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Physical Properties of Solutions

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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A solution is a homogenous mixture of 2 or more substances

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

The solvent is the substance present in the larger amount

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.

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Solid Solutions

•  Sapphire (less precious) color blu is due to the Fe and Ti impurities in the Al2O3

•  Ruby is a solid solution of Cr2O3 in Corundum (Al2O3)

•  The variety called “pigeon blood” is one of the most precious stone.

© Dario Bressanini 5

Ideal Solutions

  Ideal solutions have : ΔHsol = 0

 The formation of ideal solutions is always spontaneous:

ΔGsol < 0

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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.

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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

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“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 solvents

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)

Two driving forces are the basis of the solution formation (solubility 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 energy 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.

C7H16 – C8H18

London Forces

Chloroform-acetone

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 energies 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

IONIC CRYSTAL SOLVATED IONS

NaCl(s) → Na+(aq) + Cl-(aq)

- + + - - +

+ - + + - -

+ - + + - -

-

- +

+

-

NaCl in water:

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.

Bigger the lattice energy of ionic compounds, lower will be their solubilities 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 small dimensions (S2- PO4

3-) are generally not soluble.

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Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

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Temperature and Solubility

O2 gas solubility and temperature

solubility usually decreases with

increasing temperature

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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 = kP c is the concentration (M) of the dissolved gas

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

low c

high P

high c

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

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Henry Law

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

AAA Kxp =

Gas (in H2O) K/(10 Mpa) CO2 0.167 H2 7.12 N2 8.68 O2 4.40

  Carbon dioxide dissolves very well in water

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Henry Law

•  At high pressure molecular Nitrogen and Oxygen are dissolved in the blood.

•  Emerging too quickly can induce the formation of emboli

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.

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Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

Percent by Mass

% by mass = x 100% mass of solute mass of solute + mass of solvent

= x 100% mass of solute mass of solution

Mole Fraction (X)

XA = moles of A sum of moles of all components

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Concentration Units Continued

M = moles of solute

liters of solution

Molarity (M)

Molality (m)

m = moles of solute

mass of solvent (kg)

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Diluted solutions Addition of solvent

M1V1 = M2V2 The moles of solute are constant

Mixing of two solutions

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

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Vapor Pressure and Ideal Solutions

 The solutions following the Raoult law are called Ideal

Solutions  These solutions have

ΔsolH =0

pA = xApA*

Francois Raoult (1830-1901)

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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*

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Solution of two volatile liquids Both liquids have a vapor pressure

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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*

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PA = XA P A 0

PB = XB P B 0

PT = PA + PB

PT = XA P A 0 + XB P B 0

Ideal Solution

Vapor-Pressure of a solution

Raoult’s law

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Ideal and not Ideal Solution

ptot = χA pA* + χB pB*

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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

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: ethanol-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, acetic acid-water, hydracid-water

DISTILLATION OF IDEAL SOLUTIONS OF TWO LIQUIDS

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.

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.

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:

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.

P = P0A + P0

B

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

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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.

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 1 0 - P1 = ΔP = X2 P 1 0

P 1 0 = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

P1 = X1 P 1 0

For dilute solutions we will demostrate that:

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

They have °C/m units.

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

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

Boiling-Point Elevation ΔTb = Kb m

Freezing-Point Depression ΔTf = Kf m

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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

Boiling-Point Elevation

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From the Clapeyron equation:

dP/dT = PΔHev/RT2

and then: dP = PΔHevdT

RT2

Using both expression of dP:

-Pdxb + PΔHevdT = 0 RT2

quindi: dxb = ΔHevdT

RT2

Integrando

xb = ΔHev (T-Teb) ≅ ΔHevΔT R (TxTeb) RTeb

2

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Xb ≅ m = ΔHevΔT ns RTeb

2

then: Keb = RTeb

2

nsΔHev and: ΔT = Kebm

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Boiling-Point Elevation

ΔTb = Tb – T b 0

Tb > T b 0 ΔTb > 0

T b is the boiling point of the pure solvent

0

T b is the boiling point of the solution

ΔTb = Kb m

m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) for a given solvent

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Freezing-Point Depression

ΔTf = T f – Tf 0

T f > Tf 0 ΔTf > 0

T f is the freezing point of the pure solvent

0

T f is the freezing point of the solution

ΔTf = Kf m

m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) for a given solvent

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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.

ΔTf = Kf m

m = moles of solute

mass of solvent (kg) = 2.41 m =

3.202 kg solvent

478 g x 1 mol 62.01 g

Kf water = 1.86 oC/m

ΔTf = Kf m = 1.86 oC/m x 2.41 m = 4.48 oC

ΔTf = T f – Tf 0

Tf = T f – ΔTf 0 = 0.00 oC – 4.48 oC = -4.48 oC

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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.

dilute more concentrated

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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

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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

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High P

Low P

Osmotic Pressure (π)

π = MRT

M is the molarity of the solution R is the gas constant T is the temperature (in K)

solvent solution

time

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A cell in an:

isotonic solution

hypotonic solution

hypertonic solution

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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.

Vapor-Pressure Lowering P1 = X1 P 1 0

Boiling-Point Elevation ΔTb = Kb m

Freezing-Point Depression ΔTf = Kf m

Osmotic Pressure (π) π = MRT

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Colligative Properties of Electrolyte 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

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

number of formula units initially dissolved in soln

nonelectrolytes NaCl CaCl2

i should be

1 2 3

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Boiling-Point Elevation ΔTb = i Kb m

Freezing-Point Depression ΔTf = i Kf m

Osmotic Pressure (π) π = iMRT

Colligative Properties of Electrolyte Solutions