george mason university general chemistry 212 chapter 13

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George Mason UniversityGeneral Chemistry 212

Chapter 13Properties of Mixtures: Solutions & Colloids

Acknowledgements

Course Text: Chemistry: the Molecular Nature of Matter and Change, 7th edition, 2011, McGraw-HillMartin S. Silberberg & Patricia Amateis

The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

Chap 13: Properties of Mixtures

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Intermolecular Forces & Solubility (Intramolecular) (Bonding)

● Ionic● Covalent● Metallic

Intermolecular● Ion-Dipole (strongest)● Hydrogen bonding● Dipole-Dipole● Ion-Induced Dipole● Dipole-Induced Dipole● Dispersion (London) (weakest

Chap 13: Properties of Mixtures Types of Solutions

Liquid Solutions

Gas Solutions

Solid Solutions The Solution Process

Solvation /Hydration

Solute

Solvent

Mixing

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Chap 13: Properties of Mixtures Thermodynamic Solution Cycle, an application of

“Hess’s” LawThe Enthalpy Change Of An Overall Process Is The

Sum Of The Enthalpy Changes Of The Individual Steps

Heat of Solution

Heat of Hydration

Lattice Energy (Hsolute)

Solution Process & Change in Entropy04/21/23 4

solution solute solvent mixΔH = ΔH + ΔH + ΔH

solution solute hydrationΔH = ΔH + ΔH

hydration solvent mixΔH = ΔH + ΔH

solution lattice hydration of the ionsΔH = ΔH + ΔH

gas liquid solidΔS > ΔS > ΔS > 0

Chap 13: Properties of Mixtures Solution as an Equilibrium Process

Effect of Temperature Effect of Pressure

Concentration Molarity & Molality Parts of Solute by Parts of Solution Interconversion of Concentration Terms

Colligative Properties of Solutions Vapor Pressure Boiling Point Elevation Freezing Point Depression Osmotic Pressure

Structure and Properties of Colloids

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Intermolecular Forces

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Phases and Phase Changes (Chapter 12) are due to forces among the molecules

Bonding (Intramolecular) forces and Intermolecuar forces arise from electrostatic attractions between opposite charges Intramolecular Forces - Attractive forces within

molecules: Cations & Anions (ionic bonding) Electron pairs (covalent bonding) Metal Cations and Delocalized Valence Electrons

(metallic bonding) Intermolecular Forces - Nonbonding Attractive and

Repulsive forces involving partial charges among particles: Molecules Atoms ions

Intermolecular Forces Intramolecular (Bonding within molecules)

Relatively strong, larger charges close to each other Ionic Covalent Metallic

Intermolecular (Nonbonding Attractions between molecules) Relatively weak attractions involving smaller charges

that are further apart Types (in decreasing order of bond energy)

Ion-Dipole (strongest) Hydrogen bonding Dipole-Dipole Ion-Induced Dipole Dipole-Induced Dipole Dispersion (London, instantaneous dipoles)

(weakest)04/21/23 7

Intramolecular Forces

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Intermolecular Forces

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Intermolecular Forces & Solubility Solutions – Homogeneous mixtures

consisting of a solute dissolved in a solvent through the actions of intermolecular forces

Solute dissolves in Solvent

Distinction between Solute & Solvent not always clear

Solubility – Maximum amount of solute that dissolves in a fixed quantity of solvent at a specified temperature

“Dilute” & “Concentrated” are qualitative (relative) terms

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Intermolecular Forces & Solubility Substances with similar types of

intermolecular forces (IMFs) dissolve in each other

“Like Dissolves Like”

Nonpolar substances (e.g., hydrocarbons) are dominated by dispersion force IMFs and are soluble in other nonpolar substances

Polar substances have Hydrogen Bonding, ionic, and dipole driven IMFs (e.g., Water, Alcohols) and would be soluble in polar substances (solvents)

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Intermolecular Forces & Solubility Intermolecular Forces

Ion-Dipole forces Principal forces involved in solubility of ionic

compounds in water Each ion on crystal surface attracts

oppositely charged end of Water dipole These attractive forces overcome attractive

forces between crystal ions leading to breakdown of crystal structure

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Water

Intermolecular Forces & Solubility

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Intermolecular Forces Hydrogen Bonding – Dipole-Dipole force that arises between

molecules that have an H atom bonded to a small highly electronegative atom with lone electrons pairs (N, O, F)

Especially important in aqueous (water) solutions. Oxygen- and nitrogen-containing organic and biological

compounds, such as alcohols, sugars, amines, amino acids.

O & N are small and electronegative, so their bound H atoms are partially positive and can get close to negative O in the dipole water (H2O) molecule

H N

:

FH

:

OH:

H-Bond


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Intermolecular Forces Dipole-Dipole forces

In the absence of H bonding, Dipole-Dipole forces account for the solubility of polar organic molecules, such as Aldehydes (R-CHO) in polar solvents such as Chloroform (CHCl3)


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Intermolecular Forces Ion-induced Dipole forces

An ion-induced dipole attraction is a weak attraction that results when the approach of an ion induces a dipole in an atom or nonpolar molecule by disturbing the arrangement of electrons in the nonpolar species

The Ion increases the magnitude of any nearby dipole; thus contributes to the solubility of salts in less polar solventsEx. LiCl in Ethanol


Dipole-Induced Dipole forces Intermolecular attraction between a polar molecule

and the oppositely charged pole it induces in a nearby non-polar molecule

Also based on polarizability, but are weaker than ion-induced dipole forces because the magnitude of charge is smaller – ions vs dipoles (coulombs law)

Solubility of atmospheric gases (O2, N2, noble gases) have limited solubility in water because of these forces

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Dispersion (or London or Instantaneous Dipole) forces These very weak forces are due to the non-zero

instantaneous dipole moments of all atoms and molecules

Such polarization can be induced either by a polar molecule or by the repulsion of negatively charged electron clouds in non-polar molecules

Thus, London interactions are caused by random fluctuations in electron density in an electron cloud

Contribute to the solubility of all solutes in all solvents

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Liquid Solutions Solutions can be:

Liquid Gaseous Solid Physical state of Solvent determines physical state of

solution Liquid solutions – forces created between solute and

solvent comparable in strength (“like” dissolves “like”) Liquid – Liquid

Alcohol/water – H bonds between OH & H2O

Oil/Hexane – Dispersion forces Solid – Liquid

Salts/Water– Ionic forces Gas – Liquid

(O2/H2O) – Weak intermolecular forces (low solubility, but biologically important)

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Liquid Solutions Gas Solutions

Gas – Gas Solutions – All gases are infinitely soluble in one another

Gas – Solid Solutions – Gases dissolve into solids by occupying the spaces between the closely packed particles Utility – Hydrogen (small size) can be

purified by passing an impure sample through a solid metal like palladium (forms Pd-H covalent bonds)

Disadvantage – Conductivity of Copper is reduced by the presence of Oxygen in crystal structure (copper metal is transformed to Cu(I)2O

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Liquid Solutions Solid Solutions

Solid – Solid Solutions Alloys – A mixture with metallic properties

that consists of solid phases of two or more pure elements, solid-solid solution, or distinct intermediate (heterogeneous) phases

Alloys Types Substitutional alloys – Brass (copper & zinc),

Sterling Silver(silver & copper), etc. substitute for some of main element atoms

Interstitial alloys – atoms of another elements (usually a nonmetal) fill interstitial spaces between atoms of main element, e.g., carbon steel (C & Fe)

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Practice ProblemA solution is ____________

a. A heterogeneous mixture of 2 or more substances

b. A homogeneous mixture of 2 or more substances

c. Any two liquids mixed together

d. very unstable

e. the answer to a complex problem

Ans: b

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The Solution Process

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Elements of Solution Process Solute particles must separate from each other Some solvent particles must separate to make

room for solute particles Solute and Solvent particles must mix together Energy must be absorbed to separate particles Energy is released when particles mix and

attract each other Thus, the solution process is accompanied by

changes in Enthalpy (∆H), representing the heat energy tied up in chemical bonds and Entropy (S), the number of ways the energy of a system can be dispersed through motions of particles)

2. Entropy (S)

Solution Process - Solvation

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Solvation – Process of surrounding a solute particle with solvent particles

Hydrated ions in solution are surrounded by solvent molecules in defined geometric shapes

Orientation of combined solute/solvent is different for solvated cations and anions

Cations attract the negative charge of the solvent

Anions attract the positive charge of the solvent

Hydration # for Na+ = 6 Hydration # for Cl- = 5

Solution Process

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Coulombic - attraction of oppositely charged particles

van der Waals forces: a. dispersion (London)b. dipole-dipolec. ion-dipole

Hydrogen-bonding - attractive interaction of a Hydrogen atom and a strongly electronegative Element:

Oxygen, Nitrogen or Fluorine

Hydration Energyattraction between solvent and solute

Lattice Energysolute-solute forces

Solute v. Solvent

Practice ProblemPredict which solvent will dissolve more of the given

solute:

a. Sodium Chloride (solute) in Methanol or in 1-Propanol

Ans: Methanol

A solute tends to be more soluble in a solvent whose intermolecular forces are similar to those of the solute

NaCl is an ionic solid that dissolves throughion-dipole forces

Both Methanol & 1-Propanol contain a “Polar” -OH group

The Hydrocarbon group in 1-Propanol is longer and can form only weak dispersive forces with the ions; thus it is less effective at substituting for the ionic attractions in the solute

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Practice Problem (con’t)

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Predict which solvent will dissolve more of the given solute:

b. Ethylene Glycol (HOCH2CH20H) inHexane (CH3(CH2)4CH3) or in Water (H2O)

Ans: Very Soluble in Water

Ethylene Glycol molecules have two polar –OH groups and they interact with the Dipole water molecule through H-Bonding

The Water H-bonds can substitute for the Ethylene Glycol H-Bonds better than they can with the weaker dispersive forces in Hexane (no H-Bonds)

Practice Problem (con’t)

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Predict which solvent will dissolve more of the given solute:

c.Diethyl Ether (CH3CH2OCH2CH3) in

Water (H2O) or in Ethanol (CH3CH2OH)

Ans: Ethanol

Diethyl Ether molecules interact with each other through dipole-dipole and dispersive forces and can form H-Bonds to both H2O and Ethanol

The Ether is more soluble in Ethanol because Ethanol can form H-Bonds with the Ether Oxygen and the dispersion forces of the Hydrocarbon group (CH3CH2) are compatible with the dispersion forces of the Ether’s Hydrocarbon group

Water, on the other hand, can form H bonds with the Ether, but it lacks any Hydrocarbon portion, so it forms much weaker dispersion forces with the Ether


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Solute particles absorb heat (Endothermic) and separate from each other by overcoming intermolecular attractions

Solvent particles absorb heat (Endothermic) and separate from each other by overcoming intermolecular attractions

Solute and Solvent particles mix (form a solution) by attraction and release of energy - an Exothermic process

soluteSolute (aggregated) + Heat Solute (separated) ΔH > 0

solventSolvent (aggregated) + Heat Solvent (separated) ΔH > 0

mixSolute (separated) + Solvent (separated) Solution + Heat ΔH < 0

The Solution Process Total Enthalpy change is called the:

““Heat of SolutionHeat of Solution” (∆” (∆HHsolnsoln)

This process, called a “Thermodynamic Solution Cycle”, is an application of “Hess’s” Law

Enthalpy Change is the Sum of the

Enthalpy Changes of Its Individual Steps

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EndothermicH > 0

EndothermicH > 0

ExothermicH < 0

soln solute solvent mixΔH = ΔH + ΔH + ΔH


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Solution Cycles, Enthalpy, and the Heat of Solution

In an Exothermic Solution Process theSolute is Soluble

If (Hsolute + Hsolvent) < Hmix (Hsoln < 0)

In an Endothermic Solution Process theSolute is Insoluble

If (Hsolute + Hsolvent) > Hmix (Hsoln > 0)

Exothermic

Endothermic

(+) (+) (–)

(+) (+) (–)

The Solution Process In an Exothermic Solution Process, the Solute is

Soluble

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Note: Recall Born-Haber Process – Chap 6

ΔH < 0

The Solution Process In an Endothermic Solution Process, the solute is

Insoluble

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Note: Recall Born-Haber Process – Chap 6

ΔH > 0

Heats of Hydration

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Hsolvent & Hmix are difficult to measure individually

Combined, these terms represent the Enthalpy change during “Solvation” – The process of surrounding a solute particle with solvent particles

Solvation in “Water” is called “Hydration”

Thus:

Since:

Then:

Forming strong ion-dipole forces during Hydration (Hmix < 0) more than compensates for the braking of water bonds(Hsolvent > 0); thus the process of Hydration is “Exothermic”, i.e.,

hydrationΔH always < 0 (Exothermic)

hydration solvent mixΔH = ΔH + ΔH

solution solute solvent mixΔH = ΔH + ΔH + ΔH

solution solute hydrationΔH = ΔH + ΔH

Heat of Hydration & Lattice Energy

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The energy required to separate an ionic solute (Hsolute) into gaseous ions is:

The Lattice Energy (Hlattice)

This process requires energy input (Endothermic)

Thus, the Heat of Solution for ionic compounds in water combines the Lattice Energy (always positive) and the combined Heats of Hydration of Cations and Anions (always negative)

+ - + -M X (s) + Heat M (g) + X (g)

solute latticeΔH (always > 0) = ΔH

Heat of Hydration & Lattice Energy Hsoln can be either positive or negative

depending on the net values of:

Hlattice (always >0) and Hhydration (always <0)

Recall:

If the Lattice Energy is large relative to the Hydration energy, the ions are likely to remain together and the compound will tend to be more insoluble

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soln solute hydration of the ionsΔH = ΔH + ΔH

soln lattice hydration of the ionsΔH = ΔH + ΔH

Charge Density - Heats of Hydration

Heat of Hydration is related to the

“Charge Density” of the ion

Charge Density – ratio of ion’s charge to its volume

Coulombs Law – the greater the ion’s charge and the closer the ion can approach the oppositely charged end of the water molecule, the stronger the attraction and, thus, the greater the charge density

The higher the charge density, the more negative Hhydr (always <0), i.e., the greater the solubility

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Charge Density - Heats of Hydration

Periodic Table

Going down Group

Same Charge, Increasing Size

Charge Density & Heat of Hydration Decrease

Going across Period

Greater Charge, Smaller Size

Charge Density & Heat of Hydration Increase

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Heat of Hydration & Lattice Energy

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Hydration vs. Lattice Energies

The solubility of an ionic solid depends on the relative contribution of both the energy of Hydration (Hhydr) and the Lattice energy (Hlattice)

Lattice Energy, Hlattice (always >0)

Small size + high charge = high Hlattice

For a given charge, lattice energy (Hlattice) decreases (less positive) as the radius of ions increases, increasing solubility

Δ = Δ + ΔH H Hsoln lattice hydration

Heat of Hydration & Lattice Energy Energy of Hydration (Hhydr)

Energy of Hydration also depends on ionic radius

Small ions, such as Na1+ or Mg2+ have concentrated electric charge and a strong electric field that attracts water molecules

High Charge Density (more negative Hhydr) produces higher solubility

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Heats of Hydration

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Cation

Radius (pm)

Hhydr (kJ/Mol)

Anion

Radius (pm)

Hhydr

(kJ/Mol)

Na+ 102 -410 F- 119 -431

K+ 138 -336 OH- 119 -460

Rb+ 152 -315 NO31- 165 -314

Cs+ 167 -282 Cl- 167 -313

Mg2+ 72 -1903 CN1- 177 -342

Ca2+ 100 -1591 N31- 181 -298

Sr2+ 118 -1424 Br- 182 -284

Ba2+ 135 -1317 SH1- 193 -336

I- 206 -247

BF41- 215 -223

ClO41- 226 -235

CO32- 164 -1314

S2- 170 -1372

SO42- 244 -1059

Solubility – Solution Temperatures

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Solubilities and Solution temperatures for 3 salts

NaCl NaOH NH4NO3

The interplay between (Hlattice) and (Hhydr) leads to different values of (Hsoln)

● Hlattice (Hsolute) for ionic compounds is always positive

● Combined ionic heats of hydration (Hhydr) are always negative


Solubility – Solution Temperatures The resulting temperature of the solution is a

function of the relative value of (Hsoln)

If the mixture solution becomes colder, the water, representing the “surroundings,” has lost energy to the system, i.e. an Endothermic process

(Hlattice dominates & Hsoln > 0)

If the solution becomes warmer, the surroundings have gained energy, i.e., an Exothermic process

Hhydr dominates & Hsoln < 0

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Solubility – Solution Temperatures

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Born-Haber diagram for the dissolution of 3 salts (Hsoln) is positive for solutions that become “Colder”

(Hsoln) is negative for solutions that become “Warmer”

HotNaOH HotHsoln = - 44.5

kJ/mol(Exothermic)

Solution Process – Entropy Change

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The Heat of Solution (Hsoln) is one of two factors that determine whether a solute dissolves in a solvent

The other factor concerns the natural tendency of a system to spread out, distribute, or disperse its energy in as many ways as possible

The thermodynamic variable for this process is called:

Entropy (S, S) Directly related to the number of ways that a

system can distribute its energy Closely related to the freedom of motion of

the particles and the number of ways they can be arranged


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Solids vs. Liquids vs. Gases In a solid, particles are relatively fixed in

their positions In a liquid the particles are free to move

around each other This greater freedom of motion allows the

particles to distribute their kinetic energy in more ways

Entropy increases with increased freedom of motion; thus, a material in its liquid phase will have a higher Entropy than when it is a solid

Sliquid > Ssolid A gas, in turn, would have a higher Entropy than its liquid phase

Sgas > Sliquid > Ssolid


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Relative change in Entropy (∆S)

The change in Entropy from solid to liquid to gas is always positive

The change from a liquid to a gas is called:

“Vaporization,”

thus, ∆Svap > 0

gas liquid solidΔS > ΔS > ΔS > 0

Solution Process – Entropy Change The change from a liquid to a solid is

called:

“Fusion” (freezing)

thus, Entropy change would be negative

After the liquid has frozen there is less freedom of motion, thus the Entropy is less,

∆Sfusion < 0

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Entropy and Solutions There are far more interactions between

particles in a solution than in either the pure solvent or the pure solute

Thus, the Entropy of a solution is higher than the sum of the solute and solvent Entropies

The solution process involves the interplay between the:

Change in Enthalpy (∆H)Change in Entropy (∆S)

Systems tend toward“State of Lower Enthalpy (∆H)”“State of Higher Entropy (∆S)”

Dissolution

soln solute solventS > (S + S ) or ΔS > 0

Enthalpy Change vs Entropy Change

Does Sodium Chloride dissolve in Hexane?

NaCl and Hexane have very dissimilar intermolecular forces:

Ionic (NaCl) vs Dispersive (Hexane)

Separating the solvent (Hexane) is easy because intermolecular forces are weak ∆Hsep > 0

Separating NaCl requires supplying Lattice Energy ∆Hsep >> 0

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Mixing releases very little heat because ion-induced dipole attractions between Na+ & Cl- ions and Hexane are weak ∆Hmix 0

The sum of the Endothermic term (∆Hsep) is much larger than the Exothermic term (∆Hmix), thus ∆Hsoln is highly positive; thus “no mixing”

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Solution does not form because the Entropy increase that would accompany the mixing of the solute and solvent is much smaller than the Enthalpy increase required to separate the ions


Does Octane dissolve in Hexane?

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Both compounds (Hydrocarbons with no polarized functional groups) consist of nonpolar molecules held together by Dispersive Intermolecular Forces (IMF’s), which are relatively weak

Thus, these similar intermolecular forms would suggest solubility

HexaneOctane


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Note ∆Hsoln is close to 0, suggesting little heat is released

With little Enthalpy change in the solution process, Octane dissolves readily in Hexane because the Entropy increase (∆S) greatly exceeds the relatively small Enthalpy change as a result of the ∆Hmix process

Practice Problem

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Water is added to a flask containing solid Ammonium Chloride (NH4Cl). As the salt dissolves, the solution becomes colder.

a.Is the dissolving of NH4Cl Exothermic or Endothermic?

Ans: The solution process results in a cold solution, i.e. – the system has taken energy from the surroundings (water loses energy becoming colder), thus,

an Endothermic process (∆Hsoln > 0)

Practice Problemb. Is the magnitude of ∆Hlattice of HN4Cl larger or smaller

than the combined ∆Hhyd of the ions?

Ans: Since ∆Hsoln > 0 and ∆Hhyd is always < 0; then ∆Hlattice must be much greater than ∆Hhyd

c. Given the answer to (a), why does NH4Cl dissolve in water?

Ans: The increase in Enthalpy (∆Hsoln > 0) would suggest the solute would be insoluble

Since a solution does form, the increase in Entropy must outweigh the change in Enthalpy

Thus, Ammonium Chloride dissolves

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Solubility - Hhydr vs Hlattice

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Relative Solubilities – Alkaline Earth Hydroxides

Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2

Lattice energy decreases (becomes less positive) as the radius of alkaline earth ion increases down the group from Mg2+ to Ba2+

Lattice energy decreases down a group suggesting solubility to increase from Magnesium Hydroxide to Barium Hydroxide


Energy of Hydration becomes more negative (Exothermic) as the ions become smaller

Since the atom and ion size increase down a group, Energy of Hydration becomes more positive (less Exothermic) down the group suggesting a decrease in solubility down the group from Mg(OH)2 to Ba(OH)2

However, in the case of Alkaline Earth Hydroxides, Lattice energy (Hlattice) dominates and Ba(OH)2 is the more soluble

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Relative Solubilities - Alkaline Earth Sulfates

MgSO4 > CaSO4 > SrSO4 > BaSO4

Lattice energy depends on the sum of the anion/cation radii

Since the Sulfate (SO42-) ion is much larger

than the Hydroxide ion (244 pm vs 119 pm), the percent change in Lattice Energy (Hlattice) going from Magnesium to Barium in the Sulfates is smaller than for the Hydroxides, i.e., the solubility change is small


The Energy of Hydration (HHydr) for the cations decreases (becomes more positive) by a greater amount going from Mg to Ba in the Sulfates relative to the Hydroxides

Thus in the case of Alkaline Earth Sulfates, the energy of Hydration (HHydr) dominates and Mg(SO4) is more soluble than Ba(SO4)

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Alkaline Earth Hydroxides

Down group● Lattice energy becomes less

positive (more soluble)● Hydration energy becomes less

negative (less soluble) Lattice-energy (Hlattice)

dominates Ba(OH)2 more soluble than

Mg(OH)2

Alkaline Earth Sulfates●The Sulfate (SO4

2-) ion is much larger than the Hydroxide ion (244 pm vs 119 pm)●The percent change in lattice energy going from Magnesium to Barium in the Sulfates is smaller than for the Hydroxides●The energy of hydration for the cations decreases by a greater amount going from Mg to Ba in the Sulfates relative to the Hydroxides Hydration Energy (Hhydr) dominates

MgSO4 is more soluble than BaSO4

SolubilityLattice Energy (Hlattice)

Always Positive(Endothermic)

Hydration Energy (Hhyd)Always Negative

(Exothermic)

SolubilityLow

SolubilityHigh

Less Neg LowHigh

AlkalineEarth

Hydroxides

AlkalineEarth

Sulfates

Higher Lower

IonicSize

Mg++

Ba++

Less Pos

More Pos More Neg SolubilityLower

SolubilityHigher

Predicting Solubility

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Compatible vs Incompatible Intermolecular Forces (IMFs)

C6H14 + C10H22 compatible: both display dispersion forces non-polar molecules (soluble)

C6H14 + H2O incompatible: dispersion vs polar H-bonding (insoluble)

H2O + CH3OH compatible; both H-bonding (soluble)

NaCl + H2O compatible; ion-dipole (soluble)

NaCl + C6H6 incompatible; ion vs dispersion (insoluble)

CH3Cl + CH3COOH compatible; both polar covalent(dipole-dipole) (soluble)

Practice Problem

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Predicting Solubility

Which is more soluble in water?

a. NaCl vs C6H6

b. CH3OH vs C6H6

c. CH3OH vs NaI

d. CH3OH vs CH3CH2CH2OH

e. BaSO4 vs MgSO4

f. BaF vs MgF

Ans: NaCl

Ans: CH3OH

Ans: NaI

Ans: CH3OH

Ans: Mg(SO4)

Ans: BaF

Writing Solution Equations1. Write out chemical equation for the dissolution of

NaCl(s) in water

NaCl(s) Na+(aq) + Cl-(aq)

2. Write out chemical equation for the dissolution of Tylenol (C8H9NO2) in water

C8H9NO2(s) C8H9NO2(aq)

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H2O

H2O

Solubility as Equilibrium Process Ions disaggregate and become dispersed in

solvent

Some undissolved solids collide with undissolved solute and recrystallize

Equilibrium is reached when rate of dissolution equals rate or recrystallization

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Solute Solute(undissolved) (dissolved)€

Solubility as Equilibrium Process Saturation – Solution contains the maximum

amount of dissolved solute at a given temperature

Unsaturation – Solution contains less than maximum amount of solute – more solute could be dissolved!

Supersaturation – Solution contains more dissolved solute than equilibrium amount

Supersaturation can be produced by

slowly cooling a heated equilibrium solution

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Comparison of Unsaturated andSaturated Solutions

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Solubility Equilibrium

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Solubility Equilibrium

A(s) A(aq)

A(s) = solid lattice form

A(aq) = solvated solution form

Rate of dissolution = rate of crystallization

Amount of solid remains constant

Dynamic Equilibrium

As many particles going into

solution as coming out of solution

Gas Solubility – Temperature

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Gas particles (solute) are already separated; thus, little heat required for disaggregation

Hsolute 0

Heat of Hydration of a gaseous species is always Exothermic (Hhydration < 0)

Hsoln = Hsolute + Hhydration < 0

solute(g) + solvent(l) sat’d soln + heat

Gas molecules have weak intermolecular forces and the intermolecular forces between a gas and solvent are also weak

As temperature rises, the kinetic energy of the gas molecules also increases allowing them to escape, lowering concentration, i.e.

Solubility of a gas decreases with increasing temperature

Salt Solubility – Temperature

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Most solids are more soluble at higher temperatures

Most ionic solids have a positive Hsoln because (Hlattice) is greater than (HHydr)

Note exception - Ce2(SO4)3

Hsoln is positive

Heat must be absorbed

to form solution

If heat is added, the rate of solution should increase

Solute + Solvent + Heat saturated solution

Effect of Pressure on Gas Solubility

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At a given pressure, the same number of gas molecules enter and leave a solution per unit time – equilibrium

As partial pressure (P) of gas above solution increases, the concentration of the molecules of gas increases in solution

Gas + Solvent Saturated Solution

2 2CO COP C

Effect of Pressure on Solutions

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Pressure changes do not effect solubility of liquids and solids significantly aside from extreme P changes (vacuum or very high pressure)

Gases are compressible fluids and therefore solubilities in liquids are Pressure dependent

Solubility of a gas in a liquid is proportional ( to the partial pressure of the gas above the solution

Quantitatively defined in terms of Henry’s Law

S P S = kHP

S = solubility of gas (mol/L)Pg = partial pressure of gas (atm)

KH = Henry’s Law Constant (mol/Latm)

Practice Problem

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The partial pressure of Carbon Dioxide (CO2) gas in a bottle of cola is 4 atm at 25oCWhat is the Solubility of CO2?Henry’s Law constant for CO2 dissolved in water is:

3.3 x 10-2 mol/LatmAns:

2H COSolubility = k × P

2

-2COS = 3.3×10 mol / L •atm × 4 atm = 0.132 = 0.1 mol / L

Practice ProblemThe solubility of gas A (Mm = 80 g/mol) in water is 2.79 g/L at a partial pressure of 0.24 atm. What is the Henry’s law constant for A (atm/M)?

04/21/23 73

hK = 0.145 mol / L • atm

gasmM = 80g / mol

gas h gasS = K × P

gash

gas

g 1 mol2.79 ×S L 80 g

K = = P 0.24 atm

Practice Problem

04/21/23 74

If the solubility of gas Z in water is 0.32 g/L at a pressure of 1.0 atm, what is the solubility (g/L) of Z in water at a pressure of 0.0063 atm?

gas(2)S = 0.0020 g / L

gas(1) h gas(1)S = K × Pgas(1)

hgas(1)

SK =

P

gas(2) h gas(2)S = K × Pgas(2)

hgas(2)

SK =

P

gas(2) gas(1)

gas(2) gas(1)

S S =

P P

gas(1)ga(2) gas(2)

gas(1)

S 0.32g / LS = P = 0.0063atm

P 1atm

Concentration

04/21/23 75

DefinitionsConcentration Term

Molarity (M)amount (mol) of solutevolume (L) of solution

Molality (m)amount (mol) of

solutemass (kg) of solvent

Parts by massmass of solute

mass of solution

Parts by volumevolume of solute

volume of solution

Mole Fraction (Xsolvent)

amount (mol) of solute

moles of solute + moles of solvent

Ratio

Mole Fraction (Xsolute)

(i)amount (mol) of solute

(i)moles of solute + moles of solvent

Practice Problem

04/21/23 76

The molality of a solution is defined as _______

a. moles of solute per liter of solutionb. grams of solute per liter of solutionc. moles of solute per kilogram of solutiond. moles of solute per kilogram of solvente. the gram molecular weight of solute per

kilogram of solvent

Ans: d

Practice Problem

04/21/23 77

What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water

Ans:

Step 1 - Convert Mass to Moles

Step 2 – Compute Molality

amount (mol) soluteMolality =

mass (kg) solvent

22 2 2

2

1 mol CaClMoles CaCl = 32.0 g CaCl × = 0.280 mol CaCl

110.98 g CaCl

22

3

0.288 mol CaClmol soluteMolality = = = 1.06 m CaCl

1 kgkg solvent 271 g×10 g

Practice Problem

04/21/23 78

Fructose, C6H12O6 (FW = 180.16 g/mol, is a sugar occurring in honey and fruits. The sweetest sugar, it is nearly twice as sweet as sucrose (cane or beet sugar). How much water should be added to 1.75 g of fructose to give a 0.125 m solution?

2 2solvent H O H O

1000 gkg = 0.0777 kg × = 77.7 g

1 kg

solute

solvent

molMolality(m) =

kg

2

sucrosesucrose

solute sucrosesolvent

H O

1 mol1.75 g

mol 180.16 gkg = =

m 0.125 mol / kg

Colligative Properties of Solutions Presence of Solute particles in a solution

changes the physical properties of the solution

The number of particles dissolved in a solvent also makes a difference in four (4) specific properties of the solution known as:

“Colligative Properties” Vapor Pressure

Boiling Point Elevation

Freezing Point Depression

Osmotic Pressure04/21/23 79

Colligative Properties of Solutions

04/21/23 80

Phase Diagram showing various phases of a substance andthe conditions under which each phase exists

Triple PointOf Solution


04/21/23 81

Colligative properties deal with the nature of a solute in aqueous solution and the extent of the dissociation into ions

Electrolyte – Solute dissociates into ions and solution is capable of conducting an electric current Strong Electrolyte – Soluble salts, strong acids,

and strong bases dissociate completely; thus, the solution is a good conductor

Weak Electrolyte – polar covalent compounds, weak acids, weak bases dissociate weakly and are poor conductors

Nonelectrolyte – Compounds that do not dissociate at all into ions (sugar, alcohol, hydrocarbons, etc.) are nonconductors


04/21/23 82

Prediction of the magnitude of a colligative property Solute Formula

Each mole of a nonelectrolyte yields 1 mole of particles in the solution

Ex. 0.35 M glucose contains 0.35 moles of solute particles (glucose molecules) per liter

Each mole of strong electrolyte dissociates into the number of moles of ions in the formula unit

Ex. 0.4 M Na2SO4 contains 0.8 mol of Na+ ions and 0.4 mol of SO4

2- ions (total 1.2 mol of particles) per liter of solution

Vapor Pressure

04/21/23 83

Vapor pressure (Equilibrium Vapor Pressure):

The pressure exerted by a vapor at equilibriumwith its liquid in a closed system

Vapor Pressure increases with increasing temperature

nonvolatile nonelectrolyte (ex. sugar) & pure solvent

The vapor pressure of a solution of a nonvolatile nonelectrolyte (solute) is always lower than the vapor pressure of the pure solvent

Presence of solute particles reduces the number of solvent vapor particles at surface that can vaporize

At equilibrium, the number of solvent particles leaving solution are fewer than for pure solvent, thus, the vapor pressure is less

Vapor Pressure

04/21/23 84

The vapor pressure of the solvent above the solution (Psolvent) equals the Mole Fraction of solvent in the solution (Xsolvent) times the vapor pressure of the pure solvent (Po

solvent)(Raoult’s Law)

Psolvent = Xsolvent Po

solvent

In a solution, the mole fraction of the solvent (Xsolvent) is always less than 1; thus the partial pressure of the solvent above the solution (Psolvent) is always less than the partial pressure of the pure solvent Po

solvent

Ideal Solution – An ideal solution would follow Raoult’s law for any solution concentration

Most gases in solution deviate from ideality Dilute solutions give good approximation of Raoult’s

law

Vapor Pressure A solution consists of Solute & Solvent The sum of their mole fractions equals 1

04/21/23 85

The magnitude of P (vapor pressure lowering) equals the mole fraction of the solute times the vapor pressure of the pure solvent

solvent solute

solvent solute

X + X = 1

X = 1 - X

o osolvent solvent solvent solute solvent

o osolvent solvent solute solvent

From Raoult's law

P = X P = (1- X ) × P

P = P - X × P

o osolvent solvent solute solvent

Rearranging

P - P = ΔP = X × P

Vapor Pressure

04/21/23 86

Vapor Pressure & Volatile Nonelectrolyte Solutions

The vapor now contains particles of both a volatile solute and the volatile solvent

From Dalton’s Law of Partial Pressures

o osolvent solvent solvent solute solute soluteP = X P and P = X P

o ototal solvent solute solvent solvent solute soluteP = P + P = X P + X P

The presence of each volatile component lowers the vapor pressure of the other by making each mole fraction less than 1

Example Problem

04/21/23 87

Given: Equi-molar solution of Benzene (XGiven: Equi-molar solution of Benzene (XBB = 0.5) = 0.5) and and Toluene (X Toluene (XTT = 0.5) (Both nonelectrolytes)) (Both nonelectrolytes)o o

ben benX = 0.5 P = 95.1 torr @ 25 C

o otol tolX = 0.5 P = 28.4 torr @ 25 C

oben ben benP = X × P = 0.5 × 95.1 torr = 47.6 torr

otol tol tolP = X × P = 0.5 × 28.4 torr = 14.2 torr

Benzene lowers the vapor pressure of Toluene and Toluene lowers the vapor pressure of Benzene

Compare Vapor composition vs Solution composition

toltol

tot

P 14.2 torrX = = = 0.230 torr

P 47.6 torr +14.2 torr

Mole fractions in vapor are different

benben

tot

P 47.6 torrX = = = 0.770 torr

P 47.6torr + 14.2torr


04/21/23 88

Boiling Point (Tb) of a liquid is the temperature at which its vapor pressure equals the external pressure (Atm Press)

The vapor pressure of a solution (solvent + solute) is lower than that of the pure solvent at any temperature

Thus, the difference between the external (atmospheric) pressure and the vapor pressure of the solution is greater than the difference between external pressure and the solvent vapor pressure - Psoln > Psolvent

The boiling point of the solution will be higher than the solvent because additional energy must be added to the solution to raise the vapor pressure of the solvent (now lowered) to the point where it again matches the external pressure

The boiling point of a concentrated solution is greater than the boiling point of a dilute solution - Antifreeze in your car!!


04/21/23 89

The magnitude of the boiling point elevation is proportional to the Molal concentration of the solute particles

solution solvent

b b b

b b b

solute solvent

b

ΔT m or ΔT = K m

where : ΔT = T - T = boiling point elevation

m = solution molality (mol / kg )

K =

molal boiling point elevation constant Note the use of “Molality”

Molality is related to mole fraction, thus particles of solute

Molality also involves “Mass of Solvent”; thus, not affected by temperature


04/21/23 90

Recall: The addition of a solute to a solvent dilutes the number of solvent particles in the solution

As the solute concentration increases there is increased competition between solute and solvent particles for space at the surface preventing solvent molecules from escaping to the vapor phase, thus the vapor pressure of the solution is lower than the vapor pressure of the pure solvent

Recall: The vapor pressure lowering, ΔP, of a solution is a function of the mole fraction of solute and vapor pressure of pure solvent, that is:

Raoults Law

o osolvent solvent solute solventP - P = ΔP = X × P

Freezing Point Depression The Freezing Point of a substance (solvent) is the

temperature at which an equilibrium is established between the number of solid particles coming out of solution and the number of particles dissolving

Note: In solutions with nonvolatile solutes, only solvent molecules can vaporize; thus, only solvent molecules can solidify (freeze)

The vapor pressure of solid and liquid solvent particles in equilibrium is same

Since vapor pressure is a function of temperature, the lowering of the vapor pressure of the solution means a lower temperature at which equilibrium exists between solid & liquid solvent particles, i.e. the freezing point

04/21/23 91


04/21/23 92

Thus, as the concentration of the solute in the solution increases, the freezing point of the solution is lowered


04/21/23 93

The Freezing Point depression has the magnitude proportional to the Molal concentration of the solute

solvent solution

f f f

f f f

solute solvent

f

ΔT m or ΔT = K m

where : ΔT = T - T = freezing point depression

m = solution molality (mol / kg )

K =

molal freezing point depression constant

Molal Boiling Point Elevation and Freezing Point DepressionConstants of Several Solvents

The van’t Hoff Factor

04/21/23 94

From Slide 82, colligative properties depend on the relative number of solute to solvent “particles”

In strong electrolyte solutions, the solute formula specifies the number of particles affecting the colligative property

Ex. The BP elevation of a 0.5 m NaCl soln would be twice that of a 0.5 m Glucose soln because NaCL dissociates into “2” particles per formula unit, where glucose produces “1” particle per formula unit

The van’t Hoff factor (i) is the ratio of the measured value of the colligative property, e.g. BP elevation, in the electrolyte solution to the expected value for a nonelectrolyte solution

measured value for electrolyte solution=

expected value for nonelectrolyte solutioni

The van’t Hoff Factor

04/21/23 95

To calculate the colligative properties of strong electrolyte solutions, incorporate the van’t Hoff factor into the equation

CH3OH(l) CH3OH(aq) 1:1, i = 1

NaCl(s) Na+(aq) + Cl-(aq) 2:1, i = 2

CaCl2(s) Ca2+(aq) + 2Cl-(aq) 3:1, i = 3

Ca3(PO4)2(s) 3 Ca2+(aq) + 2 PO43-(aq) 5:1, i = 5

Ex. Freezing Point Depression with van’t Hoff factor for Ca3(PO4)2(s)

5T K cf f m T iK cf f m


04/21/23 96

Colligative Mathematical Property Relation1. Vapor Pressure

(Raoult’s Law)

2. Freezing Point Depression

3. Boiling Point Elevation

4. Osmotic Pressure

solvent solvent solute

oP = i P X

f f

T i K m

T i K mb b

i MRT

Practice Problem

04/21/23 97

What is the boiling point of 0.0075 m aqueous calcium chloride, CaCl2?

CaCl2 = 3 particles (1 Ca & 2 Cl) i = 3

CaCl2

obT = 100.1 C

b b bΔT m or ΔT = K m (i = 3)i

solution solventb b bwhere : ΔT = T - T = boiling point elevation

solution solventb b bT - T = 3 K m

CaCl H O2 2

b b bT = 3 K m + T

CaCl2

oo

b

CT = 3 0.512 0.0075 m + 100.0 C

m

Practice Problem

04/21/23 98

What is the freezing point of a 0.25 m solution of glucose in water (Kf for water is 1.86°C/m)?

a. 0.93°C b. –0.93°C c. 0.46°C d. –0.46°C e. 0.23°C

Ans: d

solution

ofT = -0.46 C

1)solvent solutionf f f fΔT = T - T = K m (i i

solution solventf f f-T = K m - T

solution solventf f fT = T - K m

fsolution

oCoT = 0.0 C - 1.86 × 0.25 mm

Practice Problem

04/21/23 99

What is the freezing point of 0.150 g of glycerol (C3H8O3) in 20.0 g of water? (Kf for water is 1.86°C/m)

solvent solutionf f f fΔT = T - T = iK m (i = 1)

solution solventf f f-T = K m - T

solution solventf f fT = T - K m

solution

oo

f

1mol0.150g

92.094gCT = 0.0 C - 1.86 ×

m 1kg20.0g ×

1000g

solution

oo

f

C molT = 0.0 C - 1.86 × 0.0814

mol kgkg

solution

ofT = -0.151 C

Practice Problem

04/21/23 100

What is the molar mass (Mm) of Butylated Hydroxytoluene (BHT) if a solution of 2.500 g of BHT in 100.0 g of Benzene (Kf = 5.065 oC/m; Tf = 5.455 oC) had a freezing point of 4.880 oC?

solvent solutionf f f fΔT = T - T = iK m (i = 1)

solvent solutionf f

f

T - Tm =

iK

o o

BHT Benzeneo

5.455 C - 4.880 Cm = = 0.1135 mol / kg

1× 5.065 C / mmBHT BHTM = Molar Mass (g / mol)

BHT BHT

BHT mBHT mBHT

Benzene BenzeneBenzene

1mol 1mol2.500 g× 25.00 g

mol M (g) M (g)m = 0.1135 = =

1kgkg 1kg100 g1000g

mBHT BHT

mBHT BHT

M g 25.00 g= = 220.3g / mol (Actual M = 220.3 g / mol)

1 mol 0.1135 mol

Colloids

04/21/23 101

Suspensions vs Mixtures vs Colloids A Heterogeneous mixture – fine sand suspended

in water – consists of particles large enough to be seen by the naked eye, clearly distinct from surrounding fluid

A Homogeneous mixture – sugar in water – forms a solution consisting of molecules distributed throughout and indistinguishable from the surrounding fluid

Between these extremes is a large group of mixtures called colloidal dispersions – Colloids

Colloid particles are larger than simple molecules, but small enough to remain in suspension and not settle out

Colloids Colloids have tremendous surface areas, which allows

many more interactions to exert a large total adhesive force, which attracts other particles Particle Size & Surface Area

Diameter – 1 to 1000 nm (10-9 to 10-6) Single macromolecule or aggregate of many

atoms, ions, or molecules Very large surface area Large surface area attracts other particles

through various intermolecular forces (IMF) Surface Area

A cube with 1 cm sides (SA – 6 cm2) if divided into 1012 cubes (size of large colloidal particles) would have a total surface area of 60,000 cm2

04/21/23 102

Colloids

04/21/23 103

Colloid Classifications: Colloids are classified according to whether

the dispersed and dispersing substances are gases, liquids, or solids

Types of Colloids

Colloids Tyndall Effect

Light passing through a colloid is scattered randomly because the dispersed particles have sizes similar to wavelengths of visible light

The scattered light beams appears broader than one passing through a solution

Brownian Motion Observed erratic change of speed and

direction resulting from collisions with molecules of the dispersing medium

Einstein’s explanation of Brownian motion further enhanced the concept of the molecular nature of matter

04/21/23 104

Colloids Stabilizing & Destabilizing Colloids

Colloidal particles dispersed in water have charged surfaces

Two major classes: Hydrophilic (Stabilized) Colloids

London force attractions between charges on dispersed phase (colloid) and the partial charges on continuous phase (water) Appear like normal solutions, no settling

outo Gelatin (protein solution) in watero Lipids & Soap form spherical micelles with

charged heads on exterior and hydrocarbon tails

o Oily particles can be dispersed by adding ions, which are absorbed on the particle surface

04/21/23 105

Colloids Hydrophobic (Unstable) Colloids

Several methods exist to overcome the attractions between the dispersed and continuous phases causing the particles to aggregate and settle out At the mouths of rivers, where salt water

meets fresh water, the ions in salt water are absorbed onto the surfaces of clay particles causing them to aggregate. The settling out process produces deltas.

In smokestack gases, electrolyte ions are absorbed onto uncharged colloidal particles, which are then attracted to charged plates removing them from the effluent

04/21/23 106

Osmosis

04/21/23 107

Osmosis, a colligative property, is the movement of solvent particles through a semipermeable membrane separating a solution of one concentration from a solution of another concentration, while the solute particles stay within their respective solutions

Many organisms have semipermeable membranes that regulate internal concentrations

Osmosis Water can be purified by “reverse osmosis”

The direction of flow of the solvent is from the solution of lower concentration to the solution of higher concentration increasing its volume, decreasing concentation

Osmotic pressure is defined as the amount of pressure that must be applied to the higher concentration solution to prevent the dilution and change in volume

04/21/23 108

Osmosis

04/21/23 109

Osmosis (2-Compartment Chamber)

04/21/23 110

• Osmosis = solvent flow across membrane

• Net movement of solvent is from 2 to 1; pressure increases in 1

• PA,1 < PA,2, osmosis follows solvent vapor pressure

• Osmotic Pressure = pressure applied to just stop osmosis

= solute molecule= solvent molecule

1 2

Semipermeable membrane

Osmotic Pressure

04/21/23 111

At equilibrium the rate of solvent movement into the more concentrated solution is balanced by the rate at which the solvent returns to the less concentrated solution.

The pressure difference at equilibrium is the Osmotic Pressure (), which is defined as the applied pressure required to prevent the net movement of solvent from the less concentrated solution to the more concentrated solution

Osmotic Pressure Osmotic Pressure () is proportional to the

number of solute particles (moles) in a given volume of solution, i.e.,

Molarity (M) (moles per liter of solution)

04/21/23 112

solute

solution

solute

solution

nΠ or Π M

V

The Proportionality Constant is :

"R" times the absolute temperature (T)

nΠ = RT = MRT

V

Osmotic Pressure

04/21/23 113

Osmotic Pressure / Solution Properties The volume of the concentrated solution will

increase as the solvent molecules pass through the membrane into the solution decreasing the concentration

This change in volume is represented as a change in height of the liquid in a tube open to the externally applied pressure

This height change (h), the solution density (d), and the acceleration of gravity (g) can also be used to compute the Osmotic Pressure ()

= g d h = g(m/s2)d(kg/m3)h(m) = pascals(kg/ms2)

1 pascal = 7.5006 x 10-3 torr(mm)= 9.8692 x 10-6 atm

Reverse Osmosis Membranes with pore sizes of 0.1 – 10 m are used

to filter colloids and microorganisms out of drinking water

In reverse osmosis, membranes with pore sizes of 0.0001 – 0.01 m are used to remove dissolved ions

A pressure greater than the osmotic pressure is applied to the concentrated solution, forcing the water back through the membrane, filtering out the ions

Removing heavy metals from domestic water supplies and desalination of sea water are common uses of reverse osmosis

04/21/23 114

Practice Problem

04/21/23 115

How much potable water can be formed per 1,000 L of seawater (~0.55 M NaCl) in a reverse osmosis plant that operates with a hydrostatic pressure of 75 atm at 25 oC?

NaCl SolnV = 179.5 L = 180 L

NaCl NaClNaCl soln NaCl NaCl soln NaClSoln

NaCl Soln NaCl Soln

n nΠ = RT M = n = M × V

V V

NaCl NaCl soln NaCl SolnNaCl Soln

n M × VV = RT = RT

Π Π

o oNaCl Soln o

mol0.55 × 1000 L

L • atmLV = 0.0821 [25 C + 273.15] K75 atm mol • K

Practice Problem

04/21/23 116

What is the osmotic pressure (mm Hg) associated with a 0.0075 M aqueous calcium chloride, CaCl2, solution at25 oC?

a. 419 mm Hg b. 140 mm Hg c. 89 mm Hg

d. 279 mm Hg e. 371 mm Hg

760 mmΠ = 0.5507576 atm = 418.5758 = 419 mm Hg

1 atm

Π = (MRT) = 3i i

oo

mol L • atmΠ = 3 × 0.0075 × 0.0821 × [25 + 273.15] K

L mol • K

Practice Problem

04/21/23 117

Consider the following dilute NaCl(aq) solutionsa. Which one will boil at a higher temperature?b. Which one will freeze at a lower temperature?c. If the solutions were separated by a semi permeable

membrane that allowed only water to pass, which solution would you expect to show an increase in the concentration of NaCl?

Ans: a. B vapor pressure of more concentrated solution is lower; thus, a higher temperature required to reach VP at boiling point

b. B vapor pressure of more concentrated solution is lower; thus, temperature at equilibrium (freezing) point is lower

c. A

The movement of solvent betweensolutions is from the solution oflower concentration to the solutionof higher concentration), thus less solvent in beaker A resulting in higher concentration

Practice Problem Consider the following three beakers that contain

water and a non-volatile solute. The solute is represented by the orange spheres.

a. Which solution would have the highest vapor pressure?

b. Which solution would have the lowest boiling point?

c. What could you do in the laboratory to make each solution have the same freezing point?

04/21/23 118

a. A More solvent particles can escape because of fewer

solute particles present

b. A Requires less change in vapor pressure to reach VP at boiling point

c. Condense “A” by 1/2

Practice Problem

04/21/23 119

Caffeine, C8H10N4O2 (FW = 194.14 g/mol), is a stimulant found in tea and coffee. A Sample of the substance was dissolved in 45.0 g of chloroform, CHCl3, to give a 0.0946 m solution. How many grams of caffeine were in the sample?

Ans:

caffeine caffeinecaffeine

194.14gMass = mol × = 0.826g

mol

3

caffeine

CHCl

molesmolality(m) =

kg

3caffeine CHClmoles = m×kg

3

3

caffeinecaffeine CHCl

CHCl

mol 1kgmoles = 0.0945 × 45.0g

kg 1000g

caffeine caffeinemoles = 0.0042525 mol

Practice Problem

04/21/23 120

A solution contains 0.0653 g of a molecular compound in 8.31 g of Ethanol. The molality of the solution is 0.0368 m. Calculate the molecular weight of the compound

solute solventmolality(m) = mol / kg

ethanol

1mol0.0653g

Mm0.0368 =1kg

8.31g1000g

Mm = 213g / mol = MolecularWeight

0.0653gMm =

mol 1kg0.0368 × 8.31g ×

kg 1000g

Practice Problem

04/21/23 121

What is the vapor pressure (mm Hg) of a solution of 0.500 g of urea [(NH2)2CO, FW = 60.0 g/mol] in 3.00 g of water at 25 oC?

What is the vapor pressure lowering of the solution?

The vapor pressure of water at 25 oC is 23.8 mm Hgurea

urea urea ureaurea

1molmol = 0.500g = 0.00833 mol

60.0g2

2 2 2

2

H OH O H O H O

H O

1molmol = 3.00 g = 0.187 mol

18.016 g

2H O

0.187Mole Fraction Water (X ) = = 0.957

0.00833 + 0.187

urea0.00833

Mole Fraction Urea (X ) = = 0.04250.00833 + 0.187

Partial Pressure of Solution, i.e., the solvent containing non - volatile electrolyte

solvent solvent solventoP = i P X solventP = 1×(23.8 mmHg×0.957) = 22.8 mmHg

VaporPressure lowering

solvent solventoP - P = ΔP = 23.8 mm Hg - 22.8 mmHg = 1.00 mm HG

solvent solvent solvento o

ureaP - P = ΔP = X × P = 0.0425 × 23.8 mm Hg = 1.01 mm Hg

or

Practice Problem

04/21/23 122

What is the vapor pressure lowering in a solution formed by adding 25.7 g of NaCl (FW = 58.44 g/mol) to 100. g of water at 25 oC? (The vapor pressure of pure water at 25 oC is 23.8 mm Hg)

NaClNaCl NaCl NaCl

NaCl

1molmol = 25.7g = 0.4398 mol

58.44g

NaCl0.4398

Mole Fraction NaCl (X ) = = 0.073420.4398 + 5.5506

2

2 2 2

2

H OH O H O H O

H O

1molmol = 100. g = 5.5506 mol

18.016 g

Vapor Pressure lowering

solvento

NaClΔP = i(X × P ) = 2 0.07432 × 23.8 mm Hg = 3.49 mm Hg

Practice Problem

04/21/23 123

A 0.0140-g sample of an ionic compound with the formula Cr(NH3)5Cl3 (FW = 243.5 g/mol) was dissolved in water to give 25.0 mL of solution at 25 oC.

The osmotic pressure was determined to be 119 mm Hg. How many ions are obtained from each formula unit when the compound is dissolved in water?

Π = i(MRT)

solute

soln

1mol0.0140 g

mol 243.5 gMolarity(M) = = = 0.0022998 M

1LL 25 ml1000 mL

oo

1atm119mmHg

Π 760 mmHgi = =

mol L •atmMRT 0.0022998 ×0.0821 ×(273.15 + 25.0) KL mol • K

i = 2.78 = 3The number of particles produced by the dissolution

of a "complex" molecule may not be readily descernible

by physical inspection of the compond formula

Equation Summary

04/21/23 124

Component Enthalpies of Heat of Solution

Component Enthalpies of Ionic Compound Heat of Soln

Relating Gas Solubility to its Partial Pressure (Henry’s Law)

Δ = Δ + Δ + Δsoln solute solvent mixH H H H

Δ = Δ + ΔH H Hsoln lattice hydr of the ions

gas gasS = PHk

Δ = Δ + Δsolute hydrationH H Hsoln

Equation Summary

04/21/23 125

Relationship between vapor pressure & mole fraction

Vapor Pressure Lowering



Osmotic Pressure

solution solventb b b bΔT = T - T = K m i

)o oP - P = ΔP = (X × Psolvent solvent solute solventi

solvent solutionf f f fΔT = T - T = K m i

solute

solution

n Π = RT = MRT

Vi i

osolvent solvent Solvent

osolute solute Solute

o ototal solvent solute solvent Solvent solute Solute

P = X × P

P = X × P

P = P + P = (X × P ) + (X × P )

Equation Summary

04/21/23 126

Concentration Term

Molarity (M)

Ratio

amount (mol) of solutevolume (L) of

solution

Molality (m)amount (mol) of

solutemass (kg) of solvent

Parts by massmass of solute

mass of solution

Parts by volumevolume of

solutevolume of solution

Mole Fraction

(Xsolute)

(i)amount (mol) of solute(i)moles of solute + moles of

solvent

Mole Fraction

(Xsolvent)

amount (mol) of solvent(i)moles of solute + moles of solvent

george mason university general chemistry 212 chapter 13

Documents

repulsive forces

intermolecuar forces

properties of mixturestypes

properties of colloids

properties of mixturessolution

solute solvent

additional material

general chemistry courses