Chapter 13

Physical Properties of Colloids and Solutions

 

There are three types of mixtures including suspensions, colloids, and solutions. One of the main features that differentiate between these mixtures is the size of particles dispersed.

 

Suspensions

 

In this case, large particles are dispersed or suspended throughout a medium. For instance, sand dispersed in a solvent like water, snow flakes dispersed in air, or precipitated particles dispersed in a solvent are all examples of suspensions. In all these cases, suspended particles are large enough to be seen by naked eye or an ordinary microscope; and will settle with time. Suspended particles can be separated from the solvent by simple filtration or through centrifugation. It is important to indicate that the physical properties of suspensions, like boiling and freezing points as well as vapor pressure, will be little affected by the suspended particles since the number of these particles is extremely small as compared to the number of solvent molecules.

 

Solutions

 

Molecules or ions are dissolved in a solvent where the molecular or ionic sizes of solutes are very small and can not be seen by excellent microscopes. Solutes form homogeneous mixtures with solvents where solute molecules are uniformly dispersed in the solvent. It is important to indicate that the physical properties of solutions will be affected by the presence of solutes.

 

Colloids

 

When the particle size of solutes is intermediate between suspensions and solutions (1-1000 nm), the mixture formed from such solutes is called a colloidal suspension or dispersion. Usually, in colloids, the terms dispersed phase and dispersing medium replace the solute and solvent, respectively. Milk, smoke and mist are examples of colloidal dispersions. Solutions containing proteins are also colloidal suspensions. An important property of colloidal dispersions is that they are stable in the colloidal state for a long time and they show Tyndall effect. Scattering of visible monochromatic light (Tyndall effect) can be used to identify the existence of a colloidal suspension as well as determine the shape of colloidal particles.

 

Stability of Colloidal Dispersions

 

For a colloidal solution to be stable, dispersed particles should be prevented from sticking together when they collide. For a liquid in liquid colloidal solution (emulsion), an emulsifying agent is used. An emulsifying agent is a molecule with polar head and non polar tail. The polar head orients toward the polar constituent of the solution (dispersion medium or dispersed phase) while the hydrophobic tail orients toward the non polar constituent.

 

 

 

 

 

Surfactant

Reversed Micelle

Normal Micelle

 

 

An example of an emulsifying agent is the casein in milk, which prevents fine fat droplets from coalescing.

 

Colloids of solids in liquids (sols) are stabilized by adsorption of ions on the surface of the solid resulting in particles carrying similar charges which tend to repel each other, thus stabilizing the solution. An example is a colloidal solution of Fe2O3.xH2O where Fe3+ ions are adsorbed on the surface of solid Fe2O3.xH2O and prevent sticking

 

Colloidal solutions of solids in gas (aerosols) are stabilized by picking static charges from air constituents, and are thus stabilized.

 

Destabilizing Colloids

 

 Countering the factors stabilizing the colloid can destabilize colloids. For example, addition of PO43- ions to the iron oxide sol will force the particles to coagulate as the excess Fe3+ ions are now surrounded with phosphate ions which neutralize the charge on the particles. Generally, two factors are effective in destabilizing colloids:

 

Ø      Heating the solution can force charged particles to collide and stick together when the kinetic energy is larger than the repulsion forces. Forming larger particles (lower surface area) results in lower energy.

Ø      Addition of a strong electrolyte will shield the charge on colloidal particles and thus they can collide and coagulate, as in the phosphate-iron oxide case.

 

Types of Solutions

 

There are several types of solutions including:

Ø      A solute dissolved in a liquid. The most common of which is dissolving a solute in water and most of our study in the coming chapters will use this type of solution.

Ø      Dissolving a solid solute in a solid can form a solid solution. An example is alloys; where two types of solid solutions can be identified:

ü      Substitutional solid solutions can be formed when an atom or molecule of a solute takes the place of an atom or molecule in the crystal lattice of a solid.

ü      Interstitial solid solutions are formed when the small solute atoms or molecules are placed within the voids or interstices of the solid. Usually, interstitial solid solutions are dense and can be very strong, like tungsten carbide.

 

Concentration Units

 

In addition to molarity (number of moles of solute dissolved in one liter of solution), other units can be defined:

 

q       Mole Fraction

The number of moles of a particular component to the overall number of moles of everything in solution is called the mole fraction of that component.

For example, a solution having 1.0 mole of acetone, 2.0 moles of ethanol, 3.0 moles of methanol, and 15.0 moles of water will have:

Xacetone = 1.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.05

Xethanol = 2.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.10

Xmethanol = 3.0/(1.0 + 2.0 + 3.0 + 14.0) = 0.15

Xwater = 14.0/(1.0 + 2.0 + 3.0 + 15.0) = 0.70

XT = 0.05 + 0.10 + 0.15 + 0.70 = 1.0

 

q       Mole Percent

The mole fraction multiplied by 100 is called the mole percent. In the above example:

Mol % acetone = 0.05 x 100% = 5%

Mol % ethanol = 0.10 x 100% = 10%

Mol % methanol = 0.15 x 100% = 15%

Mol % water = 0.70 x 100% = 70%

 

q       Weight Fraction

The weight of a particular component to the total number of grams of solution is called the weight fraction of that component.

Wcomponent = g component/g solution

For example, a solution containing 12.0 g methanol and 38.0 g water has:

wmethanol = 12.0/(12.0 + 38.0) = 0.24

wwatrer = 38.0/(12.0 + 38.0) = 0.76

 

q       Weight Percent

The weight fraction multiplied by 100 is called the weight percent. In the example above:

Wt% Methanol = 0.24 x 100% = 24%

Wt% Water = 0.76 x 100% = 76%

 

Note that summation of mole or weight percent of all components will add up to 100%. Also, summation of mole or weight fractions of all components will add up to 1.0.

 

Molality

 

Molality is defined as the number of moles of solute dissolved in one kilogram of solvent. Therefore, the volume of one molal solution can be more or less than 1.0 L. Molality is not concerned with volume. You should be able to contrast molality and molarity and be able to use molality in calculations as well as conversion to other concentration units as mentioned earlier.

 

Example

A certain aqueous solution contains 7% ethanol (C2H5OH) by mass. Calculate the mole fraction, mole percent, and the molality

Solution
  • 7% by mass would mean that we have 7g ethanol in a 100g sample (and, therefore, 93g H2O)
  • Molar mass of ethanol is (2 x 12.0) + (6 x 1.01) + (1 x 16.0) = 46.1g/mole
  • Molar mass of H2O is (2 x 1.01) + (1 x 16.0) = 18.0g/mole

Number of moles ethanol = 7/46.1 = 0.152 mol

Number of moles of water = 93/18.0 = 5.17 mol

  • Total moles in sample is (0.152 + 5.17) = 5.32
  • Xethanol = 0.152/5.32 = 0.0286

Xwater  = 1 – 0.0286 = 0.9714

Mol% ethanol = 0.0286x100% = 2.9%

Mol% water = 0.9714x100% = 97.1%

Molality:

  • In the same 100g sample above, the ethanol 0.152 mol and the H2O weighs 93g (or 0.093 kg), therefore:

molality = 0.152mol/0.093 kg = 1.63 m

 

Example

 

A MgSO4 (FW= 120.4 g/mol) aqueous solution has a weight fraction of 0.2. What is the molality of the solution?

 

Solution

 

A 0.2 weight fraction means that the solution contains 20 g MgSO4 and 80 g water (0.080 kg).

Mol MgSO4 = 20 g/(120.4 g/mol) = 0.166 mol

Molality = 0.166 mol/0.080 kg = 2.08 m

 

Example

 

The mole fraction of benzene in a benzene/chloroform solution is 0.45. What is the mole fraction of chloroform (FW = 119.4) and the weight percent of benzene (C6H6, FW = 78.1)?

 

Solution

 

Since the solution contains benzene and chloroform only, the mole fraction of chloroform can be calculated:

Xchloroform = 1 – 0.45 = 0.55

Assuming a total number of moles equal 1.0, then:

g C6H6 = 0.45 mol x (78.1 g/mol) = 35.1 g

g Chloroform = 0.55 mol x (119.4 g/mol) = 65.7 g

 

Wt% C6H6 = 35.1/(35.1 + 65.7) x 100% = 34.8%

 

Example

 

A CaCl2 (FW = 111.0 g/mol) solution is 4.57 m. Find the mole fraction of CaCl2 and water in solution.

 

Solution

 

4.57 m means 4.57 mole CaCl2 in 1000 g of water

Number of moles of water = 1000 g/(18.0 g/mol) = 55.6 mol

XCaCl2 = 4.57 mol/(4.57 + 55.6) = 0.0759

Xwater  = 55.6/(4.57 + 55.6) = 0.9421

 

Example

 

Calculate the weight percent of formic acid solution (FW = 46.03 g/mol) that is 1.099 M. The density of the solution is 1.0115 g/mL.

 

Solution

 

Assume 1000 mL of solution. We have 1.099 mol formic acid:

g formic acid = 1.099 mol * (46.03 g/mol) = 50.95 g

 

Weight of solution = 1000 mL * 1.0115 g/mL = 1011.5 g

Wt% formic acid = {50.95/1011.5}* 100% = 5.001%

 

Energy and Disorder in the Formation of Solutions

 

Two factors are important to consider when explaining the solubility of substances in liquids. These factors are:

 

v     Entropy

 

A process tends to occur spontaneously (preferred) when the disorder or randomness increase. This is why a gas confined in a container will rapidly diffuse out when the container is open and a drop of ink will diffuse in a solution till the concentration in all regions is the same. This is a natural tendency  where solutes tend to dissolve in order to increase randomness or entropy.

 

v     Intermolecular Attractions

 

Benzene does not dissolve in water while methanol is freely soluble in water. Why is this behavior? To answer this and similar questions, we should consult our knowledge about intermolecular forces. First, let us have a close look at the solubility process:

§       Solvent molecules should be expanded so that a space for solutes is created. In order to expand solvent molecules, we need to input energy in order to overcome the intermolecular forces

§       Solute molecules should also be expanded and we need to input energy to overcome intermolecular forces between solute molecules

§       The solubility of expanded solute molecules in solvent will release energy. The released energy should, at least, be of the same magnitude as the input energy for expanding both solute and solvent molecules so that a solute may dissolve in a solvent.

 

Now, consider the case of benzene and why it does not dissolve in water. A high input energy is needed to expand water molecules since there are the relatively strong hydrogen bonds. It seems that these hydrogen bonds will be broken so that dispersion forces are formed between water and benzene molecules. This is an unfavorable process since we need to input much higher energy for solute expansion but will gain very little as dispersion forces are formed. Therefore, benzene is not expected to dissolve in water and the two solutions are said to be immiscible.

On the other hand, methanol dissolves freely in water since water and methanol have same type of intermolecular attractions (hydrogen bonding) and when hydrogen bonds in the solvent are broken, new hydrogen bonds with the solute are formed. This is  a feasible process and such solutions are very stable.

When the hydrocarbon chain becomes longer and longer, the alcohol becomes less soluble (miscible) in water since the alcohol size becomes larger which requires breaking several hydrogen bonds to accommodate one solute molecule; an unfavorable process. Methanol, ethanol and propanol are all completely miscible with water while octanol is not. On the other hand miscibility decreases from butanol to heptanol.

 

It could be fairly concluded that polar solutes will dissolve in polar solvents and vice versa. In another statement, one can say that the intermolecular forces in both solute and solvent should be alike for the solute to dissolve in the solvent. A non polar solute like I2 has only dispersion forces as the intermolecular forces between I2 molecules and will thus dissolve in a non polar solvent (has dispersion forces) like CCl4 but will not dissolve appreciably in water.

 

Ionic solids are highly polar and will only dissolve in very polar solvents like water. Even polar solvents like ethanol do not have enough polarity to dissolve ionic solutes. Ions in water become hydrated where water molecules form a layer of water around each ion; thus decreasing the available charge, which adjacent ions can feel.

 

an ionic crystal dissolving in water

 

 

Heats of Solutions

 

When solutes dissolve in solvents, the solution process either releases or absorbs energy. The amount of heat that is absorbed or released when a solute is dissolved in a solvent is called the heat of solution (DHsoln).

DHsoln is negative when the solution process releases heat (exothermic, like solubility of LiCl, LiI, AlCl3, and Al2(SO4)3 in water) and is a positive value when the solution absorbs heat (as for the solution of KCl, KBr, NH4Cl, and NH4NO3 in water). DHsoln = 0 in the case where no heat is absorbed or released (Hsoln­  = Hsolutes, like the case of solubility of benzene in carbon tetrachloride). The magnitude of DHsoln provides information about relative intermolecular forces of solute, solvent, and solution. When DHsoln = 0, the solution is referred to as an ideal solution. Two points could be made:

v     The solution process will be exothermic when the intermolecular forces between solute-solvent molecules are stronger than solute-solute or solvent-solvent molecules.

v     An endothermic solution process is observed when the intermolecular forces between solute-solute and solvent-solvent molecules are stronger than solute-solvent molecules.

 

§         The solubility process of liquids or nonionized solids in liquid solvents involves the following steps:

§         Expanding solute molecules requires an input of energy equals DHsolute to overcome potential energy.

§         Expanding solvent molecules requires an input of energy equals DHsolvent to overcome potential energy.

§         Combining expanded solute and solvent molecules will release energy equals DHcombination, due to decrease in potential energy (attraction occurs) 

DHsoln = DHcombination – (DHsolvent  + DHsolute)

 

This can be schematically represented by the following graphics:

  • For the solubility of solids in liquids, imagine a path where the solid solute is first evaporated into gaseous ions, which requires an input of energy equals DH1 (which corresponds to potential energy needed to overcome lattice energy). The second step will be the hydration of the ions by water molecules, which releases energy DHhydration. Three situations can be encountered:

 

ü      When DHhydration = DH1 , this is an ideal solution

ü      When DHhydration > DH1, this is an exothermic process

ü      When DHhydration < DH1, this is an endothermic process

 

Schematically, these processes can be presented as below:

 

 

Solubility and Temperature

 

For most solids and liquids dissolved in water, the solubility increases as the temperature is increased. This means that the equilibrium shifts towards forming more concentrated solutions at higher temperatures. At any temperature, a solute dissolves to the extent where the equilibrium at that definite temperature is reached.

In the case of the solubility of gas solutes in liquids, usually the solubility decreases as the temperature is increased. This is because the solubility of gases in liquids is almost always exothermic. According to Le Chatelier principle solubility will thus be inversely related to temperature.

 

Fractional Crystallization

 

Solubility of substances is affected differently with a rise in temperature. Some substances will increase their solubility in very fast rates, like NH4NO3 while others will respond very slowly to temperature changes, like NaCl. The differential effect of temperature on solubility of different solutes is successfully used for the purification of solutes by fractional crystallization. The technique involves dissolving the sample in a small amount of a hot solvent in which the desired solute is less soluble. Upon slowly cooling the solvent, solute crystals start to separate while impurities stay in solution. The crystals of the solute are then collected by filtration.

 

Effect of Pressure on Solubility

 

Solubility of solids and liquids in liquid solvents are usually not affected by changes in pressure. However, solubility of gases in liquid solvents is highly dependent on pressure. A quantitative assessment of the effect of pressure on solubility of gases can be clearly shown by Henry's law, which states that: The solubility of a gas (Cg) in a liquid is directly proportional to the pressure of the gas above the solution.

Cg = kg Pg

kg  is the Henry's law constant. This relation is useful in calculation of the solubility of a gas at some pressure provided that the solubility at some other pressure is known:

Cg1 / Cg2 = Pg1 / Pg2

 

Example

 

At 25 oC, O2 gas collected over water at a total pressure of 1.0 atm is soluble to the extent of 0.0393 g/L. What would its solubility be if its partial pressure over water was 800 torr. Pwater = 23.8 torr at 25 oC

 

 

Solution

 

PT = PO2 + Pwater

760 torr = PO2 + 23.8

PO2 = 736 torr

 

Substitution in the equation Cg1 / Cg2 = Pg1 / Pg2 gives:

0.0393/CO2 = 736/800

CO2 = 0.0427 g/L

 

Colligative Properties of Solutions

 

A property that depends only on the relative amounts of solute and solvent is called a colligative property. Colligative properties are physical properties where, depending on the amount of solute relative to solvent, would reflect a change in some physical properties. Physical properties which will be studied will include vapor pressure, boiling point, melting point, as well as osmotic pressure.

 

v     Vapor Pressures of Solutions

 

The presence of solutes in liquid solutions has great effects on the physical properties of the solution. The boiling point, freezing point, as well as the vapor pressure of the solution will vary depending on the amount and nature of dissolved solute. Raoult's law discusses the effects of solutes on the vapor pressure of solutions provided that these solutes are non dissociable (non ionics). Raoult,s law states the vapor pressure of solution (Psoln) is directly proportional to the mole fraction of the solvent. This can be represented by the relation:

 

Psoln = xsolvent * Posolvent

Where xsolvent  is the mole fraction of the solvent and Posolvent is its vapor pressure. Therefore, it is obvious that:

 

Ø      Dissolving a nonvolatile solute will result in a decrease in the vapor pressure of the solution.

Ø      When the vapor pressure of the solute equals the vapor pressure of the solvent, no change in the vapor pressure of the solution would be observed.

Ø      When the vapor pressure of the solute is more than the vapor pressure of the solvent, the vapor pressure of the solution would be increased.

 

Example

 

10.0 g of a paraffin (FW = 282 g/mol), a nonvolatile solute, was dissolved in 50.0 g of benzene (C6H6, FW = 78.1 g/mol). At 53 oC, the vapor pressure of pure benzene is 300 torr. What is the vapor pressure of the solution at that temperature?

 

Solution

 

nsolvent = 50.0 g/(78.1 g/mol) = 0.640

nsolute = 10.0 g/(282 g/mol) = 0.035

xsolvent = 0.640/(0.640 + 0.035) = 0.948

Psoln = xsolvent * Posolvent

Psoln = 0.948 * 300 torr = 284 torr

 

Solutions with more than One Volatile Component

 

When solutions contain more than one volatile component, the vapor pressure of the solution is the sum of the partial vapor pressures of all components. Assume a situation where a solution is composed from three volatile components A, B, and C, the vapor pressure of the solution is:

Psoln = xA * PoA + xB * PoB + xC * PoC

 

Example

 

A solution containing 50.0 g of CCl4 (FW = 153.8 g/mol) and 50.0 g of CHCl3 (FW = 119.4 g/mol). At 50 oC, if the vapor pressure of pure CCl4 and CHCl3 is 317 and 526 torr, respectively, find the partial pressure of each substance and the vapor pressure of the solution.

 

Solution

 

nCCl4 = 50.0 g /(153.8 g/mol) = 0.325 mol

nCHCl3 = 50.0 g /(119.4 g/mol) = 0.419 mol

xCCl4 = 0.325/(0.325 + 0.419) = 0.437

xCHCl3 = 1 – 0.437 = 0.563

PCCl4 = xCCl4* PoCCl4

PCCl4 = 0.437 * 327 torr = 139 torr

 

PCHCl3 = xCHCl3* PoCHCl3

PCHCl3 = 0.563 * 526 torr = 296 torr

 

Psoln = PCCl4 + PCHCl3

Psoln = 139 torr + 296 torr = 437=5 torr

 

The Psoln can also be calculated directly form the relation below without calculation of partial pressures in  separate steps.

 

Psoln = xA * PoA + xB * PoB

Psoln = 0.437 * 317 torr + 0.563 * 526 torr = 435 torr

It should be emphasized here that Raoult's law is valid for dilute solutions only while concentrated solutions usually show deviations from that law. Solutions obeing Raoult's law are referred to as ideal solutions while non ideal solutions show either a negative or positive deviations from raoult's law. Three situations can be stated:

 

Ø      The situation where the intermolecular forces between solute-solute and solvent solvent are of the same strength (DHsoln = 0). No deviations from Raoult's law as this is an ideal solution (like benzene-carbon tetrachloride solution).

Ø      The situation where the intermolecular forces between solute- solvent molecules are larger than solute-solute and solvent-solvent intermolecular forces (DHsoln = -ve, exothermic process). Negative deviations from Raoult's law is observed  (like acetone-water solution).

Ø      The situation where the intermolecular forces between solute-solvent molecules are less than solute-solute and solvent-solvent intermolecular forces (DHsoln = +ve, endothermic process). Positive deviations from Raoult's law is observed  (like ethanol-hexane solution).

 

Fractional Distillation

 

The separation of mixtures of volatile components is accomplished by a technique called fractional distillation. The idea is simple where assuming a mixture of two volatile components (A and B) in a solution. The solution starts to boil when the atmospheric pressure is equal to the sum of the partial pressures of A and B.

 

fractional distillation

 

 If A is more volatile than B, we can draw a schematic between temperature and mole fraction of B, for instance.

 

Text Box: Temperature

 

As the mole fraction of the less volatile component (B) is increased, the boiling point of the mixture will also increase.

 

Example

 

Assume a mixture of A and B boils at some temperature. Find the mole fractions of A and B provided that the vapor pressures of pure A and pure B are 1140 and 570, respectively.

 

Solution

 

The mixture boils when the total vapor pressure of the solution equals 760 torr. Applying Dalton’s relation for partial pressures we get:

Psoln = xA * PoA + xB * PoB

760 torr = xA * 1140 torr + (1 – xA) * 570 torr

Solving the equation for xA we get:

 xA = 1/3

xB = 1 – xA = 1 – 1/3 = 2/3

 

Example

 

If 1.0 mole of A is mixed with 2.0 moles of B and, at some temperature, the resulting solution boils. Find the partial pressure of A and B provided that the vapor pressures of pure A and pure B are 1140 and 570, respectively.

 

Solution

 

XA = 1.0/(1.0 + 2.0) = 1/3

XB = 2.0/(1.0 + 2.0) = 2/3

 

PA  = xA * PoA

PA  = 1/3 * 1140 torr = 380 torr

               

PB  = xB * PoB

PA  = 2/3 * 570 torr = 380 torr

 

For the composition of the vapor, we may write:

PA = xA(vapor) PT(vapor)

380 torr = xA(vapor) * 760 torr

xA(vapor) = 0.500

Note that the mole fraction of A in solution is just 0.333, it was increased in the vapor.

 

Also, we can calculate xB(vapor) to be also 0.500

Therefore, it can be concluded that the vapor will be richer in the more volatile component. This can be easily seen from studying the boiling point diagram below:

 

Text Box: Temperature 

 

 

The upper curve on the boiling point diagram indicates the composition of the vapor. The horizontal line between the liquid boiling point curve and the vapor composition curve is called a tie line. To understand the diagram, we read it as follows:

 

Ø      When the composition of the mixture is x1, it boils at temperature T1, to provide a vapor that has a composition x2. The vapor is collected.

Ø      When the vapor with composition x2 is condensed and reheated, it will boil at temperature T2.

Ø      When the vapor with composition x2 boils at T2, the vapor composition will be x3.

Ø      Repetition of this process will result in vapor that is continuously richer in component A.

This procedure is a typical fractional distillation technique in which volatile liquids are separated from each other.

 

Solutions that Show Large Deviations from Raoult’s Law

 

There are some solutions that show very large deviations from Raoult’s law. Deviations can be positive or negative where positive deviations (higher vapor pressures) mean the presence of a low boiling composition like the one in the figure below. A solution with such a minimum is called a minimum-boiling azeotrope. Fractional distillation of solution compositions below or above the minimum is possible and will give either component plus the azeotrope. However, separation of the mixture at the azeotrope composition is not possible with fractional distillation. Therefore, in terms of boiling point the azeotrope behaves as if it were a new substance.

 

Low Boiling Azeotrope (like propanol-water mixture)

High Boiling Azeotrope (like aquous HCl solution)

 

 

v     Freezing Point Depression

 

Since nonvolatile solutes lower the vapor pressure of solutions, the freezing point of the solution will also be affected where depression of the freezing point is observed. This is a consequence of the effect of the nonvolatile solute on the properties of solutions. The phenomenon is clear when the phase diagram for the solution is studied:

 

From the graph, as the vapor pressure decreases the phase transfer from liquid to solid would occur at a lower temperature. The decrease in freezing point depends on the depression in vapor, due to presence of the nonvolatile solute. The relation governing the effect of amount of added nonvolatile solute on freezing point depression is:

DTf  = kf m

Where DTf  is the freezing point depression, kf is a molal freezing point depression constant and m is the molality of the solution.

 

v     Boiling Point Elevation

 

In addition to depression in freezing point, it should be obvious that the boiling point of a solution having a nonvolatile solute should be higher than the pure solvent. This is simply due to the lower vapor pressure of the solution where boiling point would be reached when the vapor pressure of solution equals the atmospheric pressure. Since the vapor pressure is lowered as a result of the nonvolatile solute, the boiling point is increased. This is very clear from the phase diagram above. The relation governing the effect of amount of added nonvolatile solute on boiling point elevation is:

DTb  = kb m

Where DTb is the boiling point elevation, kb is a molal boiling point elevation constant and m is the molality of the solution.

 

 Example

 

What is the freezing point and boiling point of a solution containing 6.50 g of ethylene glycol (FW = 62.1 g/mol) in 200 g of water? Kf = 1.86 oC/m and kb = 0.51 oC/m

 

Solution

 

m = mol solute/kg solvent

mol solute = Wt/FW = 6.50g/(62.1 g/mol) = 0.105 mol

 

DTf  = kf m

DTf  = 1.86 oC/m * {0.105 mol/0.200 kg} = 0.976 oC

 

DTb  = kb m

DTb  = 0.51 oC/m * {0.105 mol/0.200 kg} = 0.27 oC

 

Therefore, the solution will boil at 100.27 oC and will freeze at – 0.976 oC.

 

An important application of the phenomena of boiling point elevation and freezing point depression is the determination of molecular masses of nonvolatile solutes; as in the example below:

 

Example

 

A 5.50 g of a newly synthesized compound was dissolved in 250 g of benzene (kf = 5.12 oC/m) and the freezing point depression was found to be 1.2 oC. Find the molecular mass of the compound.

 

Solution

 

DTf  = kf m

1.2 oC = 5.12 oC/m * m

Molality = 0.199 m

Molality = mol solute/kg solvent

0.199 = mol solute/0.250 kg

mol solute = 0.0498 moles

mol solute = Wt solute/FW

FW = 5.50 g/0.0498 mol = 110 g/mol

It should be realized that a solvent with high kf is an advantage for such experimental calculations of molecular masses.

 

v     Osmotic Pressure

 

 Osmosis is defined as the process whereby solvent molecules pass from a dilute solution to a more concentrated one, through a semi permeable membrane. An equilibrium force which will ultimately bring both solutions to the same concentration drives this process forward. An osmometer is a device used for measurement of osmotic pressure (P). The solution is placed in the osmometer, which is inserted in a solvent. Solvent molecules pass through the osmotic membrane and the value of the height of the solution in a capillary is measured and related to the osmotic pressure. A relation for the osmotic pressure can be shown to resemble the ideal gas equation where:

P is proportional to the molarity of the solution as well as the absolute temperature, where:

P = MRT

Where R is the gas constant

 

M = n/V where M is the molarity of the solution

P = (n/V)RT

P V= nRT

This is called the van’t Hoff equation and is very similar to the common ideal gas law (PV = nRT).

 

The magnitude of the osmotic pressure, even for dilute solutions, is usually large and thus can be advantageously used for accurate determination of molecular masses. Solutions that have the same osmotic pressure are called isotonic solutions.

 

Reverse Osmosis

 

If a pressure greater than the osmotic pressure is applied at the surface of the solution, solvent will pass from the more concentrated to dilute solution. This process is called reverse osmosis and is the basis for a wide spreading technique for water purification.

 

Solutions of Electrolytes

 

All previous discussion about effect of nonvolatile solute on colligative properties was limited to non-electrolyte solutes. Electrolytes, on the other hand, are even more effective in causing lowering in vapor pressure and freezing point and more elevation in boiling point of solutions. This is simply because when electrolytes dissociate in solution, they produce more species. For instance, 1.0 m NaCl will produce 1.0 m Na+ and 1.0 m Cl- and the overall effective molality of the solute will be 2.0 m.

 

Example

 

What is the freezing point depression of a 0.15 m aqueous solution of Al2(SO4)3? (kf = 1.86 o C/m)

 

Solution

 

Al2(SO4)3 g 2 Al3+ + 3 SO42-

 

Therefore, the overall effective molality of the solution is:

m = 0.15 * 5 = 0.75

 

DTf  = {1.86 oC/m}*0.75 m = 1.4 oC

 

Example

 

A solution of CaCl2 (FW = 111 g/mol) was prepared by dissolving 25.0 g of CaCl2 in exactly 500 g of H2O. What is the expected vapor pressure at 80 oC (Pwater at 80 oC = 355 torr)? What would the vapor pressure of the solution be if CaCl2 were not an electrolyte?

 

Solution

 

Psoln = xsolvent Posolvent

Therefore, calculate the number of moles of water and CaCl2

 

nwater = 500g/{18.0 g/mol} = 27.8

nCaCl2 = 25.0 g/{111 g/mol} = 0.225

nspecies = 3 * 0.225 = 0.675

xwater = 27.8/{27.8 + 0.675} = 0.975

 

Psoln = 0.975 * 355 torr = 346 torr

 

If CaCl2 were not an electrolyte, CaCl2 will not dissociate and the number of moles of all species of solute will be 0.225:

Xwater = 27.8/{27.8 + 0.225} = 0.993

 

Psoln = 0.993 * 355 torr = 352 torr

 

It is clear from this example that the pressure will decrease if the solute is an electrolyte. In fact, electrolytic solutes will result in increased depression in vapor pressure and freezing point and increased elevation in boiling point, mainly because electrolytes dissociate in solutions.

 

Interionic Attractions and the van’t Hoff Factor

 

The above examples assume that positive and negative ions have no attraction forces and thus each ion behaves as an intact species. This is, in fact, not true since there are some attractions which result in decreasing the number of independent species. For example, the following table provides some quantitative assessment of attractions:

 

Salt

0.1 m

0.01 m

0.001 m

Theoretical factor (i)

NaCl

1.87

1.94

1.97

2

K2SO4

2.32

2.70

2.84

3

MgSO4

1.21

1.53

1.82

2

 

Note that as the electrolytic solution becomes more dilute, the effect of attraction forces is diminished. Ultimately, Attraction forces will be zero at infinite dilution. The degree to which an electrolyte behaves as if its ions were independent is quantitatively given by van’t Hoff factor, i:

i = DTf (measured)/{DTf calculated as non electrolyte}

 

It should also be observed that attractive forces are even more effective for multiply charged ions.

 

Example

 

Find the boiling point elevation of a 0.100 m MgSO4  aqueous solution (kb = 0.51 oC/m, i = 1.21).

 

Solution

 

DTb = kb * m

m = 1.21 * 0.100 m = 0.121

 

DTb = 0.51 oC/m * 0.121m = 0.062 oC

 

If MgSO4 was not an electrolyte

 

DTb = 0.51 oC/m * 0.100 m = 0.051 oC.