional to its partial pressure at any given temperature:

C = kP

Page 27: Chapter 12 Solutions

Henry’s Law Constants in Water for Various Gases (molal/atm)

Gas 0°C 20°C 40°C 60°C

CO2 7.60 x 10-2 3.91 x 10-2 2.44 x 10-2 1.63 x 10-2

C2H4 1.14 x 10-2 5.60 x 10-3 3.43 x 10-3 ---

He 4.22 x 10-4 3.87 x 10-4 3.87 x 10-4 4.10 x 10-4

N2 1.03 x 10-3 7.34 x 10-4 5.55 x 10-4 4.85 x 10-4

O2 2.21 x 10-3 1.43 x 10-3 1.02 x 10-3 8.71 x 10-4

Page 28: Chapter 12 Solutions

• Water at 20C is saturated with air that contains CO2 at a partial pressure of 8.0 torr. What is the molal concentration of CO2 in the solution?

Henry’s Law Calculation

Page 29: Chapter 12 Solutions

• Experiments show that the way solubility changes with temperature depends on the sign of the enthalpy of solution.• Solubility increases with increasing

temperature if Hsoln is positive (endothermic).• Solubility decreases with increasing

temperature if Hsoln is negative (exothermic).

Solubility and Temperature

Page 30: Chapter 12 Solutions

K2Cr2O7(s) → K2Cr2O7(aq) H = +66.5 kJ/mol

The solubility increases with increasingtemperature when Hsoln is positive.

Solubility and Hsoln

Page 31: Chapter 12 Solutions

Temperature Dependence on Solubility

Page 32: Chapter 12 Solutions

• Colligative property: Any property of a solution that changes in proportion to the concentration of solute particles.• Many colligative properties are directly

related to the lowering of solvent vapor pressure by the presence of solute particles.

Colligative Properties of Solutions

Page 33: Chapter 12 Solutions

Effect of Solute on Evaporation• The rate of evaporation of solvent in a

solution is lower than that of the pure solvent. Solute particles block opportunities for solvent particles to enter the vapor phase.

Page 34: Chapter 12 Solutions

Raoult’s Law• Raoult’s law: The vapor pressure of

solvent above a dilute solution equals the mole fraction of the solvent times the vapor pressure of the pure solvent.

Psolv = solvPsolv

• Another form of this equation gives the lowering of the vapor pressure.

Psolv = solutePsolv

Page 35: Chapter 12 Solutions

• At 27C, the vapor pressure of benzene is 104 torr. What is the vapor pressure of a solution that has 0.100 mol of naphthalene in 9.90 mol of benzene?

Example: Raoult’s Law

Page 36: Chapter 12 Solutions

Boiling Point Elevation• Because a solute lowers the vapor

pressure of the solvent, it raises the boiling point of the solution. Below, the concentration of the solution is increasing from (a) to (e).

Page 37: Chapter 12 Solutions

Boiling Point Elevation• The boiling point elevation is

Tb = mkb

where m is the molal concentration and kb is the boiling point constant for the solvent.

Solvent B.P. (°C) kb (°C/m)

Acetic acid 117.90 3.07

Benzene 80.10 2.53

Water 100.0 0.512

Page 38: Chapter 12 Solutions

Freezing Point Depression• Solute particles interfere with the ability

of solvent particles to form a crystal and freeze. Thus, it takes a lower temperature to freeze solvent from a solution than from the pure solvent. This is freezing point depression.

Page 39: Chapter 12 Solutions

Freezing Point Depression• The freezing point depression is

Tf = mkf

where m is the molal concentration and kf is the freezing point constant for the solvent.

SolventFreezing Pt.

(°C)kf (°C/m)

Acetic acid 16.60 3.90

Benzene 5.51 4.90

Naphthalene 80.2 6.8

Water 0.00 1.86

Page 40: Chapter 12 Solutions

• Benzophenone freezes at 48.1C. A solution of 1.05 g urea ((NH2)2CO, molar mass = 60.06 g/mol) in 30.0 g of benzophenone freezes at 42.4C. What is kf for benzophenone?

Example: Calculate kf

Page 41: Chapter 12 Solutions

• Benzophenone freezes at 48.1C and has a kf of 9.8C/molal. A 2.50-g sample of solute whose molar mass is 130.0 g/mol is dissolved in 32.0 g of benzophenone. What is the freezing point of the solution?

Test Your Skill

Page 42: Chapter 12 Solutions

• Semipermeable membranes allow water and small molecules to pass through them.• Osmosis is the diffusion of a fluid

through a semipermeable membrane.

Osmosis

Page 43: Chapter 12 Solutions

• When a semipermeable membrane separates a solution from the pure solvent, the net effect is for pure solvent to move through the membrane into the solution.

• The higher level of liquid produces an additional pressure, called osmotic pressure.

Osmosis (continued)

Page 44: Chapter 12 Solutions

• Osmotic pressure is a colligative property, and can be calculated by the equation

= MRTwhere:

= osmotic pressureM = molar concentration of soluteR = ideal gas law constantT = temperature in Kelvin

Osmotic Pressure

Page 45: Chapter 12 Solutions

• A 5.70 mg sample of protein is dissolved in water to give 1.00 mL of solution. Calculate the molar mass of the protein if the solution has an osmotic pressure of 6.52 torr at 20C.

Example: Molar Mass by Osmotic Pressure

Page 46: Chapter 12 Solutions

Colligative Properties - Summary

Property Symbol Conc. Unit Constant

Vapor pressure

∆P Mole fraction P°

Boiling Point ∆Tb Molal Kb

Freezing point

∆Tf Molal Kf

Osmotic pressure

Π Molar RT

Page 47: Chapter 12 Solutions

Electrolyte Solutions• The colligative properties of electrolyte

solutions are more pronounced because electrolytes separate into ions in solution.• The van’t Hoff factor, i, is defined by the

equation

lytenonelectro for value expected

property ecolligativ measuredi

Page 48: Chapter 12 Solutions

The van’t Hoff Factor• In dilute solution, the van’t Hoff factor for

salts approaches the number of ions produced by one formula unit of the substance.

NaCl → Na+(aq) + Cl-(aq) i = 2

MgBr2 → Mg2+(aq) + 2Br-(aq) i = 3

• The van’t Hoff factor generally decreases as the concentration increases.

Page 49: Chapter 12 Solutions

Example: the van’t Hoff Factor

• Arrange the following aqueous solutions in order of increasing boiling points: 0.03 m urea (a nonelectrolyte), 0.01 m NaOH, 0.02 m BaCl2, 0.01 m Fe(NO3)3.

Page 50: Chapter 12 Solutions

Mixtures of Volatile Substances• In a solution of two or more volatile

compounds, all components of the mixture are in equilibrium with their vapors.• An ideal solution is one in which all

volatile components obey Raoult’s law for all compositions.

PA = APA, PB = BPB, PC = CPC, etc.

Page 51: Chapter 12 Solutions

Ideal Solutions• Mixtures of toluene and benzene form

nearly ideal solutions.

Page 52: Chapter 12 Solutions

• At 27C, the vapor pressure of carbon tetrachloride (CCl4) is 127 torr and that of chloroform (CHCl3) is 212 torr. What is the partial pressure of each substance, and the total vapor pressure of the solution, of a solution that contains 0.40 mol of CCl4 and 0.60 mol of CHCl3?

Example: Vapor Pressure of Solutions

Page 53: Chapter 12 Solutions

Distillation

• Distillation is the separation of a mixture of components based on differences in volatility (vapor pressure) by repeated evaporation and condensation of the mixture.• The vapor always contains a larger mole

fraction of the more volatile component.

Page 54: Chapter 12 Solutions

Distillation Apparatus

Page 55: Chapter 12 Solutions

Deviations from Raoult’s Law

• Most liquid-liquid solutions deviate from the ideal behavior predicted by Raoult’s law.• Solutions have positive deviations if the

vapor pressure is higher than predicted.• Solutions have negative deviations if the

vapor pressure is lower than predicted.