f in al e x am : we dn esday , ap ri l 26 fro m 10:15 a m

Solutions

University Chemistry II

Spring 2006

Instructor: Dr. Sarah A. Green

Office: Chem Sci. 607

Phone: 487-2048

Email address: [email protected]

Office hours: Wednesday 1:00–3:00 pm

Class time: MWF 11:05-11:55 Place: DOW 641

Lab Supervisor: Lorri Reill y, Chemical Sci. 508B, lareill [email protected] ; 7-2044

Learning Center Coordinator: Lois Blau, Chem Sci. 206A [email protected]; 7-2297

Textbook: Chemistry: The Central Science, 10th edition, by Brown, LeMay, and Bursten.

Week Dates Chapter Topic

1 Jan 9-13 13 Solutions

2 Jan 16-20 14 Chemical Kinetics

3 Jan 23-27 15 Chemical Equilibrium

4 Jan 30-Feb 3 16 Acid-Base Equili bria

5 Feb 6-8 17 No class Friday: Winter Carnival

6 Feb 13-15 17, Review

EXAM 1 Feb 15, 6:00 pm 13-16 No class Friday, Feb 17

7 Feb 20-24 18 Environmental Chemistry

8 Feb 27-March 3 19 Thermodynamics

BREAK March 6-10

9 March 13-17 20 Electrochemistry

10 March 20-22 Review

EXAM II March 22, 6:00 pm 17-20 No class Friday, March 24

11 March 27-31 21 Nuclear Chemistry

12 April 3-7 22 Nonmetals

13 April 10-12 23 Metals

EXAM III April 12, 6:00 pm 21-23 No class Friday, April 14

14 April 17-21 25 Organic/Biochem

FINAL EXAM: Wednesday, April 26 from 10:15 am - 12:15 pm

This schedule is subject to modification. Any changes will be announced in class and posted to

the class via WebCT.

Solutions

Chapter 13

Properties of Solutions

Adapted by SA Green from:

John D. Bookstaver

St. Charles Community College

St. Peters, MO

2006, Prentice Hall, Inc.

Chemistry, The Central Science, 10th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

Solutions

Solutions

• Solutions are homogeneous mixtures of two

or more pure substances.

• In a solution, the solute is dispersed uniformly

throughout the solvent.

Solutions

Solutions

How does a solid dissolve into a liquid?

What „drives‟ the dissolution process?

What are the energetics of dissolution?

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How Does a Solution Form?

1. Solvent molecules attracted to surface ions.

2. Each ion is surrounded by solvent molecules.

3. Enthalpy ( H) changes with each interaction broken or formed.

Ionic solid dissolving in water

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How Does a Solution Form?

1. Solvent molecules attracted to surface ions.

2. Each ion is surrounded by solvent molecules.

3. Enthalpy ( H) changes with each interaction broken or formed.

Solutions

How Does a Solution Form

The ions are solvated

(surrounded by

solvent).

If the solvent is water,

the ions are

hydrated.

The intermolecular

force here is ion-

dipole.

Solutions

Energy Changes in Solution

To determine the enthalpy

change, we divide the

process into 3 steps.

1. Separation of solute

particles.

2. Separation of solvent

particles to make

„holes‟.

3. Formation of new

interactions between

solute and solvent.

Solutions

Enthalpy Changes in Solution

The enthalpy

change of the

overall process

depends on H for

each of these steps.

Start

End

EndStart

Solutions

Enthalpy changes during dissolution

The enthalpy of

solution, Hsoln, can

be either positive or

negative.

Hsoln = H1 + H2 + H3

Hsoln (MgSO4)= -91.2 kJ/mol --> exothermic

Hsoln (NH4NO3)= 26.4 kJ/mol --> endothermic

Solutions

Why do endothermic processes

sometimes occur spontaneously?

Some processes,

like the dissolution

of NH4NO3 in water,

are spontaneous at

room temperature

even though heat is

absorbed, not

released.

Solutions

Enthalpy Is Only Part of the Picture

Entropy is a measure of:

• Dispersal of energy in the system.

• Number of microstates (arrangements) in the system.

b. has greater entropy, is the favored state

(more on this in chap 19)

Solutions

Entropy changes during dissolution

Each step also involves a

change in entropy.

1. Separation of solute

particles.

2. Separation of solvent

particles to make

„holes‟.

3. Formation of new

interactions between

solute and solvent.

Solutions

SAMPLE EXERCISE 13.1 Assessing Entropy Change

In the process illustrated below, water vapor reacts with excess solid sodium

sulfate to form the hydrated form of the salt. The chemical reaction is

Does the entropy of the system increase or decrease?

Solutions

Dissolution vs reaction

• Dissolution is a physical change—you can get back the

original solute by evaporating the solvent.

• If you can‟t, the substance didn‟t dissolve, it reacted.

Ni(s) + HCl(aq) NiCl2(aq) + H2(g) NiCl2(s)dry

Solutions

Degree of saturation

• Saturated solution

Solvent holds as much

solute as is possible at

that temperature.

Undissolved solid

remains in flask.

Dissolved solute is in

dynamic equilibrium

with solid solute

particles.

Solutions


• Unsaturated Solution

Less than the

maximum amount of

solute for that

temperature is

dissolved in the

solvent.

No solid remains in

flask.

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• Supersaturated

Solvent holds more solute than is normally

possible at that temperature.

These solutions are unstable; crystallization can

often be stimulated by adding a “seed crystal” or

scratching the side of the flask.

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Unsaturated, Saturated or Supersaturated?

How much solute can be dissolved in a solution?

More on this in Chap 17

(solubility products, p 739)

Solutions

Factors Affecting Solubility

• Chemists use the axiom

“like dissolves like”:

Polar substances tend to

dissolve in polar solvents.

Nonpolar substances tend

to dissolve in nonpolar

solvents.

Solutions


The stronger the

intermolecular

attractions between

solute and solvent,

the more likely the

solute will dissolve.Example: ethanol in water

Ethanol = CH3CH2OH

Intermolecular forces = H-bonds; dipole-dipole; dispersion

Ions in water also have ion-dipole forces.

Solutions


Glucose (which has

hydrogen bonding)

is very soluble in

water.

Cyclohexane (which

only has dispersion

forces) is not water-

soluble.

Solutions


• Vitamin A is soluble in nonpolar compounds

(like fats).

• Vitamin C is soluble in water.

Solutions

Which

vitamin is

water-soluble

and which is

fat-soluble?

Solutions

Gases in Solution

• In general, the

solubility of gases in

water increases with

increasing mass.

Why?

• Larger molecules

have stronger

dispersion forces.

Solutions

Gases in Solution

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

Solutions

Gases in Solution

• The solubility of

liquids and solids

does not change

appreciably with

pressure.

• But, the solubility of

a gas in a liquid is

directly proportional

to its pressure.

Increasing

pressure

above

solution

forces

more gas

to dissolve.

Solutions

Henry‟s Law

Sg = kPg

where

• Sg is the solubility of

the gas;

• k is the Henry‟s law

constant for that gas in

that solvent;

• Pg is the partial

pressure of the gas

above the liquid.

Solutions

Henry‟s Law

Sg = kPg

k for N2 at 25°

=6.8 x 10-4 mol/L atm

Solutions

Temperature

Generally, the

solubility of solid

solutes in liquid

solvents increases

with increasing

temperature.

Solutions

Temperature• The opposite is true of

gases. Higher temperature drives gases out of solution.

Carbonated soft drinks are more “bubbly” if stored in the refrigerator.

Warm lakes have less O2 dissolved in them than cool lakes.

Solutions

Chap 13:

Ways of Expressing

Concentrations of

Solutions

Solutions

Mass Percentage

Mass % of A =mass of A in solution

total mass of solution100

Solutions

Parts per Million and

Parts per Billion

ppm =mass of A in solution


Parts per Million (ppm)

Parts per Billion (ppb)

ppb =mass of A in solution


Solutions

moles of A

total moles in solutionXA =

Mole Fraction (X)

• In some applications, one needs the

mole fraction of solvent, not solute—

make sure you find the quantity you

need!

Solutions

mol of solute

L of solutionM =

Molarity (M)

• You will recall this concentration

measure from Chapter 4.

• Because volume is temperature

dependent, molarity can change with

temperature.

Solutions

mol of solute

kg of solventm =

Molality (m)

Because neither moles nor mass

change with temperature, molality

(unlike molarity) is not temperature

dependent.

Solutions

Solutions

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations

(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass

percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+

What is the concentration of Zn2+ in parts per million?

PRACTICE EXERCISE(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A

commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl

in a bottle containing 2500 g of bleaching solution?

PRACTICE EXERCISEA commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole

fraction of NaOCl in the solution.

Solutions

Colligative Properties

• Colligative properties depend only on

the number of solute particles present,

not on the identity of the solute

particles.

• Among colligative properties are

Vapor pressure lowering

Boiling point elevation

Melting point depression

Osmotic pressure

Solutions

Vapor Pressure

As solute molecules are

added to a solution,

the solvent become

less volatile

(=decreased vapor

pressure).

Solute-solvent

interactions contribute

to this effect.

Solutions

Vapor Pressure

Therefore, the vapor

pressure of a solution

is lower than that of

the pure solvent.

Solutions

Raoult‟s Law

PA = XAP A

where

• XA is the mole fraction of compound A

• P A is the normal vapor pressure of A at

that temperature

NOTE: This is one of those times when you

want to make sure you have the vapor

pressure of the solvent.

Solutions

SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering

Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor

pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of

pure water at 25°C is 23.8 torr (Appendix B).

PRACTICE EXERCISEThe vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor

pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene

glycol in the solution?

Solutions

Boiling Point Elevation and

Freezing Point Depression

Solute-solvent

interactions also

cause solutions to

have higher boiling

points and lower

freezing points than

the pure solvent.

Solutions

Boiling Point Elevation

The change in boiling

point is proportional to

the molality of the

solution:

Tb = Kb m

where Kb is the molal

boiling point elevation

constant, a property of

the solvent.Tb is added to the normal

boiling point of the solvent.

Solutions


• The change in freezing

point can be found

similarly:

Tf = Kf m

• Here Kf is the molal

freezing point

depression constant of

the solvent.Tf is subtracted from the normal

freezing point of the solvent.

Solutions

Boiling Point Elevation and


In both equations,

T does not depend

on what the solute

is, but only on how

many particles are

dissolved.

Tb = Kb m

Tf = Kf m

Solutions

Colligative Properties of

ElectrolytesBecause these properties depend on the number of

particles dissolved, solutions of electrolytes (which dissociate in solution) show greater changes than those of nonelectrolytes.

e.g. NaCl dissociates to form 2 ion particles; its limiting van‟t Hoff factor is 2.

Solutions

Colligative Properties of

ElectrolytesHowever, a 1 M solution of NaCl does not show

twice the change in freezing point that a 1 M

solution of methanol does.

It doesn‟t act like there are really 2 particles.

Solutions

van‟t Hoff Factor

One mole of NaCl in

water does not

really give rise to

two moles of ions.

Solutions

van‟t Hoff Factor

Some Na+ and Cl−

reassociate as hydrated ion pairs, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

Solutions

The van‟t Hoff Factor

• Reassociation is

more likely at higher

concentration.

• Therefore, the

number of particles

present is

concentration

dependent.

Solutions

The van‟t Hoff Factor

We modify the

previous equations

by multiplying by the

van‟t Hoff factor, i

Tf = Kf m i

i = 1 for non-elecrtolytes

Solutions

Osmosis

• Semipermeable membranes allow some

particles to pass through while blocking

others.

• In biological systems, most

semipermeable membranes (such as

cell walls) allow water to pass through,

but block solutes.

Solutions

OsmosisIn osmosis, there is net movement of solvent from the area of higher solvent concentration (lowersolute concentration) to the are of lower solvent concentration (highersolute concentration).

Water tries to equalize the concentration on

both sides until pressure is too high.

Solutions

Osmotic Pressure

• The pressure required to stop osmosis,

known as osmotic pressure, , is

n

V= ( )RT = MRT

where M is the molarity of the solution

If the osmotic pressure is the same on both sides

of a membrane (i.e., the concentrations are the

same), the solutions are isotonic.

Solutions

Osmosis in Blood Cells

• If the solute

concentration outside

the cell is greater than

that inside the cell, the

solution is hypertonic.

• Water will flow out of

the cell, and crenation

results.

Solutions

Osmosis in Cells

• If the solute

concentration outside

the cell is less than

that inside the cell, the

solution is hypotonic.

• Water will flow into the

cell, and hemolysis

results.

Solutions

Solutions

Molar Mass from

Colligative Properties

We can use the

effects of a colligative

property such as

osmotic pressure to

determine the molar

mass of a compound.

Solutions

Colloids:

Suspensions of particles larger than

individual ions or molecules, but too small to

be settled out by gravity.

Solutions

Tyndall Effect

• Colloidal suspensions

can scatter rays of light.

• This phenomenon is

known as the Tyndall

effect.

Solutions

Colloids in Biological Systems

Some molecules have

a polar, hydrophilic

(water-loving) end and

a nonpolar,

hydrophobic (water-

hating) end.

Solutions


Sodium stearate

is one example

of such a

molecule.

Solutions


These molecules

can aid in the

emulsification of fats

and oils in aqueous

solutions.

Solutions

END Chap 13

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