Physical Properties of

1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A is a homogenous of 2 or more substances

The solute is(are) the substance(s) present in the smaller amount(s)

The is the substance present in the larger amount

2 We can have:

Gaseous Solutions: gases can spontaneously mix in any proportion.

Liquid Solutions: liquid solutions are the most common and they can be obtained by dissolution of a gaseous, liquid or solid solute.

Solid Solutions: they are generally called alloys, where two or more metals are present. Mercury alloys are called amalgams, and they can be liquid or solid. Solid Solutions

• Ruby is a of

Cr2O3 in Corundum (Al2O3) • The variety called “pigeon blood” is one of the most precious stone.

• Sapphire (less precious) color blu is due to the Fe and Ti

impurities in the Al2O3

4 Ideal Solutions

 Ideal solutions have :

ΔHsol = 0  The formation of ideal solutions is always spontaneous:

ΔGsol < 0

© Dario Bressanini 5 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.

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

6 Three types of interactions in the solution process: • solvent-solvent interaction • solute-solute interaction • solvent-solute interaction

Molecular view of the formation of solution

ΔHsoln = ΔH1 + ΔH2 + ΔH3

7 “like dissolves like”

Two substances with similar intermolecular forces are likely to be soluble in each other.

• non-polar molecules are soluble in non-polar

CCl4 in C6H6 • polar molecules are soluble in polar solvents

C2H5OH in H2O • ionic compounds are more soluble in polar solvents

NaCl in H2O or NH3 (l)

8 Two driving forces are the basis of the solution formation ( of two substances):

1. Disorder tendency (entropic factor). It is the only factor presents in the case of (ideal) gases, which can be mixed in any proportion.

2. Intermolecular attraction among the molecules of the two substances (energetic factor).

If the attractions A-A and B-B are stronger than those A-B, the two substances will not have the tendency to mix.

The solubility of a solute in a solvent depends on the balance among these two factors. Molecular Solutions In this case the solute (liquid or solid) is a molecular species, characterized by low intermolecular forces. In the case of liquids the two compounds are soluble if these forces are similar, or if the heterogeneous interactions are stronger than the homogeneous one. -acetone

C7H16 – C8H18

London Forces

Hydrogen Bonds Ionic Solutions

In this case the solute is a ionic solid, and polar solvents are able to overcome the strong ionic bonds. The factors influencing the solution process of a ionic solid are:

- the lattice energy of the solid (sum of the attraction of anions and cations): bigger this energy, lower the tendency of the solid to dissolve in the liquid phase

- the attraction energy of dipole-ion among the ions of the solute and the electric dipoles of the solvent molecules: bigger this energy, bigger the tendency to the solution formation NaCl in water: NaCl(s) → Na+(aq) + Cl-(aq)

+

- - - + + + - + - + - - - + - + - + - + - +

IONIC CRYSTAL SOLVATED IONS NaCl(s) → Na+(aq) + Cl-(aq)

This process is called hydration and the + electrostatic interaction energy of the ion with water molecules is called hydration energy.

The solubility of a ionic solid - depends on the balance among lattice and hydration energy.

SOLVATED IONS Bigger the lattice energy of ionic compounds, lower will be

their and vice versa. The lattice energy depends on both the ionic charges and their distances :

- bigger the ionic charges, bigger will be the lattice energy - higher the ionic distances (bigger ions), lower will be the lattice energy

The situation is complicated by the fact that the hydration energy is also bigger for small and highly charged ions. In general the lattice energy prevails, so we can predict that:

- solids formed by ions with a single charge and big + + dimensions (K , NH4 ) are generally soluble - solids formed by ions with two a three charges and 2- 3- small dimensions (S PO4 ) are generally not soluble. Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

15 Temperature and Solubility

O2 gas solubility and temperature

solubility usually decreases with increasing temperature

16 Pressure and Solubility of Gases

The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).

c is the (M) of the dissolved gas c = kP P is the pressure of the gas over the solution k is a constant for each gas (mol/L•atm) that depends only on temperature

low P high P

low c high c 17 Molecular view: bigger the partial pressure of the gas, higher the number of the gas molecules colliding the liquid surface and consequently dissolved into the liquid phase Henry Law

pA = xAK A

• Knowing the Henry constants is important for a wide range of applications

Gas (in H2O) K/(10 Mpa) CO2 0.167  Carbon dioxide dissolves H2 7.12 very well in water N2 8.68 O2 4.40 19 Henry Law

• At high pressure molecular Nitrogen and Oxygen are dissolved in the blood. • Emerging too quickly can induce the formation of emboli

20 Blood Solubility

 At high pressure molecular Nitrogen and Oxygen are dissolved in the blood.  Molecular Oxygen can be consumed, but Nitrogen will remain in the blood. Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass mass of solute % by mass = x 100% mass of solute + mass of solvent mass of solute = x 100% mass of solution

Mole Fraction (X) moles of A XA = sum of moles of all components 22 Concentration Units Continued Molarity (M)

moles of solute M = liters of solution

Molality (m)

moles of solute m = mass of solvent (kg)

23 Diluted solutions Addition of solvent

M1V1 = M2V2 The moles of solute are constant

Mixing of two solutions

M1V1 + M2V2 = MfVf = Mf(V1 + V2)

24 Vapor Pressure and Ideal Solutions

pA = xApA*

 The solutions following the Raoult law are called Ideal Solutions Francois Raoult (1830-1901)  These solutions have

ΔsolH =0

25 Raoult law • Assuming that the solvent- solvent and solute-solute interactions are identical to those solute-solvent, we can conclude that the vapor pressure is:

pA = xApA*

26 Solution of two volatile liquids

Both liquids have a vapor pressure

27 Raoult law for volatile liquids – The vapor pressure of each component can be calculated by the Raoult law – The total pressure is the sum of both partial pressures.

pA = χA pA* pB = χB pB*

ptot = pA + pB = χA pA* + χB pB*

28 Vapor-Pressure of a solution

P = X P 0 A A A 0 PB = XB P B

PT = PA + PB 0 0 PT = XA P A + XB P B

Raoult’s law

29 Ideal and not Ideal Solution

ptot = χA pA* + χB pB* 30 Not Ideal Solutions

• The not ideal behavior is the common case for real solutions • A and B interactions are different from those AA and BB

Positive deviation Negative deviation 31 POSITIVE DEVIATION FROM THE RAOULT LAW

The solution shows a vapor pressure higher than that calculated by the Raoult law; The cohesion forces A-A or B-B are stronger than the adhesion forces A-B

Examples of these solutions are: -water, isopropanol-water, propanol-water NEGATIVE DEVIATIONS FROM RAOULT LAW

The solution shows a vapor pressure lower than that calculated by the Raoult law; The cohesion forces A-A or B-B are weaker than the adhesion forces A-B

Examples of these solutions are: pyridine-water, -water, hydracid-water DISTILLATION OF IDEAL SOLUTIONS OF TWO LIQUIDS

Distillation: separation process based on the vaporization of a liquid phase by heat application and subsequent condensation by cooling the vapor in a container different from that where the vaporization was carried out.

When an ideal solution of two liquids is heated to the boiling point it is possible to obtain a curve similar to that reported below, where for each composition the vapor is more rich of the more volatile component, while the liquid is more rich of the other component.

These diagrams are in the isobaric (constant pressure) form. The two curves represent the composition of the liquid (blue) and the vapor (red) phases, in equilibrium at a certain temperature. The blue line is called ebullition curve, the red one condensation curve. Fractional Distillation

In this technique a column (fractioning column) is posed among the vaporization and condensation site. In this column the liquid-vapor phase change is repeated several times, thanks to the condensation over the cold surface of the column and the heat transferred by the hot vapor fluxing on the column. These liquid/vapor equilibria increase the efficiency of the liquids separation. From the figure it is possible to observe this phenomenon. DISTILLATION OF NON IDEAL SOLUTIONS OF TWO LIQUIDS - 1

Solutions having positive deviations from the Raoult law show the diagram reported below; In this case we observe a solution vapor pressure higher than that expected.

The distillation diagram shows a minimum called minimum azeotrope.

In this point liquid and vapor have the same composition and the azeotrope behaves as a pure liquid.

In this case it is not possible to separate the two components by distillation, because we obtain always the azeotrope as the distillate (and one of the two components as the residue). DISTILLATION OF NON IDEAL SOLUTIONS OF TWO LIQUIDS - 2

Solutions having negative deviations from the Raoult law show the diagram reported below; In this case we observe a solution vapor pressure higher than that expected.

The distillation diagram shows a minimum called maximum azeotrope.

In this point liquid and vapor have the same composition and the azeotrope behaves as a pure liquid.

In this case it is not possible to separate the two components by distillation; we will obtain always the azeotrope as the residue (and one of the two components as the distillate). DISTILLATION OF A MIXTURE OF TWO NOT SOLUBLE LIQUIDS

Two not soluble (ot only partially soluble) liquids form a heterogeneous mixture. In the limit case of a totally not soluble liquids, no interactions are present among the two components, and the vapor pressure is the sum of the vapor pressure of the two pure liquids:

0 0 P = P A + P B

When this mixture is heated, the vapor pressure increases and the mixture boils when the external pressure is reached. The boiling temperature of the mixture is consequently lower than that of the pure components and it does not depend on the mixture composition; the distillation produces a vapor having a constant composition (heteroazeotrope), until one component is totally consumed. This process is exploited in the steam distillation, where a high-boiling liquid not soluble in water can be distilled at a temperature lower than than the boiling point of water. Distillation under vacuum

The vacuum application allows to decrease the temperature of the boiling point; this technique is exploited for compound having boiling point over 200 °C.

The effect of the boiling point decrease can be extimated by using the diagram reported below. DISTILLATION OF SOLUBLE AND NOT SOLUBLE LIQUIDS

Soluble Liquids:

Ptotal = PA° NA + PB°NB Raoult’s law

Distillation Techniques: simple distillation; fractioned distillation

Not soluble Liquids:

Ptotal = PA° + PB° Dalton’s law

Distillation Techniques: steam distillation

The boiling point occurs when Pext = Ptot

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering 0 P 0 = vapor pressure of pure solvent P1 = X1 P 1 1

Raoult’s law X1 = of the solvent

If the solution contains only one solute:

X1 = 1 – X2 0 0 P - P1 = ΔP = X2 P X2 = mole fraction of the solute 1 1 41

Boiling-Point Elevation ΔTb = Kb m

Freezing-Point Depression ΔTf = Kf m

For dilute solutions we will demostrate that:

ΔTb= Tb(solution) - Tb(solvent) = Kb m

ΔTf= Tf(solvent) - Tf(solution) = Kf m

Kb, ebullioscopic constant, and Kf, cryoscopic constant, depend only on the solvent.

They have °C/m units. Boiling-Point Elevation The lowering of vapor pressure, due to the addition of a non volatile solute, makes necessary an increase of T, to allow the solution to reach again the boiling point. The first factor, ΔP (vapor pressure lowering), is obtained from the Raoult law, while the second, ΔT (boiling point elevation), is obtained from the Clapeyron equation. From the Raoult law:

P = P°xa

lnP = lnP° + lnxa

dlnP = dlnxa = dln(1-xb) ≅ -dxb

dlnP = dP/P and then dP = -Pdxb 44 From the Clapeyron equation:

2 dP/dT = PΔHev/RT

and then: dP = PΔHevdT 2 RT Using both expression of dP:

-Pdxb + PΔHevdT = 0 RT2 quindi: dxb = ΔHevdT RT2 Integrando

xb = ΔHev (T-Teb) ≅ ΔHevΔT R (TxT ) RT 2 45 eb eb

Xb ≅ m = ΔHevΔT n RT 2 s eb then:

2 Keb = RTeb nsΔHev and:

ΔT = Kebm

46 Boiling-Point Elevation 0 ΔTb = Tb – T b 0 T b is the boiling point of the pure solvent

T b is the boiling point of the solution 0 Tb > T b ΔTb > 0

ΔTb = Kb m

m is the of the solution

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

for a given solvent 47 Freezing-Point Depression

0 ΔTf = T f – Tf 0 T f is the freezing point of the pure solvent

T f is the freezing point of the solution 0 T f > Tf ΔTf > 0

ΔTf = Kf m

m is the molality of the solution

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

for a given solvent 48 49 What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The of ethylene glycol is 62.01 g.

o ΔTf = Kf m Kf water = 1.86 C/m 1 mol 478 g x moles of solute 62.01 g m = = = 2.41 m mass of solvent (kg) 3.202 kg solvent

o o ΔTf = Kf m = 1.86 C/m x 2.41 m = 4.48 C 0 ΔTf = T f – Tf 0 o o o Tf = T f – ΔTf = 0.00 C – 4.48 C = -4.48 C

50 Osmotic Pressure (π) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (π) is the pressure required to stop osmosis.

more dilute concentrated

51 Osmotic Pressure

When the equilibrium is reached, if a mole of solvent passes through the membrane, the variation of the free energy ΔG should be zero,and consequently both term influencing the free energy, (concentration and pressure variations) should be equal:

ΔGconc + ΔGpress = 0

then: ΔGconc= RTlnP = RTlnxa P° ΔGpress = VmΔP

Join both terms: VmΔP + RTlnxa = 0 52

VmΔP = -RTlnxa

Considering that ΔP is the osmotic pressure π:

Vmπ = -RTlnxa = -RTln(1-xb) = RTxb

πVm = RT n ns

From the expression we can obtain the osmotic pressure equation:

πV = nRT or

π = MRT 53

Osmotic Pressure (π)

time solvent solution

High Low P P

π = MRT

M is the molarity of the solution R is the gas constant T is the temperature (in K) 54 A cell in an:

isotonic hypotonic hypertonic solution solution solution 55 Colligative Properties

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

0 Vapor-Pressure Lowering P1 = X1 P 1

Boiling-Point Elevation ΔTb = Kb m

Freezing-Point Depression ΔTf = Kf m

Osmotic Pressure (π) π = MRT

56 Colligative Properties of Solutions

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

0.1 m NaCl solution 0.2 m ions in solution

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

i should be nonelectrolytes 1 NaCl 2

57 CaCl2 3 Colligative Properties of Electrolyte Solutions

Boiling-Point Elevation ΔTb = i Kb m

Freezing-Point Depression ΔTf = i Kf m

Osmotic Pressure (π) π = iMRT

58