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Chapter 12Solutions

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Solution Formation

1. Types of Solutions

2. Solubility and the Solution Process

3. Effects of Temperature and Pressure on Solubility

Contents and Concepts


Colligative Properties

4. Ways of Expressing Concentration

5. Vapor Pressure of a Solution

6. Boiling-Point Elevation and Freezing-Point Depression

7. Osmosis

8. Colligative Properties of Ionic Solutions

Colloid Formation

9. Colloids


Solution Formation

1. Types of Solutions

a. Define solute and solvent.

b. Define miscible fluid.

c. Provide examples of gaseous solutions, liquid solutions, and solid solutions.

Learning Objectives


2. Solubility and the Solution Process

a. List the conditions that must be present to have a saturated solution, to have an unsaturated solution, and to have a supersaturated solution.

b. Describe the factors that make one substance soluble in another.

c. Determine when a molecular solution will form when substances are mixed.

d. Learn which conditions must be met to create an ionic solution.


3. Effects of Temperature and Pressure on Solubility

a. State the general trends for the solubility of gases and solids with temperature.

b. Explain how the solubility of a gas changes with temperature.

c. Apply Henry’s law.


Colligative Properties

4. Ways of Expressing Concentration

a. Define colligative property.

b. Define molarity.

c. Define mass percentage of solute.

d. Perform calculations with the mass percentage of a solute.

e. Define molality.

f. Calculate the molality of a solute.

g. Define mole fraction.

h. Calculate the mole fractions of components.


i. Convert molality to mole fraction.

j. Convert mole fraction to molality.

k. Convert molality to molarity.

l. Convert molarity to molality.

5. Vapor Pressure of a Solution

a. Explain vapor-pressure lowering of a solvent.

b. State Raoult’s law.

c. Calculate vapor-pressure lowering.

d. Describe an ideal solution.


6. Boiling-Point Elevation and Freezing-Point Depression

a. Define boiling-point elevation and freezing-point depression.

b. Calculate boiling-point elevation and freezing-point depression.

c. Calculate the molecular mass of a solute from molality.

d. Calculate the molecular mass from freezing-point depression.


7. Osmosis

a. Describe a system where osmosis will take place.

b. Calculate osmotic pressure.

8. Colligative Properties of Ionic Solutions

a. Determine the colligative properties of ionic solutions.


Colloid Formation

9. Colloids

a. Define colloid.

b. Explain the Tyndall effect.

c. Give examples of hydrophilic colloids and hydrophobic colloids.

e. Describe coagulation.

f. Explain how micelles can form an association colloid.


A solution is composed of two parts: the solute and the solvent.

SoluteThe gas (or solid) in a solution of gases (or solids), or the component present in the smaller amount.

SolventThe liquid in the case of a solution of gases or solids, or the component present in the larger amount.


Fluids that mix with or dissolve in each other in all proportions are said to be miscible (left).

Fluids that do not dissolve in each other are said to be immiscible (right).


a.The 5 g of Mo is the solute; the 80 g of Cr is the solvent.

b.MgCl2 is the solute; H2O is the solvent.

c. O2 and N2 are the solutes; Ar is the solvent.

Concept Check 12.1Identify the solute(s) and solvent(s) in the following solutions.a. 80 g of Cr and 5 g of Mob. 5 g of MgCl2 dissolved in 1000 g of H2Oc. 39% N2, 41% Ar, and the rest O2


A saturated solution is in equilibrium with respect to the amount of dissolved solute.

The rate at which the solute leaves the solid state equals the rate at which the solute returns to the solid state.


The solubility of a solute is the amount that dissolves in a given quantity of solvent at a given temperature.

An unsaturated solution is a solution not in equilibrium with respect to a given dissolved substance and in which more of the substance can be dissolved.


A supersaturated solution is a solution that contains more dissolved substance than a saturated solution does. This occurs when a solution is prepared at a higher temperature and is then slowly cooled. This is a very unstable situation, so any disturbance causes precipitation.


Solubility can be understood in terms of two factors:

1. The natural tendency toward disorder favors dissolving.

2. The relative forces between and within species must be considered.

Stronger forces within solute species oppose dissolving.

Stronger forces between species favor dissolving.


For molecular solutions, this can be summarized as “Like dissolves like.” In other words, solutes dissolve in solvents that have the same type of intermolecular forces.

An immiscible solute and solvent are illustrated at right.


Concept Check 12.2You have two molecular compounds, X and Y. Compound X has stronger intermolecular forces than compound Y. X has no dipole-dipole bonding of any kind, and Y exhibits hydrogen bonding. Predict the relative solubility of compounds X and Y in water and in a nonpolar solvent. Explain your reasoning.


When considering ionic solutes in water, we need to examine the hydration energy and the lattice energy.


The stronger ion-dipole force between the ion and the solvent—that is, hydration energy—favors dissolving.

A stronger force between ions—that is, lattice energy—opposes dissolving.


The force of attraction between water and both a cation and an anion is illustrated to the left with lithium fluoride, LiF.


The process of dissolving occurs at the surfaces of the solid. Here we see water hydrating (dissolving) ions.


The hydration energy for AB2 must be greater than the hydration energy for CB2.

Concept Check 12.3The hypothetical ionic compound AB2 is very soluble in water. Another hypothetical ionic compound, CB2, is only slightly soluble in water. The lattice energies for these compounds are about the same. Provide an explanation for the solubility difference between these compounds.


In general, solubility depends on temperature.


In most cases, solubility increases with increasing temperature. However, for a number of compounds, solubility decreases with increasing temperature.

The difference is explained by differences in the heat of solution.


When dissolving absorbs heat (is endothermic), the temperature of the solution decreases as the solute dissolves. The solubility will increase as temperature increases.

When dissolving releases heat (is exothermic), the temperature of the solution increases as the solute dissolves. The solubility will decrease as temperature increases.


Hot packs use an exothermic solution process.

Cold packs use an endothermic solution process.


Henry’s law describes the effect of pressure on gas solubility: The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution.

This is expressed mathematically in the equation

S = kHPwhereS = gas solubility

kH = Henry’s law constant for the gasP = partial pressure of the gas over the solution


In general, pressure has little or no effect on the solubility of solids or liquids in water.

The solubility of a gas increases as pressure increases, as illustrated at right.

?


Helium–oxygen mixtures are sometimes used as the breathing gas in deep-sea diving. At sea level (where the pressure is 1.0 atm), the solubility of pure helium in blood is 0.94 g/mL. What is the solubility of pure helium at a depth of 1500 feet?

Pressure increases by 1.0 atm for every 33 feet of depth, so at 1500 feet the pressure is 46 atm. (For a helium–oxygen mixture, the solubility of helium will depend on its initial partial pressure, which will be less than 1.0 atm.)


P1 = 1.0 atm P2 = 46 atm

S1 = 0.94 g/mL S2 = ?

1

212

P

PSS

atm 1.0

atm 46

mL

g 0.942

S

mL

g 432S


At high elevations, the partial pressure of oxygen decreases, decreasing the solubility of oxygen in water. Fish require a certain minimum level of dissolved oxygen to survive.

Concept Check 12.4Most fish have a very difficult time surviving at elevations much above 3500 m. Howcould Henry’s law be used to account for this fact?


The concentration of a solute can be quantitatively expressed in several ways:

1. Molarity

2. Mass percentage of solute

3. Molality

4. Mole fraction


Molarity is the moles of solute per liter of solution. It is abbreviated as M.

solutionofliters

soluteofmolesM


Mass percentage of solute is the percentage by mass of solute in a solution.

100% solutionofgrams

soluteofgrams

solute of percentage Mass

?


An experiment calls for 36.0 g of a 5.00% aqueous solution of potassium bromide. Describe how you would make up such a solution.


A 5.00% aqueous solution of KBr has 5.00 g KBr per 100. g solution. The remainder of the 100. g is water: 95 g.

We can use this ratio to determine the mass of KBr in 36.0 g solution:

Since 1.80 g KBr is required for 36.0 g of solution, the remainder consists of 34.2 g water.

We make the solution by mixing 1.8 g KBr in 34.2 g water.

KBr g 1.80 solution100.g

KBrg 5.00solution g 36.0


Molality is the moles of solute per kilogram of solvent. It is abbreviated as m.

solventofkilograms

soluteofmolesm

?


Iodine dissolves in a variety of organic solvents. For example, in methylene chloride, it forms an orange solution. What is the molality of a solution of 5.00 g iodine, I2, in 30.0 g of methylene chloride, CH2Cl2?


Mass of solute = 5.00 g I2

Mass of solvent = 30.0 g CH2Cl2

kg 1

g 10

I g 253.8

I mol 1

solventg 30.0

I g 5.00 3

2

22 m

kg

mol 0.657m


Mole fraction is the moles of component over the total moles of solution. It is abbreviated .

solution of moles total

soluteofmolesΧ

?


A solution of iodine, I2, in methylene chloride, CH2Cl2, contains 5.00 g I2 and 56.0 g CH2Cl2. What is the mole fraction of each component in this solution?


Mass of solute = 5.00 g I2

Mass of solvent = 56.0 g CH2Cl2

mol 0.01970I g 253.8

I mol 1I g 5.00

solute Moles

2

2

2

mol 0.6594ClCH g 84.93

ClCH mol 1I g 56.0

solvent Moles

22

222

mol 0.6791

mol 0.6594mol 0.01970moles Total


0.02902I

Χ

total mol 0.6791

I mol 0.01970 2I2

Χ

0.97122ClCH Χ

2 2

2 2CH Cl

0.6594 mol CH Cl

0.6791 mol totalΧ

?


A bottle of bourbon is labeled 94 proof, meaning that it is 47% by volume of alcohol in water. What is the mole fraction of ethyl alcohol, C2H5OH, in the bourbon? The density of ethyl alcohol is 0.80 g/mL.

One liter of bourbon contains 470 mL of alcohol and 530 mL of water. To solve this problem, we will convert the volume of ethyl alcohol to mass using density, and then convert to moles using molar mass.


mol 8.16 g 46.08

mol 1

mL 1

g 0.80 OHHC mL 470

52

0.22ethanol Χ

mol 29.4g 18.02

mol 1

mL 1

g 1.00 OH mL 530 2

mol 37.6mol 29.4mol 8.16moles Total

total mol 37.6

OHHC mol 8.16 52ethanol Χ

?


A 3.6 m solution of calcium chloride, CaCl2, is used in tractor tires to give them weight. The addition of CaCl2 also prevents water in the tires from freezing at temperatures above –20°C. What are the mole fractions of CaCl2 and water in such a solution?

The 3.6 m solution contains 3.6 mol CaCl2 in 1.0 kg of water. To solve this problem, we will convert the mass of water to moles, and then compute the mole fractions.


Moles solute = 3.6 mol CaCl2

0.0612CaCl Χ

OH mol 55.5g 18.02

mol 1

kg 1

g 10 OH kg 1.0

solvent Moles

2

3

2

mol 59.1mol 55.5mol 3.6moles Total

total mol 59.1

mol 3.62CaCl Χ

0.94OH2Χ

total mol 59.1

mol 55.52H OΧ


Converting between molality and mole fraction is relatively simple, because you know the masses or moles of both the solute and the solvent.

?


A solution contains 8.89 × 10-3 mole fraction of I2 dissolved in 0.9911 mole fraction of CH2Cl2 (methylene chloride). What is the molality of I2 in this solution?

We will assume we have 1 mole of the solution, so we begin with 8.89 × 10-3 mol I2 and 0.9911 mol CH2Cl2. Next, we will convert the moles of solvent into grams, and then into kilograms. Finally, we will compute the molality of the solution.


Moles solute = 0.00889 mol I2

kg

mol 0.106m

22

3

ClCH kg 0.08417

g 10

kg 1

mol 1

g 84.93 mol 0.9911 solvent Kilograms

kg 0.08417

mol 0.00889m


Converting between molality and molarity requires knowing the density of the solution. This enables you to calculate the mass or volume of the solution. You can then distinguish the amount of solute from the amount of solvent, or combine them to find the volume of solution.

These types of conversions are illustrated in the following problems.

?


Citric acid, HC6H7O7, is often used in fruit beverages to add tartness. An aqueous solution of citric acid is 2.331 m HC6H7O7. What is the molarity of the solution? The density of the solution is 1.1346 g/mL.

A 2.331 m solution contains 2.331 mol solute in 1.000 kg solvent. We will use this relationship first to convert the moles of citric acid to grams and then to find the mass of solution. Using density, we can then find the volume of solution. Finally, we will compute the molarity of the solution.


Moles solute = 2.331 mol HC6H7O7

L

mol 1.827

g 447.88mol 1

g 192.14 mol 2.331 solute Mass

L 1.2761

mL

L10

g 1.1346

mL 1 g 1447.88solution Liters

3

g 1447.88 g 1000.00g 447.88solution of Mass

L 1.2761

mol 2.331 M

?


An aqueous solution of ethanol is 14.1 M C2H5OH. The density of the solution is 0.853 g/cm3. What is the molality of ethanol in the solution?

We will work with 1.00 L of solution. First, we will convert volume to mass using the density. Then, we will find the masses of the solute and the solvent. Finally, we will compute the molality.


g 649.7mol 1

g 46.08 mol 14.1 solute of Mass

g 853

mL

g 0.853

L 10

mL 1 L 1.000solution of Mass

3-

kg

mol 69.4

kg 0.2033g 10

kg 1 g 203.3

g 649.7g 853solvent of Mass

3

kg 0.2033

mol 14.1 m


Colligative properties of solutions are properties that depend on the concentration of the solute molecules or ions in solution but not on the chemical identity of the solute.

1. Vapor-pressure lowering

2. Boiling-point elevation

3. Freezing-point lowering

4. Osmotic pressure


The vapor pressure of a solution, P, is less than the vapor pressure of the pure solvent, P°.

When the solute is nonvolatile, the vapor pressure of a solution is the mole fraction of the solvent times the vapor pressure of pure solvent.

osolventsolventsolution PXP

oPP solventsolution


We can explain this relationship by recalling that evaporation occurs at the surface of the solution. When a solute is present, less of the surface is occupied by solvent molecules.


To establish an equilibrium, the gaseous solvent will condense in the more concentrated solution until the vapor pressures and concentrations are equal.


To find the vapor-pressure lowering (a colligative property), we rearrange the following equation:

solutiono

solvent PP P

o

solventsolute

solventsolute 1- Since

P X P

XX

osolventsolvent

osolvent PP P X

solvento

solvent 1- XP P


Note that vapor-pressure lowering is directly proportional to the solute concentration, the definition of a colligative property.

osolventsolutesolvent PXP

?


Eugenol, C10H12O2, is the chief constituent of oil of clove. This pale yellow liquid dissolves in ethanol, C2H5OH; it has a boiling point of 255°C. As a result, we know eugenol’s vapor pressure is very low; it can be considered nonvolatile. What is the vapor-pressure lowering at 20.0°C of a solution containing 8.56 g of eugenol in 50.0 g of ethanol? The vapor pressure of ethanol at 20.0°C is 44.6 mmHg.


mol 0.0521 g 164.22

mol 1 g 8.56 solute Moles

mmHg 2.04Δ P

mol 1.085 g 46.08

mol 1 g 50.0solvent Moles

mmHg 44.6 0.04581 Δ

Δ 0

solventsolvent

P

PΧP

mol 1.137

mol 1.085 mol 0.0521 moles Total

0.04581 mol 1.137

mol 0.0521solute

Χ


When the solute is nonvolatile, it has no appreciable vapor pressure itself and forms an ideal solution.

When the solute is volatile, a nonideal solution results.


The phase diagram on the next slide shows the changes in freezing and boiling points when a nonvolatile solute is added to a solvent. The blue line shows the pure solvent; the purple line shows the solution.


The boiling point of a solution is higher than the boiling point of pure solvent. The boiling-point elevation, Tb, is given by the following equation:

Tb = mKb


The freezing point of a solution is lower than the freezing point of pure solvent. The freezing-point depression, Tf, is given by the following equation:

Tf = mKf

?


A solution is made up of eugenol, C10H12O2, in diethyl ether (“ether”). If the solution is 0.575 m eugenol in ether, what are the freezing point and the boiling point of the solution? The freezing point and the boiling point of pure ether are –116.3°C and 34.6°C, respectively. The freezing-point depression and boiling-point elevation constants are 1.79°C/m and 2.02°C/m, respectively.


m = 0.575 m

Kf = 1.79°C/m Kb = 2.02°C/m

Tf° = –116.3°C Tb° = 34.6°C

Tf = m × Kf Tb = m × Kb

Tf = 0.575 m × 1.79°C/m

Tf = 1.03°C This is the freezing-point depression.

Tf = –116.3 – 1.03 = –117.3°C

Tb = 0.575 m × 2.02°C/m

Tb = 1.16°C This is the boiling-point elevation.

Tb = 34.6 + 1.16 = 35.8°C

?


In a freezing-point depression experiment, the molality of a solution of 58.1 mg anethole in 5.00 g benzene was determined to be 0.0784 m. What is the molar mass of anethole?

First, we will use the freezing-point depression data to find the molality of the solution. Next, we will use the molality and mass of solvent to find the moles of solute. Finally, we will use the moles of solute and mass of solute to find the molar mass.


Solute mass = 58.1 mgSolvent mass = 5.00 g

m = 0.0784 mol/kg

mol 10 3.92

kg 0.00500 m 0.0784

solvent kg solute mol

4-

m

solvent kg

solute mol m

g/mol. 148 is mass molar The

g/mol 148 mol 10 3.92

g 10 58.1mass Molar

4-

-3

?


An 11.2-g sample of sulfur was dissolved in 40.0 g of carbon disulfide. The boiling-point elevation of carbon disulfide was found to be 2.63°C. What is the molar mass of the sulfur in the solution? What is the formula of molecular sulfur? The boiling-point elevation constant, Kb, for carbon disulfide is 2.40°C/m.


Solute mass = 11.2 gSolvent mass = 40.0 g

Tb = 2.63°C

Kb = 2.40°C/m

First, we will use the definition of the colligative property to calculate the molality. Next, we will use the definition of molality to calculate the moles of solute. Finally, using the mass and moles of solute, we will find the molar mass and the molecular formula of the sulfur.


m

mK

Tm 1.096

C2.40

C2.63

Δ

f

b

fb Δ KmT

mol 0.04383

kg 0.0400 1.096

solvent kg solute mol

m

m

solvent kg

solute mol m


.8Sformula molecular The

g/mol 255.5 mol 0.04383

g 11.2mass Molar

g/mol. 32.065 is mass formula The

S. is sulfur for formula empirical The

8 32.065

255.5n


Osmosis is the phenomenon of solvent flow through a semipermeable membrane to equalize the solute concentration on both sides of the membrane.

A semipermeable membrane allows solvent molecules to pass through but not solute molecules.


Water is the solvent.

Water will flow from the left to the right.


Water is the solvent. It flows from the beaker into the thistle tube.


Osmotic pressure, , is equal to the pressure that, when applied, just stops osmosis. Osmotic pressure is a colligative property of a solution.

= MRT

?


Dextran, a polymer of glucose units, is produced by bacteria growing in sucrose solutions. Solutions of dextran in water have been used as a blood plasma substitute. At 21°C, what is the osmotic pressure (in mmHg) of a solution containing 1.50 g of dextran dissolved in 100.0 mL of aqueous solution, if the average molecular mass of the dextran is 4.0 108 amu?


T = 21°C + 273 = 294 K

Mass solute = 1.50 g

Liters solution = 100.0 10-3 L

Molecular mass = 4.0 108 amu

Molar mass = 4.0 108 g/mol

mol 10 3.75

g 10 4.0


8

9

mol/L 10 3.75 L 10 100.0

mol 10 3.75 Molarity 8-

3-

-9


8 mol L atm 3.75 10 0.08206 294 K

L mol Kπ

MRTπ

4 6.9 10 mmHgπ

atm 1

mmHg 760 atm 10 9.05 7


One can show that 760.0 mmHg is equivalent to the pressure exerted by a column of water 10.334 m high. Thus each 1.00 mmHg of pressure is equivalent to the pressure of a 1.36-cm column of water. If the density of this dextran solution is equal to that of water, what height of solution would exert a pressure equal to its osmotic pressure?

4 9.4 10 cmπ

4 1.36 cm 6.9 10 mmHg

1.00 mmHgπ


Ionic solutes dissolve to form more than one particle per formula unit. We alter the colligative property equations to account for this fact by including i, the number of ions per formula unit:

P = iP°AXB

Tb = iKbm

Tf = iKfm

= iMRT

?


What is the osmotic pressure at 25.0°C of an isotonic saline solution (a solution having an osmotic pressure equal to that of blood) that contains 0.900 g NaCl in 100.0 mL of aqueous solution? Assume that i has the ideal value (based on the formula).


T = 25°C + 273 = 298 K

Mass solute = 0.900 g NaCl

Volume solution = 100.0 mL = 100.0 10−3 L

Molar mass solute = 58.44 g/mol

i = 2

mol 0.01540 g 58.44


2

3

1.540 10 molMolarity 0.1540 mol/L

100.0 10 L


K 298 K mol

atm L0.08206

L

mol0.1540 2

π

iMRTπ

atm 7.53 π


A colloid is a dispersion of particles of one substance (the dispersed phase) throughout another substance or solution (the continuous phase). The dispersed particles range from 1000 pm to 200,000 pm in size—much larger than single molecules or single ions.

Fog is an example of a colloid. In fog, water droplets are dispersed through air.


Colloids exhibit the Tyndall effect.

The path of the light is visible through a colloid because the light is reflected by the relatively larger-sized particles in the dispersed phase.


Colloid Solution


Colloids are characterized according to the state (solid, liquid, or gas) of the dispersed phase and the state of the continuous phase.

• Fog and smoke are aerosols, which are liquid droplets or solid particles dispersed throughout a gas.

• An emulsion consists of liquid droplets dispersed throughout another liquid (for example, particles of butterfat dispersed through homogenized milk). • A sol consists of solid particles dispersed in a

liquid.


Colloids in which the continuous phase is water are categorized into two major classes: hydrophilic colloids and hydrophobic colloids.

Hydrophilic colloidA colloid in which there is a strong attraction between the dispersed phase and the continuous phase (water).

Hydrophobic colloid A colloid in which there is a lack of attraction between the dispersed phase and the continuous phase (water).


Coagulation is the process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase. It is analogous to precipitation from a solution.

Curdled milk is an example of coagulation.


When molecules or ions that have both a hydrophobic end and a hydrophilic end are dispersed in water, they associate, or aggregate, to form colloidal-sized particles called micelles.

A colloid in which the dispersed phase consists of micelles is called an association colloid.


Soap consists of compounds such as sodium stearate. Sodium stearate is an example of a molecule with hydrophobic and hydrophilic ends.

Hydrophobicend

Hydrophilicend


In water solution, the stearate ions associate to form micelles in which the hydrocarbon ends point inward toward one another and away from the water, and ionic carboxyl groups are on the outside of the micelle facing the water.

The cleansing action of soap occurs because oil and grease can be absorbed into the hydrophobic centers of soap micelles and washed away.


Far left:

Vegetable oil floating on water (dyed green).

Left:

When the mixture is shaken with soap, an emulsion forms as the oil droplets are absorbed into soap micelles.


Synthetic detergents also form association colloids. Sodium lauryl sulfate is a synthetic detergent present in toothpastes and shampoos.


The detergent molecules we have discussed so far are classified in the trade as anionics, because they have a negative charge at the hydrophilic end.

Other detergent molecules are classified as cationics, because they have a positive charge at the hydrophilic end.


Many cationic detergents also have germicidal properties and are used in hospital disinfectants and in mouthwashes.

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