Solutions

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1414SolutionsSolutions

EndothermicEndothermic

ExothermicExothermic

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Chapter GoalsChapter GoalsThe Dissolution Process

Spontaneity of the Dissolution Process•Dissolution of Solids in Liquids •Dissolution of Liquids in Liquids

(Miscibility)•Dissolution of Gases in Liquids •Rates of Dissolution and Saturation •Effect of Temperature on Solubility •Effect of Pressure on Solubility •Molality and Mole Fraction

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Chapter GoalsChapter GoalsColligative Properties of Solutions

• Lowering of Vapor Pressure and Raoult’s Law

• Fractional Distillation • Boiling Point Elevation • Freezing Point Depression • Determination of Molecular Weight by

Freezing Point Depression or Boiling Point Elevation

• Colligative Properties and Dissociation of Electrolytes

• Osmotic Pressure

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Chapter GoalsChapter GoalsColloids

• The Tyndall Effect • The Adsorption Phenomena • Hydrophilic and Hydrophobic

Colloids

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The Dissolution ProcessThe Dissolution Process 溶解過程溶解過程

• Solutions are homogeneous mixtures of two or more substances. – Dissolving medium is called the solventsolvent.– Dissolved species are called the solutesolute.

• There are three states of matter (solid, liquid, and gas) which when mixed two at a time gives nine different kinds of mixtures.– Seven of the possibilities can be

homogeneous.– Two of the possibilities must be

heterogeneous.

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The Dissolution ProcessThe Dissolution ProcessSeven Homogeneous Possibilities

Solute Solvent Example• Solid Liquid salt water• Liquid Liquid mixed drinks• Gas Liquid carbonated beverages• Liquid Solid dental amalgams• Solid Solid alloys• Gas Solid metal pipes • Gas Gas air

Two Heterogeneous Possibilities• Solid Gas dust in air • Liquid Gas clouds, fog

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Spontaneity of the Spontaneity of the Dissolution ProcessDissolution Process

Many solid do dissolve in liquids by endothermic processes.

A large increase in disorder of the solute during the dissolution process

Many solid do dissolve in liquids by endothermic processes.

A large increase in disorder of the solute during the dissolution process

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•As an example of dissolution, let’s assume that the solvent is a liquid.–Two major factors affect dissolution of solutes

1.Change of energy content or enthalpy of solution, Hsolution – If Hsolution is exothermic (< 0) dissolution is

favored – If Hsolution is endothermic (> 0) dissolution is not

favored 2.Change in disorder, or randomness, of the

solution Smixing – If Smixing increases (> 0) dissolution is favored – If Smixing decreases (< 0) dissolution is not

favored

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• Thus the best conditions for dissolution are :– For the solution process to be exothermic .

Hsolution < 0

– For the solution to become more disordered. Smixing > 0

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• Disorder in mixing a solution is very common. Smixing is almost always > 0.

• What factors affect Hsolution?– There is a competition between several

different attractions.• Weak solute-solute attractions favor solubility• Weak solvent-solvent attractions favor solubility• Strong solvent-solute attractions favor solubility

• Solute-solute attractions such as ion-ion attraction, dipole-dipole, etc.– Breaking the solute-solute attraction requires

an absorption of Energy.

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Dissolution of Solids in Dissolution of Solids in LiquidsLiquids

• The energy released (exothermic) when a mole of formula units of a solid is formed from its constituent ions (molecules or atoms for nonionic solids) in the gas phase is called the crystal lattice energycrystal lattice energy. (always negative)

•The crystal lattice energy is a measure of the attractive forces in a solid.•These attractions are strong, a large amount of energy is released as the solid forms

•The crystal lattice energy increases as the charge density increases.

Ionic solid:M+

(g)+ X-(g) M+X-

(s) + crystal lattice energy

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•This process can be considered the hypothetical first step in forming a solution of a solid in a liquid•It always endothermic •The smaller the magnitude of the crystal lattice energy, the more readily dissolution occurs

MX(s) + energy M+(g)+ X-

(g)

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• Solvent-solvent attractions such as hydrogen bonding in water.– This also requires an absorption of Energy.

• Solvent-solute attractions, solvationsolvation, releases energy.– If solvation energy is greater than the sum of

the solute-solute and solvent-solvent attractions, the dissolution is exothermic, Hsolution < 0.

– If solvation energy is less than the sum of the solute-solute and solvent-solvent attractions, the dissolution is endothermic, Hsolution > 0.

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Spontaneity of the Dissolution Spontaneity of the Dissolution ProcessProcess

• Solvent-solute attractions, solvationsolvation, releases energy.– When the solvent is water, the more specific

term is hydration– Hydration energy: the energy change involved in

the hydration of one mole of gaseous ions Mn+

(g)+ xH2O(l) M(OH2)xn++ energy (for cation)

Xy-(g)+ rH2O(l) X(H2O)r

y-+ energy (for anion)– Hydration is usually highly exothermic for ionic

or polar covalent compounds, because the polar water molecules interact very strongly with ions and polar molecules

– Nonpolar solids do not dissolve appreciably in polar solvent, because they do not attract each other

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亂度增加

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•Dissolution is a competition between:1.Solute -solute attractions• crystal lattice energy for ionic

solids

2.Solvent-solvent attractions

• H-bonding for water

3.Solute-solvent attractions• Solvation or hydration (the

solvent is water) energy

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• Solvation is directed by the water to ion attractions as shown in these electrostatic potentials.

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• In an exothermic dissolution, energy is released when solute particles are dissolved.– This energy is called the energy of

solvation or the hydration energy (if solvent is water).

• Let’s look at the dissolution of CaCl2.

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8or 7ely approximat is where

OHCl2)Ca(OHCaCl 2-2

62OH

(s)22

x

x

Ca

OH2

OH2

OH2

OH2

H2O

H2O

2+

Cl-

OH H H

OH

HO H

H H O

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• The energy absorbed when one mole of formula units becomes hydrated is the molar energy of hydrationmolar energy of hydration.

-y

n22-y

+nn22

+n

Xfor Ehydration O)X(HOH )(X

Mfor Ehydration )M(OHOH +(g)M

y

x

ng

x

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Dissolution of Liquids Dissolution of Liquids in Liquids (Miscibility)in Liquids (Miscibility)

• Most polar liquids are miscible in other polar liquids.

• In general, liquids obey the “like dissolves like” rule.– Polar molecules are soluble in polar

solvents.– Nonpolar molecules are soluble in

nonpolar solvents.• For example, methanol, CH3OH, is

very soluble in water

Sulfuric acid is always diluted by adding the acid slowly and carefully to waterto water.

Water should never never be added to the acid

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50ml H2SO4

50ml H2O

50ml H2SO4 +50ml H2O

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Dissolution of Liquids in Dissolution of Liquids in Liquids (Miscibility)Liquids (Miscibility)

• Nonpolar molecules essentially “slide” in between each other.– For example, carbon tetrachloride and

benzene are very miscible.

C Cl

Cl

Cl

Cl

C

CC

C

CC

H

H

H

H

H

H

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Dissolution of Gases in Dissolution of Gases in LiquidsLiquids

• Polar gases are more soluble in water than nonpolar gases.– This is the “like dissolves like” rule in

action.• Polar gases can hydrogen bond with water• Some polar gases enhance their solubility

by reacting with water.

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Dissolution of Gases in Dissolution of Gases in LiquidsLiquids

• A few nonpolar gases are soluble in water because they react with water.

• Because gases have very weak solute-solute interactions, gases dissolve in liquids in exothermic processes.

acidk very wea

HCOOHCOHOHCO 33

OH

3222

2

aqaqg

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Rates of Dissolution and Rates of Dissolution and SaturationSaturation

• Finely divided solids dissolve more rapidly than large crystals.– Compare the dissolution of granulated sugar and sugar

cubes in cold water.• The reason is simple, look at a single cube of NaCl.

• The enormous increase in surface area helps the solid to dissolve faster.

NaCl

Breaks

upmany smaller crystals

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• Saturated solutions Saturated solutions have established an equilibrium between dissolved and undissolved solutes– Examples of saturated solutions include:

•Air that has 100% humidity.• Some solids dissolved in liquids.

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• Symbolically this equilibrium is written as:

• In an equilibrium reaction, the forward rate of reaction is equal to the reverse rate of reaction.

-aqaqs XM MX

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Rates of Dissolution and Rates of Dissolution and SaturationSaturation• The solubilities of many solids The solubilities of many solids

increase at higher increase at higher temperaturetemperature SupersaturatedSupersaturated solutions have higher-than-saturated concentrations of dissolved solutes.– Metastable (temporarily

stable)

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Effect of Temperature on Effect of Temperature on SolubilitySolubility

•According to LeChatelier’s Principle–when stress is applied to a system at equilibrium, the system responds in a way that best relieves the stress.–Since saturated solutions are at equilibrium, LeChatelier’s principle applies to them.

•Possible stresses to chemical systems include:

1.Heating or cooling the system.2.Changing the pressure of the system. 3.Changing the concentrations of reactants or products.

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•What will be the effect of heating or cooling the water in which we wish to dissolve a solid?– It depends on whether the dissolution is exo-

or endothermic.• For an exothermic dissolution, heat can be

considered as a product.

– Warming the water will decrease solubility and cooling the water will increase the solubility.

• Predict the effect on an endothermic dissolution like this one.

LiBr Li Br kJ / molsH O + -2 aq aq 48 8.

KMnO kJ / mol K MnO4 sH O +

4-2 43 6. aq aq

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• For ionic solids that dissolve endothermicallyendothermically dissolution is enhanced by heatingheating.

• For ionic solids that dissolve exothermicallyexothermically dissolution is enhanced by coolingcooling.

• Be sure you understand these trends.

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Effect of Pressure on Effect of Pressure on SolubilitySolubility

• Pressure changes have little or no effect on solubility of liquids and solids in liquids.– Liquids and solids are not compressible.

• Pressure changes have large effects on the solubility of gases in liquids.– Sudden pressure change is why

carbonated drinks fizz when opened.– It is also the cause of several scuba

diving related problems including the “bends”.

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• The effect of pressure on the solubility of gases in liquids is described by Henry’s Law.

Cgas = kPgas

Where Cgas =the concentration of gas k= is constant for a particular gas and solvent at a particular temperature Pgas= the pressure of the gas

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MolalityMolality 重量莫耳濃度重量莫耳濃度 and and Mole FractionMole Fraction 莫耳分率莫耳分率

•In Chapter 3 we introduced two important concentration units.

1.% by mass of solute

%100solution of mass

solute of mass = w/w%

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2.Molarity

solution of Liters

solute of moles = M

• We must introduce two new concentration units in this chapter.

Molality and Mole Molality and Mole FractionFraction

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• Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.


m=kg of solvent

moles of solute

In dilute aqueous solutions molarity and molality are nearly equal

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Example 14-1: MolalityWhat is the molality of a solution that contain 128g of

CH3OH in 108g of water?

The molecule weight of CH3OH is 32? mole CH3OH=128/32=4 mole4mole/108x10-3 kg =37m CH3OH

Example 14-1: MolalityWhat is the molality of a solution that contain 128g of

CH3OH in 108g of water?

The molecule weight of CH3OH is 32? mole CH3OH=128/32=4 mole4mole/108x10-3 kg =37m CH3OH

Example 14-2: MolalityHow many grams of H2O must be used to dissolve

50.0g of sucrose to prepare a 1.25m solution of sucrose, C12H22O11?

The molecule weight of C12H22O11is 3421.25m=(50/342)/xkgx=0.117kg H2O

Example 14-2: MolalityHow many grams of H2O must be used to dissolve

50.0g of sucrose to prepare a 1.25m solution of sucrose, C12H22O11?

The molecule weight of C12H22O11is 3421.25m=(50/342)/xkgx=0.117kg H2O

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Molality and Mole FractionMolality and Mole FractionExample 14-1: Calculate the molarity and the

molality of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g

?mol C6H12O6

kg H2O 90.0gx10-3 H2O=

10.0g C6H12O6

180.0g C6H12O6=0.617 m C6H12O6

This is the concentration in molalitymolality

?mol C6H12O6

L solution (100g/1.04)x10-3 sol’n=

10.0g C6H12O6

180.0g C6H12O6

=0.587 M C6H12O6

This is the concentration in molaritymolarity

10% means: 10g C6H12O6 in 90g H2O

m=

M=

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Example 14-2: Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 2.00 x 102 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

?mol C6H5COOHkg C6H6

=

7.25g C6H5COOH122g C6H5COOH

=0.338 m C6H5COOH

2.0x102 ml x0.879 C6H6

1000

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• Mole fraction is the number of moles of one component divided by the moles of all the components of the solution– Mole fraction is literally a fraction using moles of one

component as the numerator and moles of all the components as the denominator.

• In a two component solution, the mole fraction of one component, A, has the symbol XA.

B of moles ofnumber +A of moles ofnumber

A of moles ofnumber AX

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• The mole fraction of component B - XB

1.00. equalmust fractions mole theall of sum The

1 that NoteB of moles ofnumber +A of moles ofnumber

B of moles ofnumber

A

B

B

XX

X

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Example 14-3: Mole FractionWhat are the fractions of CH3OH and H2O in the

solution described in Example 14-1? It contains 128g of CH3OH and 108g of H2O?

? mole CH3OH=128/32= 4 mol? Mole H2O=108/18= 6 molXCH3OH= 4mol/(4mol+6mol) = 0.4

XH2O= 6mol/(4mol+6mol) = 0.6

Example 14-3: Mole FractionWhat are the fractions of CH3OH and H2O in the

solution described in Example 14-1? It contains 128g of CH3OH and 108g of H2O?

? mole CH3OH=128/32= 4 mol? Mole H2O=108/18= 6 molXCH3OH= 4mol/(4mol+6mol) = 0.4

XH2O= 6mol/(4mol+6mol) = 0.6

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• Example 14-3: What are the mole fractions of glucose and water in a 10.0% glucose solution (Example 14-1)?

Let the solution equal 100g 10.0g of glucose, 90.0g of water

?mole C6H12O6 = 10.0g/180=0.0556 mol C6H12O6 ?mole H2O = 90.0g/18=5.0 mol H2O

XH2O =5.0 mole H2O

5.0 mole H2O + 0.0556 mol C6H12O6

= 0.989

XC6H12O6 =0.0556 mole C6H12O6

5.0 mole H2O + 0.0556 mol C6H12O6

= 0.011

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The sulfuric acid in a car battery has a density of 1.225 g/cm3 and is 3.75 M. What is the molality, mole fraction and percentage of sulfuric acid by mass in this solution?

Let the solution volume is 1.0L3.75 M means that 3.75 moles of H2SO4

?g H2SO4=3.75 mole x 98=367.5 g H2SO4 Mass of solute?g solution=1.00L x 103 x 1.225g/ml = 1225 g solution

Mass of solutionMass of solvent = 1225-367.5 = 857.5g water

m= moles of solute / kg of solvent = 3.75 moles/0.8575kg = 4.37 m

% = g of solute / g of solution = 367.5 g/ 1225 g = 30%

XH2SO4 =3.75 mol H2SO4

(857.5g/18) mole H2O + 3.75 mol H2SO4 = 0.073

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Colligative Properties of Colligative Properties of SolutionsSolutions

• Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution.– Colligative properties do not depend on

the kinds of particles dissolved.• Colligative properties are a physical

property of solutions.

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Colligative Properties of Colligative Properties of SolutionsSolutions

• There are four common types of colligative properties:1.Vapor pressure lowering2.Freezing point depression3.Boiling point elevation4.Osmotic pressure

Vapor pressure lowering is the key to all four of the colligative properties.

54Lowering of vapor pressure

Lowering of Vapor Lowering of Vapor Pressure and Raoult’s Pressure and Raoult’s

LawLaw

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LawLaw• Addition of a nonvolatile solute to a solution

lowers the vapor pressure of the solution.– The effect is simply due to fewer solvent

molecules at the solution’s surface.– The solute molecules occupy some of the

spaces that would normally be occupied by solvent.

• Raoult’s Law models this effect in ideal solutions .•溶液的組成決定了溶液的蒸汽壓。•根據拉午耳定律，溶質為非揮發性非電解質的稀薄溶液，其蒸氣壓和溶劑的莫耳分率成正比。

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LawLaw• Derivation of Raoult’s Law.

Psolvent = Xsolvent P0solvent

Where psolvent = vapor pressure of solvent in solution P0

solvent = vapor pressure of pure solvent Xsolvent = mole fraction of solvent in solution

Lowering of vapor pressure, Psolvent, is defined as: Psolvent = P0

solvent - Psolvent

= P0solvent –(Xsolvent)(P0

solvent) = (1-Xsolvent) P0

solvent

Xsolute = 1- Xsolvent

Psolvent = XsoluteP0solvent

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LawLawExample 14-4: Vapor Pressure of a Solution

of Nonvolatile SoluteSucrose is a nonvolatile, nonionizing solute in water.

Determine the vapor lowering, at 25oC, of 1.25 m sucrose solution in Example 14-2. Assume that the solution behaves ideally. The vapor pressure of pure water at 25oC is 23.8 torr.

1.25m sucrose= 1.25 mol/1kg solvent Xsucrose= 1.25mol/(1.25mol+(1000/18)mol) = 0.0220Psolvent = XsoluteP0

solvent

= (0.022)(23.8 torr) = 0.524 torr

Example 14-4: Vapor Pressure of a Solution of Nonvolatile Solute

Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor lowering, at 25oC, of 1.25 m sucrose solution in Example 14-2. Assume that the solution behaves ideally. The vapor pressure of pure water at 25oC is 23.8 torr.

1.25m sucrose= 1.25 mol/1kg solvent Xsucrose= 1.25mol/(1.25mol+(1000/18)mol) = 0.0220Psolvent = XsoluteP0

solvent

= (0.022)(23.8 torr) = 0.524 torr

•若溶質為揮發性高蒸氣壓物質–溶液的蒸汽壓等於溶液中每一成份的蒸汽壓的總和 (包含溶質和溶劑 )–每一成份的蒸汽壓等於此純物質的蒸汽壓和它在溶液中的莫耳分率 (mole fraction)的乘積

•溶液的蒸汽壓符合拉午耳定律的溶液稱為「理想溶液 (Ideal solution)」–理想溶液中不論是溶質－溶質、溶劑－溶劑或溶質－溶劑間的分子之間的引力相似–此時溶液的體積具有加成性，符合1 + 1 = 2原則–因為分子間的引力相近，混合時無能量改變，理想溶液在混合時沒有能量的變化，不會吸熱或放熱。譬如苯 (Benzene)與甲苯 (Toluene)混合溶液，兩者的分子大小相近、化學結構和極性相似，可形成理想溶液。

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volatile Solutevolatile Solute• When a solution consists of two components

that are very similar– Such as heptane C7H16 and Octane C8H18

– Experiences nearly the same intermolecule force

– Ptotal =PA+ PB Ptotal =XAP0

A+ XBP0B

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Example 14-5: Vapor Pressure of a Solution of volatile Components

At 40oC, the vapor pressure of pure heptane 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.0 mole of hepatne and 4.0 mole of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution.

Xheptane= 1.0mol/(1.0mol+ 4.0mol) = 0.2Xcctane= 1- 0.2 = 0.8Pheptane = XheptaneP0

heptane= (0.2)(92.0)=18.4 torr

Poctane = XoctaneP0octane= (0.8)(31.0)=24.8 torr

Ptotal = Pheptane + Poctane= 18.4 +24.8 =43.2 torr

Example 14-5: Vapor Pressure of a Solution of volatile Components

At 40oC, the vapor pressure of pure heptane 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.0 mole of hepatne and 4.0 mole of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution.

Xheptane= 1.0mol/(1.0mol+ 4.0mol) = 0.2Xcctane= 1- 0.2 = 0.8Pheptane = XheptaneP0

heptane= (0.2)(92.0)=18.4 torr

Poctane = XoctaneP0octane= (0.8)(31.0)=24.8 torr

Ptotal = Pheptane + Poctane= 18.4 +24.8 =43.2 torr

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Example 14-6: Composition of VaporCalculate the mole fractions of heptane and octane

in the vapor that is equilibrium with the solution in Example 14-5.

The mole fraction of a component in a gaseous mixture equals the ratio of its partial pressure to the total pressure

Xheptane = Pheptane/Ptotal= 18.4/43.2= 0.426Xoctane = Pocptane/Ptotal= 24.8/43.2= 0.574

Example 14-6: Composition of VaporCalculate the mole fractions of heptane and octane

in the vapor that is equilibrium with the solution in Example 14-5.

The mole fraction of a component in a gaseous mixture equals the ratio of its partial pressure to the total pressure

Xheptane = Pheptane/Ptotal= 18.4/43.2= 0.426Xoctane = Pocptane/Ptotal= 24.8/43.2= 0.574

• Some solution do not behave ideally over the entire concentration range.

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Positive deviation Negative deviation

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Boiling Point ElevationBoiling Point Elevation• Addition of a nonvolatile solute to a solution

raises the boiling point of the solution above that of the pure solvent.– This effect is because the solution’s vapor

pressure is lowered as described by Raoult’s law.

– The solution’s temperature must be raised to make the solution’s vapor pressure equal to the atmospheric pressure.

• The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

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Boiling Point ElevationBoiling Point Elevation• Boiling point elevation relationship is: Tb = Kbm Where Tb = boiling point elevation m= molality concentration of solute Kb = boiling point elevation constant for

solvent

• Example 14-4: What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

Tb = Kbm = (0.512 oC/m)(2.5m) =1.28 oCBoiling point of the solution=100+1.28=101.28oC

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Freezing Point Freezing Point DepressionDepression

• Addition of a nonvolatile solute to a solution lowers the freezing point of the solution relative to the pure solvent.

• See table 14-2 for a compilation of boiling point and freezing point elevation constants.

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• Relationship for freezing point depression is:Tf = Kfm Where Tf = freezing point depression of solvent

m= molality concentration of solute Kb = freezing point depression constant for

solventExample 14-5: Calculate the freezing point of a 2.50 m aqueous glucose solution.

Tf = Kfm = (1.86 oC/m)(2.5m) =4.65 oCFreezing point of the solution=0-4.65=-4.65oC

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• Notice the similarity of the two relationships for freezing point depression and boiling point elevation.

• Fundamentally, freezing point depression and boiling point elevation are the same phenomenon.– The only differences are the size of the

effect which is reflected in the sizes of the constants, Kf & Kb.

• This is easily seen on a phase diagram for a solution.

mm bbff K T vs.KT

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Example 14-6: Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.

molality = mole of benzoic acid / kg of benzene

= (8.50/122)/(75.0/1000) = 0.929 m

Tf = Kfm = (5.12 oC/m)(0.929m) =4.76 oCFreezing point of the solution=5.48-4.76=0.72oC

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Determination of Molecular Determination of Molecular Weight by Freezing Point Weight by Freezing Point

DepressionDepression•The size of the freezing point depression depends on two things:

1.The size of the Kf for a given solvent, which are well known.

2.And the molality concentration of the solution which depends on the number of moles of solute and the kg of solvent.

•If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

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Determination of Molecular Weight Determination of Molecular Weight by Freezing Point Depressionby Freezing Point Depression

Example 14-7: A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound? (Kf =1.86)Tf = Kfm m = Tf /Kf

= (0-(5.58))(1.86) =3.0m3.0=mole of compound/0.2kg solvent Mole of compound = 0.6 mole0.6mole =37/molecular weight The compound molecular weight = 37/0.6 =61.7 g/mol

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Colligative Properties Colligative Properties and Dissociation of and Dissociation of

ElectrolytesElectrolytes• Electrolytes have larger effects on boiling

point elevation and freezing point depression than nonelectrolytes.– This is because the number of particles

released in solution is greater for electrolytes

• One mole of sugar dissolves in water to produce one mole of aqueous sugar molecules.

• One mole of NaCl dissolves in water to produce two moles of aqueous ions:– 1 mole of Na+ and 1 mole of Cl- ions

75


ElectrolytesElectrolytes• Remember colligative properties depend on

the number of dissolved particles.– Since NaCl has twice the number of particles

we can expect twice the effect for NaCl than for sugar.

• The table of observed freezing point depressions in the lecture outline shows this effect.

76


ElectrolytesElectrolytes• Ion pairing or association

of ions prevents the effect from being exactly equal to the number of dissociated ions

77


ElectrolytesElectrolytes• The van’t Hoff factor, symbol i, is used to

introduce this effect into the calculations.• i is a measure of the extent of ionization or

dissociation of the electrolyte in the solution.

=Tf (actual)

Tf (if nonelectrolyte)i

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ElectrolytesElectrolytes

• i has an ideal value of 2 for 1:1 electrolytes like NaCl, KI, LiBr, etc.

• i has an ideal value of 3 for 2:1 electrolytes like K2SO4, CaCl2, SrI2, etc.

unit formulaions 2 ClNaClNa -

aq+aq

OH-+ 2

unit formulaions 3 Cl 2CaClCa -

aq+2aq

OH-2

+2 2

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ElectrolytesElectrolytes• Example 14-8: The freezing point of 0.0100 m NaCl

solution is -0.0360oC. Calculate the van’t Hoff factor and apparent percent dissociation of NaCl in this aqueous solution.

meffective = total number of moles of solute particles/kg solvent First let’s calculate the i factor.

=Tf (actual)

Tf (if nonelectrolyte)i =

Kfm effective

Kfm stated

=m effectivem stated

Tf (actual)= Kf meffective =Tf (actual)

Kf meffective =

0.0360oC1.86oC/m

meffective= 0.0194m =m effectivem stated

i =0.0194m0.0100m

=1.94

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ElectrolytesElectrolytes• Next, we will calculate the apparent percent

dissociation.• Let x = mNaCl that is apparently dissociated.

NaCl Na+ + Cl-H2O

(0.01-x)m xm xmmeffective= [0.01-x+x+]m =[0.01+x]m =0.0194m x=0.094m

=mapp dissm stated

Apparent % dissociation x100%

= 0.0094m0.0100m

x100%

= 94%

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ElectrolytesElectrolytes• Example 14-9: A 0.0500 m acetic acid

solution freezes at -0.0948oC. Calculate the percent ionization of CH3COOH in this solution.CH3COOH H+ + CH3COO-

(0.05-x)m xm xmmeff= [0.05-x+x+x]m=[0.05+x]m

Tf = Kf meff =Tf

Kf

meff =0.0948oC1.86oC/m

=0.051m

meff= [0.05+x]m =0.051m x= 0.001m

=m ionizedm original

% ionized x100%

= 0.0010m0.0500m

x100%

= 2.0% ionized and 98.0% unionized

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Osmotic PressureOsmotic Pressure•Osmosis is the net flow of a solvent

between two solutions separated by a semipermeable membrane.– The solvent passes from the lower

concentration solution into the higher concentration solution.

•Examples of semipermeable membranes include:1.Cellophane and saran wrap2.skin3.cell membranes

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Osmotic PressureOsmotic Pressure

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net solvent flow

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Osmotic PressureOsmotic Pressure• Osmosis is a rate controlled phenomenon.

– The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium.

• The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.

M

M

RT

where: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821L atmmol K

T = absolute temperature

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Osmotic PressureOsmotic Pressure For very dilute aqueous solutions, molarity

and molality are nearly equal. M m

m

for dilute aqueous solutions only

RT

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Osmotic PressureOsmotic Pressure•Osmotic pressures can be very large.

– For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i.

•Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as:1.Polymers2.Biomolecules like

• proteins• ribonucleotides

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Osmotic PressureOsmotic PressureExample 14-18: A 1.00 g sample of a biological

material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.

M M

M M

RT RT

atm = 2.80 torr1 atm

760 torr atm =

= atm

0.0821 KL atmmol K

? .

..

0 00368

0 00368

298150 10 4

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M M

M M

M

RT RT

atm = 2.80 torr1 atm

760 torr atm =

= atm

0.0821 K

g

mol

1.00 g

0.100 L

L

typical of small proteins

L atmmol K

gmol

? .

..

?

..

0 00368

0 00368

298150 10

1

150 106 67 10

4

44

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Osmotic PressureOsmotic PressureWater Purification by Reverse Osmosis

• If we apply enough external pressure to an osmotic system to overcome the osmotic pressure, the semipermeable membrane becomes an efficient filter for salt and other dissolved solutes.– Ft. Myers, FL gets it drinking water from

the Gulf of Mexico using reverse osmosis.– US Navy submarines do as well.– Dialysis is another example of this

phenomenon.


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Hypotonic solutionIsotonic solutionHypertonic solution

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ColloidsColloids•Colloids are an intermediate type of mixture that has a particle size between those of true solutions and suspensions.–The particles do not settle out of the solution but they make the solution cloudy or opaque.

•Examples of colloids include:1.Fog 2.Smoke 3.Paint 4.Milk 5.Mayonnaise6.Shaving cream7.Clouds

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The Tyndall EffectThe Tyndall Effect• Colloids scatter light when it is shined upon

them.– Why they appear cloudy or opaque.– This is also why we use low beams on

cars when driving in fog.

Distilled water Tap water500ml distilled water1 drop milk

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The Adsorption The Adsorption PhenomenonPhenomenon

• Colloids have very large surface areas– They interact strongly with substances

near their surfaces.• One of the reasons why rivers can carry so

much suspended silt in the water.

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2 3 3

Fe Cl H O Fe O H O + 6 H + Cl

A colloidal particle contains many Fe O H O units with Fe ions

bound to its surface. The + charged particles repel each other and

keep the colloid from precipitating.

2+ -2 2 3 2

+ -

2 3 23+

y y

y

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Hydrophilic and Hydrophilic and Hydrophobic ColloidsHydrophobic Colloids

• Hydrophilic colloids like water and are water soluble. – Examples include many biological proteins like

blood plasma.• Hydrophobic colloids dislike water and are

water insoluble. – Hydrophobic colloids require emulsifying

agents to stabilize in water. • Homogenized milk is a hydrophobic colloid.

– Milk is an emulsion of butterfat and protein particles dispersed in water

– The protein casein is the emulsifying agent.

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• Mayonnaise is also a hydrophobic colloid.– Mayonnaise is vegetable oil and eggs in a

colloidal suspension with water.– The protein lecithin from egg yolk is the

emulsifying agent.• Soaps and detergents are excellent

emulsifying agents.– Soaps are the Na or K salts of long chain

fatty acids.– Sodium stearate is an example of a typical

soap.

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• Sodium stearate

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micelle

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• So called “hard water” contains Fe3+, Ca2+, and/or Mg2+ ions– These ions come primarily from minerals

that are dissolved in the water.• These metal ions react with soap anions and

precipitate forming bathtub scum and ring around the collar.Ca soap anion Ca(soap anion)

insoluble scum

22(s)

• Synthetic detergents were designed as soap substitutes that do not precipitate in hard water• Detergents are good emulsifying agents.• Chemically, we can replace COO- on soaps with

sulfonate or sulfate groups

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• The use of detergents containing phosphates is now discouraged because they cause eutrophication in rivers

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• Linear alkylbenzenesulfonates are good detergents.

Solutions

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