Lecture Notes 112 Weeks 1 and 2 Solutions Chapter 12 Chemistry by Chang

Jan. 13-20, 2005

Connections: Review

Chemistry 111:

Chapter 4 The nature and behavior of solutions and types of reactions that

occur in aqueous solutions

Chapter 9 and 10 Lewis Dot Structures of Molecules

Chapter 11 Intermolecular Attractions/Forces and Physical Properties

Chemistry 111Lab: Projects I and II

Polar and non-polar solvents

“Like Dissolves Like”

Molar Mass by Freezing Point Depression

Concentration Units

Reactions in Aqueous Solution

The Physical Properties of Solutions and Their Behavior

What is a solution? A solution is a homogeneous mixture of two or more substances.

Solute is the substance present in the smaller amount.

Solvent is the substance present in the larger amount.

Types of solutions: The types of mixtures are “defined” by the phase of the solute(s) and the solvent::

Describing Solutions:

A Solution may be:

saturated which indicates that the solution contains the maximum amount of

solute that will dissolve in the solvent at that temperature.

unsaturated which indicates that more solute may be dissolved in the solvent at

at that temperature.

supersaturated which indicates that the solution contains more solute than a

saturated solution of that solute in solvent at that temperature.

Supersaturated solutions are unstable and the “extra” solute is likely

to precipitate when the solution is disturbed.

Factors influencing the dissolving process: How do substances dissolve and what factors influence dissolving of a solute by a solvent?

1. From Chemistry 111 and 111 Lab, you know the general rule of “Like Dissolves Like”

2. Intermolecular/Interparticle attractions hold particles together in liquids and solid samples. These same interactions govern the dissolving process. There are three different attractions to consider:

A. Solvent-solvent attractions that must break to allow solution formation.

B. Solute-solute attractions that must break to allow solution formation.

C. Solvent-solute attractions that must form to allow solution formation.

Each of these steps involves energy. Energy is required for the breaking of attractions in the first two steps and energy is released in the formation of attractions. The overall change in energy can be determined by considering the DH for each step and, by Hess’ Law, the DH for the overall process. An overall exothermic change is one of two factors that drive physical or chemical changes to occur.

DH_solution = DH₁ + DH₂ + DH₃

The second factor that drives reactions to occur is the tendency of nature to move toward disorder. Mixing substances increases the disorder of the system and favors the formation of a mixture of the two materials.

Some terms that are used to describe solutions:

Miscible describes two liquids that are completely soluble in all proportions.

Solvation/hydration is the process in which solvent molecules surround the solute molecules or ions in a cage-like pattern. When the solvent is water, the process is called

hydration.

The general rule “Like Dissolves Like” is quite useful.

Example 12.1 from text p. 490 Predict the relative solubilities for bromine (Br₂) in

benzene(C₆H₆) and in water. [In the textbook, the problem provides the dipole moment

(m = 0 for benzene indicating that it is non-polar; and m = 1.87 for water indicating that it

is polar.)] Answer: The non-polar molecules of bromine will be more soluble in the non-

polar solvent benzene than in the polar solvent water.

Quick Review: “Like Dissolves Like” General Behavior:

1. Non-polar molecules are more soluble in non-polar solvents.

2. Polar molecules are more soluble in polar solvents.

3. Ionic compounds are more likely to dissolve in polar solvents.

Note: 112 students are expected to be able to predict whether compounds are ionic,

non-polar molecular, or polar molecular using information mastered in Chemistry 111. If

you have questions on these topics, please see me, check your notes from Chem 111,

and check the early chapters of the textbook.

The composition of a solution is given in units of concentration. There are several ways to express the concentration of a solution by giving information about the proportion of solute to solvent.. Concentration units one needs to know:

1. Percent(%) by mass or weight = mass of solute X 100%

Total mass of mixture

2. Mole Fraction (X) = # moles of component A

Total # moles in the mixture

3. Molarity (M) = # moles of solute

# liters of solution

4. molality(m) = # moles of solute

# kilograms of solvent

5. volume percent = volume of solute X 100%

Total volume of mixture

6. ppm (parts per million) = # grams of solute per 1,000,000 grams of solvent

Problems: Knowing the definitions of these concentration units will allow you to solve simple problems expressing the concentrations of solutions.

More complex type of problem: It is sometimes necessary to be able to translate information from one expression of concentration to another.

Example: The concentration of a solution of ethanol is given as 5.86 M and its density is listed as 0.927 g/mL. What is the molality of the solution?

Example: The concentration of a solution of ethanol(C₂H₅OH) is given as 5.86 M and its density is listed as 0.927 g/mL. What is the molality of the solution?

Hints:

Pick a sample, if one is not given. Here one could choose to evaluate a 1.000 L.

Draw a picture of that sample and record all information possible about that sample using

the information given. For this problem:

Volume= 1.000 liter

The sample contains 5.86 moles ethanol

Determine the # grams ethanol: _____________

The mass of this sample is 927 grams based on the density.

Determine the # grams of the solvent: _______________

Determine the # moles of solvent(always assume water unless told

otherwise.): ______________

Since you know the number of moles of solute in this sample and the mass of the solvent in this sample, it is possible to determine the molality.

molality ) = # moles of solute

# kilograms of solvent

The Effect of Temperature on Solubility

1. Solids: Solubility of Solids is often increased by increasing temperature but not always. It is important to have empirical information such as that given in this graph:

On the graph, look at KNO₃. How is the solubility of the compound affected by an increase in temperature? _____

On the graph, look at Na₂SO₄. How is the solubility of the compound affected by an increase in temperature? _____

One of the major ideas in chemistry is that, if there is a difference in a property of two materials,one may be able to use that difference to separate or identify the two. You used that idea in Chemistry 111 laboratory in Projects I and II to identify and/or separate materials. One can use the differences in solubility to separate materials, if they have significantly different solubilities at any given temperature. This process is called Fractional Crystallization.

Example: USE THE GRAPH ABOVE>

Suppose you have a mixture of 90 grams of KNO₃ and 10 grams NaCl. Use the graph above and note that, if you place the mixture in 100 mL of water and heat to about 60 degrees C, all of both compounds will dissolve since the solubility of KNO₃ at that temperature is about 125g and the solubility of NaCl at that temperature is about 40 g.

What happens if you cool the mixture to around 0 degrees C?

The solubility of NaCl is 35 g and all of the 10 g of NaCl will still be dissolved in

the solution.

The solubility of KNO₃ is only about 12 grams and that amount will remain in

solution. However, the rest of the compound, 90 g- 12 g, will precipitate

which will form 78 grams of pure KNO₃.

2. The Effect of Temperature on the Solubility of Gases

An increase in temperature usually reduces the solubility of a gas.

Example: Oxygen gas dissolves in water. It is this dissolved O₂ that is used by aquatic

life. If the temperature of a body of water is raised, the amount of

dissolved oxygen is reduced. This Thermal Polution may cause death

to aquatic life in streams, lakes, the ocean if the amount of dissolved O₂,

falls below that needed to support life.

The Effect of Pressure on Solubility:

Solubility of Solids: External pressure has no effect on the solubilities of solids and

liquids.

Solubility of Gases: External pressure has a major effect of the solubility of a gas.

The solubility of a gas is proportional to the pressure of the gas over the solution.

One of the best examples is the solubility of carbon dioxide is carbonated drinks.

The carbonation in drinks is due to dissolved CO₂ . The soda is bottled

under very high CO₂ pressure which causes the a lot of CO₂ to dissolve

in the drink. When the drink is opened, the concentration of CO₂ in the

atomosphere is low and, therefore, the solubility of the carbon dioxide is

much lower and the excess carbon dioxide bubbles out of solution.

This is quantitatively expressed by Henry’s Law:

(Concentration of dissolved gas) = k (Pressure of that gas)

c= kP

where c = concentration of dissolved gas

P = pressure of the gas over the solution

k = a constant (called Henry’s constant) for the gas that depends

on the gas and on the temperature

Examples to consider:

a. Carbonated Drinks:

b. Dissolved gases in bodies of water

c. Dissolved CO₂- Lake Nyos(see p. 500)

d. The “Bends”- Nitrogen gas under pressure dissolving in blood

Colligative Properties

These are properties in which a solution differs from the pure solvent. They are properties of a solution that depend on the number of solute particles present in the mixture and not on the nature of the solute particles.

1. Vapor Pressure

Define: The pressure of the vapor when it is in equilibrium with its liquid.

Vapor pressure of a pure substance depends on:

a. the nature of the liquid which governs the strength of the bonds holding the

particles to each other (intermolecular attractions). The stronger the IMA,

the lower the vapor pressure of the substance.

b. the temperature. The higher the temperature, the higher the vapor pressure.

Vapor Pressure of solutions: The presence of a solute affects the vapor pressure of a solvent.

a. The vapor pressure of a solution is always lower that the vapor pressure of

the pure solvent at the same temperature

b. The value of the vapor pressure of a solution depends on the number of

particles of dissolved solute present in the mixture, i.e., it is a colligative

property. The vapor pressure(P₁) above the solution for a volatile component

depends on the mole fraction(X₁) and the vapor pressure of a pure sample of

that component(P^o₁) : Raoult’s Law

P₁ = X₁P₁^o

A. Vapor Pressure of a solution with one non-volatile solute:

P₁ = X₁P₁^o

X₁ + X₂ = 1 [Since there are only two components, X₁ = mole fraction solvent and X₂ = mole fraction of the solute ]

X₁ = (1 - X₂)

Raoult’s Law says: P₁ = X₁P₁^o

P₁ = (1 - X₂)P₁^o

Rearrange terms :

P₁ = P₁^o - X₂P₁^o

Therefore, the change in vapor pressure = P₁^o - P₁ = X₂P₁^o = DP

B. Vapor Pressure of a solution with two volatile liquids: both components

contribute to the vapor pressure over the mixture.

P_total = P_A + P_B

If P_A = X_AP_A^o and P_B = X_BP_B^o

Then: P_total = X_AP_A^o + X_BP_B^o

For an ideal solution, the vapor pressure of mixtures of different proportions of A and B would be the values along the solid line in this graph. Each point would be the addition of the values on the dotted lines

for the mixture.

Colligative Properties Continued

2.Boiling Point Elevation and Freezing Point Lowering:

Chemistry 111 Review:

For each pure substance, one can draw a Phase Diagram which indicates the phase of the substance under a wide range of temperatures and pressures. Of course, the temperatures and pressures at which substances will exist in each phase differ drastically. However, the general shape of a phase diagram is:

Region A: solid, Region B: liquid, Region C: gas

GENERAL PHASE DIAGRAM:

Vapor pressure and boiling: The boiling point of a material is defined as the temperature at which the vapor pressure is equal to the atmospheric pressure. From Chem 111 laboratory, you know that a solution has a higher boiling point than the pure solvent. The phase diagram for the solution indicates the following changes: Effect of solute particles on vapor pressure expressed graphically

Vapor Pressure and freezing: The freezing/melting point of a solution is lower than that of the pure solvent. This is one of the main ways to check for purity of a substance as was used in Project 1 of Chemistry 111.

The variation in boiling point and freezing point of a solution can be expressed:

Change in boiling point:

DT_b = K_bmi

DT_b = change in the boiling point

K_b = molal boiling-point elevation constant that depends on the solvent

m = the concentration of the solution in molality

i = the van’t hoff factor = # particles of solute/one mole of solute

Change in freezing/melting point:

DT_f = K_fmi

DT_f = change in the freezing/melting point

K_f = molal freezing-point lowering constant that depends on the solvent

m = the concentration of the solution in molality

i = the van’t Hoff factor = # particles of solute/one mole of solute

Example: What is the freezing point of a solution containing 478 grams of ethylene glycol

dissolved in 3202 grams of water? The molar mass of ethylene glycol is

62.01 g/mol To check yourself: answ.= DT = 4.48 degees

and the freezing point is -4.48^oC.

The van’t Hoff factor, i, depends on whether the solute is ionic or not and on how many

ions are produced per mole of dissolved solvent. For electrolyes, an estimate of

the value of i is the number of moles of ions/one mole of solute;

NaCl i = ~2

Na₂SO4 i = ~ 3

3. Colligative Properties: Osmosis

Osmosis is the selective passage of particles through a membrane. In considering what will occur when solutions are separated by a semi-permeable membrane:

1. Determine which particles can pass through the membrane.
2. Remember that materials will move (if they can pass through the membrane) from where they are in high concentration to where they are in low concentration.
3. Evaluate the consequences when particles move through the membrane until they are in equal concentrations on either side of the membrane.

Example; Two solutions are separated by a membrane that is permeable by water only. The solution on the left is 0.500 M Urea and the solution on the right is 1.0 M Urea.

a. The membrane is permeable to water only.

b. The solution on the left has a high % water and not much solute; the solution on the right has a lower % water and more solute. Water will move from where it (the water) is in higher proportion to where it is in lower proportion.

c. Therefore, the water will move from the left to the right à. This will cause the container on the left to go down and the level of the solution on the right to go up. It will also cause the solution on the left to have a lower % of water and so become more concentrated in the solute. Whereas, the solution on the right will gain water and become more dilute. The process will occur until the two solutions have the same % water and the same concentration of solute.

Osmotic Pressure is the pressure that must be applied to the system to keep the process from occurring. If a pressure greater than the osmotic pressure of the solution is applied, it is possible to obtain Reverse Osmosis. Reverse Osmosis is used as a method to purify water (p. 516-517).

Osmotic pressure = p = MRT

M = molarity

R = the gas law constant

T = temperature in degrees Kelvin

What happens to a cell in different solutions:

An Isotonic solution: the solution has the same solute concentration and % water as the cell.

A Hypotonic: the solution has a lower solute concentration and a higher % water compared to the

cell. If the cell is placed in the solution, water will move through the membrane and

into the cell since the % water outside the cell is greater than inside,. The cell will

expand and may even burst.

A Hypertonic: the solution has a higher solute concentration and a lower % water compared to

thecell. If the cell is placed in the solution, water will move through the membrane and

into the cell since the % water outside is lower than the % water inside the cell. The cell will shrink in size.

Example 1:

1. The surrounding solution is isotonic with the solution inside the cell membrane.
2. The surrounding solution is hypotonic compared to the solution inside the cell.
3. The surrounding solution is hypertonic compared to the solution inside the cell

Example 2: It is possible to remove the shell of an egg by soaking it in vinegar. What remains is

essentially a very large cell surrounded by a cell membrane that is permeable to water. What will happen to the egg?

Use Colligative Properties to Determine Molar Mass of a Solute:

1. Use Freezing Point Depression: Chemistry 111 Lab Project I, Exercise F

Assume the solute is a non-electrolyte. Therefore, i = 1.

DT_f = K_fm

DT_f/K_f = m = molality

m = # moles of solute/#kg solvent

#moles solute = # grams solute/Molar Mass solute

2. Use Osmotic pressure:

p = MRT

M = molarity = # moles solute/ #liters of solution

#moles solute= #grams solute/Molar Mass solute

Another Type of Mixture: Colloids

What makes a mixture a colloid rather than a solution? A colloid is a mixture in which

the particles of one material are dispersed throughout another material but are

not completely homogeneous. Colloids are likely to form when particles are just

too large to make a solution.

What is the Tyndall Effect ? One way to detect colloids is the Tyndall Effect which is that

when a beam of light is passed through a colloid, the path of the light beam is

visible This does not happen in solutions because the particle size is too small to

cause the scattering of the light beam. Examples of the Tyndall Effect are seen

on a foggy night when you can see the beam of headlights in the mist or when

the path of a beam of sunshine in a room shows up because of the dust particles

suspended in the air.

What is meant by the terms hydrophobic and hydrophyllic and which is which?

Hydrophobic colloids are ones that are not likely to form in water due to the

non-polar nature of the colloidal material.

Hydrophilic colloids are ones that are likely to form in water due to the ability of

the material to form intermolecular attractions to water.

Explain briefly how soaps and detergents are emulsifying agents? Materials that act as soaps or detergents are likely to have a non-polar part that is attracted to grease and other non-polar materials and a polar end that is attracted to water. An Example: