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Study Guide SOLIDS & LIQUIDS and SOLUTIONS N.B. This document is intended as a supplement to the textbook – not a replacement. You still have to read the book. There is no replacement, substitute, or shortcut for reading the textbook. These are only my notes. Feel free to add your own notes on the following pages.

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Page 1: sg.Solids & Liquids - docfish.comdocfish.com/wp-content/uploads/2012/04/sg.Solids__Liquids.pdf · i. ability to flow: liquids (and gases); not solids. ii. b/c: attractive forces are

Study Guide SOLIDS & LIQUIDS

and SOLUTIONS

N.B. This document is intended as a supplement to the textbook – not a replacement. You

still have to read the book. There is no replacement, substitute, or shortcut for reading the textbook. These are only my notes. Feel free to add your own notes on the following pages.

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Chemistry LIQUIDS & SOLIDS p.1

TABLE OF CONTENTS I. LIQUIDS & SOLIDS .......................................................................................................................... 2

A. Overview of the States of Matter .................................................................................................. 2 B. Physical Properties of Liquids & Solids ....................................................................................... 3

II. CHANGES OF STATE ...................................................................................................................... 5

A. Overview ...................................................................................................................................... 5 B. Equilibrium & Le Châtelier’s Principle ........................................................................................ 5 C. Equilibrium Vapor Pressure of a Liquid ....................................................................................... 6 D. Phase Diagrams ............................................................................................................................ 7

III. SOLUTIONS ....................................................................................................................................... 8

A. Terminology ................................................................................................................................. 8 B. Processes ..................................................................................................................................... 10 C. Concentration Units .................................................................................................................... 10

1. Percent by Mass (a.k.a., percent by weight) ............................................................................ 10 2. Molarity (M) (by far the most common concentration unit) ................................................... 11 3. Molality (m) ............................................................................................................................. 11

D. Effects of Temperature and Pressure on Solubility .................................................................... 12 1. Temperature and Solubility (Solids and Gases). ..................................................................... 12 2. Pressure and Solubility (Gases). .............................................................................................. 13

IV. COLLIGATIVE PROPERTIES ........................................................................................................ 15

A. Overview .................................................................................................................................... 15 B. Vapor-pressure lowering (Raoult’s law) ................................................................................. 15 C. Boiling-pressure elevation ....................................................................................................... 16 D. Freezing-point depression ....................................................................................................... 17 E. Osmotic pressure ..................................................................................................................... 17 F. Application – Determining Molar Mass .................................................................................. 17

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Chemistry LIQUIDS & SOLIDS p.2

I. LIQUIDS & SOLIDS A. Overview of the States of Matter The three common states of matter are gases, liquids and solids. § Liquids assume the shape of their containers but possess a definite volume. § Solids possess their own shape and volume. § Gases assume the shape and volume of their containers.

o Gases are the most compressible and expandable of the states of matter o Gases mix evenly and completely when confined in the same container o Gases have much lower density than liquids or solids.

For our purposes, the strength of ionic and covalent bonds is the same. The strength of the intermolecular bonds – the bonds holding together molecules – is less but still important. There are essentially three types of intermolecular bonds:

· Hydrogen bonds – hydrogen atom is attracted to a highly electronegative atom (N, O, or F) on an adjacent molecule,

· Dipole-dipole interaction – dipole (equal but opposite charges separated by a small distance) attraction between polar molecules,

· London dispersion forces1 – result from the constant motion of electrons and creation of instantaneous dipoles.

The strongest is the H-bonds; the weakest, London dispersion. Together dipole-dipole interactions and London dispersion forces are called collectively van der Waals forces. The strength of attraction between particles determines the temperature at which a substance melts (solid becomes liquid) and boils (liquid becomes gas). The stronger the attractions, the high the melting and boiling points. For example, ionic compounds have strong attractions and are mostly solid at room temperature. This is also illustrated for molecules by comparing the state at room temperature of water, ammonia (H3N) and methane (H4C). Hydrogen bonding (the strongest of the intermolecular bonds) results from one hydrogen atom’s attraction to a lone pair of electrons on an adjacent molecule – in a 1-to-1 relationship. Water (H2O) has two lone pairs of electrons and two hydrogen atoms per molecule, ammonia (H3N), one and three; methane (H4C), none and four. Thus, water can form the most number of H-bonds, ammonia some but not as many because is doesn’t have enough lone pairs to balance the number of hydrogen atoms (remember – it’s a 1-to-1 relationship). This results in methane being a gas at room temperature; ammonia, a volatile liquid; and, water a relatively nonvolatile liquid at room temperature (Figure 1).

A. Methane (CH4)

B. Ammonia (H3N)

C. Water (H2O) D. Melting points

E. Boling points

Figure 1. Relationship between intermolecular bonding and state of matter. 1 Fritz London (in 1930), a German-born American physicist.

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Chemistry LIQUIDS & SOLIDS p.3

In a solid, the particles are held in fixed in 3-dimensional space because the attractive forces are the strongest and are much greater than the kinetic energy necessary to separate them. In these fixed positions, the particles have only vibrational motion (e.g., quivering). As the solid melts, the particles gain enough energy, relative to the attractive forces, for rotational motion (roll and tumble around) while still in contact with each other. When the substance boils, kinetic energy of the particles overwhelms the attractive forces and it becomes a gas. These gas particles have translational motion (movement) (Figure 2.)

Figure 2. Arrangement of particles in a gas (left), liquid (center), and solid (right). B. Physical Properties of Liquids & Solids The Kinetic-Molecular Theory was described by the Scottish physicist James Clerk Maxwell and the Austrian physicist Ludwig Georg Boltzman around 1860. An important contribution was made later by Albert Einstein in his 1905 paper on Brownian motion (the random motion exhibited by gas particles and dust in water seen under a microscope). The kinetic-molecular theory and the concept of the ideal gas, on which it is based, was described in the study guide for Gases. The list below presents different characteristics of matter and explains the behavior based on the kinetic-molecular theory.

A. Expansion i. don’t significantly expand ii. b/c: particles are attracted to each other and hold each other in relative space (solids more

than liquids) B. Fluidity

i. ability to flow: liquids (and gases); not solids. ii. b/c: attractive forces are sufficient to hold particles close together in liquids but not strong

enough to hold in fixed 3-D space. Attraction within solids is sufficiently strong to hold fixed in space.

C. Relative Density i. density of gas is ~ 1/1000th that of the corresponding liquid or solid; density of liquid is only

about 1/10th of the solid ii. b/c: particles are close together

D. Relatively Incompressibility

i. particles are about as close together as they will get. The volume of water (20oC) compressed by 1000 atm is decreased by only ~4%. (Contrast this to gas ≈ 1000x’s larger than liquid)

ii. b/c: particles are closely packed

E. Ability to Diffuse i. liquids gradually diffuse ii. b/c: particles are (1) closer together, and (2) attractive forces between particles slow their

movement. As temperature increases, kinetic energy increases, particles move faster and diffuse more readily.

F. Summary

Compare the general properties of solid, liquid and gas (using ice, water and steam, respectively) in the following table.

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Chemistry LIQUIDS & SOLIDS p.4

Table 1. Comparison of States of Matter. Shape* Solid Liquid Gas

Volume* has own (definite) has own (definite) fills container (indefinite)

Extent & Type of Particle Movement

Relative Density**

Compare Strength of Attraction (A) versus Kinetic Energy (KE)

* E.g., does it take on the shape (volume) of its container or have its own definite shape (volume)? ** E.g., very low, low, medium, etc.

Other Properties G. Surface Tension (liquids) ≡ force that tends to pull adjacent particles on a

liquid’s surface area to the smallest possible size a. The molecules in water are held in place by hydrogen bonds. At

the surface the molecules are drawn together and toward the body of the liquid because they are missing in one direction (e.g., top of the liquid). This results in the lowest surface area-to-volume ratio, which is a sphere. i) example – water drops are spherical

b. b/c: H-bonding (H & N, O, F) is a strongly attractive force c. Is it strong?

i) belly-flop = breaking H-bonds that create the surface tension of water; only separating the molecules, not breaking them into hydrogen & oxygen

ii) keeps “water skeeters” afloat

H. Capillary Action ≡ attraction of the surface of a liquid to the surface of a solid a. explains why when the edge of a paper towel is placed in water, the water

rises up the paper towel b. explanation: The water moves upward as a result of its adhesion (attraction

between different molecules) of the water to the walls of the vessel. The water rising up the walls of the vessel produce the notable meniscus of water. The surface tension holds the water intact so, instead of just the edges moving up along the vessel, the entire liquid is dragged upward2.

I. Shapes - Solids Recall that solids had definite shapes. At the molecular level, the definite shape results from two basic classifications:

a. crystalline solids i) ≡ consist of crystals – in which the particles are arranged in an orderly,

geometrically-repeating pattern ii) most solids are crystalline solids – e.g., salt (an ionic compound) and sugar (a

molecule) are easily arranged as crystalline solids readily identifiable as crystalline to

2 source: http://hyperphysics.phy-astr.gsu.edu/hbase/surten.html#c4

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Chemistry LIQUIDS & SOLIDS p.5

the naked eye or a low-power microscope. A diamond is an example of a covalent network crystal.

b. amorphous solids

i) ≡ the particles are randomly arranged randomly ii) example: glass

II. CHANGES OF STATE A. Overview

The state in which matter exists can change between solid, liquid and gas (Table 2).

Table 2. Changes of State Change of State Process Example

solid ⇒ liquid melting ice ⇒ water solid ⇒ gas sublimation dry ice (CO2) ⇒ CO2 gas liquid ⇒ solid freezing* water ⇒ ice liquid ⇒ gas vaporization** liquid bromine ⇒ bromine vapor gas ⇒ liquid condensation steam ⇒ water gas ⇒ solid deposition steam ⇒ ice * All liquids will freeze, given a low enough temperature and high enough pressure, e.g., Neon

freezes at -459.7°F (-272.2°C, 0oK) under 367 psia pressure.

** Vaporization includes both boiling and evaporization: · Evaporation = particles escape from the surface of a non-boiling liquid and enter the gas phase. · Boiling = bubbles of vapor appear throughout the liquid as well as at the surface; occurs when

the vapor pressure within the liquid equals the atmospheric pressure B. Equilibrium & Le Châtelier’s Principle

Equilibrium (or dynamic equilibrium) is a dynamic condition in which two opposing changes occur at equal rates in a closed system.

Equilibrium is a special condition that exists during a change of state. An example of equilibrium is cans on a shelf at a grocery store: as cans are removed, more are replaced – the number of cans on the shelf may stay the same over time but any given does not remain on the shelf for long. Another example is a sealed flask of water (Figure 3).

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Chemistry LIQUIDS & SOLIDS p.6

(A) (B) (C) Figure 3. Vapor-liquid equilibrium. (A) Only liquid phase is present. (B) Evaporation occurs initially as particles begin to leave the liquid phase. (C) Particles in the gase phase begin to condense until an equilibrium is established when the rate of evaporation equals the rate of condensation. An equation representing the equilibrium between a solid and a liquid is: solid + energy liquid. Volatile liquids evaporate readily (e.g., perfume); nonvolatile liquids do not. Again, this property can be explained by the extent of intermolecular bonding. Systems in equilibrium will respond to outside changes. For example, more cans are removed from the shelf; more cans will be stocked. If a closed bottle of water is heated, there will be more steam but also the rate at which the vapor condenses will also increase. This is stated in Le Châtelier’s principle: Le Châtelier’s principle = when a system is at equilibrium is disturbed by application of a stress, the system will reach a new equilibrium that minimizes the stress. Typical stresses are changes in (1) concentration, (2) pressure (and volume) and (3) temperature. These are covered in more detail in a separate chapter (Kinetics). C. Equilibrium Vapor Pressure of a Liquid

Equilibrium vapor pressure is the pressure exerted by a vapor in equilibrium with its corresponding liquid at a give temperature. The boiling point of a liquid is the temperature at which the vapor pressure of the escaping liquid molecules equals (or becomes greater than) the atmospheric pressure pressing down on the surface of the liquid. For example, at an atmospheric pressure of 760 mm Hg, the boiling points of CHCl3, C2H5OH, and C2H3O2 are 61.7oC, 78.3oC and 117.9oC (Figure 4). At higher elevations, where the atmospheric pressure decreases, the boiling points decrease. This is why it takes longer to cook rice in Denver than in Ridgefield – the water is not as hot when the water boils. This also explains why it takes less time to cook using a pressure cooker.

Vapor Pressure v. Temperature

02004006008001000120014001600

0 10 20 30 40 50 60 70 80

Temp (oC)

Vapo

r Pre

ssur

e (m

m H

g)

Figure 4. Vapor pressures of CHCl3, C2H5OH, and C2H3O2 as a function of temperature.

CHCl3 C2H5OH

C2H3O2

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Chemistry LIQUIDS & SOLIDS p.7

When a liquid is heated, it boils. Where does the heat go? Once the liquid begins to boil (e.g., 100oC for water), the temperature does not continue to rise but stays constant throughout the boiling process. The additional heat goes to increasing the kinetic energy of the molecules sufficiently enough to overcome the attractive forces holding the molecules in the liquid state. Addition of salt (e.g., NaCl) will raise the temperature at which water boils. Why? D. Phase Diagrams A phase diagram shows the relationship between the pressure and temperature for the phases of a given substance.

Phase Diagram for Water Phase Diagram for Water

Figure 5. Phase diagram for water. The diagram at the left displays the six phase changes between solid, liquid and gas. Curve AB (right) indicates the temperature and pressure conditions at which ice and water vapor coexist. Curve AC indicates the temperature and pressure conditions at which liquid water and water vapor coexist. Curve AD indicates the temperature and pressure conditions at which ice and liquid water coexist. Notice that this curve (AD) has a negative slope: because ice is less dense than liquid water, an increase in pressure lowers the melting point. The curve (AD) for most substances has a positive slope.

Two important points are (1) the triple point and (2) the critical point.

(1) Triple Point: the temperature and pressure at which the substance can coexist as gas, liquid and solid.

(2) Critical Point: indicates the critical temperature and critical pressure. a. Critical Temperature (Tc): The temperature above which a substance cannot exist

as a liquid (374oC for water). b. Critical Pressure (Pc): The highest pressure at which a substance can exist as a

liquid (217.75 atm for water).

Figure 6. Phase diagram for carbon dioxide. Notice how the curve between solid and liquid has a (slightly) positive curve. This is typical because generally as one increases the pressure, the melting point also increases.

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Chemistry LIQUIDS & SOLIDS p.8

III. SOLUTIONS A. Terminology Recall that a solution is a homogeneous mixture (a blend of two or more kinds of matter in which each retains its own chemical identity and properties) (Error! Reference source not found. and Table 3). The two components of a solution are (1) the solute, the substance being dissolved, and (2) the solvent, the substance doing the dissolving. When the solute settles out of a solution (e.g., muddy water), it is a suspension. A colloid (or colloidal suspension) is when the solute particles are sufficiently large to disrupt the path of light but not large enough to settle out of solution. Examples of colloids are emulsions (e.g., mayonnaise and shaving cream) and foams (e.g., remaining on the beach by retreating waves). Many colloids appear to the eye to be homogenous at first. However, they scatter light such as the headlight beam from an approaching automobile or rays of light through the clouds (Figure 77).

Figure 7. The Tyndall effect resulting from a colloidal suspension of water vapor and dust in the air.

Table 3. Some Common Terms Used To Describe Solutions and Solution Components Term Definition/Description Example Soluble capable of being dissolved sugar in water

Insoluble does not dissolve in the solvent sand and water

Miscible liquid solutes and solvents that dissolve freely in one another

rubbing alcohol and water

Immiscible liquid solutes and solvents that are not soluble in each other

oil and water

Solution homogeneous mixture of two or more substances in a single phase, each retaining their own identity and chemical properties

soda

Solute the substance in a solution that is being dissolved salt in salt water

Solvent the substance in a solution that does the dissolving

water in salt water

Saturated Solution a solution that contains the maximum amount of dissolved solute

adding sugar to coffee until no more dissolves

Unsaturated Solution

a solution that does not contain the maximum amount of dissolved solute

coffee when only a little sugar is added

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Chemistry LIQUIDS & SOLIDS p.9

Supersaturated Solution

a solution that contains more dissolved solute than a saturated solution would contain under the same circumstances

honey

(also sodium acetate in the lab)

Electrolyte solute that enables a solution to conduct electricity

table salt (and other ionic compounds)

Nonelectrolyte solute that does not enable a solution to conduct electricity

sugar (and other water-soluble molecules)

The solubility of a solute in a solution is clearly identified when an unsaturated solution quickly becomes a saturated solution (Figure 8). Figure 7. Graph showing the solubility of NaC2H3O2 producing unsaturated and saturated solutions.

Solutions can exist in a variety of states of solvents and solutes (Table 4). Table 4. Examples of Solvent-Solute Solution Combinations. Solvent State Solute State Example Gas Gas oxygen in nitrogen (air) Gas Liquid water vapor in air (humidity) Liquid Gas carbon dioxide in water (carbonated water) Liquid Liquid alcohol in water (medical thermometer) Liquid Solid sugar in water (preparation for rock candy) Solid Liquid mercury in silver & tin (dental amalgam = filling) Solid Solid brass (Cu/Zn); 18 carat gold (Ag/Au)

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Chemistry LIQUIDS & SOLIDS p.10

B. Processes The dissolution of a solvent in a liquid involves a process whereby solute particles at the edges are carried away by the solvent. Thus, dissolution occurs at the interface between the solute and the solvent (Figure 9). A.

B.

C.

Figure 9. Dissolution process. A. Solvent molecules (e.g., water) are attracted to the solute particles (e.g., NaCl – the positively-charged sodium cation forms an electrostatic attraction with the negatively-charged oxygen on water). B. One solute particle (e.g., Cl–) is pulled from the surface into the solution as the solvent particle moves away. Another particle of solute (e.g., Na+) is attracted to the solvent. C. The second solute particle moves into the body of the solution and the process is repeated until no more solute remains or the solution is saturated. There are three major ways to alter the rate at which a solute dissolves in a solution: (1) agitation, (2) temperature, and (3) surface area. Agitation increases the number of solvent particles that come in contact with the solute. Increasing the temperature increases the kinetic energy of the particles, thereby increasing the frequency of collisions between solute and solvent. Similarly, increasing the surface area-to-volume ratio increases the number of solute particles in potential contact with the solvent. Realize that dissolution is a dynamic process – as more solid dissolves in a solution, and the concentration of the solute molecules increases, the collisions between dissolved particles increases and results in reformation of the solid crystal. There becomes equilibrium between the dissolving and crystallization process. When the opposing rates are equal, the solution reaches solution equilibrium. When water is the solvent, the process of dissolving is referred to as hydration and the ions are said to be hydrated (Figure 80). Generally, ionic compound are not soluble in nonpolar solvents (e.g., NaCl in oil). Figure 80. Hydrated copper(II) sulfate, CuSO4⋅5H2O (copper(II) sulfate pentahydrate). The water molecules trapped in the crystal structure are released when the substance is heated and forms the anhydrous form, CuSO4.

C. Concentration

Concentration Units Concentration of a solute in a solution can be expressed in a variety of ways. As you will see, each is useful.

1. Percent by Mass (a.k.a., percent by weight)

100% x solution of masssolute of mass 100% x

solvent of mass solute of masssolute of mass massby percent =

+= (Eq. 1)

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Chemistry LIQUIDS & SOLIDS p.11

2. Molarity (M) (by far the most common concentration unit)

solution of L 1solute of moles molarity = (Eq. 2)

3. Molality (m)

solvent of kg 1

solute of moles molality = (Eq. 3)

To prepare a solution with a given total volume (i.e., molarity), follow the following procedure:

1. Accurately mass the solute and transfer it to a volumetric flask through a funnel. (You’ll have to convert from the number of moles required to the number of grams to mass.)

2. Through the funnel carefully and slowly add water to the flask, which is carefully swirled, to dissolve the solvent and mix the solution.

3. After all the solute has dissolved, add more water to bring the level of the water to exactly the volume mark.

Example. Calculate the percent by mass and molality of a solution containing 24.4 g sulfuric acid in 198 g of water.

ANS. Percent by Mass:

mass of solute 24.4 gmass of solution 222.4 g

Molality (m ):moles solute 24.4 gkg solvent 222.4 g

moles solute: 24.4 g H2SO4 1 mol H2SO4 0.249 mol H2SO4

98.07 g H2SO4

0.249 mol H2SO4 1.26 m

0.198 g H2SO4

11.0%

= * 100% = 11.0%

=

m= =

= * 100% =

The choice for the units of concentration is based on the use at hand. Molarity is the most common because it is the easier to measure volume of solvent (e.g., using a volumetric flask) than its mass. But molarity of a solution can be affected by temperature. For example, a 1.0 M solution at 25oC may become 0.97 M at 45oC due to an increase in volume. However, percent by mass and molality are independent of temperature and therefore useful when temperature is a factor (e.g., colligative effects, q.v., below).

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Chemistry LIQUIDS & SOLIDS p.12

Dilution

It is common practice in chemistry laboratories to make, for any given substance, a single stock solution – e.g., 1 M NaCl. From this stock solution, dilutions are made for the desired concentration of the experiment at hand. The usual equation used to solve these problems is given below.

M1V1 = M2V2 (Eq. 4) Where M = concentration (mole/Liter) and V = volume for solutions 1 and 2, respectively. Example How many mL of a 1 M NaCl solution must be transferred to a second flask to make 100 mL of a 0.05 M NaCl solution?

M1 = 1 M

V1 = ? mL M1 V1 = M2 V2

M2 = 0.05 M

V2 = 100 mL

D. Effects of Temperature and Pressure on Solubility

1. Temperature and Solubility (Solids and Gases). Solids. In most cases, but not all, solubility of a solid increases with temperature (Figure 12). (A notable exception is sodium sulfate which displays a direct proportion with solubility and temperature until approximately 35oC and then an uncommon inverse relationship above that temperature.) The dissolution process can be endothermic (e.g., ammonium nitrate) or exothermic (e.g., calcium chloride). The absorption or release of heat is unrelated to the relationship between solubility and temperature.

Gases. Typically, the solubility of a gas in water usually decreases with increasing temperature (Figure 10). An example of this is the maximum temperature at which some fish can survive: trout, a relatively cold-water fish requires more dissolved oxygen in the water than catfish.

Figure 9. Solubilities of several gases in water (left) and sodium sulfate (right).

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Chemistry LIQUIDS & SOLIDS p.13

0

2

4

6

8

10

12

14

16

0 10 20 30 40 50

Temperature (oC)

O2

(mg)

/ 10

00 g

Wat

er

Max. Water Temp. for Fish Fish oC Trout 24 Lake Herring 25 Pike 30 Bass (LM) 34 Catfish 35

0

5

10

15

20

25

30

35

40

0 1 2 3 4 5 6

Pressure (atm)

O2

(mg)

/ 10

00 g

Wat

er

Figure 10. Solubility of oxygen gas in water for temperature (left) and the maximum water temperature at which several species of fish can survive (middle). The relationship between solubility and temperature for a gas (oxygen) is graphed (right).

2. Pressure and Solubility (Gases). When a gas is dissolved in a liquid, such as carbonated water, equilibrium is eventually established between the rate that the gas leaves the solution and the rate of gas entering the liquid:

gas + solvent solution. Increasing the pressure above the solution puts stress on the equilibrium and results, eventually, in an increase in the solubility of gas in the solution (Figure 10). When the pressure is reduced, the gas becomes less soluble in the solution and comes out. This is the basis for Decompression Sickness (“the Bends”), a great concern for SCUBA divers. This condition results when certain gases, such as nitrogen, are inhaled and enter the bloodstream. As one descends into the water, the pressure of the nitrogen increases 1 atm per 10 m (33 feet). This makes the nitrogen more soluble in the blood which, in turns, makes more nitrogen dissolved in the tissues. When one ascends, the pressure decreases (until reaching again 1 atm at the surface). If the diver ascends too quickly, there is insufficient time for the nitrogen to be carried out of the tissues to the blood and eliminated from the body by exhalation. The nitrogen will come out of the tissues and blood as gas bubbles. This can increase in severity from general malaise and mild joint pains to excruciating pain and possible fatal injury to the nervous and other systems. An illustrative example is the effervescence of carbonated soda such as when it is left in a hot place (e.g., the beach during summer) or shaken before being opened. Henry’s law: the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution: c = kP (Eq. 5) where c = the concentration (mole/L) of gas; P = pressure (atm); and, k is a constant that depends on temperature. Example. The solubility of nitrogen gas at 25oC and 1 atm is 6.8 x 10–4 mol/L. What is the concentration of nitrogen dissolved in water under atmospheric conditions? (Partial pressure of N2 in the atmosphere is 0.78 atm)

ANS. The first step is to determine k, Henry’s law constant: c = kP

6.8 x 10–4 mol/L = k (1 atm) k = 6.8 x 10–4 mol/(L⋅atm)

Therefore, the solubility of N2 in water is: c = (6.8 x 10–4 mol/(L⋅atm)) (0.78 atm) c = 5.3 x 10–4 mol/L

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Chemistry LIQUIDS & SOLIDS p.14

Henry’s law works for most gases but not all – especially when they react chemically with water. Exceptions include ammonia (NH3 + H2O NH4

+ + OH–) and carbon dioxide (CO2 + H2O H2CO3). Another interesting exception is oxygen gas dissolved in the bloodstream. It reversibly binds to hemoglobin (Hb + 4O2 Hb(O2)4). When the oxy-hemoglobin complex reaches a tissue with an lower partial pressure of oxygen, the oxygen is delivered to the cell.

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Chemistry LIQUIDS & SOLIDS p.15

IV. COLLIGATIVE PROPERTIES A. Overview Colligative properties (or collective properties) are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. The classic example of colligative properties is application of salt on snow and ice (freezing point elevation). Colligative properties are:

1. vapor-pressure lowering 2. boiling point elevation 3. freezing point depression 4. osmotic pressure

B. Vapor-pressure lowering (Raoult’s law) As solute dissolves in a solution, the solvent particles are attracted to the solute particles. This increased attraction decreases the amount of molecules of solvent that are able to escape by vapor-pressure equilibrium. This is a direct result of the kinetic-molecular theory. The relationship between solution vapor pressure and solvent vapor pressure depends on the concentration of the solute in the solution. This is given by Raoult’s law: o

111 P X P = (Eq. 6) where 1P is the partial pressure of the solvent over the solution; 1X is the mole fraction of the solvent in

the solution; and, o1P is the partial pressure of the pure solvent. Realize that if a solute is nonvolatile (i.e.,

essentially no vapor pressure), the vapor pressure of the solution containing it is always less than that of the pure solvent. Example. Calculate the vapor pressure of a solution made by dissolving 218 g of glucose (180.2 g/mol molar mass) in 460 mL of water at 30.oC. The vapor pressure of water at this temperature is 31.82 mmHg. What is the decrease in vapor pressure? (Assume the density of water at this temperature is 1.00 g/mL.) ANS. Using Raoult’s law ( o

111 P X P = ), we first need to calculate the mole fraction of the solution: nGLU = 218 g glucose 1 mol glucose = 1.21 mol glucose

180.2 g glucose

nH2O = 460 mL water 1 g water 1 mol water = 25.5 mol glucose1 mL water 18.02 g water

25.5 mol = 0.955nGLU + nH2O 26.71 mol

Partial pressure of the glucose solution (P1) is:P1 = 0.955 x 31.82 mmHg = 30.4 mmHg

Lastly, the vapor-pressure lowering is:ΔP = 31.82 mmHg - 30.4 mmHg = 1.4 mmHg

nGLU =mole

fraction water:

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Chemistry LIQUIDS & SOLIDS p.16

CHECK: Because the mole fraction of glucose is (1 – 0.955) 0.045, the vapor pressure lowering is (0.045)(31.82 mmHg) = 1.4 mmHg. C. Boiling-pressure elevation A solution boils when the vapor pressure of the solution equals the atmospheric pressure. Solute particles in a solution hold the solvent particles in place, thereby requiring more energy for them to escape the liquid. Thus the boiling point becomes elevated by addition of a solute to a solution. Using a phase diagram, adding a solute lowers curve and thereby raises the temperature at which a solution boils (Figure 11). Boiling point elevation (ΔTb) is calculated using the following equation: ΔTb = i Kb m (Eq. 7) where Kb is the molal boiling point elevation constant, m is the molality of the solution, and i is the van’t Hoff factor which is the number of moles of particles per mole of solute (e.g., i = 1 for glucose; i = 2 for NaCl; and i = 3 for CaCl2 because glucose, as a molecule, does not dissociate in water and remains as a single particle; NaCl → Na+ + Cl–, which are two moles of particles per mole of solute; and, CaCl2 → Ca2+ + 2Cl–, which are three moles of particles per mole of solute). Note: the boiling-point elevation constant and freezing-point depression constants are for the solvents – the identity of the solute is only important in as far as the van’t Hoff constant. The boiling-point elevation and freezing point depression constants are given for some common liquids (Table 5). A common but incorrect assumption is that salt added to water will raise its boiling point, thereby making the pasta in the pot cook quicker. The amount of salt added to water the kitchen is not enough to elevate its boiling point. Contrast this with sea water, which tastes quite salty, has a boiling point of 100.6°C. The salt is added simply to season the food and prevent the pasta from sticking.

Figure 11. Phase diagram for a solution illustrating freezing point depression and boiling point elevation.

Note the curve is shifted down with the addition of the solute thereby lowering freezing point (Tf) and raising the boiling point (Tb).

Table 5. Boiling Point Elevation and Freezing Point Depression Constants for Selected Solvents.

Solvent Normal Freezing

Point* (oC) Kf (oC/m)

Normal Boiling Point* (oC) Kb (oC/m)

Water 0.0 1.86 100.0 0.52 Benzene 5.5 5.12 80.1 2.53 Acetic Acid 16.6 3.90 117.9 2.93 Cyclohexane 6.6 20.0 80.7 2.79 * measured at 1 atm

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Chemistry LIQUIDS & SOLIDS p.17

D. Freezing-point depression Similar to boiling point elevation, the freezing point of a solution can be lowered by the concentration of solute: ΔTf = i Kf m (Eq. 8) where Kf is the molal freezing point depression constant (Table 5). E. Osmotic pressure Osmosis is the selective passage of solvent molecules through a semi-permeable membrane from an area of higher concentration of solute (lower concentration of solvent) to lower concentration of solute (higher concentration of solvent).

Figure 124. Osmosis and osmotic pressure. Osmotic pressure (π) is the pressure exerted to stop osmosis. π = MRT (Eq. 9) where M is the molarity, R = the gas constant (0.0821 L-atm/K-mol), and T is the temperature (K). Thus, the osmotic pressure is expressed in atm. Because osmotic pressure experiments are carried out at a constant temperature, we can use molarity instead of molality (as we must do in boiling point elevation and freezing point depression). The two solutions in an osmotic experiment (Figure 12), the solution with less solute is hypotonic to the more concentrated hypertonic solution. If both solutions have the same concentration, e.g., at equilibrium, the solution is isotonic. F. Application – Determining Molar Mass One application of boiling point elevation (or the other colligative properties) is the determination of molar mass. Example. A solution of 0.85 g of an unknown molecule is dissolved in 100. g of benzene. The resulting solution has a freezing point of 5.16oC. What is the molality and molar mass of the unknown compound? ANS. freezing-point depression → molality → number of moles → molar mass

1. molality: ΔTf = i Kf m

i = 1 because it is a molecule (does not dissociate in water like an ionic compound would) m = ? Kf = 5.12oC/m m = ΔTf = 5.5 - 5.16 = 0.34

oC = 0.066 m

Kf 5.12oC/m

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2. Because there are 0.066 mol of unknown in 1 kg of solvent, the number of moles of solute in 100.

g (= 0.100 kg) of solvent is:

0.100 kg solvent 0.066 mol unknown = 6.6E-03 mol unknown

1 kg solvent 3. Molar mass:

molar mass = g unknown = 0.85 g unknown = 129 g/molmol unknown 0.0066 mol unknown