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SOLUTIONS II

• COLLIGATIVE PROPERTIES

• Chapter 13.6 Silberberg

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The Properties of Mixtures:

Solutions and Colloids

Chapter 13

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The Properties of Mixtures: Solutions and Colloids

13.1 Types of Solutions: Intermolecular Forces and Solubility

13.2 Intermolecular Forces and Biological Macromolecules

13.3 Why Substances Dissolve: Understanding the Solution Process

13.5 Concentration Terms

13.6 Colligative Properties of Solutions

13.7 Structure and Properties of Colloids

13.4 Solubility as an Equilibrium Process

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Goals & Objectives

• See the following Learning Objectives on pages 543-544.

• Understand these Concepts:• 13.15-20.

• Master these Skills:• 13.5-10.

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Colligative Properties

• Colligative properties depend on the number, rather than the kind, of solute particles.

• Colligative properties are physical properties of solutions.

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Colligative Properties

• Examples of colligative properties include:– change in vapor pressure

if solute is volatile– lowering of vapor pressure

if solute is not volatile– freezing point depression– boiling point elevation

if solute is not volatile– osmotic pressure

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Colligative Properties of Solutions

Colligative properties are properties that depend on the number of solute particles, not their chemical identity.

An electrolyte separates into ions when it dissolves in water. Strong electrolytes dissociate completely while weak electrolytes dissociate very little.

A nonelectrolyte does not dissociate to form ions. Each mole of dissolved compound yields 1 mole of particles in solution.

The number of particles in solution can be predicted from the formula and type of the solute.

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Figure 13.22 Conductivity of three types of electrolyte solutions.

strong electrolyte weak electrolyte nonelectrolyte

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Volatile Nonelectrolyte Solutions

For a volatile nonelectrolyte, the vapor of the solution contains both solute and solvent.

The presence of each volatile component lowers the vapor pressure of the other, since each one lowers the mole fraction of the other.

For such a solution, the vapor will have a higher mole fraction of the more volatile component.The vapor has a different composition than the solution.

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Colligative Properties of Electrolyte Solutions

For vapor pressure lowering: P = i(Csolute x Posolvent)

For boiling point elevation: Tb = i(bm)

For freezing point depression: Tf = i(fm)

For osmotic pressure: Π = i(MRT)

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Figure 13.26

Nonideal behavior of strong electrolyte solutions.

Ions in solution may remain clustered near ions of opposite charge, creating an ionic atmosphere. The ions do not act independently, and the effective concentration of dissolved particles is less than expected.

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Vapor Pressure Lowering

The vapor pressure of a solution containing a nonvolatile nonelectrolyte is always lower than the vapor pressure of the pure solvent.

The vapor pressure lowering is proportional to the mole fraction of the solute present.

Psolvent = Xsolvent x P°solvent

Raoult’s law states that the vapor pressure of the solvent above the solution is proportional to the mole fraction of the solvent present:

DPsolvent = Xsolute x P°solvent

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Figure 13.23 Effect of solute on the vapor pressure of solution.

Equilibrium is reached with a given number of particles in the vapor.

Equilibrium is reached with fewer particles in the vapor.

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Sample Problem 13.6 Using Raoult’s Law to Find ΔP

PROBLEM: Find the vapor pressure lowering, DP, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.°C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

PLAN: We are given the vapor pressure of pure H2O, so to calculate DP we just need the mole fraction of glycerol, Xglycerol.

multiply by density

volume (mL) of glycerol

mass (g) of glycerol

mole fraction (X) of glycerol

multiply by M

amount (mol) of glycerol

vapor pressure lowering

multiply by P°glycerol

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Sample Problem 13.6

SOLUTION:

= 0.137 mol C3H8O3

DP = 0.00498 x 92.5 torr

= 27.4 mol H2O500.0 mL H2O x 0.988 g H2O

1 mL H2O

1 mol H2O

18.02 g H2Ox

10.0 mL C3H8O3 x1.26 g C3H8O3

1 mL C3H8O3

1 mol C3H8O3

92.09 g C3H8O3

x

= 0.461 torr

0.137 mol C3H8O3

0.137 mol C3H8O3 + 27.4 mol H2OXglycerol = = 0.00498

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Vapor Pressure & Raoult's Law

• Raoult’s Law states that the vapor pressure of a volatile component in an ideal solution decreases as its mole fraction decreases.– Psolution = (XA)(P

A)+(XB)(PB)+(XC)(P

C)…

• When a nonvolatile solute is dissolved in a liquid, the vapor pressure of the liquid is always lowered.– Psolution = (Xsolvent)(P

solvent) + (Xsolute)(Psolute)

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Vapor Pressure & Raoult's Law

• When an ionic solute is dissolved in a liquid, the vapor pressure of the liquid is lowered in proportion to the number of ions present.

NaCl (s) + excess H2O Na1+(aq) + Cl1(aq)

• The vapor pressure is lower for a 1.0M solution of NaCl than for a 1.0 M solution of glucose due to the number of ions.

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Vapor Pressure & Raoult's Law• Determine the vapor pressure of a solution prepared by

dissolving 18.3 g of sucrose in 500g of water at 70C. The vapor pressure of water at 70 C is 233.7 torr. The molar mass of sucrose is 342.3g/mole.

• What is the vapor pressure of a mixture that is 25% by mass acetone in water at 25 C? The vapor pressure of water at 25 C is 23.8 torr, and the vapor pressure of acetone at that temperature is 200 torr. The molar mass of acetone is 58.0 g/mole.

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Boiling Point Elevation

A solution always boils at a higher temperature than the pure solvent.This colligative property is a result of vapor pressure lowering.

DTb = Kbm

The boiling point elevation is proportional to the molality of the solution.

Kb is the molal boiling point elevation constant for the solvent.

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Boiling Point Elevation

• kb = the molal boiling point elevation constant

kb for water, H2O 0.512C/m

kb for benzene, C6H6 2.53C/m

kb for camphor 5.95C/m

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Boiling Point Elevation

• Determine the normal boiling point of a 2.50 m glucose, C6H12O6, solution.

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Figure 13.24 Boiling and freezing points of solvent and solution.

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Freezing Point Depression

A solution always freezes at a lower temperature than the pure solvent.

DTf = Kfm

The freezing point depression is proportional to the molality of the solution.

Kf is the molal freezing point depression constant for the solvent.

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*At 1 atm.

SolventBoilingPoint (oC)* Kb (°C/m) Kf (°C/m)

MeltingPoint (°C)

Acetic acid

Benzene

Carbon disulfide

Carbon tetrachloride

Chloroform

Diethyl ether

Ethanol

Water

117.9

80.1

46.2

76.5

61.7

34.5

78.5

100.0

3.07 16.6 3.90

2.53 5.5 4.90

2.34 -111.5 3.83

5.03 -23 30.

3.63 -63.5 4.70

2.02 -116.2 1.79

1.22 -117.3 1.99

0.512 0.0 1.86

Table 13.6 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents

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Sample Problem 13.7 Determining Boiling and Freezing Points of a Solution

PROBLEM: You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in your car’s radiator. What are the boiling and freezing points of the solution?

PLAN: We need to find the molality of the solution and then calculate the boiling point elevation and freezing point depression.

divide by M

mass (g) of solute

amount (mol) of solute

molality (m)

divide by kg of solvent

DTb DTfTb(solution)Tf(solution)

DTb = KbmTb + DTbDTf = Kfm Tf - DTf

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Sample Problem 13.7

SOLUTION:

1.00x103 g C2H6O2 x 1 mol C2H6O2

62.07 g C2H6O2

= 16.1 mol C2H6O2

16.1 mol C2H6O2

4.450 kg H2O= 3.62 m C2H6O2molality =

DTb = 3.62 m x 0.512°C/m = 1.85°C

Tb(solution) = 100.00 + 1.85 = 101.85°C

DTf = 3.62 m x 1.86°C/m = 6.73°C

Tb(solution) = 0.00 - 6.73 = -6.73°C

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Freezing Point Depression

• DTf = kfm

• kf = the molal freezing point depression constant

kf for water, H2O 1.86C/m

kf for benzene, C6H6 5.12C/m

kf for t-butanol, (CH3)3COH 8.09C/m

kf for camphor 37.7C/m

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Freezing Point Depression

• Calculate the freezing point of 2.50m glucose solution.

• Calculate the freezing point of a solution that contains 8.50g of benzoic acid, C6H5COOH, in 75.0g of benzene, C6H6.

kf for benzene, C6H6 = 5.12C/m

MW for benzoic acid = 122 g/mole

Freezing point of benzene = 5.5C

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Determination of Molar Masses by Freezing Point Depression

• A 37.0g sample of a new covalent compound, a nonelectrolyte, was dissolved in 200g of water. The resulting solution froze at 5.58C.

• Determine the molecular weight of the compound. (V-3)

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Osmotic Pressure

Osmosis is the movement of solvent particles from a region of higher to a region of lower concentration through a semipermeable membrane.

Π = MRT

Solvent will always flow from a more dilute solution to a more concentrated one.

Osmotic pressure is the pressure that must be applied to prevent the net flow of solvent.

M = molarity

R = 0.0821 atm·L/mol·K

T = Kelvin temperature

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Figure 13.25 The development of osmotic pressure.

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Osmotic Pressure

• Determine the osmotic pressure of a 1.00 M solution of glucose at 25OC.

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Sample Problem 13.8 Determining Molar Mass from Osmotic Pressure

PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries O2. A physician dissolves 21.5 mg of one variety in water to make 1.50 mL of solution at 5.0°C. She measures an osmotic pressure of 3.61 torr. What is the molar mass of the protein?

PLAN: We convert Π to atm and T to degrees K and calculate molarity from osmotic pressure. We can then determine the molar mass using the number of moles and the known mass.

Π = MRT

Π (atm)

M (mol/L)

M (g/mol)amount (mol) of solute

divide mass (g) by molesconvert to mol

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Sample Problem 13.8

SOLUTION:

1 atm

760 torrM = Π

RT=

3.61 torr x

(0.0821 L·atm/mol·K)(278.15 K)

= 2.08 x10-4 M

= 6.89x104 g/mol

2.08 x10-4 mol1 L 103 mL

1 L = 3.12x10-7 molx

21.5 mg x 1 g103 mg

= 0.0215 g

M =0.0215 g

3.12x10-7 mol

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Osmotic Pressure

• A 1.00 g sample of a protein was dissolved in enough water to give 100 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25C. Calculate the molarity and the approximate molecular weight of the material.

• 2.80 torr/ 760 torr = 0.00368 atm

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Figure 13.29a Osmotic pressure and cell shape.

A red blood cell in an isotonic solution has its normal shape.

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Figure 13.29b Osmotic pressure and cell shape.

A hypotonic solution has a lower concentration of particles than the cell. A cell in a hypotonic solution absorbs water and swells until it bursts.

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Figure 13.29c Osmotic pressure and cell shape.

A hypertonic solution has a higher concentration of dissolved particles than the cell. If a cell is placed in a hypertonic solution, water moves out of the cell, causing it to shrink.

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Table 13.7 Types of Colloids

Dispersed Substance

Dispersing MediumColloid Type Example(s)

Aerosol

Aerosol

Foam

Solid foam

Emulsion

Solid emulsion

Sol

Solid sol

Liquid Gas Fog

Solid Gas Smoke

Gas Liquid Whipped cream

Gas Solid Marshmallow

Liquid Liquid Milk

Liquid Solid Butter

Solid Liquid Paint; cell fluid

Solid Solid Opal

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Figure 13.30 Light scattering and the Tyndall effect.

The narrow, barely visible light beam that passes through a solution (left), is scattered and broadened by passing through a colloid (right).

Sunlight is scattered by dust in air.

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Figure 13.31 The Nile delta (reddish-brown area).

At the mouths of rivers, colloidal clay particles coalesce into muddy deltas.

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Figure B13.1

Figure B13.1 The typical steps in municipal water treatment.

Chemical Connections

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Figure B13.2 Ion exchange to remove hard-water cations.

Chemical Connections

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Figure B13.3 Reverse osmosis to remove ions.

Chemical Connections