1

Physical Properties of Solutions

Chapter 13

Outline of the Chapter

• The fundamental properties of Solutions– Types– Energetics

• Working with solutions– Concentration Units

• Saturated Solutions and Equilibrium

• Colligative Properties– Freezing point depression– Boiling point elevation

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Definitions to Know

Solution

Solute

Solvent

Saturated, unsaturated, supersaturated solutions

Solubility

Dynamic Equilibrium

Types of solutions

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Solutions

• Solutions are homogeneous mixtures of two or more pure substances.

• Solute is dispersed uniformly throughout the solvent ( we will deal with water solutions).

• Liquids that mix in all proportions are called miscible

12.1 Some Types Of Solutions

12.1

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12.1 Types of Solutions

• Saturated Solvent holds as much solute as

is possible (maximum) at that temperature.

Dissolved solute is in dynamic equilibrium with solid solute particles.

Dynamic equilibrium: rate of crystallization = rate of dissolving

Solubility: concentration of saturated solution.

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12.1 Types of Solutions

• UnsaturatedLess than the

maximum amount of solute for that temperature is dissolved in the solvent.

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12.1 Types of Solutions

• Supersaturated Solvent holds more solute than is normally possible at that

temperature. Solutions are unstable; crystallization can usually be

stimulated by adding a “seed crystal” or scratching the side of the flask.

• Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate.

A Molecular View of the Solution Process

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1. In the following processes differentiate between physical process that leads to solution formation and chemical reaction

a) Ni(s) + 2HCl → NiCl2(aq) + H2(g)

b) NaCl (s) → Na+(aq) + Cl-(aq)

c) CH3OH(l) → CH3OH(aq)

Intermolecular Forces In Solution Formation

Intermolecular Forces In Solution

Solute-solute

Solvent-solvent

Solute-solvent

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Energy needed to break the:

solute-solute interaction:

Δ H1 > 0

Solvent-solvent interactions:

Δ H2 > 0

Energy released when:Solvent-solute interaction

Δ H3 < 0

Hsoln = H1 + H2 + H3

Formation Of Solution: Energy Considerations

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Energy Changes in Solution

The enthalpy change of the overall process depends on H for each of these steps.

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Why Do Endothermic Processes Occur?

Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

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Why Do Endothermic Processes Occur?

Yet we know that in some processes, like the dissolution of NH4NO3 in water, heat is absorbed, not released.

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Enthalpy Is Only Part of the Picture

The reason is that increasing the disorder or randomness (known as entropy) of a system tends to lower the energy of the system.

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Enthalpy Is Only Part of the Picture

So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

Intermolecular Forces In Solution Formation

• Ideal Solution: All intermolecular forces are of comparable strength Δ Hsoln = 0.

(N2 and O2; gasoline ; benzene + toluene)

Enthalpy driven dissolution:• Intermolecular forces between solute and solvent molecules

are stronger than other intermolecular forces,

|Δ H3| > Δ H1 + Δ H2, Δ Hsoln < 0 (exothermic)

MgSO4, ΔHsolution = -92. kJ/mol Sometimes, volume of solution is smaller than volume of

individual components (C2H5OH + H2O)

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Intermolecular Forces In Solution Formation

Entropy driven dissolution:• Intermolecular forces between solute and solvent

molecules are weaker than other intermolecular forces,

|Δ H3| < Δ H1 + Δ H2, Δ Hsoln > 0 Endothermic. Entropy driven. NH4NO3, ΔH = 26.4 kJ/mol

• Intermolecular forces between solute and solvent are much weaker ( ΔH1 + ΔH2 << |ΔH3| ) than other intermolecular forces, the solute does not dissolve in the solvent. The compound is relatively insoluble in the solvent.

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Example

2. Predict the relative solubilities in the following case:

a) Bromine in benzene (μ = 0 D)

b) Bromine in water (μ = 1.87 D)

c) KCl in carbon tetrachloride (μ = 0 D)

d) KCl in liquid ammonia (μ = 1.46 D)

e) Formaldehyde (CH2O) in carbon disulfide CS2, μ = 0 D)

f) Formaldehyde (CH2O) in water (μ = 1.87 D)

g) Urea, (NH2)2CO in carbon disulfide

h) Urea in water

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Factors Affecting Solubility

• Chemists use the axiom “like dissolves like”:Polar substances tend to dissolve in polar solvents.Nonpolar substances tend to dissolve in nonpolar

solvents: CCl4 in C6H6

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Factors Affecting Solubility

Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane (which only has dispersion forces) is not.

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Liquid-Liquid Solutions

• Miscible: completely soluble• Immiscible: do not dissolve in each other• Solvent: component with greater

concentration

• Like dissolves like

• Acetic acid dimers

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Solutions of Solids in Liquids

Things to know:

• aqueous solutions– hydration

• polar-polar solutions– Solvation

• Crystallization

• Solubility of metals

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• Two forces

- Break ionic attractions (positive and negative ions) in the solid

- Form ion-dipole forces between solute particles and solvent particles (form hydration shell).

Solubility determined by the strength of the solid-solid IMfs and solute-solvent interactions

Network and metallic + water: does not happen

Aqueous Solutions

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Ion-Dipole Forces in Dissolution

12.4

Hydration shell

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Example

3. Predict whether each of the following is likely to be a solution or a heterogeneous mixture.

a) Ethanol, CH3OH, and water, HOH

b) Pentane, CH3(CH2)3CH3, and octane, CH3(CH2)6CH3

c) Sodium chloride, NaCl, and carbon tetrachloride, CCl4d) 1-decanol, CH3(CH2)8CH2OH, and water, HOH

12.4

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Examples

4. Some solution processes are endothermic and some are endothermic. Provide a molecular interpretation for these differences.

5. Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible? Give examples.

Answers:• (IMFs: solid-solid; liqud-liquid, solid-liquid; Energy diagrams)• (miscible: completely soluble in all proportions; IMFs comparable

in strength, butane in octane; methanol in water, describe the IMFs))

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Ways of Expressing Concentrations of Solutions

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Concentration Units

• Things to know:• Percent by mass• Mole fraction• Molarity• Molality• Comparison of Concentration Units• Parts per million (ppm)• Parts per billion (ppb)

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Mass Percentage

Mass % of A =mass of A in solutiontotal mass of solution

100

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Parts per Million andParts per Billion

ppm = mass of A in solutiontotal mass of solution

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Parts per Million (ppm), 1 ppm = 1 mg/L

Parts per Billion (ppb), 1 ppb = 1 µg/L

ppb =mass of A in solutiontotal mass of solution

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moles of Atotal moles in solution

XA =

Mole Fraction (X)

• In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

• xi < 1; x1 + x2 + x3 + … = 1

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mol of soluteL of solution

M =

Molarity (M)

• Volume is temperature dependent.

Therefore

• molarity can change with temperature.

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mol of solutekg of solvent

m =

Molality (m)

• Moles and mass do not change with temperature.

Therefore,• Molality (unlike molarity) is not

temperature dependent.

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Concentration Units: Examples

6. A sample of 0.892 g KCl dissolved in 54.6 g H2O. Calculate % by mass.

7. A solution is prepared by mixing 200.4 g C2H5OH to 143.9 g of H2O. Calculate mole fractions of both ingredients of the solution.

(Xalcohol 0.3539; xH2O = 0.647

8. Calculate molality of H2SO4 solutions containing 24.4 g H2SO4 in 198 g of H2O.

(1.26 m)

9. How many milliliters of water (d = 0.998 g/mL) are required to dissolve 25.0 g of urea and thereby produce a 1.65 m solution of urea, CO(NH2)2?

(253 mL of H2O)

10. The density of a 2.45 M aqueous methanol solution is 0.976 g/mL. What is the molality of the solution?

(2.73 m)

11. Calculate the molality of a 34.5 % (by mass) aqueous solution of phosphoric acid, H3PO4. The molar mass of phosphoric acid is 98.00 g/mol.

(5.37 m)

Example

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12. What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? (8.92 m)

m =moles of solute

mass of solvent (kg)M =

moles of solute

liters of solution

Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)

mass of solvent = mass of solution – mass of solute

= 927 g – 270 g = 657 g = 0.657 kg

m =moles of solute

mass of solvent (kg)=

5.86 moles C2H5OH

0.657 kg solvent= 8.92 m

Example (H@P, 516)

13. An aqueous solution of ethylene glycol used as an automobile engine coolant is 40.0% HOCH2CH2OH by mass and has a density of 1.05 g/mL. What are the

(a) molarity,

(b) molality, and

(c) mole fraction of HOCH2CH2OH.

(a. 6.77 M b. 10.7 m c. 0.162)

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Changing Molarity to Molality

If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

Effect of Temperature on Solubility

• Solubility: the amount of solute that will dissolve in a given amount of solvent at a given temperature

• Solubilities of ionic compounds increase significantly with increasing temperature (About 95% of compounds). Rest do not change or very slightly (NaCl).

• A very few have solubilities that decrease with increasing temperature.

• A supersaturated solution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute.

12.6

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Temperature and Solubility

Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

Exceptions?

Fractional Crystallization

When KNO3(s) is crystallized from an aqueous solution of KNO3 containing CuSO4 as an impurity,

CuSO4 remains in the solution.

12.6

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Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.

Suppose you have 90 g KNO3 contaminated with 10 g NaCl.

Fractional crystallization:

1. Dissolve sample in 100 mL of water at 600C

2. Cool solution to 00C

3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

12.6

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Solubility: Examples

15. The solubility of ammonium formate, NH4HCO2, in 100.0 g of water is 102.0 g at 0ºC and 546 g at 80ºC. A solution is prepared by dissolving NH4HCO2 in 200.0 g of water until no more can dissolve at 80ºC. The solution is then cooled to 0ºC. What mass of ammonium formate precipitates? (Assume that no water evaporates and the solution is not supersaturated.

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Gases in Solution

• In general, the solubility of gases in water increases with increasing mass.

• Larger molecules have stronger dispersion forces.

Gas Solubility and Temperature

12.6

Effect of Pressure on Solubility of Gases

Things to know:Henry’s lawMolecular Interpretation of Henry’s LawCheck online tutorial

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Gases in Solution

• The solubility of liquids and solids does not change appreciably with pressure.

• The solubility of a gas in a liquid is directly proportional to its pressure.

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Pressure and Solubility of Gases

Explain the pictures.

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Pressure and Solubility of Gases

HENRY’S LAW : the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.

c = kPc is the concentration (mol/L) of the dissolved gas

P is the pressure of the gas over the solution

k is a constant (mol/L•atm) that depends onlyon temperature (for a specific solvent/solute)

low P

low c

high P

high c12.7

Solubility of Gas in Water as a Function Of Gas Pressure

12.7

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Pressure and Solubility of Gases: Henry’s Law

16. Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.00 atm over the liquid at 25.0 ºC. The Henry's Law constant for CO2 in water at this temperature is 3.1 x 10-2 mol/L-atm. (0.12 M)

17. Calculate the concentration of CO2 in a soft drink after the bottle is opened and it equilibrates at 25.0 ºC under a CO2 partial pressure of 3.0 x 10-4 atm. (9.3 x 10-6 M)

18. The solubility of pure N2 at 25ºC and 1 atm is 6.8 x 10-4 mol/L. What is the concentration of N2 dissolved in H2O under atmospheric conditions? The partial pressure of nitrogen gas in the atmosphere is 0.78 atm. (5.3 x 10-4 mol/L)

• NOTE: gases that interact (dissolve) with the solvent do not obey Henry’s Law.

Colligative properties of Nonelectrolyte Solutions

Colligative Properties

Colligative Properties – physical properties of solutions that depend on the number of solute particles present but not on the identity (nature) of the solute.

• Freezing point depression• Boiling point elevation• Vapor Pressure of a Solution• Osmosis

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Vapor Pressure

Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.

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Vapor Pressure

Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

Vapor Pressures of Solutions

• For a solution with a nonvolatile solute ( vapor pressure = 0), the vapor pressure of its solution is always less than that of the pure solvent.

• Raoult’s law: the partial pressure of the solvent above a solution (P1) is the product of the vapor pressure of the pure solvent (Po

1) and the mole fraction of the solvent in the solution (x1):

P1 = x1 . Po1

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Raoult’s Law Vapor-Pressure Lowering: Raoult’s law

If the solution contains only one solute (nonvolatile):

X1 = 1 – X2

P 10 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

P1 = X1 P 10

01

ΔP = X2

01P

P1 = vapor pressure of solution

X2 < 1

1

01

01 )1( PP 21

011

011

01 x x- P x- P P - P P

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Example: Raoult’s Law

19. At 25 ºC, the vapor pressure of pure water is 23.76 mm Hg and that of aqueous urea solution is 22.98 mm Hg. Estimate the molality of the solution.

(1.8 m)

20. The vapor pressure of glucose, C6H12O6 solution is 17.01 mm Hg at 20.0 ºC, while that of pure water is 17.25 mm Hg at the same temperature. Calculate the molality of the solution. (0.77 m)

21. How many grams of urea (H2NCONH2) must be added to 450 g of water to give a solution with vapor pressure 2.50 mm Hg less then the pure water at 30.0 ºC? The partial pressure of water at 30.0 ºC is 31.80 mm Hg.

(126 g)

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Boiling Point Elevation and Freezing Point Depression

Nonvolatile solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

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Tb = Tb – T b0

Tb > T b0 Tb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

Tb = Kb m

m is the molality of the solution

Kb is the molal boiling-point elevation constant (0C/m)

Boiling Point Elevation

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Freezing-Point Depression

Tf = T f – Tf0

T f > Tf0 Tf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

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Colligative Properties of Nonelectrolyte Solutions

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

12.8

Vapor-Pressure Lowering P1 = X1 P 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

Molar Mass Calculations

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What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf water = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

12.8

Example: Colligative Properties

22. What mass of sucrose, C12H22O11, should be added to 75.0 g H2O to raise the boiling point to 100.35 0C?

23. Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile electrolyte. Calculate the b.p. and f.p. of a 25.0 mass percent solution of ethylene glycol in water. ( 102.8 ºC; -10.0 ºC)

24. List the following aqueous solutions in order of their expected freezing points (lowest to highest):

0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22 O11

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Determination of Molar Mass

25. A 7.85 g sample of a compound with empirical formula C5H4 is dissolved in 301 g of benzene. The freezing point of the solution is 1.05 ºC below that of pure benzene. What are the molar mass and molecular formula of this compound?

(127 g/mol, C10H8)

26. A solution of an unknown nonvolatile compound was prepared by dissolving 0.250 g of the substance in 40.0 g CCl4. The B.P of the resultant solution was 0.357 ºC higher than of the pure solvent. Calculate the molar mass of the solute. (88.0 g/mol)

Colligative Properties of Electrolyte Solutions

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Colligative Properties of Electrolytes

Since these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

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Colligative Properties of Electrolytes

However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does.

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van’t Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions.

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van’t Hoff Factor

Some Na+ and Cl− reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

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The van’t Hoff Factor

• Reassociation is more likely at higher concentration.• Therefore, the number of particles present is

concentration dependent.

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The van’t Hoff Factor

We modify the previous equations by multiplying by the van’t Hoff factor, i

Tf = Kf m i

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

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Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

Colligative Properties of Electrolyte Solutions

12.9

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12.8 Osmotic Pressure ()

12.8

Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution (high solvent concentration) to a more concentrated one (low solvent concentration).

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () is the pressure required to stop osmosis.

dilutemore

concentrated

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HighP

LowP

Osmotic Pressure ()

= RT = MRT

M is the molarity of the solution

R is the gas constant: 0.0821 L-atm/mol-K

T is the temperature (in K)12.8

V

n

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A cell in an:

Isotonic Solution: same conc.

hypotonicLess conc.

hypertonicMore conc.

12.8

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Osmosis in Blood Cells

• If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic.

• Water will flow out of the cell, and crenation results.

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Osmosis in Cells

• If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic.

• Water will flow into the cell, and hemolysis results.

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Osmotic Pressure: Example

27. The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine its molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml solution. The osmotic pressure of the solution at 25.0 ºC was found to be 1.54 torr. Calculate the molar mass of the protein. (8.45 x 103 g/mol)

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A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.

Colloid versus solution

• colloidal particles are much larger than solute molecules

• colloidal suspension is not as homogeneous as a solution

12.8

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The Cleansing Action of Soap

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Tyndall Effect

• Colloidal suspensions can scatter rays of light.

• This phenomenon is known as the Tyndall effect.

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Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (water-hating) end.

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Colloids in Biological Systems

Sodium stearate is one example of such a molecule.

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Colloids in Biological Systems

These molecules can aid in the emulsification of fats and oils in aqueous solutions.

The End