Unit 1: Review and Chapter 12: Liquids and Solids Chapters 1-12

Outline

- polyatomic ions - liquids and fluids -phase diagrams

- naming - solids - crystals, amorphous -calculations

- sig figs - changes of state -balancing

- graphing - equilibrium -Lewis structures

- lab safety - Le Châtelier's principle -e− configuration

- equations: M, #p, conc, gas, light - vapour pressure -organic chem

- stoichiometry - evaporation and boiling -mass percent

-empirical/molecular formulas - molar heats of vapourizaion and fusion

Liquids and Fluids p.363

A liquid is a substance that has a definite volume and takes the shape of its container. A fluid is a

substance that can flow and therefore take the shape of its container. Gases can be considered

fluid because they can flow. According to the kinetic-molecular theory, particles of solids, liquids

and gases are in constant motion.

increasing kinetic energy gas (dipole-dipole,

↑ | hydrogen bonding, and

liquid London dispersion)

| ↓ solid increasing intermolecular forces

Liquids have higher densities than gases and are nearly incompressible. But because their

particles are in motion, liquids will diffuse through other liquids. Gases will diffuse faster than

liquids due to their particles moving more quickly. Solids will also diffuse, but millions of times

more slowly than in liquids.

Solids p.367

A solid is a substance that has a definite volume and shape. They are generally more dense than

liquids as the particles are packed even closer together, and are considered incompressible. Solids

can be classified as crystalline or amorphous. Crystalline solids are made of crystals and have an

ordered geometric pattern. When pieces of crystal break off, they retain the geometric pattern.

Two examples are salt and diamond.

There are 4 types of crystals:

� Ionic crystals - composed of positive and negative ions - are hard, brittle, good insulators and

have high melting points (e.g. NaCl)

� Covalent network crystals - composed of atoms covalently bonded - essentially giant

molecules - are very hard and brittle, have high melting points, and are nonconductors or

semiconductors (e.g. diamond)

� Covalent molecular crystals - composed of molecules covalently bonded - low melting

points, soft, good insulators (e.g. ice)

� Metallic crystals - metal atoms in ordered pattern with sea of electrons - are able to move

throughout the crystal, are good conductors - varied melting points (e.g. gold)

Amorphous solids (Gk. without shape) have no geometric pattern, and if pieces break off, they

can have varied structures. Although their particles are arranged randomly, these particles are not

constantly changing position, as in a liquid. Two examples are glass and plastic.

Changes of State p.372

In this course, we will focus on the 3 states in which all matter on Earth can exist:

melting vapourization

least energy solid î liquid î gas most energy

freezing condensation

sublimation→←deposition

Equilibrium p.372

An equilibrium is a dynamic condition in which the forward rate of a process balances the

reverse rate in a closed system. Imagine two closed containers of gas are connected and the valve

separating them is opened. Each gas would flow into the other until the rate of each gas entering

and leaving becomes equal. This is an equilibrium. When you put the lid on a bottle of water, the

particles evaporating and condensing will reach equilibrium. (fig 12-10, p.373)

liquid + heat î vapour Equilibrium is shown by a double headed arrow

which means that the process occurs in both

H2(g) + I2(g) Ö 2HI(g) directions.

Le Châtelier's Principle p.374

Because equilibria are dynamic, any change to the reactants or products will change the

equilibrium. Le Châtelier's principle states that when an equilibrium is stressed, it will shift to

relieve the stress. For example: if this system is stressed by adding more heat,

the equilibrium will shift to the right, which means

liquid + heat î vapour that it will produce more vapour until it reaches a

↓−−−−J−−−−

↑ new equilibrium.

Stresses include changes in concentration, pressure, temperature, volume, etc. (table 12-3, p.375)

Vapour Pressure p.376

Vapour pressure is the pressure exerted by the particles in vapour phase above a liquid. The

weaker the intermolecular bonds, the more easily the liquid will evaporate, and the higher the

vapour pressure. Liquids that evaporate easily (or have a higher % of liquid particles that can

evaporate) are said to be volatile. (e.g. ether, acetone) Note that the equilibrium vapour pressure

is only dependent on temperature and is not the same as vapour pressure in an open system.

Evaporation and Boiling p.365, 378

Evaporation and boiling are both forms of vapourization. Evaporation is when particles escape

from the surface of a non-boiling liquid and enter the gas state. This happens because some of the

particles on the surface of the liquid have higher than average kinetic energies, which enables

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them to break the intermolecular forces holding them in the liquid phase. Boiling occurs when

the liquid particles throughout the liquid have enough energy to break the intermolecular bonds

→ liquid and gas phases exist in equilibrium. This occurs when the liquid's equilibrium vapour

pressure equals the atmospheric pressure, which is known as the normal boiling point. Increasing

the temperature of a liquid increases its equilibrium vapour pressure. For water:

101.32547.3419.917.332.330.61Vapour Press (kPa)

100806040200Temperature (°C)

At sea level, where the atmospheric pressure is on average 101.325 kPa, (or 1 atm or 760 torr)

water will boil at 100°C. In Banff (alt 1383 m) water will boil at 95°C and on Everest (8850 m),

it will boil at 71°C. (the b.p. drops approx. 1°C per 300 m) Increasing the pressure above a liquid

also increases the equilibrium vapour pressure, which allows the liquid to reach a higher

temperature to boil - such as in a pressure cooker.

Molar Heat of Vapourizaion and Fusion p.379

As water boils, the temperature remains constant. The energy taken in by the water is used to

break the intermolecular bonds and release vapour particles. The amount of heat required to

vapourize 1 mole of a liquid at its boiling point is called the molar heat of vapourization. At

1 atm, water has a molar heat of vapourization of 40.79 kJ/mol.

The molar heat of fusion is the amount of heat required to melt 1 mole of solid at its melting

point. Freezing and condensation work the same way, with the same energies, except energy

must be removed. At low temperatures and pressures, liquids can not exist. In that case, solids

can be in equilibrium with vapours, and sublimation and deposition occurs.

Phase Diagrams p.381

These are pressure vs. temperature graphs showing all the phases of a substance. (See fig 12-14,

p381) Note that water expands when it freezes and therefore increasing pressure lowers the

melting point. This is unusual in nature. The triple point is the temperature and pressure that a

substance can exist as a solid, liquid, and gas at the same time. The critical point is the

temperature above which the substance can not exist as a liquid, regardless of pressure. The

critical pressure is the lowest pressure at the critical point.

Calculations p.386

ex) How much heat is absorbed when 235 g of water boils?

∆Hvap = q where: q = quantity of heat (kJ)

n n = number of moles (mol)

∆Hvap = molar heat of vapourization (kJ/mol)

n = m = 235 g = 13.041... mol heat = 13.041... mol • 40.79 kJ= 531.945.. kJ

M 18.02 g/mol mol

q = 532 kJ of heat

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ex 2) What is the molar heat of fusion for ethanol if 1.50 x 1024 molecules release 12.3 kJ of

energy when they freeze?

∆Hfus = q n = #p = 1.50 x 1024 = 2.4908... mol

n 6.022 x 1023 6.022 x 1023

∆Hfus = q = 12.3 kJ = 4.93804 = 4.94 kJ/mol

n 2.4908... mol

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Unit 2: Solutions Chapters 13, 14

Outline

-Types of mixtures -Molality: m = n/kg -Changing B.P. and F.P.

-solutions -% conc, ppm, ppb -∆tb = Kbm

-suspensions -Dissociation and Ionization -∆tf = Kf m

-colloids -Precipitation Reactions -Electrolytes and ∆t

-Solubility -Overall Ionic Equation

-Henry's Law -Net Ionic Equation

-Heat of Solution (∆Hsolution) -Electrolytes: strong and weak

-Concentration -Nonelectrolytes: volatile or not

-Molarity: C = n/V -Lowering Vapour Pressure

Types of Mixtures p.395

Solutions: A solution is a homogeneous mixture of two or more substances in a phase (e.g. salt

water). Solutions are usually used to describe liquid mixtures, but solutions can be gas mixtures

(like air) or solid mixtures (like alloys). The dissolving medium is called the solvent and the

substance being dissolved is called the solute.

Suspensions: A suspension is a heterogeneous mixture where the particles in the solvent are

comparatively large, and will settle out if unagitated (e.g. muddy water). Suspensions can be

separated by filtration.

Colloids: A colloid is similar to a suspension, but the particles are smaller and therefore do not

settle and cannot be filtered (e.g. milk). Again, the particles being dispersed can be solid, liquid

or gas, and the dispersing medium can also be any of the three. See table 13-2, p.398. Colloids

and suspensions will scatter light, whereas solutions will not.

Solubility p.402

Solubility describes the amount of solute per amount of solvent required to produce a saturated

solution at a given temperature. It is usually measured in g of solute per g solvent. A saturated

solution has the maximum amount of solute dissolved in that solvent at that temperature. For

example, if you add sugar to water until some remains undisolved at the bottom, it is saturated.

(sucrose has a solubility of 203.9 g/100 g water at 20°C) The solution has reached solution

equilibrium, where dissolution and crystallization occur at the same rate.

A supersaturated solution contains more solute than a similar saturated solution under the same

conditions. This is usually done by heating the solution, adding the solute and cooling it slowly.

Dissolution - Hydration is the process of dissolving where water is the solvent. Like dissolves

like. Water is a polar molecule, and therefore polar and solids dissolve well in water, whereas

non polar solids do not dissolve as well. Likewise, non polar solvents dissolve non polar solutes

better than polar solutes. Liquids that do not mix (such as oil and water) are immiscible, whereas

liquids that mix in any proportion (such as ethanol and water) are miscible.

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Henry's Law p.407

The solubility of a gas in a liquid is proportional to the partial pressure of the gas above the

solution. In other words, if you increase the pressure of a gas above a liquid, you increase the

solubility of the liquid, and it will dissolve more gas. (such as in soft drinks)

solvent + gas î solution Le Châtelier's principle also explains this change

Increasing the temperature of liquids decreases the solubility of gases, as their particles of greater

energy form an equilibrium using fewer particles. Increasing the temperature of liquids is less

predictable for solids, but it usually increases their solubilites.

Heats of Solution p.409

When solutes mix with solvents, the solvation of the particles results in a release or absorption of

heat. Heat or energy is required to break the bonds holding the solute particles together, (and

solvent particles together) causing an absorption of heat. Also, heat is released when bonds are

formed between the solute and the solvent. If the amount of heat absorbed is greater than the heat

released, the ∆Hsolution = positive (gets cold) ; If the amount of heat released is greater, the

∆Hsolution = negative (gets hot). Recall that ∆H = q / n.

Concentration p.412

Concentration can be expressed in many ways. We will study molarity, molality, percent and

parts per million/billion.

Molarity is most frequently used, and describes the number of moles of solute per litre of

solution (not solvent). C = concentration in mol/L or M

C = n V = volume of solution in L

V n = number of moles in mol

Ex 1 ) Find the molarity of a solution made by dissolving 254 g of KCl in enough water to make

1450 mL.

C = ? n = m = 254 g = 3.407109... mol

n = M 74.55 g/mol

V = 1450 mL = 1.45 L

m = 254 g C = n = 3.407... mol = 2.3497..... = 2.35 mol/L or M

M = 74.55 g/mol V 1.45 L

Molality is used when dealing with vapour pressure and temperature changes, as it does not

change with temperature. Molality describes the number of moles of solute per kilogram of

solvent (not solution). m = molality in mol/kg

m = n msolvent = mass of solvent in kg

msolvent n= number of moles in mol

You may also wish to use these formulas: M = n/V m = n/kg

C = n/L

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Ex 2) Find the molality of a solution if 284 g of CuSO4 • 5H2O is dissolved in 6.7 kg of water.

m = ? n = m = 284 g = 1.1372... mol

n = M 249.72 g/mol

msolvent = 6.7 kg

msolute = 284 g m = n = 1.137... mol = 0.1697... = 0.17 mol/ kg or m

M = 249.72 g/mol kg 6.7 kg

Concentration can also be expressed as percent, parts per million (ppm), or parts per billion

(ppb). 1 percent means that there is 1 unit of solute per 100 units of solution. This is usually

described as weight of solute / weight of solution (w/w) or weight of solute / volume of solution

(w/v). For example: sodium fluoride in toothpaste is 0.243 % (w/w), or the concentration of

chlorine in Saskatoon's water in mg/mL is 0.17 % (w/v).

To calculate the percent concentration, you can use the following ratio:

percent means out of 100

% conc. = amount of solute

100 amount of solution 1 % is like 1 part per hundred

One part per million means that - out of 1 million particles of solution, there is 1 particle of

solute and 999 999 particles of solvent. Note that the units for the amounts of solute and solution

should be the same. For these calculations, you may assume that 1.00 mL of water = 1.00 g.

To calculate ppm or ppb, use the same ratio as above:

ppm = amount of solute = ppb

1 000 000 amount of solution 1 000 000 000

Note: 1 ppm = 1 mg/kg = 1 mL/L = 1 mg/L (for water) = 1 second/280 hours

1 ppb = 1µg/kg

Ex 1) How many grams of HCl are there in 120 g of a 6.0 % (w/w) solution?

6.0 % = x x = 7.2 g

100 120 g

Ex 2) Find the concentration of chlorine in Saskatoon's tap water in ppm if 425 g of chlorine is

found in 250 000 L of water.

note 1 L = 1000 g

ppm = 425 g

1 000 000 250 000 000 g conc = 1.7 ppm

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Dissociation and Ionization p.425

When ionic compounds dissolve, they dissociate into their ions. (assume 100 % dissociation)

CaCl2(s) –—H2

O—→ Ca2+(aq) + 2Cl–(aq) Some examples of dissociation eqns

Ni2(SO4)3(s) –—H2

O—→ 2Ni3+(aq) + 3SO42–(aq)

Ionization is different in that ions are produced from molecular substances. Ionization occurs

when the intramolecular force in the solute is weaker than the attraction to the solvent.

The attraction of the HCl molecule to water is

HCl –—H2

O—→ H+(aq) + Cl–(aq) stronger than the H-Cl bond.

Precipitation Reactions p.427

When a product of a reaction is insoluble, it precipitates out to form a solid. To predict whether a

precipitate will form, use a solubility chart. (such as table 14-1, p.427)

Zn(NO3)2(aq) + (NH4)2S(aq) → ZnS( ) + 2NH4NO3( )

3CuCl2(aq) + 2Na3PO4(aq) → Cu3(PO4)2( ) + 6NaCl( )

The overall ionic equation is

3Cu2+(aq) + 6Cl−(aq) + 6Na+(aq) + 2PO43−(aq) → Cu3(PO4)2(s) + 6Na+(aq) + 6Cl−(aq)

The spectator ions are not involved in the precipitation:

6Cl−(aq) + 6Na+(aq) → 6Na+(aq) + 6Cl−(aq)

The net ionic equation includes only the ions that make the precipitate:

3Cu2+(aq) + 2PO43−(aq) → Cu3(PO4)2(s) Note: this eqn applies to other reactions

Electrolytes and Nonelectrolytes p.399

An electrolyte is a substance that when dissolved in water, forms ions which conduct an electric

current (e.g. NaCl). Ionic compounds, acids and bases are electrolytes. Some polar molecules

will also form ions in solution. A nonelectrolyte is a substance that when dissolved in water, does

not produce ions and will not conduct an electric current (e.g. sucrose).

Increasing surface area of the solute, agitating the mixture and increasing the temperature all

increase the rate of dissolution.

Strong and Weak Electrolytes p.432

A strong electrolyte is a compound that breaks up (dissociates or ionizes) entirely or almost

entirely and exists as ions in solution. For example, HCl is 100 % ionized in water. Note that this

is independent of solubility, and even slightly soluble compounds can be strong electrolytes

(what does dissolve is in ion form)

A weak electrolyte has only a small amount of the dissolved compound existing as ions.

HCN(aq) + H2O(l) î H3O+(aq) + CN−(aq) weak electrolyte - only small amount ionizes

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Lowering Vapour Pressure p.436

A colligative property is one that depends on how many particles of solute are present, but not on

their identity. When dealing with colligative properties, use molality for concentration. When a

nonvolatile (does not readily change to gas) solute is placed in a solvent, it lowers its vapour

pressure. This is because the new solute takes the place of solvent particles on the surface,

keeping them from evaporating. This lowering depends on the concentration of solute and not the

type of solute. Therefore, equal concentrations of sugar or antifreeze (or any nonvolatile

nonelectrolyte) in water lower the vapour pressure by the same amount.

Changing the Boiling and Freezing Points p.438

Lowering vapour pressure will raise the boiling point because more heat is needed to make the

vapour pressure equal the atmospheric pressure. Similarly, adding a nonvolatile nonelectrolyte to

a solvent will lower the freezing point. Because the change in b.p. and f.p. is the same for all

nonvolatile nonelectrolytes, we can determine how much a b.p. or f.p. will change given a molal

concentration. ∆tb = change in b.p. in °C ∆tf = change in f.p. in °C

∆tb = Kbm Kb = molal b.p. constant in °C/m

∆tf = Kf m Kf = molal f.p. constant in °C/m

m = molality in mol/kg

Table 14-2 on p.438 has some molal constants. Water has a Kf of -1.86°C/m and a Kb of

0.51°C/m. That means that a 2 m solution with any nonvolatile nonelectrolyte will have a b.p. of

101.02°C and a f.p. of -3.72°C.

Ex 1) Calculate the change in boiling point by adding 85 g of glucose (C6H12O6 ) to 500. mL of

water. (Glucose is a nonelectrolyte)

∆tb = Kbm m = n n = 85 g = 0.471... mol = 0.943...mol/kg

kg 180.18 g/mol 0.500 kg

∆tb = 0.51 °C/m x 0.945... m = 0.4811... = 0.48°C the b.p. increases by 0.48°C

Electrolytes and these Changes p.443

Adding an nonvolatile electrolyte will have a similar effect to a nonelectrolyte, only to a greater

degree. For example, 1 mol of NaCl will dissociate into 2 mol of ions, which will lower the f.p.

about twice as much as 1 mol of sugar. The effects on the b.p. are similar. CaCl2 will lower the

f.p. of water by about 3 times more than sugar.

The observed change of the f.p. or b.p. differs slightly from the expected values due to attractive

forces between the dissolved ions. As the solution becomes more dilute, the observed values

approach the expected. Also, ions of greater charge will also attract each other more strongly, and

will not change the b.p. or f.p. as much as expected. For example, MgSO4 will not depress the

f.p. of water as much as does KCl.

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Unit 3: Acids and Bases Chapters 15, 16

Outline

-Acid names

-Arrhenius acids and bases -neutralization reactions

-strength of acids and bases -ionization constant of water:

-Brønsted-Lowry acids and bases -Kw = [H3O+] [OH–] = 1.0 % 10–14 M2

-polyprotic acids -calculating concentrations

-Lewis acids and bases -pH, pOH

-conjugate acids and bases -titration

-amphoteric -calculations with pH and titration

-amphiprotic

Acid Nomenclature p.454

Binary acids contain hydrogen and one of the more electronegative elements. Oxyacids contain

hydrogen, oxygen and a third element, usually a nonmetal. Some common acids:

Binary acids Oxyacids

HF hydrofluoric acid CH3COOH acetic acid see table 15-2 p.455

HCl hydrochloric acid H2CO3 carbonic acid

HBr hydrobromic acid HNO3 nitric acid

HI hydriodic acid H3PO4 phosphoric acid

H2S hydrosulfuric acid H2SO4 sulfuric acid

The prefixes and suffixes refer to the number of oxygen atoms on the central atom:

hypochlorous acid HClO hypo____ous means "2 less oxygen"

chlorous acid HClO2 ____ous means "1 less oxygen"

chloric acid HClO3 ____ic is the most common form or first discovered

perchloric acid HClO4 per____ic means "1 more oxygen"

Arrhenius Acids and Bases p.459

According to Arrhenius, an acid is a compound that will increase the [H+] in solution; A base is a

compound that will increase the [OH−]. Compounds will ionize or dissociate into these ions, or

produce them from water. For example, HCl will ionize into H+ and Cl− (acid). Actually, the

polar water molecule attracts the H+ ion and forms H3O+, the hydronium ion.

HCl(g) + H2O(l) → H3O+(aq) + Cl−(aq) acid

NaOH(s) –—H2

O—→ Na+(aq) + OH–(aq) base

Strong acids and bases ionize or dissociate completely and are also strong electrolytes. The more

polar the bond with hydrogen, the easier it is to break and form ions → strong acid. Weak acids

and bases only ionize or dissociate partially, and are weak electrolytes.

HF(g) + H2O(l) î H3O+(aq) + F−(aq) weak acid (do not confuse w/ corrosiveness)

NH3(g) + H2O(l) î NH4+(aq) + OH−(aq) weak base

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The solubility of the compound also affects the strength of acids and particularly bases. For

example, if a metal hydroxide is insoluble, it will not produce many OH− ions.

Limitations of Arrhenius definition:

-it predicted that compounds that did not produce H+ or OH− ions would be neutral

-it did not account for acids or bases not in aqueous solution

Brønsted-Lowry Acids and Bases p.464

In 1923 Brønsted and Lowry independently expanded Arrhenius' definition. A Brønsted-Lowry

acid is a molecule or ion that is a proton donor (H+ is a proton). As well, a base is a proton

acceptor.

HCl + NH3 → NH4+ + Cl−

HCl is the proton donor, NH3 is the acceptor

Arrhenius hydroxide bases such as NaOH are not Brønsted-Lowry bases, as they do not accept a

proton. The OH− ion would be the base.

HCl monoprotic acid (can donate 1 proton/molecule)

H2SO4 diprotic acid \ (also called polyprotic)

H3PO4 triprotic acid /

ionization series:

(weak acid) H3PO4 + H2O î H3O+ + H2PO4

− (greatest conc.) note: protons are

(weaker acid) H2PO4−+ H2O î H3O

+ + HPO42−

(lower conc.) donated one at a time

(weaker still) HPO42−+ H2O î H3O

+ + PO43−

(lowest conc.)

Limitations of Brønsted-Lowry definition:

-compounds could only be acids if they contained hydrogen

Lewis Acids and Bases p.467

In 1923 also, a yet broader definition of acids and bases was proposed by Gilbert Lewis. A Lewis

acid is an atom, molecule, or ion that accepts an electron pair to form a covalent bond. (Acid

Accepts) A Lewis base is an electron pair donor. (some dots are missing)

: F : . . : F : −

: F : B + : F :− → : F : B : F : : F : . . : F : Lewis acid Lewis base adduct

co-ordinate covalent bond

(where both e− come from

the same atom)

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Note that H+ acts as a Lewis acid, because it can accept an e− pair. We will use the

Brønsted-Lowry definition of acids and bases unless otherwise stated.

Summary: Acid Base Table 15-5 p.468

Arrhenius H+ (H3O+) producer OH− producer

Brønsted-Lowry H+ (proton) donor H+ (proton) acceptor

Lewis e− pair acceptor e− pair donor

Conjugate Acids and Bases p.469

When an acid donates a proton (H+), it creates a conjugate base, which is what is left over after

loosing a proton.

acid - - - - - - - - - - - - - - - - - - conjugate base A conjugate base can accept H+ to

HF(g) + H2O(l) î H3O+(aq) + F−(aq) reform that acid

base - - conjugate acid

A conjugate acid can release H+ to

NH3(g) + H2O(l) î NH4+(aq) + OH−(aq) reform that base

base acid conj acid conj base

The stronger the acid, the more it ionizes and the less likely the reverse reaction will occur.

Therefore, the stronger the acid or base, the weaker the conjugate base or acid will be. See table

15-6 on p.471.

Amphoteric p.471

In the above examples, water can act as either an acid or a base, and is called amphoteric.

donates H+ accepts H+ (water is amphoteric)

H3O+ ←H2O→OH−

acts as acid acts as base

This is easily confused with amphiprotic, which means that it can donate or accept H+. All

amphiprotic substances are amphoteric, but there are substances that can act as either acid or base

that do not have hydrogen, and can't be called amphiprotic. (fr Gk, amphi - both)

Neutralization p.473

In aqueous solutions, neutralization involves the reaction of hydronium and hydroxide ions to

form water and a salt, which is an ionic compound made from the cation from a base and the

anion from an acid.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

ionization/dissociation

HCl(g) + → H3O+(aq) + Cl−(aq) NaOH(s) –—H

2O—→ Na+(aq) + OH–(aq)

overall ionic eqn

H3O+(aq) + Cl−(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl−(aq) + 2H2O(l)

net ionic eqn

H3O+(aq) + OH–(aq) → 2H2O(l)

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Ionization of Water p.481

Pure water will actually ionize on its own. This makes water a weak electrolyte.

H2O(l) + H2O(l) î H3O+(aq) + OH–(aq)

But the concentration of its ions [H3O+], [OH–] is extremely small. [X] means the concentration

of X in mol/L. At room temperature, both [H3O+] and [OH–] = 1.0 % 10–7 M. As the [H3O

+]

increases, the [OH–] decreases, and their product is constant. This is known as the ionization

constant of water. (Kw)

Kw = [H3O+] [OH–] = 1.0 % 10–14 M2 (at 25°C)

Calculating Concentrations p.483

Because [H3O+] = [OH–] for water, it is neutral. However, if a solution has a greater [H3O

+] than

[OH–], it is acidic. For bases, [OH–] > [H3O+].

Ex 1) Calculate the [H3O+] and [OH–] for a 5.0 % 10–2 M solution of HI.

HI(l) + H2O(l) → H3O+(aq) + I–(aq)

M = 5.0 % 10–2 M M = 5.0 % 10–2 M

[H3O+] = 5.0 % 10–2 M (every mole of HI will produce 1 mole of H3O

+)

[OH–] = Kw = 1.0 % 10–14 M2 = 2.0 % 10–13 M

[H3O+] 5.0 % 10–2 M

Ex 2) What is the [H3O+] for a 3.0 M solution of H2SO4? (we will assume 100% ionization)

H2SO4(l) + 2H2O(l) → 2H3O+(aq) + SO4

−(aq) ([OH–] = 1.7 % 10–15 M)

M = 3.0 M M = 6.0 M

pH p.485

Depending on whom you believe, pH stands for the potential or power of hydrogen. The pH scale

is a way to describe the [H3O+] in a convenient way. Mathematically p means −log10.

pH = −log [H3O+] also: pOH = −log [OH−]

The pH scale generally ranges from 0 - 14, with 7 being neutral. pH below 7 is acidic, and above

is basic. Note that pH + pOH = 14. See table 16-3 p.486 for examples.

Ex 1) calculate the pH of a 2.5 % 10–2 M solution of HCl.

pH = −log [H3O+] = −log [2.5 % 10–2 M] = 1.60205999... pH = 1.60

(2 sig figs)

Significant figures with pH are a little different. The number of digits after the decimal

determines the number of sig figs in a pH value.

ex 2) Calculate the [H3O+] of a solution with a pOH of 8.0.

pH = 14 − pOH = 14 − 8.0 = 6.0

pH = −log [H3O+] −log [H3O

+] = 6.0 log [H3O+] = −6.0

[H3O+] = antilog(−6.0) [H3O

+] = 1 % 10–6 M (1 sig fig)

12

Titration p.497

So far, we have only been dealing with strong acids and bases (complete ionization/dissociation)

Weak acids and bases do not completely break into their ions, and can't be calculated this way.

An approximation of the pH can be found experimentally using and indicator or more accurately

by titration. Titration is the process of mixing an acid with a base to find the equivalence point -

when the concentration of [H3O+] = [OH–]. If you know the concentration and volume of one

reactant (the standard solution), you can determine the concentration of the other, knowing the

volume required to "neutralize" it. For a strong acid (like HCl) and a strong base (like NaOH),

the equivalence point occurs around pH = 7. However, when you titrate a weak acid (like acetic

acid) with a strong base like (NaOH), the equivalence point will be around pH = 9. It is important

to use the appropriate indicator for each titration - one that has an end point (point where it

changes colour) near the equivalence point.

Ex 1) In a titration, 50.0 mL of 2.00 M HCl is required to reach the endpoint with 25.0 mL of

NaOH. Find the concentration of sodium hydroxide.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

n = 0.1000 mol n = 0.1000 mol

C = 2.00 M C = ?

V = 0.0500 L V = 0.0250 L

n = CV = 2.00 M % 0.0500 L C = n = 0.1000 mol = 4.00 M

= 0.1000 mol V 0.0250 L

Ex 2) What is the concentration of HBr if 38.7 mL of 0.0500 M Ca(OH)2 will reach the

equivalence point with 20.0 mL?

Ca(OH)2(aq) + 2HBr(aq) → CaBr2(aq) + 2H2O(l)n = 0.001935 mol n = 0.00387 mol

C = 0.0500 M C = ? use stoich to find missing value

V = 0.0387 L V = 0.0200 L

n = CV = 0.001935 mol C = n/V = 0.00387 mol/0.0200 L = 0.1935 = 0.194 M

13

Unit 4: Reaction Energy and Kinetics Chapter 17

Outline

-exo/endothermic -Gibbs free energy

-q = m cp ∆T -∆G ° = ∆H ° −T∆S °-enthalpy -reaction mechanisms

-heat of reaction (∆H) -collision theory

-heat of formation (∆H°f) -activation energy (Ea)

-heat of combustion (∆H°c) - factors affecting reaction rate

-Hess's Law -rate = ∆ [ ]

-entropy (S) t

-∆H°reaction = �∆H°products − �∆H°reactants also for G and S

Thermochemistry p.511

Thermochemistry is the study of the transfer of heat energy in chemical reactions. All chemical

reactions have these changes in energy and can absorb heat (endothermic) and release heat

(exothermic). We cannot directly measure heat, but we can measure temperature. Temperature is

the average kinetic energy of the particles in a sample, and is measured in °C or K. Heat is energy

transferred between samples due to a temperature difference, and is measured in J. To calculate

the heat transferred, we can use the equation where: q = energy gained (or lost) in J

m = mass of sample in g

q = m cp ∆T cp = specific heat capacity at const. press. in J/g • °C

∆T = temperature change in °C (or K)*

Ex) How much heat is gained by 245g of iron that is heated from 22.0°C to 40.0°C?

q = ? q = 245 g % 0.449 J/g K % 18.0 °C

m = 245 g = 1980.09 J table 17-1 p.513 has cp values

c = 0.449 J/g K q = 1980 J

∆T = 40.0-22.0 = 18.0 °C

This equation can be rearranged to solve for any of the variables. If heat is lost, than both the ∆T

and q values will be negative.

Heat of Reaction p.514

The heat of reaction is the amount of heat released or absorbed in a chemical reaction. This is the

difference in chemical potential energy between the reactants and products. If the products have

less stored energy than the reactants, heat is released in an exothermic reaction.

2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ or ∆H = – 483.6 kJ

In this thermochemical reaction, heat is given off (exothermic). The amount of heat produced

depends on the moles of reactants. Use half as much, produce half the heat. Another way to show

heat change is by including the enthalpy value (∆H). Enthalpy is the heat of a system, and ∆H

shows the amount of energy gained or lost. (see also energy graphs p.516)

∆H = + (heat absorbed)

∆H = H products − H reactants ∆H = − (heat released)

14

The reverse reaction would be endothermic. (note: include energy or enthalpy, but not both)

2H2O(g) + 483.6 kJ → 2H2(g) + O2(g) or ∆H = +483.6 kJ

Heat of Formation p.517

In a synthesis reaction where a compound is made from its elements, heat is absorbed or released.

The amount of heat absorbed or released to form 1 mole of a substance is called the molar heat of

formation. For example, ∆Hf of H2O(g) in the above example is – 241.8 kJ/mol. The standard

molar heat of formation occurs at room temperature (25°C/298.15 K) and 1 atm. The standard

states ( °) at that temperature are used (water is a liquid at 25°C). For water: ∆H°f = -285.8 kJ/mol

The more heat released in formation, the more stable the compound. If the ∆H°f is positive (or

only slightly negative), the compound can decompose into its elements. The ∆H°f = 0 for

elements. Some examples: Fe2O3 = – 1118.4 kJ/mol C2H2 (acetylene) = +228.2 kJ/mol

See table A-14 p.902.

Heat of Combustion p.519

The heat of combustion is the heat released from the combustion of 1 mole of that substance. For

example, propane has a ∆H°c = of – 2219.2 kJ/mol.

Hess's Law p.519

Hess's law states that the overall enthalpy change in a reaction is equal to the sum of the enthalpy

changes for the steps in the reaction. Sometimes we cannot measure the heat of a reaction

because it takes too long (like rusting) or it is not possible to separate from other reactions (like

burning C - forms CO2 and CO). We can use known steps to find what we cannot measure.

For example, find the heat of reaction for the burning of atomic sulfur to form sulfur trioxide.

2S(s) + 3O2(g) → 2 SO3(g) ∆H°= ?

Known steps

S(s) + O2(g) → SO2(g) ∆H°c = – 297 kJ/mol

2SO2(g) + O2(g) → 2SO3(g) ∆H°c = – 198 kJ/mol

Look for formulas you want to keep, make sure the number of moles will match ( × if needed)

2S(s) + 2O2(g) → 2SO2(g) ∆H°c = – 594 kJ/mol ← times 2

2SO2(g) + O2(g) → 2SO3(g) ∆H°c = – 198 kJ/mol

Add the steps and cancel to get net equation

2S(s) + 2O2(g) → 2SO2(g) ∆H°c = – 594 kJ/mol

2SO2(g) + O2(g) → 2SO3(g) ∆H°c = – 198 kJ/mol

2S(s) + 3O2(g) →2 SO3(g) ∆H° = – 594 +(–198) = – 792 kJ/mol

15

Ex 2 - (harder) Calculate the heat of reaction for the combustion of propane given the following:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H°= ?Given steps: (full steps may not be given)

« 3C(s) + 4H2(g) → C3H8(g) » ∆H°f = – 103.9 kJ/mol

2H2(g)+ O2(g) → 2H2O(l) ∆H°f = – 571.6 kJ/mol

≈ C(s) + O2(g) → CO2(g) ∆ ∆H°f = – 393.5 kJ/molRearrange and multiply

C3H8(g) → 3C(s) + 4H2(g) ∆H°f = +103.9 kJ/mol ← reversed

4H2(g) + 2O2(g) → 4H2O(l) ∆H°f = – 1143.2 kJ/mol ← times 2

3C(s) + 3O2(g) → 3CO2(g) ∆H°f = – 1180.5 kJ/mol ← times 3

Cancel

C3H8(g) → 3C(s) + 4H2(g) ∆H°f = +103.9 kJ/mol

4H2(g) + 2O2(g) → 4H2O(l) ∆H°f = – 1143.2 kJ/mol

3C(s) + 3O2(g) → 3CO2(g) ∆H°f = – 1180.5 kJ/mol

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H°= – 2219.8 kJ/mol

Entropy p.526

Entropy is a measure of the disorder or randomness in a substance. Entropy has the symbol S and

its units are kJ/mol• K. Entropy increases as a substance is heated from a solid→liquid→gas.

Forming solutions, mixing gases, and dissolving all increase entropy (∆S = +). At absolute zero,

the entropy is zero.

In nature, processes drive towards the lowest enthalpy and highest entropy. When these two

oppose each other, Gibbs free energy can be used to determine which will prevail and what will

happen to the reaction. Only the change in free energy can be measured.

Where: ∆G° = free energy change in kJ/mol

∆G ° = ∆H ° −T∆S ° ∆H° = enthalpy change in kJ/mol

T = temperature in K

(note: standard states ° ) ∆S° = entropy change in kJ/mol• K

This change in free energy can be used to predict whether a reaction will occur spontaneously.

∆G ° = negative → reaction will occur (spontaneous)

∆G ° = zero → reaction at equilibrium

∆G ° = positive → reaction will not occur

Ex) Predict whether the following reaction will occur at 25°C : NH4Cl(s) → NH3(g) + HCl(g) if

∆H° = 176 kJ/mol and ∆S0 = 0.285 kJ/mol K.

∆G ° = ∆H° −T∆S° ∆G ° =176 kJ/mol − 298K • 0.285 kJ/mol•K

∆G ° = 176 kJ/mol − 84.93 kJ/mol = 91 kJ/mol not spontaneous (∆G°= +)

16

If given thermochemical data, you can calculate the ∆ H, S or G of a reaction by using these eqns

∆H°reaction = �∆H°products − �∆H°reactants ∆G°reaction = �∆G°products − �∆G°reactants

∆S°reaction = �∆S°products − �∆S°reactants � means "the sum of"

Ex) Use thermochemical data to calculate the change in entropy in the following reaction:

2Na(s) + Cl2(g) → 2NaCl(s)

∆S° (in J/mol•K) = 51.2 223.1 72.1

∆S°reaction = �∆S°products − �∆S°reactants

= 2(72.1) − 223.1 + 2(51.2)

= 144.2 - 325.5 = −181.3 J/mol• K (losing entropy, gaining order, not spontaneous)

This process can be used for enthalpy, entropy, and free energy.

Collision Theory p.532

This theory states that in order for particles to react, they must collide with enough energy and at

the right orientation to form new species. See fig 17-9 p.533 for examples. When molecules

collide, some of the higher kinetic energy is converted into potential energy, which breaks bonds.

This allows new bonds to form, creating an activated complex - a temporary structure with

partial bonding of all reactants. Depending on which bonds break, the activated complex may

reform the reactants or form new products. Energy is required to create these complexes, and is

called activation energy: Ea for the forward and Ea' for the reverse.

17

Reaction Rates p.538

A reaction rate is the change in concentration of the reactants per unit of time. Factors that

influence the rate include:

� nature of reactants - i.e. the reactivity of the reactants under those conditions

� surface area - increase surface area → increase rate (more particles in contact)

� temperature - increase temp. → increase rate (more collisions, higher energy)

� concentration - increase conc. → increase rate (more particles in contact)

� adding a catalyst → increase rate (new reaction pathway with lower E a )

Given the reaction rate of a compound, you can use mole ratios to find the rate of other

compounds.

Ex 1) If the reaction rate for CuCl2 is −1.5 M/min, what is the rate of formation of Cu3(PO4)2?

3CuCl2(aq) + 2Na3PO4(aq) → Cu3(PO4)2(s) + 6NaCl(aq)−1.5 M/min rate = + 0.5 M/min

3 = 1

−1.5 M/min x Note that rates for reactants will be (−) and products will be (+)

because reactants are being used up, while products are formed.

Reaction Mechanisms p.531

A reaction mechanism is the step-by-step sequence of reactions in a chemical change.

For I2(g) + H2(g) î 2HI(g) An example of a homogeneous reaction - where the

reactants and products are in the same phase.

I2 î 2I step 1 fast (homogeneous chemical system - all steps same too)

I + H2 î H2I step 2 fast Note I and H2I are intermediates

H2I +I î 2HI step 3 slow (a step but not in net eqn)

I2 + H2 î 2HI

Some reaction mechanisms have several steps, each with its own activation energy. The step with

the largest Ea takes the longest and is the rate determining step. See fig 17-11 p.535 - H2 and H2I

are called intermediates and are more stable than activated complexes.

In the above mechanism, steps 1 and 2 have smaller activation energies, and are faster steps.

Step 3 has the largest Ea and is therefore the rate determining step.

18

Unit 5: Equilibrium Chapter 18

Outline:

-favoured reactions -Ka and pKa

-the equilibrium constant Keq -Kb and pKb

-calculating Keq -calculations of Ka, pKa, Kb, and pKb

-equilibrium shift -buffers

-press. conc. temp. -hydrolysis

-factors affecting Keq -Ksp and calculations

-ICE method -Ksp and solubility

-common ion effect -predicting precipitates

Favoured Reactions p.554

Recall that chemical equilibrium involves a reversible reaction in which the forward rate equals

the reverse rate, and therefore the concentrations of reactants and products remain constant. So

far, we have assumed that neither forward nor reverse reactions were favoured, and the

concentrations of reactants and products are roughly equal at equilibrium.

If however, the forward reaction is favoured, the forward reaction is nearly complete before the

reverse reaction establishes equilibrium. In this case, there is a higher concentration of products

than reactants at equilibrium.

HBr(aq) + H2O(l) ___î H3O

+(aq) + Br−(aq) forward favoured

If the reverse reaction is favoured, the forward reaction is just starting when the reverse reaction

establishes equilibrium. Here, there is a higher concentration of reactants than products at

equilibrium.

H2CO3(aq) + H2O(l) ì___

H3O+(aq) + HCO3

−(aq) reverse favoured

The Equilibrium Constant p.556

The equilibrium constant (K or Keq) describes the extent to which the reaction has been carried

out by the time it reaches equilibrium. In a reaction: Where: [A]a = the conc. of A to the

aA + bB → cC +dD Keq = [C]c [D]d power of its coefficient

[A]a [B]b Keq = eq. const. (units vary)

Keq = [products] In general

[reactants]

If Keq = >1 products favoured (more products around at eq.)

Keq = 1 equal conc. of reactants and products

Keq = < 1 reactants favoured (more reactants around at eq.)

The value of Keq is found experimentally and is dependent on temperature. If gases are used,

pressure takes the place of concentration, however your text uses concentration. The conc. of

pure solids and liquids essentially remain constant and are left out of equilibrium expressions.

19

Ex 1) Calculate the equilibrium constant for the reaction at constant temperature -

N2 + O2 î NO if the concentrations at equilibrium are: [N2] = 6.4 × 10−3 mol/L, [O2] = 1.7 × 10−3

mol/L, [NO] = 1.1 × 10−5 mol/L.

Keq = [NO]2 = (1.1 × 10−5M)2 = 1.1 × 10−5

N2(g) + O2(g) î 2NO(g) [N2][O2] (6.4 × 10−3M)(1.7 × 10−3M)

Ex 2) Given: H2(g) + Br2(g) î 2HBr(g), Keq = 3.5 × 104, and constant temp.

a) Calculate the concentration of H2 if [HBr] = 9.8 × 10−2 M, and [Br2] = 4.3 × 10−3 M.

Keq = [HBr]2 [H2] = [HBr]2 [H2] = (9.8 × 10−2 M)2 = 0.000063813...

[H2][Br2] Keq [Br2] (3.5 × 104)(4.3 × 10−3 M)

= 6.4 × 10−5 M

b) Calculate the concentration of HBr if [H2]= 0.0356 M and [Br2] = 0.0298 M

Keq = [HBr]2 [HBr]2 = Keq [H2][Br2]= (3.5 × 104)(0.0356 M)(0.0298 M)

[H2][Br2]

[HBr]2 = 37.1308 M2 [HBr] =√ 37.1308 M2 = 6.0935... = 6.1 M

Equilibrium Shift p.562

Recall that Le Chatelier’s principle states that an equilibrium will shift to relieve a stress.

Pressure: Increasing the pressure will shift the eq. toward the side which has fewer moles of gas

particles. If they are equal, or there are no gases, there is no change.

N2(g) + 3H2(g) î 2NH3(g) (Haber process) increase pressure, shift to right

Note: adding a non-reacting gas to increase the pressure does not affect the eq.

Concentration: Increasing the concentration of one reactant or product will shift the eq. toward

the other side.

HBr(aq) + H2O(l) î H3O+(aq) + Br−(aq) increase [HBr], shift to right

increase [H2O], no effect on eq.

Temperature: Increasing the temperature shifts the eq. away from the energy term.

N2(g) + 3H2(g) î 2NH3(g) + 92 kJ increase temp, shift to left

CaCO3(s) + 556 kJ î CaO(s) + CO2(g) increase temp, shift to right

Note: Keq = [CO2(g)]

NOTE: Changing the pressure, concentration, or adding a catalyst does not change the Keq. For

example, more product may be formed (as in a shift to the right), but the ratio of their

concentrations remains the same.

Changing the temperature does change the Keq. For an exothermic reaction, increasing the temp

decreases the Keq; For an endothermic reaction, increasing the temp increases the Keq.

Reactions run to completion when a product in not available for a reverse reaction. Examples of

this include producing a gas that escapes, a precipitate that is insoluble, or a product that does not

ionize.

20

The ICE methodWhen reactants are converted into products, their concentrations decrease and the concentrations

of the products increase according to molar ratios. When calculating Keq, we must use

equilibrium concentrations, not initial concentrations. To find equilibrium concentrations, it is

sometimes useful to use the ICE method.

Ex 1) Given a 1.0 M HCl solution produces [H3O+] of 0.9 M at equilibrium, what is the

equilibrium concentration of HCl?

HCl(aq) + H2O(l) î H3O+(aq) + Cl−(aq)

Initial 1.0 M 0

Change −X +X

Equilibrium ? 0.9 M

Determine the change of the known: X = 0.9

Apply the change to the unknown: [HCl] = 1.0 M −X = 1.0 − 0.9 = 0.1 M

1b) Note that you can now find the Keq because the equilibrium concentrations of H3O+ and Cl−

will be the same.

Keq = [H3O+][Cl−] = (0.9) (0.9) = 8.1

[HCl] 0.1

Ex 2) If you mix 5.0 mL of 0.060 M Co2+ with 5.0 mL 0.4 M Cl–, and the concentration of

CoCl42– at eq is 0.008 M, what are [Co2+] and [Cl–] at equilibrium?

Co2+ + 4Cl– î CoCl42–

I 0.030 M 0.2 M 0 X = 0.008

C -X -4X +X

E ? ? 0.008 M

[Co2+]eq = 0.030 – X = 0.030 – 0.008 = 0.022 M

[Cl–]eq = 0.2 – 4X = 0.2 – 0.032 = 0.168 M with sig figs, = 0.022 M and 0.2 M

Common Ion Effect p.567

Introducing an ion that is common to 2 solutes will produce a precipitate or reduce ionization.

For example, placing HCl into a saturated solution of NaCl will precipitate out NaCl.

Adding HCl increases the [Cl−] and will

HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) combine with Na+ to produce NaCl(s).

NaCl(s) î Na+(aq) + Cl−(aq) Higher [Cl−] but lower [Na+] → same Keq

CH3COOH(aq) +H2O(l) î H3O+(aq) + CH3COO−(aq) Adding NaCH3COO will reduce the

ionization of CH3COOH and lower [H3O+] and raise pH.

21

The Acid Ionization Constant Ka and pKa p569

Ka is the equilibrium constant for an acid as it ionizes in water. Take the above example:

CH3COOH(aq) +H2O(l) î H3O+(aq) + CH3COO−(aq)

Ka = [H3O+][CH3COO−] Note that H2O is left out because it is constant

[CH3COOH] as are all pure solids and liquids: Ka = K[H2O]

The larger the Ka, the stronger the acid. An easier way to compare this strength is to use pKa. This

works the same way as pH. pKa has no units and the lower it is, the stronger the acid.

pKa = –log[Ka]

Kb and pKb Not in text

Kb is the equilibrium constant for a base as it reacts with water. In the reaction:

NH3(aq) + H2O (l) î OH−(aq) + NH4+(aq) the Kb = [OH−][NH4

+]

[NH3]

The larger the Kb, the stronger the base. As with pKa, pKb = –log[Kb] and the lower the pKb,

the stronger the base. Note that Ka • Kb = Kw.

Buffers p.570

A buffer is a solution that resists changes in pH when a small amount of acid or base is added. It

contains a weak acid (or base) and its conjugate base (or acid). Because of the common ion

effect, adding small amounts of acid or base to a buffered solution will not greatly affect the pH.

CH3COOH(aq) +H2O(l) î H3O+(aq) + CH3COO−(aq) a weak acid

adding sodium acetate increases the

NaCH3COO(aq) +H2O(l) î Na+(aq) + CH3COO−(aq) acetate ion and slightly raises the pH

Any added acid (H3O+) will react with the acetate ion and form more CH3COOH.

Added base (OH−) will react with H3O+ to form water, and CH3COOH will ionize to form more

H3O+, or it will react with CH3COOH to form water and CH3COO−. In all cases, the pH is only

slightly affected by small amounts of acid or base. A solution of a weak base and its conjugate

acid works in a similar way to buffer pH.

Hydrolysis p.572

Hydrolysis is any reaction with water. A salt is made in the neutralization of an acid and a base -

the cation from the base and the anion from the acid (ex NaCl from HCl and NaOH). When salts

dissolve in water, their ions hydrolyze (react with water) to produce H3O+ or OH−.

HA + H2O î H3O+ + A− A− + H2O î OH− + HA

weak acid conj. base (anion) stronger A−, more basic

22

The weaker the acid, the stronger the conjugate base (anion). This will pull H+ from water to

form OH− and make the solution basic. CH3COO− is the conjugate base of a weak acid, and will

hydrolyze: This is why adding

CH3COO−(aq) + H2O(l) î OH−(aq) + CH3COOH(aq) NaCH3COO raises the pH

Strong acids produce weak anions that do not hydrolyze. For example, HCl (strong acid)

produces Cl− which won't appreciably hydrolyze. (nor will Na+)

Bases work in a similar way. Strong bases produce weak cations that will not hydrolyze much.

(For ex Na+ from NaOH) Weak bases produce strong cations that will hydrolyze and lower the

pH by producing H3O+.

B + H2O î OH− + BH+ BH+ + H2O î H3O+ + B

weak base conj. acid (cation) stronger BH+, more acidic

NH4+(aq) + H2O (l) î H3O

+(aq) + NH3(aq) This is why adding NH4Cl lowers pH

Salt Cation from Anion from Hydrolysis resultant pH

NaCl NaOH (strong base) HCl (strong acid) neither ions neutral

NaCH3COO NaOH (strong base) CH3COOH (weak acid) anion basic

NH4Cl NH3 (weak base) HCl (strong acid) cation acidic

NH4CH3COO NH3 (weak base) CH3COOH (weak acid) both ions neutral (if = )

Hydrolysis helps explain why the endpoints of some acid-base reactions are not pH 7.00.

The Solubility Product Constant Ksp p.577

Ksp is equilibrium constant of a saturated ionic solution. In the equilibrium:

AgCl(s) î Ag+(aq) + Cl−(aq) K = [Ag+][Cl−] the amount of solid

[AgCl] remains constant

K[AgCl] = Ksp = Ksp = [Ag+][Cl−]

Don't forget that the concentrations are raised to the power of their coefficient:

MgF2(s) î Mg2+(aq) + 2F−(aq) Ksp = [Mg2+][F−]2

Ex 1) Calculate the Ksp of CuCl at 25°C if its solubility is 0.0108 g/100. g H2O.

n = 0.0108 g = 1.09... × 10−4 mol = 0.00109... M Convert the solubility into M

99.00 g/mol 0.100 L (assume Dwater = 1.000 g/ml)

CuCl(s) î Cu+(aq) + Cl−(aq) Because the solubility tells you the conc. of ions,

and the ratio is 1:1, just find the product of their [ ].

Ksp = [Cu+][Cl−] = (0.00109...)(0.00109...) = 1.19 × 10−6 (no units)

23

Ex 2) Calculate the Ksp for PbCl2 if it has a solubility of 1.0 g/100.0 g at a particular temperature.

n = 1.0 g = 0.00359... mol = 0.0359... M Convert the solubility into M

278.10 g/mol 0.100 L

Be careful! There are twice as many Cl−, so

PbCl2(s) î Pb2+(aq) + 2Cl−(aq) we must × 2 and then square the [ ].

Ksp = [Pb2+][Cl−]2 = (0.0359...)(2 × 0.0359...)2

= (0.0359...)(0.0719...)2

= (0.0359...)(0.00517...) = 1.9 × 10−4

Using Ksp to Calculate Solubilities p.581

Recall that a large K means that the concentrations of the products are higher than the reactants.

(or the reaction goes to completion) A small K means that the reaction does not produce much

product, and most remains unreacted. The Ksp works the same way. A large Ksp means that there

are many ions in solution (i.e. it is very soluble) A low Ksp means that there are very few ions in

solution (i.e. it is not very soluble or insoluble). The higher the Ksp, the greater the solubility. For

example, PbCl2 has a Ksp of 1.6 × 10−5, (large) and is soluble at room temp. In contrast, HgS has a

Ksp of 1.6 × 10−52, (small) and is insoluble at room temp.

Given a Ksp, you can find the solubility of a sparingly soluble salt.

Ex) Calculate the solubility of BaCO3 if its Ksp is 5.1 × 10−9.

BaCO3(s) î Ba2+(aq) + CO32−(aq) Ksp = [Ba2+][CO3

2−] =

(x)(x) = 5.1 × 10−9 x = √ 5.1 × 10−9 = 7.1 × 10−5 M

Predicting Precipitates p.582

If the product of the [ions] for a slightly soluble salt exceeds the Ksp, a precipitate forms.

Ex) Will a precipitate form if 20.0 mL of 0.010 M BaCl2 is mixed with 20.0 mL of 0.0050 M

Na2SO4 if the Ksp for BaSO4 is 1.1 × 10−10? NaCl is soluble, but BaSO4 is sparingly

BaCl2 + Na2SO4 î BaSO4 + 2NaCl soluble. So if [Ba2+][SO42−] is >1.1 × 10−10

a ppt will form.

[Ba2+] = C1V1 = C2V2 = (0.010M)(0.0200L) = C2(0.0400L) = 0.00500 M

[SO42−] = C1V1 = C2V2 = (0.0050M)(0.0200L) = C2(0.0400L)= = 0.00250 M

[Ba2+][SO42−] = (0.00500)(0.00250) = 1.2 × 10−5

1.2 × 10−5 is larger than the Ksp for BaSO4 of 1.1 × 10−10, so yes, a ppt will form.

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Unit 6: Oxidation and Reduction Chapter 19

Outline: -hydrogen peroxide

-oxidation numbers -strength of oxidizing and reducing agents

-oxidation-reduction reactions -electrochemistry

-LEO GER -voltaic cells -wet cells, dry cells

-half reactions -An Ox, Red Cat

-redox and covalent bonds -Batteries

-balancing redox reactions -electrolytic cells

-oxidation number method -electroplating

-half reaction method -electrode potentials

-Eº = Eºcathode – Eºanode

Oxidation Numbers p.591

Recall from Chem 20:

The distribution of electrons in a molecule can be described by oxidation numbers, which are

usually similar to the charge an atom will make. Oxidation numbers differ from charges in that

they are not physically real, but only a description of attraction for bonding electrons.

Examples: Li+1 , Ba+2 oxidation states

Li 1+ , Ba2+ charges

Electronegativity is the ability of an atom to attract a shared pair of electrons. (See p. 151)

Rules for Assigning Oxidation Numbers

1. Uncombined elements have an Ox # of 0.

ex) Na, C, O2, P4, S8 all have Ox # of 0

2. Elements are assigned the Ox # equal to the charge they would have as an ion. The most and

least electronegative elements get their group charges.

ex) NaOH Na is +1, O is −2 (H is +1)

3. Fluorine in a compound is always −1.

4. Oxygen is almost always −2. Exceptions H2O2 (O = −1), OF2 (O = +2)

5. Hydrogen is +1 when bonded to a more electronegative element. When bonded to metals,

hydrogen is −1.

6. Ox #s add up to 0 in a neutral compound. In a polyatomic ion, they add up to equal the charge.

7. These rules can also be applied to ionic compounds. ex) NaCl Na is +1, Cl is −1.

Use these rules to assign Ox #s to each element in a compound:

NaF Na = +1 F = − 1

H2SO4+1H2 S ? O4

− 2 2(1) +___+ 4(−2) = 0 S is +6

PO43− ? PO4

− 3 ____ + 4(−2) = −3 P is +5

Oxidation and Reduction p.593

Red-ox reactions involve transfer of electrons. Oxidation originally described reactions of metal

and oxygen to form oxides: Fe + O2 → Fe2O3

Reduction originally described reactions where the metal ore was purified to metal:

Fe2O3 → Fe + O2

Now, we use a broader definition. An oxidation reaction is a reaction where an ion or atom

increases its oxidation state. A reduction reaction decreases the oxidation state of an element.

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0 0 +1 −1 In this example, sodium goes from an Ox # of 0 to +1.

Ex) 2Na + Cl2 → 2NaCl Chlorine goes from 0 to −1.

0 +1

Na → Na+ + 1e− (oxidation half reaction) Sodium is oxidized because it loses an electron.

0 −1

Cl2 + 2e− → 2Cl

− (reduction half reaction) Chlorine is reduced because it gains an electron.

says

LEO GERLoss of Electrons is Oxidation Gain of Electrons is reduction

Na → Na+ + e−

Cl2 + 2e− → 2Cl

−

sodium goes from 0 to +1 chlorine goes from 0 to −1

loss of 1 e− → oxidized gain of 1 e

− → reduced

increase in Ox # = oxidation decrease in Ox # = reduction

Oxidation produces electrons while reduction acquires them. These processes must happen

together and the number of e− produced must equal the number acquired. The substance that is

oxidized is also called a reducing agent - it allows another substance to be reduced. The

substance that is reduced is also called an oxidizing agent (also oxidant or oxidizer) - it allows

another substance to be oxidized.

Some Oxidizers at St. Joseph Common Oxidizers

NaNO3 (NH4)2Cr2O7 Chemical Group Chemical Formula NH4NO3 (NH4)2S2O8 peroxides O2

2−

Sr(NO3)2 KMnO4 nitrates NO3−

AgNO3 K2Cr2O7 nitrites NO2−

Ba(NO3)2 K2CrO4 perchlorates ClO4−

KNO3 KClO3 chlorates ClO3−

Ni(NO3)2 KIO3 chlorites ClO2−

Mg(NO3)2 Na3BO3 hypochlorites ClO−

Pb(NO3)2 MnO2 dichromates Cr2O72−

LiNO3 H2O2 permanganates MnO4−

Cu(NO3)2 Ca(NO3)2 persulfates S2O82−

Co(NO3)2 Fe(NO3)3

Ex 2) Silver nitrate acts on copper to form copper(II) nitrate and silver. Identify the species that is

oxidized and reduced.

0 +1 +5 −2 +2 +5 −2 0 Ag is reduced (ox # decreases)

Cu + AgNO3 → Cu(NO3)2 + Ag Cu is oxidized (ox # increases)

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Ex 3) Identify the reducing and oxidizing agents.

0 +1 +5 −2 +2 +5 −2 +4 −2 +1 −2

Zn + HNO3 → Zn(NO3)2 + NO2 + H2O (unbalanced)

Element Ox # change e− change process agents

Zn 0 to +2 loss of 2 oxidation Zn is a reducing agent

N +5 to +4 gain of 1 reduction HNO3 is an oxidizing agent

(note: Nitrates are oxidizers, Zn(NO3)2 too is a strong oxidizer)

Pennies will react with nitric acid to produce NO in the following redox reaction:

3Cu + 2NO3− + 8H+ → 3Cu2+ + 2NO + 4H2O (you write in the Ox #s)

0 +2 +5 −2 +1 +2 −2 +1 −2

3Cu → 3Cu2+ + 6e− 2NO3− + 6e− + 8H+ → 2NO + 4H2O

oxidation half reaction reduction half reaction (nitrate is reduced to NO)

2NO + O2 → 2ΝΟ2 2ΝΟ2 + 2Η2Ο → Η3Ο+ + ΝΟ3−+ ΗΝΟ2

brown gas in air (smog) light brown in water nitric acid (acid rain)

Note that the following is not a redox reaction because there is no change to oxidation numbers:

+1 −1 +1 +5 −2 +1 +5 −2 +1 −1

NaCl + AgΝO3 → NaNO3 + AgCl

Redox and Covalent Bonds p.594

Oxidation-reduction occurs in covalently bonded molecules as well.

0 0 +1 −1 Even though there is no loss or gain of e−, they are not

H2 + Cl2 → 2HCl equally shared. Here, +1 describes a partial transfer of an e−

In other words, the shared pair of e−s hangs around Cl more than the H.

Balancing Redox Reactions p.597

There are many methods developed to balance redox reactions. Often they are difficult to balance

by inspection because you must balance both the number of atoms and oxidation states.

Oxidation Number Method: This method is basically inspection, but we get a starting

Ex 1) Cu + Ag+ → Cu2+ + Ag position from the Ox #s.

0 +1 +2 0 Start by writing out the Ox #s

Cu + Ag+ → Cu2+ + Ag

0 +2 Isolate the electron changes

Cu → Cu2+ + 2e−

+1 0

Ag+ + e− → Ag x 2 = Make the number of e− lost and gained equal by multiplying

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2Ag+ + 2e− → 2Ag Cu → Cu2+ + 2e−

Cu + 2Ag+ → Cu2+ + 2Ag Use the numbers of the redox components to balance rest

Ex 2) NH3 + O2 → NO2 + H2O

−3 +1 0 +4 −2 +1−2

NH3 + O2 → NO2 + H2O

−3 +4 0 −2 Note: we start with 2 oxygen, so x 2 here

N → N + 7e− O + 2e− → O

2O + 4e− → 2O

N → N + 7e− x 4 This multiplier will help you balance the eqn.

2O + 4e− → 2O x 7 Apply it to the oxygen you were counting.

There is some trial and error here.

4NH3 + 7O2 → NO2 + H2O

4NH3 + 7O2 → 4NO2 + 6H2O Now balance the rest as usual.

Ex 3) Zn + HNO3 → Zn(NO3)2 + NO2 + H2O

0 +1 +5−2 +2 +5 −2 +4 −2 +1 −2 Write the Ox #s

Zn + HNO3 → Zn(NO3)2 + NO2 + H2O

0 +2 +5 +4

Zn → Zn + 2e− N + e− → N multiply N by 2 to balance electrons

Zn → Zn + 2e− There are 3 possible N to apply this 2. Ignore the Zn(NO3)2, as it

2N + 2e− → 2N contains the oxidized Zn, and its N remains +5. So use either the

HNO3 or the NO2. If you use 2HNO3 you already have more than

2N on the right, so try the NO2. (this is the trial and error part)

Zn + HNO3 → Zn(NO3)2 + 2NO2 + H2O

Zn + 4HNO3 → Zn(NO3)2 + 2NO2 + H2O Balance the N first.

Zn + 4HNO3 → Zn(NO3)2 + 2NO2 + 2H2O Always double check by counting.

Balancing Redox Reactions in Acidic Conditions

Without going into the half reaction method, here is a short-cut method to balance these:

1) balance as before

2) If the H and O don't balance, balance the O first by adding H2O to one side

3) Add H+ to the needed side to balance H

Ex) Balance MnO4− + Fe2+ → Mn2+ + Fe3+ , which takes place in acidic conditions.

+7-2 +2 +2 +3 +7 +2 +2 +3

MnO4− + Fe2+ → Mn2+ + Fe3+ Mn + 5e

− → Mn Fe2+ → Fe3+ +1e

− x 5

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MnO4− + 5Fe2+ → Mn2+ + 5Fe3+

MnO4− + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O Now add H2O to balance O

8H+ + MnO4− + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O Add H+ to balance H

Balancing Redox Reactions in Basic Conditions

Balance the equation as if it were in acidic condition. If there is H+ left over, then add OH− to

both sides to cancel the H+. Finally, combine the H+ and OH− to form H2O.

Ex) Balance Cr(OH)3 + BrO3−→ CrO4

2− + Br− in basic conditions.

+3-2+1 +5-2 +6-2 -1 +3 +6 +5 -1

Cr(OH)3 + BrO3−→ CrO4

2− + Br−

Cr → Cr +3e−

x 2 Br + 6e− → Br

−

2Cr(OH)3 + BrO3−→ 2CrO4

2− + Br−+ H2O add H2O

2Cr(OH)3 + BrO3−→ 2CrO4

2− + Br− + H2O +4H+ add H+ (if acidic, this would be done)

4OH− + 2Cr(OH)3 + BrO3

−→ 2CrO4 2− + Br

−+ H2O + 4H+ + 4OH

−add OH

−to both sides

4OH− + 2Cr(OH)3 + BrO3

−→ 2CrO4 2− + Br

− + 5H2O Combine H2O

Strength of Oxidizing and Reducing Agents p.602

Recall that:

substance Ox # electrons category example

oxidized goes up loss of e−

reducing agent Li → Li+

reduced goes down gain of e−

oxidizing agent Ag+ → Ag

The strength of oxidizing and reducing agents can be compared based on their abilities to give

and take electrons. The easier a substance gives up electrons, the stronger a reducer it is. The

more strongly a substance attracts electrons, the stronger an oxidizer it is. These properties are

summarized by the activity series and are related to electronegativity. (See table 19-3, p.603) Any

reducing agent can be oxidized by the oxidizers below it, and vice versa. The farther apart the

two, the more likely a redox reaction will take place.

−1 −2 0

Hydrogen Peroxide 2HOOH → 2HOH + O2 p.604

The peroxide ion O22- can easily be both oxidized and reduced. This is autooxidation -self red-ox.

Electrochemistry p.606

Electrochemistry involves changes in chemical and electrical energy. We will study two

electrochemical cells: voltaic cells and electrolytic cells.

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Voltaic Cells p.606

A voltaic electrochemical cell (also called galvanic cell) spontaneously converts chemically

stored potential energy into moving electrons through a redox reaction. (Ex - battery) If this

reaction occurs in contact, such as in a beaker, there is an increase in temperature. However, if

the electrodes are separated by a conductor, useful electrical energy can be produced. An

electrochemical cell is made of two electrodes; one is oxidized (the anode), the other is reduced

(the cathode). The classical example of this is a zinc-copper cell. Because Zn loses its electrons

more easily than Cu, Zn is oxidized, while Cu is reduced.

Zn → Zn + 2e− Cu + 2e−→ Cu2+

Oxidation Reduction

Anode (−) Cathode (+)

An Ox Red Cat

Anode = Reduction at

Oxidation Cathode

sd

A voltaic wet cell:

Notice that the electrodes are in an

electrolytic solution with the same

metal as an ion. In order for the e− to

flow, we must complete the circuit with

a salt bridge - a path for ions to migrate

in either direction, but keeping the

metal atoms from doing the same.

Zn loses e−s and forms Zn2+ in solution

Cu2+ gains e−s from the copper and grows

Zn disintegrates, while copper grows larger

With excess Zn 2+, the reaction would stop

but the salt bridge allows more NO3− to

cross over, keeping it electrically neutral.

K+ performs the same function on the

other side.

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You may encounter voltaic cells represented this way:

Zn(s) | Zn(NO3)2(aq) §§Cu(NO3)2(aq) | Cu(s) anode always on left (a b/f c)

(The §§is the salt bridge, and the | a phase boundary)

Batteries p.608

Dry cells are like wet cells, but the electrolytic component is a paste, rather than a solution. Here

are some of the batteries we use: (See p.609 for diagrams)

Primary (non-rechargeable) Secondary (rechargeable)

Zn-MnO2 (alkaline - AAA, AA, 9V) NiOOH-Cd or NiCd (reg. rechargeable AA, etc.)

Zn-C (C and D) Ni-MH (M is an alloy of La, Ce, Nd, Pr, Ni, Co, Mn

HgO-MnO2 (watch battery) C-LiCoO2 (Li+ ion)

Pb-PbO2 (car battery)

Electrolytic Cells p.610

Electrolytic cells are cells where electricity is used to reverse a spontaneous redox process. Two

examples of electrolysis are recharging batteries and electroplating. Because electrolysis is

non-spontaneous, it requires an external voltage source to reverse the reaction. A good example

of this is the car battery. This battery is made of 6 cells of Pb and PbO2. Under oxidation of the

Pb and reduction of PbO2 in sulfuric acid, both electrodes are coated with PbSO4 as the battery

runs down. In this case, the battery acts as a voltaic cell, as it produces 12V. However, the car's

alternator (and a rectifier) produce slightly more than 12V in the opposite direction. This reverses

the redox reaction, and converts the PbSO4 back to Pb and PbO2. Here, the battery is acting as an

electrolytic cell.

Electroplating p.611

In a voltaic cell, the anode disappears, while the cathode is plated and grows larger. If a DC

voltage is applied (negative terminal connected to the former anode) the process is reversed.

Electroplating uses electricity to change the anode to a cathode, which now accepts ions from the

solution causing metal deposition on its surface.

The solution contains a salt of the metal

so that its cations will be available to plate

on the surface of the object.

Again, you would need a porous barrier

(not shown) to keep the reaction going.

Electroplating can be used to coat a reactive

metal with a less reactive one to prevent

corrosion. It is also used in jewelry and

coins. Until 1996, pennies were mostly

copper, but are now only copper plated.

These older coins contain ~2¢ of copper.

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Electrode Potentials p.613

Metals tend to lose e−s in reactions and are therefore oxidized. Some metals lose e−s more easily

than others as was seen in the activity series. The tendency of any half reaction to occur as a

reduction reaction is called the reduction potential. We can also quantify this characteristic by

measuring the potential difference (emf or E) in volts produced between any two electrodes. This

is called an electrode potential. To standardize these half reaction voltages, we set the half

reaction of hydrogen to zero volts:

2H+ 2e− → H2 Eº = 0.00 V (recall: º = standard states: 25ºC, 1 atm, 1M)

Any substance that holds on to its electrons more strongly than hydrogen will be reduced and

hydrogen will be oxidized. If however, the substance loses electrons more easily than hydrogen,

that substance will be oxidized and hydrogen will be reduced. The difference in the ease of e−

loss is measured in volts. If a substance holds on to e−s more strongly than hydrogen, the voltage

is (+); If it loses electrons more easily than hydrogen, the voltage is (−).

A table of these reduction half reactions shows the standard electrode potentials relative to the

standard hydrogen electrode (SHE) See table 19-3, p.615 for values. Of the substances listed, Li

most easily loses e−s, while F2 most easily gains them. One way to look at the table is that all the

reactions with positive voltages will be spontaneous, while the ones with negative voltages will

be spontaneous in the opposite direction.

Sample reduction half reactions:

Cu2+ + 2e−→ Cu Eº = +0.34 V Eº = +0.34 V spontaneous

Cu2+ is reduced, H is oxidized

Zn2+ + 2e−→ Zn Eº = − 0.76 V Eº = −0.76 V not spontaneous

Zn is oxidized, H is reduced

(all reactions are written as reductions)

To tell if a cell will produce useful electricity, we can calculate the potential difference for the

cell at standard states (Eºcell)

If the Eºcell is (+), the reaction is spontaneous

Eºcell = Eºcathode – Eºanode If the Eºcell is (–), the reaction is not spontaneous

Note: this is opposite to ∆H and ∆G, but same as ∆S

When creating a voltaic cell with any two metals, the metal with the more positive electrode

potential will be the cathode, and the more negative will be the anode.

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Ex 1) Determine the voltage of a cell made from Mg | Mg2+ and Cu | Cu2+. Will this process be

spontaneous?

Mg2+ + 2e−→ Mg Eº = − 2.37 V The reaction with the more negative value

Cu2+ + 2e−→ Cu Eº = +0.34 V will be oxidized.

(or for the spontaneous direction: more + = cathode | more - = anode)

Eº = Eºcathode – Eºanode = 0.34 - (-2.37) = +2.71 V

Note that the reaction will always be spontaneous in one direction.

Cu as cathode:

Cu2+ + Mg → Cu + Mg2+ Eº = +2.71 V (spontaneous)

Mg as cathode:

Mg2+ + Cu → Mg + Cu2+ Eº = -2.71 V (non spontaneous) -could occur as electrolysis

Ex 2) Determine the voltage of the following reaction and state whether it is spontaneous.

2Al3+ + 3Sn → 2Al + 3Sn2+

red cathode

|--------------------| Eº = -1.66 V the more neg will be the anode in the

2Al3+ + 3Sn → 2Al + 3Sn2+ spontaneous direction, so no – this

|___________| Eº = -0.14 V is not spontaneous

ox anode

Also Eº = Eºcathode – Eºanode = -1.66 – (-0.14) = -1.52 V

This reaction would be non-spontaneous. However, if we applied at least 1.52 V with the

negative terminal connected to Al, this reaction would occur as electrolysis.

(When the cell is not under standard state conditions, we must use the Nernst equation, but we

will save that for another day.)

End Course Notes

You did it!

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