Блог Анкара

In This Chapter

^ Concentrating on the most important concepts ^ Listing key kinds of calculations

^ Breaking down exam questions into their component parts

M Curing the course of the AP exam, you’ll be asked to answer a variety of types of chem-V^^istry questions. Some of these questions will be explicit and straightforward; in other words, you’ll know for certain just what you’re being asked. Other questions will be implicit and indirect; in other words, you’ll have to unpack the question to determine just what you’re being asked. Whether explicit or implicit, no matter what clothes they wear, the questions will fall into one or another of a limited number of categories. Here’s an overview of those categories.

Periodic Table

Knowledge of the patterns of the periodic table is essential to do well on the AP chemistry exam; both explicitly and implicitly, questions will require you to refer to this basic chemical knowledge. Here are some of the contexts in which you might be called to consult the table:

What is the subatomic composition of a particular atom?

What is the ground state or excited electronic structure of an atom?

What kind of compound, with what formula, will be formed if two elements react?

Is a particular chemical species likely or unlikely?

Will two compounds react or not?

What kind of bond forms between the atoms of two elements? Is a particular atom likely to form an ion and if so, which ion? How many covalent bonds will a particular atom form?

Composition and Structure

Knowing the rules by which atoms assemble into compounds is key. Know how to do the following:

Determine a percent composition for a compound, given the compound’s formula

Determine a compound’s empirical formula, given its percent composition (and a molecular formula, given the compound’s molar mass)

Move easily between compound names and formulas

Given a compound’s formula, be able to

• Determine whether the compound is ionic or molecular

• Propose a likely covalent structure for a molecular compound, including possible double or triple bonds, and draw its Lewis dot structure

• Propose a likely molecular shape, especially by using VSEPR principles

• Explain the polarity of a molecule with respect to molecular shape and bond polarity

Solubility and Colligative Properties

Solubility is important because it can control the phases of reactants or products, and even determine whether a given reaction is possible. Almost all soutions on the AP exam have water as the solvent. Whether or not a substance dissolves in a given solvent is intimately connected to the structure and properties of both the substance and the solvent. In other words, be familiar with the following:

I How molecular structure, especially polarity, controls solubility and miscibility in different solvents

I How temperature affects the solubility of solid and gaseous solutes I How pressure affects the solubility of gaseous solutes

Different chemistry problems require you to use different units of concentration. Some problems require you to convert between different units. So, you must be familiar with these different units, when it is appropriate to use each unit, and how to calculate dilutions:

Molarity: Moles solute per liter solution — by far the most used

Molality: Moles solute particles per kilogram solvent

• To be used in colligative properties calculations

Partial pressure of an ideal gas: Equals the mole fraction of the gas multiplied by the total pressure

Percent solution: Mass percent, ppm, ppb, and volume percent — relatively less common

Colligative properties, the properties of a solution that depend on the number of solute particles, vary with molality. Molality can be used to calculate three key properties with which you should be familiar:

I Boiling point elevation

Freezing point depression I Molar mass of a solute

In addition, you should be able to describe the effects of added nonvolatile solute on the vapor pressure of solvent over a solution.

Gas Behavior and Phase Behavior

Don’t enter the exam without knowing ideal gas behavior thoroughly and effortlessly. Here are the key relationships that may crop up in questions:

Charateristics of an ideal gas: For instance, molecules have no volume, molecules are in random motion, molecules neither attract nor repel each other, molecules collide elastically

I Meaning of gas temperature: As the temperature of a gas increases, the distribution of kinetic energies of the particles gets broader, and the average kinetic energy increases.

Ideal Gas Law: PV = NRT, The most complete statement of ideal gas behavior — forget this one at your peril.

Combined Gas Law: P1V1/T1 = P2V2/T2, A useful combination of other laws that allows you to calculate responses to changes in pressure, volume, and temperature

• Boyle’s Law: PV = K, Which leads to P1V1= P2V2

• Charles’s Law: V/T = k, which leads to V1/T1= V2/T2

• Gay-Lussac’s Law: P/T = k, which leads to P1/T1= P2/T2

Dalton’s Law: Ptotal=P1+P2…+Pn

I Graham’s Law: The ratio of the square roots of the molar masses equals the inverse of the rates of effusion or diffusion

The same concepts of kinetic energy and particle motion (in other words, kinetic molecular theory) that inform our model of gas behavior can also be used to help explain the movement of substances between gas, liquid, and solid phases. Be comfortable with the following:

I The microscopic description of solid, liquid, and gas phases, especially as it relates to particle motions and kinetic energy

I How the microscopic details of each phase correspond to macroscopic properties such as density, viscosity, and heat capacity

I How molecular structure and interparticle forces (like dipole-dipole and dispersion forces) factor into phase behavior, especially in condensed phases

Stoichiometry and Titration

Knowing the numerical relationships between reactants and products is the starting point for countless questions. So, be sure you can do the following:

I Balance reaction equations

Use balanced reaction equations to construct conversion factors

I Move easily between units of moles and other units (mass, volume, and particles)

Apply balanced reaction equations to determine the identity of a limiting reagent and calculate the theoretical, actual, and percent yield of a reaction

Apply knowledge of stoichiometry to acid-base reactions, especially in accounting for acid/base equivalents within titration reactions

Make predictions about whether a given titration reaction is at its equivalence point (equal amonts of acid and base reaction according to H+ + OH- H2O)

Move easily between molar concentrations of acid or base and related quantities of pH, pOH, Ka, Kb, And Kw

Understand the concept of a buffered solution, when buffered solutions are effective, and be able to use the Henderson-Hasselbach equation

Equilibrium

The concept of equilibrium is core to understanding how chemical systems respond to change, and it figures prominently in a large percentage of questions. Here are some of the ways in which equilibrium might rear its head:

Determine the direction in which a given reaction will shift, in response to the following

• increasing or decreasing the concentration of a reactant

• increasing or decreasing the concentration of a product

• increasing or decreasing temperature

• increasing or decreasing pressure or volume

Given concentrations of reactants or products, calculate an equilibrium constant and vice versa

Using an equilibrium constant or a reaction quotient and the concentrations of reac-tants or products to

• Determine whether a system is at equilibrium

• Determine the direction in which a nonequilibrium system will proceed

Manipulate equilibrium constant expressions and values for coupled equilibria and for forward versus reverse reactions

Apply equilibrium concepts to questions of solubility and acid-base behavior, especially with respect to Ksp, Ka, Kb And Kw

Thermodynamics and Thermochemistry

Thermodynamics deals quantitatively with differences in energy between reactants and products and states of substances, and thermochemistry deals specifically with thermal energy transfer as heat during chemical reactions. Reaction equilibrium can be predicted from knowledge of thermodynamic parameters enthalpy and entropy. Both of these topics are fundamental to chemical change. Have a solid grasp on the concepts and calculations associated with them.

Move easily between equilibrium concentrations and the free energy change for a reaction by using AG° = – RT ln Keq. and for nonequilibrium conditions using AG = AG°-RT Ln Q.

Understand the interrelationships of free energy, enthalpy, entropy, and temperature in determining the spontaneity of a chemical reaction

• Know and be able to use the Gibbs equation, AG = AH - TAS

• Be able to connect the concepts of enthalpy and entropy with behavior at the atomic and molecular level such as bonding and motion

Understand and be able to differentiate the concepts of heat capacity, molar heat capacity, and specific heat capacity

Be able to apply the concept of heat capacity and calorimetry to reactions in which heat is released or absorbed, especially by using Q = MCAT

Apply Hess’s Law to coupled systems of chemical reactions and determine a missing quantity from a Hess’s Law cycle

Kinetics

Just as a grasp of thermodynamics is key to understanding spontaneity, having a grasp of kinetics is key to understanding the rates of reactions. Knowing one concept without the other is like walking on one stilt — you go nowhere and you fall down a lot.

Understand the concept of reaction rate and how questions of rate differ from questions of equilibrium

Given a set of experimental data, be able to formulate and use rate laws and understand the distinct roles of

• Rate constants

• Reactant concentrations

• Reaction orders

Be able to apply the concepts of molecularity, elementary steps, and rate-determining steps to the mechanism of a chemical reaction

Interpret reaction energy diagrams with regard to the kinetics and mechanism of a reaction

Understand the concept of activation energy, Ea, And be able to apply it to questions of rate, especially by using the Arrhenius equation

Redox and Electrochemistry

Charges can change during chemical reactions, and when they do, they become critical players, altering the stoichiometry and energetics of the reaction. Oxidation-reduction reactions and electrochemistry are must-know categories for the exam, especially with respect to the following skills:

Determining the oxidation number of all atoms in a balanced reacton equation and using those numbers to determine

• Whether or not a given reaction involves redox

• The identity of both oxidizing and reducing agents

Decomposing redox reactions into oxidation and reduction half-reactions

Using half-reactions to balance overall redox reaction equations (under acidic and basic conditions)

Applying redox chemistry to electrochemical and electrolytic cells in order to

• Determine the identity of anodes, cathodes

• Determine the direction in which electrons and ions flow

• Determine if and and at which electrode electroplating occurs

• Calculate standard cell potentials, especially with regard to standard reduction potentials of half-reactions, electrical current and the Faraday, and free energy changes under standard conditions

• Calculate cell potential under nonstandard conditions from concentrations and the standard cell potential by using the Nernst equation

Descriptive and Organic Chemistry

These topics are sometimes given short shrift in general Chemistry classes, but that doesn’t mean they’ll be absent from the AP exam — up to one-sixth of the material will test this information directly or indirectly. Be sure to focus on these items:

Given the reactants, make reasonable predictions about the products of the most common kinds of reactions, to include

• Single and double replacement/displacement

• Synthesis/combination and decomposition

• Combustion Be familiar with scenarios most likely to result in

• Precipitation

• Gas release

• Redox

• Formation of a colored compound

Be able to move easily between the names for organic compounds and their structures, to include

• Alkanes, alkenes, and alkynes

• Cyclic and aromatic compounds, including existence of isomers and resonance structures

• Geometric isomers

• Compounds with functional groups, including halides, alcohols, ethers, and carboxylic acids

Be able to make reasonable predictions about the properties of simple organic compounds (such as solubility, volatility, and reactivity)

Be familiar with common organic reaction types, to include substitution, condensation, hydrolysis, addition, and elimination

In This Chapter

^ Recognizing and balancing oxidation-reduction (redox) reactions ^ Flowing through electrochemical cells

^ Understanding and using standard cell potentials and the Nernst potential

/n chemistry, electrons get all the action. Among other things, electrons might transfer between reactants during a reaction. Reactions like these are called Oxidation-reduction Reactions, or Redox Reactions for brevity. Redox reactions are as important as they are common. But they’re not always obvious. In this chapter, you discover how to recognize redox reactions and how to balance the equations that describe them. All combustion, combination, decomposition, and single replacement reactions are redox. You’ll also learn how to figure out what goes on inside electrochemical cells, chemical systems that use redox reactions to generate electrical current.

Using Oxidation Numbers to Recognize Redox Reactions

If electrons move between reactants during redox reactions, then it should be easy to recognize those reactions just by noticing changes in charge, right?

Sometimes.

The following two reactions are both redox reactions:

Mg(s) + 2H+(a<7) — Mg2+(aq) + H2(g) 2H2(g) + O2(g) — 2H2O(g)

In the first reaction, it’s obvious that electrons are transferred from magnesium to hydrogen. In the second reaction, it’s not so obvious that electrons are transferred from hydrogen to oxygen. We need a way to keep tabs on electrons as they transfer between reactants. In short, we need Oxidation numbers.

VjjjABEfl Oxidation numbers are tools to keep track of electrons. Sometimes an oxidation number

Describes the actual charge on an atom. Other times an oxidation number describes an imagi-Lifll ) Nary sort of charge — the charge an atom would have if all its bonding partners left town,

Taking their own electrons with them. In ionic compounds, oxidation numbers come closest to describing actual atomic charge. The description is less apt in covalent compounds, in which electrons are less clearly "owned" by one atom or another. The point is this: Oxidation numbers are useful accounting tools, but they are not direct descriptions of physical reality. And remember that oxidation numbers apply to individual atoms, so it is possible that different atoms of the same element can have different oxidation numbers within the same compound.

Here are some basic rules That are never broken For figuring out an atom’s oxidation number:

Atoms in elemental form have an oxidation number of zero. So, the oxidation number of both Mg(s) And O2(g) is zero.

Single-atom (monatomic) ions have an oxidation number equal to their charge. So, the oxidation number of Mg2+(aq) is +2 and the oxidation number of Cl-(aq) is -1.

In a neutral compound, oxidation numbers add up to zero. In a charged compound, oxidation numbers add up to the compound’s charge.

Here are some basic rules that That have exceptions For figuring out an atom’s oxidation number:

In compounds, oxygen usually has an oxidation number of -2. An annoying exception is the peroxides, like H2O2, in which oxygen has an oxidation number of -1.

In compounds, hydrogen has an oxidation number of +1 when it bonds to nonmetals (as in H2O), and an oxidation number of -1 when it bonds to metals (as in NaH).

In compounds

• Group IA atoms (alkali metals) have oxidation number +1.

• Group IIA atoms (alkaline earth metals) have oxidation number +2.

• Group IIIA atoms have oxidation number +3.

• Group VIIA atoms (halogens) usually have oxidation number -1.

By applying these rules to chemical reactions, you can discern who supplies electrons to whom. The chemical species that loses electrons is Oxidized, And acts as the Reducing agent (or Reductant). The species that gains electrons is Reduced, And acts as the Oxidizing agent (or Oxidant). All redox reactions have both an oxidizing agent and a reducing agent.

Oxidation may or may not involve bonding with oxygen, breaking bonds with hydrogen or losing electrons — but oxidation always means an increase in oxidation number.

Reduction may or may not involve bonding with hydrogen, breaking bonds with oxygen or gaining electrons — but reduction always means a decrease in oxidation number.

For example, consider the following reaction:

Cr2O3(s) + 2Al(s) — 2Cr(s) + Al2O3(s)

Which are the oxidizing and reducing agents? To answer this question, you must assign oxidation numbers to each atom and then see how those numbers change between the reactant side and the product side. Recall that atoms in elemental form (like solid Al and solid Cr) have oxidation number zero, and that oxygen typically has oxidation number -2 in compounds. The oxidation number breakdown (shown in Figure 19-1) reveals that Al(s) is oxidized to Al2O3(s), and Cr2O3(s) Is reduced to Cr(s). So, Al(s) Is the reducing agent and Cr2O3(s) Is the oxidizing agent.

Oxidizing reducing agent agent

Figure 19-1:

Oxidation numbers reveal redox in a reaction.

Cr n

2 3

+ 2Al — 2Cr +

Sum = 0

2 x (+3) 3 x (-2) 2 x (0)

Al2O3

2 x (0) 2 x (+3) 3 x (-2)

Sum = 0

Respecting the Hyphen: Oxidation-Reduction Half-Reactions

When you balance a chemical reaction equation, the primary concern is to obey the principle of Conservation of Mass — the total mass of the reactants must equal the total mass of the products. This is done by conserving and accounting for atoms. See Chapter 9 if you need to review this process. In redox reactions, you must obey a second principle as well: the Conservation of Charge. The total number of electrons lost must equal the total number of electrons gained. In other words, you can’t just leave electrons lying around. The universe is finicky about that type of thing. People hire accountants to keep the universe straight, so AP chemistry is just Accounting 101!

Sometimes simply balancing a redox reaction with an eye to mass results in a charge-balanced equation as well. In other words, tracking atoms also tracks all the electrons. Like a string of green lights, that’s a lovely thing when it occurs, but you can’t count on it. So, it’s best to have a fail-safe system for balancing redox reactions. The details of that system depend on whether the reaction occurs in acidic or basic condtions — in the presence of excess H+ or excess OH-. Both variations use Half-reactions, Incomplete parts of the total reaction that reflect either oxidation or reduction alone.

Balancing under acidic conditions

To properly conserve mass and charge as you balance a redox reaction under acidic conditions, you often need to add water or hydrogen ion to the reactants or products. You are allowed to add water because, as the solvent, water is always available to participate in the reaction. You are allowed to add hydrogen ion because that species is present at higher concentration in acidic solutions.

Here is a summary of the method for balancing a redox reaction equation for a reaction under Acidic conditions:

1. Separate the reaction equation into the oxidation half-reaction and the reduction half-reaction. Use oxidation numbers to identify these component half-reactions.

2. Balance the half-reactions separately, temporarily ignoring O and H atoms.

3. Balance the half-reactions separately, using H2O to add O atoms and using H+ to add H atoms.

4. Balance the half-reactions separately for charge by adding electrons (e ).

5. Balance the charge of the half-reactions with respect to each other by multiplying the reactions such that the total number of electrons is the same in each half-reaction.

6. Reunite the half-reactions into a complete redox reaction equation.

7. Simplify the equation by canceling items that appear on both sides of the arrow.

For example, here is an unbalanced redox reaction: NO2- + Br- — NO + Br2

To balance the redox reaction, assuming that it occurs under acidic conditions (the problem will typically tell you whether this is the case), simply go through the steps, 1 through 7:

1. Divide the equation into half-reactions for oxidation and reduction:

2. Balance half-reactions, temporarily ignoring O and H:

2Br- — Br2 NO2- — NO

3. Balance half-reactions for O and H by adding H+ and H2O, respectively:

2Br- — Br2

2H+ + NO2- — NO + H2O

4. Balance the charge within each half-reaction by adding electrons (e"):

2Br- — Br2 + 2e-

1e – + 2H+ + NO2- — NO + H2O

5. Balance the charge of the half-reactions with respect to each other:

2Br- — Br2 + 2e-

2e- + 4H+ + 2NO2- — 2NO + 2H2O

6. Add the half-reactions, reuniting them within the total reaction equation:

2e – + 4H+ + 2Br – + 2NO2 – — Br2 + 2NO + 2H2O + 2e -

7. Simplify by canceling items that appear on both sides of the equation:

4H+ + 2Br – + 2NO2 – — Br2 + 2NO + 2H2O

Fine, you say, but what if the reaction occurs under basic conditions? Do I have to learn a whole different set of rules? No — the process for balancing redox equations under basic conditions is 90 percent identical to the one used for balancing under acidic conditions. In other words, master one and you’ve mastered both. Instead of using water and hydrogen ion as balancing tools, you are allowed to use water and hydroxide ion, the ion present in greater concentration in basic solutions.

Br – — Br2 NO2 – — NO

(oxidation) (reduction)

Balancing under basic conditions

Here’s how easy it is to adapt your balancing method for basic conditions:

Perform Steps 1 through 7 as described for balancing under acidic conditions.

Observe where H+ is present in the resulting equation. Add an identical amount of OH-to both sides of the equation such that all the H+ is "neutralized," becoming water.

Cancel any amounts of H2O that appear on both sides of the equation. That’s it. Really.

For example, here is another unbalanced redox reaction: Cr2+ + Hg — Cr + HgO

Assuming the reaction takes place under basic conditions (again, problems typically tell you whether acidic or basic conditions apply), begin balancing as if the reaction occurs under acidic conditions, and then neutralize any H+ ions by adding OH – equally to both sides. Finally, cancel any excess H2O molecules.

Follow Steps 1 through 7 as under acidic conditions:

H2O + Cr2+ + Hg — Cr + HgO + 2H+ Neutralize H+ by adding sufficient and equal amounts of OH- To both sides:

2OH- + H2O + Cr2+ + Hg — Cr + HgO + 2H2O Simplify by canceling H2O as possible from both sides:

2OH – + Cr2+ + Hg — Cr + HgO + H2O

Redox reactions have a bad reputation among chemistry students due the the perceptions that they’re difficult to understand and even more difficult to balance. But the perception is only a perception.

The whole process boils down to the following principles:

Balancing redox reaction equations is exactly like balancing other equations; you simply have one more component to balance — the electron.

Use oxidation numbers to discern the oxidation and reduction half-reactions.

Under acidic conditions, balance O and H atoms by adding H2O and H+.

Under basic condtions, balance as you do with acidic conditions, but then neutralize any H+ by adding OH -.

There are other methods for balancing redox reaction equations, so feel free to use another method if you’re more comfortable with it!

Keeping Current: Electrochemical Cells

Redox reactions are responsible for some not-so-useful things, like rust. But they’re also responsible for some very useful things, like Electrochemical cells, Devices that convert the chemical energy of a redox reaction into electric energy of flowing current, or vice versa. Voltaic cells Are electrochemical cells that convert chemical energy into electrical energy.

Batteries are examples of voltaic cells. A device called an Electrolytic cell Converts electrical energy into chemical energy. Keeping track of the relationships between different reaction mixtures and the direction and magnitude of current flow can be tricky. In this section, we describe the fundamentals of electrochemical and electrolytic cells. In addition, we give you the basic math you need to relate redox reactions to current flow in these kinds of cells.

Navigating voltaic cells

The energy provided by batteries is created in a unit called a Voltaic cell, The kind of electrochemical cell that converts chemical energy to electrical energy. Many batteries use a number of voltaic cells wired in series, and others use a single cell. Voltaic cells harness the energy released in a redox reaction and transform it into electrical work. A voltaic cell is created by connecting two metals called electrodes in solution with an external circuit. In this way, the reactants are not in direct contact, but are able to transfer electrons to one another through an external pathway, allowing the redox reaction to occur. Obviously, to be useful, a battery involves reactions that are far from equilibrium and are just waiting to power your Mp3 player by sending electrons through it so the reaction can reach equilibrium. At equilibrium, the battery is "dead." The electrode that undergoes oxidation is called the Anode.

This is easily remembered if you can burn the phrase "an ox" (for anode oxidation) into your memory. The phrase "red cat" is equally useful for remembering what happens at the other electrode, called the Cathode Where the reduction reaction occurs.

Electrons created in the oxidation reaction at the anode of a voltaic cell flow along an external circuit to the cathode, where they fuel the reduction reaction taking place there. We will use the spontaneous reaction between zinc and copper as an example of a voltaic cell here, but it is important to realize that there are many powerful redox reactions that power many types of batteries, so they are not limited to reactions between copper and zinc.

Zinc metal will react spontaneously with an aqueous solution of copper sulfate when they are placed in direct contact with one another. Zinc, being a more reactive metal than copper, displaces the copper ions in solution. The displaced copper deposits itself as pure copper metal on the surface of the dissolving zinc strip. It may at first appear to be a simple single replacement reaction, but it is also a redox reaction. The oxidation of zinc proceeds as follows:

Zn(s) — Zn2+(ag) + 2e-

The two electrons created in this oxidation of zinc are consumed by the copper in the reduction half of the reaction:

Cu2+(a<7) + 2e – — Cu(s)

These two half-reactions make up the total reaction:

Cu2+(a< ) + 2e – + Zn(s) — Cu(s) + Zn2+(a< ) + 2e -

The electron duo appears on both sides of this combined reaction and therefore cancels, leaving

Cu2+(a< ) + Zn(s) — Cu(s) + Zn2+(a< )

This is the reaction that happens when the two reactants are in direct contact, but recall that a voltaic cell is created by connecting the two reactants by an external pathway. The electrons created at the anode in the oxidation reaction travel to the reduction half of the reaction along this external pathway. This flow of traveling electrons is electrical current. A voltaic cell utilizing this same oxidation-reduction reaction between copper and zinc is shown in Figure 19-2, which we will examine piece by piece.

Zn

Anode

Figure 19-2:

A voltaic cell that uses a redox reaction between Cu and Zn.

Cu

Cathode

Zn (s)-*~ Zn 2+ (aq) + 2e-

Cu2+ (aq) + 2e – ■

Cu(s)

Movement of cations

Movement of anions

First, note that zinc is being oxidized at the anode, which is labeled with a negative sign. It is important to note that this does not mean that the anode is negatively charged. Rather, it is meant to indicate that electrons are being released there. The oxidation of zinc releases Zn2+ cations into the solution as well as two electrons that flow along the circuit to the cathode. This oxidation thus results in an increase of Zn2+ ions into the solution and a decrease in the mass of the zinc metal anode.

The electrons released by the oxidation of zinc at the anode carry out the reduction of Cu at the cathode. This pulls Cu2+ from solution and deposits more Cu metal on the cathode. The result is the exact opposite of the effect occurring at the anode: The solution becomes less concentrated as Cu2+ ions are used up in the reduction reaction, and the electrode gains mass as Cu metal is deposited.

No doubt you have noticed that the diagram also contains a U-shaped tube connecting the two solutions. This is called the Salt bridge And it permits the system to adjust the charge imbalance created as the anode releases more and more cations into its solution (resulting in a net positive charge) and the cathode uses up more and more of the cations in its solution (leaving it with a net negative charge). The salt bridge contains an electrolytic salt (in this case NaNO3). A good salt bridge is created with an electrolyte whose component ions will not react with the ions already in solution. The salt bridge functions by sucking up the extra NO3 – ions in the cathode solution and depositing NO3 – ions from the other end of the bridge into the anode solution. It also sucks up the excess positive charge at the anode by absorbing Zn2+ ions and depositing its own cation, Na+, into the cathode solution. These actions of the salt bridge are necessary because the solutions must be neutral in order for the redox reaction to continue. The bridge also completes the cell’s electrical circuit by allowing for the

Flow of charge back to the anode. Instead of a salt bridge, electrochemical cells can be set up with a porous barrier separating (and in direct contact with) the anode and cathode solutions. Whether a salt bridge or a porous barrier is used, anions travel to the anode, while cations travel to the cathode.

Flowing through electrolytic cells

Another kind of electrochemical cell is an Electrolytic cell, A cell in which electrical energy is converted into chemical energy. An electrolytic cell is essentially a voltaic cell run in reverse. Instead of using the potential energy of a spontaneous chemical reaction to drive current, electrolytic cells use an external energy source (like a battery) to drive nonspontaneous chemical reactions. Because electrolytic cells are essentially the opposite of voltaic cells, you can understand both with the same set of electrochemical principles.

For example, an electrolytic cell can be used to drive the following reaction, in which liquid (melted) sodium chloride decomposes into liquid sodium and chlorine gas:

2NaCl(l) — 2Na(l) + Cl2(g)

Like any other redox reaction, this one can be divided into half-reactions:

2Na+(l) + 2e – — 2Na(l) 2Cl -(l) — Cl2(g) + 2e -

The former half-reaction occurs at a cathode and latter occurs at an anode. Figure 19-3 depicts an electrolytic cell in which an external voltage source is being used to drive this nonspontaneous reaction forward.

Battery —>- e + -

As in a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode. Thus our "an ox" and "red cat" mnemonics are still valid.

Calculating EMF and standard cell potentials

Voltaic cells connect two halves of a redox reaction with a current-carrying conductor. But what exactly is it that causes the flow of charge across the conductor? To answer this question, you need to understand the concept of electromotive force. To be able to calculate the diriving force behind the flow of current, you need to be able to use standard cell potentials. In this section we describe the connection between electromotive force and standard cell potential, and show you how to use them in the context of a specific redox reaction.

Electromotive force and standard cell potentials have to do with potential energy. When a difference in potential energy is established between two locations, an object has a natural tendency to move from an area of higher potential energy to an area of lower potential energy. When Wile E. Coyote places a large circular boulder at the top of a hill overlooking a road on which he knows his nemesis the Roadrunner will soon be traveling, he takes it for granted that when he releases the boulder it will simply roll down the hill. This is due to the difference in gravitational potential energy between the top and bottom of the hill, which have high and low potential energies respectively. In a similar, though less diabolical, manner, the electrons produced at the anode of a voltaic cell have a natural tendency to flow along the circuit to a location with lower potential: the cathode.

It is this potential difference between the two electrodes that causes the Electromotive force, Or EMF, of the cell. EMF is also often referred to as the Cell potential And is denoted ECell. The cell potential varies with temperature and concentration of products and reactants and is measured in Volts (V). A Volt is a derived unit — one Volt equals one Joule per Coulomb (V = J C-1); in other words, a Volt is an amount of energy held by an amount of charge. As in many other parts of chemistry, chemists have found it much easier to track relative changes rather than absolute ones. So it has been found to be convenient to have standards by which cells can be compared. The Јcell that occurs when concentrations of solutions are all at 1mol L1 and the cell is at STP (1atm pressure, 0°C) is given the special name of Standard cell potential, Or Ј°cell.

You can calculate electrochemical (voltaic or electrolytic) cell potentials by taking the difference between the standard potentials of the oxidation and reduction half-reactions. Once upon a time, a group of chemists sitting in a small, dark room decided that all half-reaction potentials should be listed as the potential for reduction (not oxidation). As such, these potentials are often referred to as Standard reduction potentials. They are calculated using the formula

Ј°cell = Ј°red(cathode) – Ј°red(anode)

Note that this equation will not be provided to you on the AP exam — you must memorize it! Standard reduction potentials are reported relative to a reference reduction potential, one that sets the "zero" for the spectrum of other reduction potentials. The reference potential is that for the reduction of 2 moles of hydrogen ion in solution, specifically at 1 molar concentration to 1 mole of hydrogen gas at 1 atmosphere of pressure:

2H+(ag) + 2e – — H2(g) Ј°red = 0 V

Here’s a happy consequence of the fact that Ј°red values are reported in Volts: Because Volts are already standardized per unit of charge (per Coulomb, specifically), the value of E°red stays the same, even if the stoichiometry of the half-reaction changes. So, you get the same potential for H*(aq) + E~ — LA H2(g).

Table 19-1 lists some standard reduction potentials along with the reduction half-reactions associated with them. The table is ordered from the most negative Ј°red (least-favorable reduction) to the most positive E°red (most-favorable reduction). Note that different tables

May list items in reverse order, so be sure to orient yourself to the table you’re using. The reactions with negative E°red are therefore reactions that happen at anodes, while those with a positive E°red occur at cathodes.

Put another way, the reactants in half-reactions with the most negative E°red are the strongest reducing agents (like Li+). The reactants in half-reactions with the most positive Ј°red are the strongest oxidizing agents (like F2).

Note that not all of the reactions in Table 19-1 show the oxidation or reduction of solid metals as in our examples so far. Note that there are liquids and gases in the mix. Not every voltaic cell is fueled by a reaction taking place between the metals of the electrodes. Although the cathode itself must be made of a metal to allow for the flow of electrons, those electrons could be passed into a gas or a liquid in order to complete the reduction half-reaction. Don’t panic! You do not need to memorize this table. The folks at the College Board do give you a table of reduction potentials on the AP exam!

Using Concentration and Standard Cell Potentials to Predict Results

Electrochemistry occurs in situations other than in beakers connected by salt bridges. Electrochemistry occurs anywhere in the universe where you find a suitable set of half-reactions. By connecting the lessons you’ve learned about electrochemical cells with other chemical wisdom about the concentrations of reactants and products, you get a powerful set of tools to predict whether redox reactions will occur in a given situation, or to explain why they did occur. In this section, we show you how to use standard cell potentials and concentrations to make predictions about redox reactions. We also introduce you to the Faraday constant, a key equation that ties electricity to chemistry by relating charge and the mole. To

Allow you to make predictions about redox reactions under nonstandard, real-life conditions, we add another tool to your mathematical toolbelt, the Nernst equation.

Predicting spontaneity with standard cell potentials

The painstaking measurement of myriad standard reduction potentials serves a greater purpose than simply making Impressive Tables of Official Data even more impressive. These potentials can be used to predict whether a redox reaction is Spontaneous Or Nonspontaneous — whether it occurs on its own or whether it requires an input of energy to go forward. Remember though, that spontaneity is defined relative to equilibrium. No obvious reaction will occur if a redox system is at equilibrium.

Recall that you can calculate the standard cell potential in a voltaic or electrolytic cell by taking the difference in the standard reduction potentials of the half-reactions occurring at the cathode and anode:

E°cell = E°red(cathode) – E°red(anode)

In the same way, you can calculate the electromotive force of any particular redox reaction occurring outside of an electrochemical cell:

E°= E°red(reduction half-reaction) – E°red(oxidation half-reaction)

Here’s the kicker: If E Is positive, the reaction is spontaneous; if E° is negative, the reaction is nonspontaneous.

Note that this sign convention is different from that typically used in discussions of energy and spontaneity. The Gibbs free energy change For a reaction (AG), for example, is negative for a spontaneous process. If you pay close attention to the equation for the relationship between Gibbs free energy and cell potential given below, you have a strong reminder about them being opposite in sign.

You can use calculated electromotive force values to predict whether or not a displacement reaction will occur. For example, here is a candidate displacement reaction:

Cu(s) + CaCl2(aq) — CuCl2(aq) + Ca(s)

Will this reaction occur spontaneously? Because it is a redox reaction as well as a displacement reaction, we can decide by examining the redox half-reactions:

Cu(s) — Cu2+(a<7) + 2e- (oxidation half-reaction)

Ca2+(a<7) + 2e – — Ca(s) (reduction half-reaction)

The relevant calculation of electromotive force is

E°= E° Red(Ca2+ — Ca) – E° Red(Cu2+ — Cu) Plugging in values from Table 19-1, we get

E° = ( -2.76 V) – (+0.34 V) = -3.10 V

Because the calculated E° Is negative, we know that the reaction is nonspontaneous — copper will not displace calcium from calcium chloride without the input of additional energy.

Tying electricity to chemistry: The Faraday constant

Using standard reduction potentials, you can calculate the spontaneity of a redox reaction under standard conditions. In other words, you have to assume that the reaction takes place with the reactants at 1mol L-1 concentration. Most real-life reactions don’t take place under standard conditions. Wouldn’t it be nice if we could use standard reduction potentials to make predictions about real-life reactions? We can, once we have a solid mathematical connection between the flow of current and the individual atoms involved in redox reactions. That connection is provided by Faraday’s constant.

As noted briefly above (and discussed at greater length in Chapter 21), spontaneous processes are those that occur with a decrease in Gibbs free energy; the products have lower free energy than the reactants, so the overall change in free energy (AG) Is negative. What we really need, then, is a way to relate electromotive force for a real-life reaction, E, To the free energy change for that reaction, AG. Here is where the work of Michael Faraday helps us out.

Faraday was a high-powered English chemist, physicist, and by most accounts a generally swell guy. Among other things he spent quite a bit of time pondering the relationship between chemistry and electricity. Studying the properties of electrolytic cells, he noticed two important things:

The mass of substance produced at an electrode (an anode or cathode) is directly proportional to the number of electrons transferred through the cell.

The amount of the substances produced by electrolysis depends on the atomic masses of the substances as well as the "excess charge" of the substances (in other words, the ionic charge of the atoms).

The upshot of these observations was that Faraday had discovered how to relate current to chemistry at the level of individual atoms. To make this connection explicit and useful for predictions, we now use a number called the Faraday constant, F. The Faraday constant is simply the product of Avogadro’s number, VVA (that is, one mole) and the charge of a single electron, E:

F = WA-e = (6.022 x 1023 mol-1)(1.602 x 10-19 C) = 9.649 x 104 C mol-1

The Faraday constant serves as a bridge to allow us to relate the free energy change for a redox reaction to the electromotive force for that reaction:

AG = -n F E

In this simple but powerful equation, N Is the number of molar equivalents of electrons that are transferred during the reaction. In other words, N Is the number of electrons written in a balanced redox reaction equation. The negative sign on the right side of the equation reflects the fact that spontaneous redox reactions have opposite signs in AG And E.

Bringing it all together with the Nernst equation

To make predictions about behavior in real-life redox reactions, where concentrations aren’t at the standard of 1mol L-1, we need to involve those real-life, nonstandard concentrations in our calculations. This section introduces the right tool for that job: the Nernst equation. This useful equation combines nonstandard concentration information in the form of the reaction quotient, Q, with the electricity-chemistry bridge of the Faraday constant, F.

Nernst was a German chemist and Nobel Prize-winner who was born shortly before Faraday died. So, we can imagine Faraday as sort of handing off the electrochemical baton to Nernst. Nernst ran with it. Nernst knew that the free energy change for a real-life reaction, AG, Could be related to the standard free energy change for that reaction (AG, when reactants and products are all at 1mol L-1) in the following way:

AG = AG + RT Ln Q

In this equation, R Is the gas constant, TIs temperature, and Q Is the Reaction quotient. The reaction quotient is the key component for our purposes because it describes the actual concentrations of reactants and products in a reaction. So, for the general, reversible reaction

AA + bB cC + dD

The reaction quotient is [ C ]c [ D ]d

Q

By combining this relationship with Faraday’s observation that AG = – N F E, We obtain the very useful Vernst e<uation:

E = E °- ■R-T Ln Q nF

Armed with the Nernst equation and a table of standard reduction potentials, you can predict the electromotive force of real life reactions, ones that occur at odd temperatures and with odd concentrations of reactants and products. Or, given an observed electromotive force, you can calculate unknown concentrations or temperatures. Thanks to Faraday and Nernst, you have a quantitative, scientifically deep explanation for why your car battery fails you on a cold winter morning.

The 5th Wave_By Rich Tennant

"Your formula Јor a carbohydrate is close, but not entirely accurate. I’m pretty sure carbohydrates consist oЈ carbon, hydrogen, oxygen, sour cream, and bacon bits.*

In this part. . .

/n earlier times, chemists relied extensively on high-tech instruments like "eyes" and "noses" and so forth. Chemistry used to be largely a descriptive science, without the extensive calculations and precise measurements that characterize it today. Qualitative observations are still a useful part of chemistry, and understanding qualitative patterns in things like color and solubility is a big part of developing an elusive and prized skill called "chemical intuition." This part gives an overview of descriptive chemistry as it applies to the AP Chemistry exam. Organic chemistry is another branch of chemical science with deep roots and time-honored, qualitative traditions. But beyond its history, organic chemistry is a huge modern enterprise with massive application in medicine and industry, among other things. We close this part by giving you a glimpse into organic chemistry, the chemistry of carbon-based compounds, preparing you for its inevitable appearance on the exam.

In This Chapter

► Taking or embracing electrons

► Naming compounds

► Getting a glimpse of ionic bonds

► Cozying up to covalent bonds

► Arming yourself with info on polarity and electonegativity

Chemistry textbooks would be much thinner if it weren’t for bonding. As appealing as thinner textbooks may be, they’d come at the cost of other appealing things, such as cars, chocolate (Heaven help us!), and being alive. Bonds of different types between different elements and in different arrangements underlie the stunning diversity and complexity of matter. Although that complexity may sound daunting, we make taking a look at bonding a breeze by breaking down just how bonding builds things up.

Bringing Atoms Together: Bonding Basics

Sometimes atoms hang out together. Sometimes they form close friendships, sometimes only loose alliances. Close mutual attractions between atoms, ones strong enough to hold them persistently together, are called Bonds. Isn’t that sweet?

Why do atoms buddy up and bond with one another? In short, pairs of atoms seek stability. In particular, atoms prefer to dwell in states with more stable electron configurations. (Chapter 3 describes what makes for a stable electron configuration in case you need to check it out. . . and you do for the exam.) Atoms just tend to be most stable when their valence shells are completely filled with electrons so each atom does its best to fill its Valence shell, Its outermost shell.

Just like the different types of bonds you form in life — bonds between buddies, sisters and brothers, your children, your coworkers (okay, maybe that’s a stretch) — atoms form different types of bonds as well, and we’ll give you the skinny on those types of bonds in the following sections.

Attracting the opposite: Ionic bonds

Ionic bonds Form between oppositely charged ions. Ions Themselves form when an atom either gains or loses valence electrons in pursuit of filling its valence shell. Positively charged ions are electrostatically attracted to negatively charged ions, so the two nestle together to form an ionic bond. In a sense, atoms transfer whole electrons to create ions and form these bonds (more on cations and anions in Chapter 3). Sometimes, an ion consists of more than one kind of atom. These multiatom ions are called Polyatomic ions, Which have won many awards for their extreme inconvenience to chemistry students. Table 5-1 lists the most

Common polyatomic ions, grouped by ionic charge. They’ll pop up frequently. They’ll annoy you until you simply buckle down and memorize them. You can find out more about ionic bonds later in the chapter in the section, "Eyeing Ionic Bonds."

Sharing electrons: CoValent bonds

Covalent bonds Form between atoms that share electrons, rather than transfer them. These shared electrons are in orbits that surround both bonded atoms. Sharing may or may not be equal, depending on which atom "wants" the electrons more. The chemical term for electron greed is Electronegativity. Differences in electronegativity between atoms largely determine what kind of bond forms between them. As described more completely in the section, "Getting a grip on electronegativity," differences in electronegativity determine whether a bond is ionic or covalent.

Keeping it solid with metals: Metallic bonding

Okay, we’re not referring to being groupies with a heavy metal band. You can do that after the AP exam. Instead, Metallic bonding Occurs within metallic solids (big surprise, we know). Metallic bonding helps to explain the unique properties of metals:

The Electron-sea model Of metallic bonding describes metallic solids as lattices of metal cations immersed within a fluid "sea" of mobile electrons. Although the electrons are constrained to the cation lattice, they can move freely within it. This model provides a satisfying intuitive explanation for metals’ electrical conductivity because the electrons within the lattice aren’t constrained to orbitals around individual atoms, but are in delocalized orbitals.

For example, a sample of solid copper holds itself together with metallic bonds. The electrons in this sample don’t clearly "belong" to any particular copper atom, but can move through the sample. So, if you draw out the sample into a wire, you can use the mobile electrons to conduct electric current.

So what do you get when you bond? In life, you get families and friends. In chemistry, you get compounds. Chemical Compounds Are the consequence of bonding. A compound consists of bonded atoms, and has properties that are different from a simple mixture of the identical, nonbonded atoms. In other words, the structure of a compound confers properties to that compound. The relationship between structures and properties is a major theme of chemistry. That theme revolves around the concept of the chemical bond. In the next section we begin to deal with the basics of compounds, especially how to recognize and interpret their names and formulas.

Naming Names: Nomenclature, Formulas, and Percent Composition

Okay, so our parents always told us not to call people names, but in chemistry, there are exceptions (and we don’t mean your chemistry teacher). There are more known chemical compounds than there are residents of New York City so something must be done to keep these compounds straight. In the early days of chemistry, far fewer compounds were known, so chemists assigned those compounds Common names, Like "ammonia." Calling a compound by its common name is essentially like calling it "Reggie." Eventually, it became clear that common names, while charming, made a mess of things because these names told you nothing about the compound. So, modern chemists prefer Systematic names, Ones assigned by certain rules, which tell you a great deal about a compound. The process of using rules to assign names to compounds is called chemical Nomenclature.

In the next sections you’ll learn to move easily back and forth between the chemical formula for a compound (H2O) and its systematic name (dihydrogen oxide). For common compounds, there is always the possibility of a charmingly messy common name (water). Finally, you’ll learn to crunch the numbers of percent composition, one of the common ways chemists begin to define an unknown compound.

Knitting names to formulas: Nomenclature

Different naming rules apply to inorganic compounds (ones not based on multi-carbon structures) versus organic compounds (ones that Are Based on multi-carbon structures). The system presented in this chapter applies to inorganic compounds; organic compound nomenclature is described in Chapter 25.

The main idea behind systematic naming is that the poor, addled chemist, drowning in compounds, can move easily between the name of a compound and that compound’s Formula.

Formulas can be one of two types:

Empirical: An empirical formula lists the kinds of atoms in a compound and gives the ratio of atoms to each other — not necessarily their actual numbers.

Molecular: A molecular formula also lists the kinds of atoms in a compound, but gives the actual number of each kind of atom within one molecule of that compound. The compounds nitrogen monoxide and dinitrogen tetroxide make clear the difference between empirical and molecular formulas; just check out Table 5-2.

KBEft When you write out the molecular formula, if the number of atoms within a molecule of the compound is one (1), you don’t write the numeral one (1), as it’s implied. Nitrogen monoxide only has one atom within each molecule of that compound so you write its molecular formula simply as NO.

A great number of chemical compounds are binary compounds, ones that are built of only two kinds of elements. Binary compounds can be either ionic (held together by ionic bonds) or molecular (held together by covalent bonds). Ionic compounds have only empirical formulas because they are not composed of distinct molecules. Molecular compounds have both empirical and molecular formulas.

Naming binary compounds

Many chemical substances are binary compounds, consisting of only two kinds of elements. The generic formula for a binary compound is XAYB, where X and Y are the different elements, and A and B are the relative (and sometimes actual) amounts of each element within one representative unit of the compound. A representative unit might be a molecule or it might be a Formula unit, The smallest repeating unit of an ionic compound.

Figure 5-1 provides a visual summary of the method for assigning systematic names to inorganic compounds, but we also give you the steps for naming a binary compound below.

Is X hydrogen? Yes^^ no

X. Y„ is An acid

A B

Is X a metal? Yes no

Can X have variable charge? Yes no

XAYB is a Molecular compound,

AB

Prefixes are required

Figure 5-1:

Flowchart for assigning systematic names to inorganic compounds.

XAYB is an Ionic compound,

Roman numerals are required

XAYB is an

Ionic compound

Is Y a polyatomic ion?

Yes

Ending depends on anion

No

Ending is -ide

To name a binary compound, ask yourself these questions:

1. Is X hydrogen? Compounds that contain hydrogen cations (H+) often fall into a class of compounds called Acids. Sadly, many common acids are still saddled with common names. Table 5-3 summarizes many of these names. Although these are common names (and are therefore inconvenient), some patterns exist that help make sense of things:

• Acids composed with monoatomic (single-atom) anions use the prefix Hydro – And the suffix -ic.

• Acids composed with polyatomic anions with names ending in -ate Use the suffix -ic.

• Acids composed with polyatomic anions with names ending in -ite Use the suffix -ous.

• If a polyatomic anion includes the prefix Per- Or Hypo-, That prefix transfers to the acid name.

• Be careful. If a compound like HCl is in a gaseous state, it is Not Called hydrochloric acid.

2. Is X a nonmetal or a metal? If X is a nonmetal, you’re dealing with a molecular compound. As such, you’ll need to use prefixes to describe the number of each type of element, X and Y. Table 5-4 summarizes the prefixes you’ll most commonly encounter. If X is a metal, then you’re dealing with an ionic compound — proceed to step 3.

The name of a molecular compound usually lists the element furthest to the left within the periodic table as the first. If both elements occur in the same column, the lower one is usually listed first. The second element in the name of the compound receives the suffix -ide.

Prefixes are always given on both elements unless there is only one of the first element; in that case, the "mono" prefix is dropped.

Nitrogen monoxide (NO) and dinitrogen tetroxide (N2O4) are examples of molecular compounds named according to these rules.

3. Is X a metal that can form cations of different charges? Although alkali metals and alkaline earth metals reliably form only one kind of cation, metals in the center of the periodic table (especially the transition metals) can sometimes form different charges of cations. If a compound contains one of these metals, you must specify the charge of the cation within the compound name. To do so, use Roman numerals within parentheses after the name of the metal; the Roman numeral corresponds to the size of the positive charge. For example, the element iron, Fe, frequently occurs as either a +2 cation or a

+3 cation. The Roman numeral system describes Fe2+ as iron (II) and Fe3+ as iron (III).

Unlike the case with molecular compounds, ionic compound names do not include prefixes to describe the number of each kind of atom within the ionic compound. Why not? Because ionic compounds always occur in the combination of ions that results in zero overall charge. Because ionic compounds have zero overall charge, you can use the charges of the individual ions to determine the formula of the ionic compound — and vice versa. Binary ionic compounds add the suffix -ide To the name of the second element.

Here are some examples of binary ionic compound formulas and names: •Na+ and Cl – combine to form NaCl, known as sodium chloride. •Fe2+ and O2- combine to form FeO, known as iron (II) oxide. •Fe3+ and O2- combine to form Fe2O3, known as iron (III) oxide.

4. Is Y a polyatomic ion? If Y is a polyatomic ion, you have permission to be momentarily annoyed. After the moment has passed, either recall the name of the polyatomic ion from memory (as you’ll have to do on the AP Chemistry exam), or refer to a handy table like Table 5-1.

For example, if Y is SO42 , then you’re dealing with the polyatomic ion called Sulfate. If your compound is Na2SO4, then the name is Sodium sulfate.

Calculating percent composition

Formulas and names emphasize the numbers of different kinds of AtomS within a compound. Percent composition Emphasizes the Mass Of different kinds of atoms within a compound. Naming by percent composition simply means that you make a list of each kind of atom in a compound accompanied by the percent of the compound’s total mass contributed by that kind of atom.

To calculate percent composition you

1. Determine the mass for each mole of the compound. Do this by counting the number of each kind of element within the compound, finding the elements’ atomic masses from the periodic table, and then adding up the individual masses of all the elements.

2. Determine the mass for each mole of each atom. Again, find the elements’ atomic masses on the periodic table.

3. Multiply the mass of the atom by however many of that atom the compound contains.

4. Divide the mass contributed by an individual element by the total mass of the compound.

5. Multiply the result of step 4 by 100%.

6. Repeat steps 2 through 5 for each element type in the compound.

7. Write out the percent composition by listing each element type alongside that element’s percent mass.

Consider the compound water, H2O.

Each mole (6.022 x 1023) of water molecules has mass 18.02g.

Each mole of water molecules contains one mole of oxygen atoms and two moles of hydrogen atoms.

Each mole of oxygen atoms has mass 16.00g. Each mole of hydrogen atoms has mass 1.008g.

So, you calculate the percent composition of water

Oxygen: (16.00g mol-1 / 18.02g mol-1) x 100.0% = 88.79% Hydrogen: [(2 x 1.008g mol-1) / 18.02g mol-1] x 100.0% = 11.21% More briefly, the percent composition is O 88.79%, H 11.21%.

So, by sheer number of atoms, water is mostly hydrogen. But by mass — in other words, by percent composition — water is overwhelmingly oxygen.

Eyeing Ionic Bonds

Ionic bonds form between cations (positively charged ions) and anions (negatively charged ions). The strength of an ionic bond derives from electrostatic attraction between ions of opposite charge. Ionic compounds form extended, three-dimensional lattices, such as the one shown in Figure 5-2. The exact geometrical arrangement of ions in an ionic lattice results from an interplay of factors, all conspiring to maximize the favorable (as in attractive) interactions between the ions.

Figure 5-2:

A lattice of Na+ and Cl-ions within the ionic compound sodium chloride, NaCl.

The NaCl lattice shown in Figure 5-2 consists of repeating, two-atom units of Na+ and Cl-. You can imagine that each sodium atom has donated an electron to a neighboring chlorine atom. Thinking about ionic bonding in this way makes clear why ionic compounds typically form between a metal and a nonmetal. Metals tend to lose valence electrons easily. Nonmetals tend to gain extra valence electrons avidly. Ions scratch each other’s backs.

The strength of ionic bonds within an ionic compound is expressed by Lattice energy. Lattice energy represents the amount of energy required to completely separate the component ions of one mole of an ionic compound into gaseous ions. Larger positive lattice energies correspond to stronger ionic bonds.

Electrostatic attraction increases not only with the magnitude of opposing charges, but also as the distance between those charges decreases. In other words, greater quantity of charge attracts more strongly than less quantity of charge, and closer charges attract more strongly than distant charges. The overall charge of an ion effectively acts as a point charge at the atom’s center. The centers of bigger-sized atoms can’t nestle as closely together as those of smaller-sized atoms. As a result, ionic bonds tend to be stronger between ions with greater magnitude of charge and between ions of smaller size. So, the strongest ionic bonds are between two small, highly charged ions.

Considering Covalent Bonds

Unlike ionic bonds, where atoms either lose or gain electrons, covalent bonds are a a kinder, gentler bond. Covalent bonds form when atoms share valence electrons. Atoms do this kind of thing because it helps them to fill their valence shells. Covalent bonds tend to form between atoms that do not completely give up electrons. In other words, covalent bonds tend to form between nonmetals.

Sharing electrons… or not

The attractive force of a covalent bond arises from the attraction of the shared electrons to the positively charged nuclei of the bonded atoms. Within bonds, electrons don’t act as truly distinct particles, but are distributed into "clouds" of varying density. The shared electrons of a covalent bond distribute with higher density in the region directly between nuclei, as shown in Figure 5-3.

Each single covalent bond houses two shared electrons. In a standard covalent bond, each bonded atom contributes one electron. So, each atom gains one electron (that of its bonding partner) in the bargain.

Sometimes one atom donates both electrons to a covalent bond, with the other atom contributing no electrons. This kind of bond is called a Coordinate covalent bond. Atoms with Lone pairs Of electrons often engage in coordinate covalent bonding. A Lone pair Consists of two electrons that are not used in bonding paired within the same orbital.

Even though covalent bonding usually occurs between nonmetals, metals can engage in coordinate covalent bonding. Usually, the metal receives electrons from an electron donor called a Ligand.

Figure 5-3:

Distribution of electron density

Within the + +

Electron cloud of a

Covalent bond.

Getting to know structural formulas

Atoms can share more than a single pair of electrons. When atoms share two pairs of electrons, they form a double bond, and when they share three pairs of electrons they form a triple bond. A solid line drawn between element symbols serves as shorthand in structural formulas to indicate that atoms are covalently bonded. So, the covalently bonded atoms of water, carbon dioxide, and dinitrogen can be indicated as shown in Table 5-5.

Rules for determining the number of covalent bonds between atoms and for estimating the geometric arrangement of covalently bonded atoms are described in Chapter 7.

Measuring the strength of covalent bonds

Bond enthalpy Describes the strength of covalent bonds. The bond enthalpy (AH) Is an estimate of the amount of energy required to break the bond. In the case of diatomic molecules with single covalent bonds (such as Cl2), the bond enthalpy is a very good estimate. In the case of polyatomic molecules (such as CH4, which contains four distinct C-H bonds), the bond enthalpy is an Average bond enthalpy, An estimate averaged over the four bonds of the molecule.

Larger bond enthalpies correspond to stronger covalent bonds. Typically, atoms held together with more bonds and/or with stronger bonds approach each other more closely than do atoms held together with fewer and/or with weaker bonds.

Bond enthalpies are easily measured, so chemists frequently use them to help determine the strength of bonds and to estimate bond distances. On the AP exam, you might be asked to make the same kinds of estimates from a set of bond enthalpy data.

Separating Charge: Polarity and Electronegativity

Polarity Refers to an uneven distribution of charge. In chemical bonds, polarity arises from a difference in Electronegativity Between bonded atoms. More electronegative atoms draw greater electron density toward themselves. You might think that atoms with more protons (more positively charged nuclei) are always more electronegative, but this isn’t the case. Why? The electronegativity of an atom derives from the positive charge of its nucleus And From the extent to which that positive charge is offset or "shielded" by the successive layers of electron density that surround that nucleus. Because atoms with many protons also tend to have a greater number of electron shells, the large positive charge of these atoms’ nuclei is partially offset by the negatively charged electron shells.

Getting a grip on electronegativity

Within a given row of the periodic table, electronegativity tends to increase from left to right. Within a given column of the table, electronegativity tends to increase from bottom to top. These trends are only overall patterns because electronegativity is influenced by more subtle factors. The electronegativities of the elements are shown in Figure 5-4.

Figure 5-4:

Electronegativities of the elements.

If-*~

OO

CO

The greater the difference in electronegativity between bonded atoms, the more polar is the bond between those atoms. Covalent bonds that are very polar more closely resemble ionic bonds than do covalent bonds that are less polar. In fact, no real physical distinction exists between ionic and covalent bonds — ionic bonds are simply so polar that it becomes useful to imagine that one atom has emerged entirely victorious from the tug-of-war between competing nuclei for electron density.

Although different sources use slightly different numbers to make the split between polar and nonpolar, usually, differences in electronegativity are interpreted with the following categories:

A difference in electronegativity between bonded atoms of less than about 0.3 leads to a description of the bond as "nonpolar."

A difference in electronegativity ranging from 0.3 to about 1.7 leads to a description of "polar."

A difference in electronegativity greater than about 1.7 leads to the description "ionic."

Digging into dipoles

Within a polar covalent bond, electrons are distributed unevenly between the two atoms. The more electronegative atom is surrounded by greater electron density and assumes a Partial negative charge. The less electronegative atom draws correspondingly less electron density and assumes a Partial positive charge. Partial negative and partial positive charges are indicated by the symbols 5- and 8+, respectively. Separation of charge along the line connecting two bonded atoms (the Bond axis) Creates a Bond dipole. Bond dipoles are often indicated in one of two different ways, as shown in Figure 5-5:

Figure 5-5:

Two different depictions of a

Bond dipole in the HCl molecule.

8+ 8-H — Cl

H->

H — Cl

The size of a bond dipole is measured quantitatively by the Dipole moment, U.. A dipole moment measures the polarity of a bond by taking into account two key factors:

IU How much charge is separated along the bond axis IU How far apart the charge is separated

Polar bonds have a larger dipole moment than nonpolar bonds. Dipole moments are vector quantities, which means that they have both size and direction. Within a molecule, different bonds may point in different directions, and these differences can be important.

Individual bonds’ dipoles sum over all the bonds of a molecule, resulting in a Molecular dipole. In addition to the Permanent dipoles Created by polar bonds, Instantaneous dipoles Can flicker into and out of existence within nonpolar bonds and molecules. Both kinds of dipoles play important roles in the interactions between molecules. Permanent dipoles lead to Dipole-dipole interactions And to Hydrogen bonds. Instantaneous dipoles lead to attractive London forces.

In addition to ion-ion interactions, these are the forces that must be overcome in order to turn a liquid into a gas or a solid into a liquid. Different compounds have different boiling points and different melting points because they engage in different collections of interactions. Here is a summary of the intermolecular forces that contribute to boiling and melting points, and examples of compounds in which each kind of force dominates.

Force Compound

Ion-ion NaCl

Hydrogen bonding H2O

Dipole-dipole H2CO

London forces CH4

Melting point Boiling point

1074 K 1738 K

273 K 373 K

156 K 254 K

91 K 112 K

In This Chapter

^ Checking out the nomenclature of organic compounds

^ Taking in testable trends in the properties of organic compounds

^ Discovering isomerism and chirality

^ Running through common organic reactions

M W Rganic chemistry Is a huge subfield, and the AP chemistry exam merely scratches the

Surface of it. Still, a good grasp of the concepts and naming rules covered in this chapter could get you those few extra points needed to push your score up one grade on exam day.

Don’t spend too much time here if you’re still struggling with the big-point-gaining chapters earlier in this book. Your time is better spent polishing your knowledge of the concepts in those earlier chapters, if you’re struggling, because so few questions on the AP exam address organic chemistry. But if you feel comfortable with the big-point-gaining chapters, then tackle this chapter with vigor and get ready to file lots of new information away in that chemistry-packed brain of yours — organic chemistry is heavy on memorization.

In addition to the few questions on each AP exam directly addressing organic chemistry, a basic knowledge of the information in this chapter, including basic structure and naming. can aid you in answering many other questions on the exam. Organic compounds are sometimes used in questions testing concepts such as kinetics, stoichiometry, and colligative properties.

Naming Names: Nomenclature and Properties of Hydrocarbons

Any study of organic chemistry begins with the study of Hydrocarbons. Hydrocarbons are some of the simplest and most important organic compounds and they contain only hydrogen and carbon atoms. Organic compounds Are based on carbon skeletons. Carbon skeletons can be modified — you can dress them up with chemically interesting atoms like oxygen, nitrogen, halogens, phosphorous, silicon, or sulfur. This cast of atomic characters may seem like a rather small subset of the more than 100 elements in the periodic table. It’s true: Organic compounds typically use only a very small number of the naturally occurring elements. Yet these molecules include the most biologically important compounds in existence. In this section you will learn how to recognize and name all of the types of hydrocarbons that appear on the AP exam, including alkenes, alkanes, alkynes, and cyclic hydrocarbons.

Starting with straight-chain alkanes

Starting with straigi linear hydrocarbons

The simplest of the hydrocarbons fall into the category of Alkanes. Alkanes are chains of carbon atoms connected by single covalent bonds. Chapter 5 describes how single covalent bonds result when atoms share pairs of valence electrons. Carbon molecules have four valence electrons. So, carbon atoms are eager to donate their four valence electrons to cova-lent bonds so that they can receive four donated electrons in turn, filling their valence shell. In other words, carbon really likes to form four bonds. In alkanes, each of these is a single bond with a different carbon or hydrogen partner.

The simplest of the alkanes, called Continuous – Or Straight-chain alkanes, Consist of one straight chain of carbon atoms linked with single bonds. Hydrogen atoms fill all the remaining bonds. Other types of alkanes include closed circles and branched chains.

We begin with straight-chain alkanes because they make clear the basic strategy for naming hydrocarbons. From the standpoint of naming, the hydrogen atoms in a hydrocarbon are more or less "filler atoms." Alkanes’ names are based on the largest number of consecutively bonded carbon atoms. So, the name of a hydrocarbon tells you about that molecule’s structure. To name a straight-chain alkane, simply match the appropriate chemical prefix with the suffix -ane. The prefixes relate to the number of carbons in the continuous chain, which we list in Table 25-1.

The naming method in Table 25-1 tells you how many carbons are in the chain. Because you know that each carbon has four bonds and because you are fiendishly clever, you can deduce the number of hydrogen atoms in the molecule as well. Any carbon at the end of a chain must be bonded to three hydrogens and all interior carbons must be bonded to two. Consider the carbon structure of pentane, for example, shown in Figure 25-1.

Figure 25-1: I I I I I

Pentane’s I I I I I

Carbon — C-C-C-C-C—

Skeleton.

Only four carbon-carbon bonds are required to produce the five-carbon chain of pentane. This leaves many bonds open — two for each interior carbon and three for each of the terminal carbons. These open bonds are satisfied by carbon-hydrogen bonds, thereby forming a hydrocarbon, as shown in Figure 25-2.

Figure 25-2:

Pentane’s hydrocarbon

Structure.

H HHH H

CCC C H

HHHHH

H

C

If you add up the hydrogen atoms in Figure 25-2, you get 12. So, pentane contains 5 carbon atoms and 12 hydrogen atoms.

As the organic molecules you study get more and more complicated, it will become more and more important to draw the molecular structure to visualize the molecule. In the case of straight-chain alkanes, the simplest of all organic molecules, you can remember a convenient formula for calculating the number of hydrogen atoms in the alkane without actually drawing the chain:

Number of hydrogen atoms = (2 x number of carbon atoms) + 2

You can refer to the same molecule in a number of different ways. For example, you can refer to pentane by its name (ahem. . . Pentane), By its molecular formula, C5H12, or by the complete structure in Figure 25-2. Clearly, these different names include different levels of structural detail. A Condensed structural formula Is another naming method, one that straddles the divide between a molecular formula and a complete structure. For pentane, the condensed structural formula is CH3CH2CH2CH2CH3. This kind of formula assumes that you understand how straight-chain alkanes are put together. Carbons on the end of a chain, for example, are only bonded to one other carbon, so they have three additional bonds to be filled by hydrogen and are labeled as CH3 in a condensed formula. Interior carbons are bonded to two neighboring carbons and have only two hydrogen bonds, and so are labeled CH2.

Going out on a limb: Making branched alkanes by substitution

Not all alkanes are straight-chain alkanes. That would be too easy. Many alkanes are so-called Branched alkanes. Branched alkanes differ from continuous-chain alkanes in that carbon chains substitute for a few hydrogen atoms along the chain. Atoms or other groups (like carbon chains) that substitute for hydrogen in an alkane are called Substituents.

Naming branched alkanes is slightly more complicated, but you need only to follow these simple set of steps to arrive at a proper (and often lengthy) name:

1. Count the longest continuous chain of carbons. Tricky chemistry problems often show branched alkanes with the longest chain snaking through a few branches instead of obviously lined up in a row. Consider the two carbon structures shown in Figure 25-3. The two are actually the same structure, drawn differently! Yikes. In either case, the longest continuous chain in this structure has eight carbons. That wasn’t hard — and yippee, now you already know the basic name for the compound. It is an octane.

Cc

C1 C2 C3 C4 C5 C

C6

B

C7

7

C8

8

C

Figure 25-3′ Ci – c2-c3 - c4- c5- cb – c7 – c

One carbon i i i

Structure | | |

Drawn two c c c

Different I

Ways. I

2. Number the carbons in the chain Starting with the end which is closest to a branch.

You can always check to be sure you have done this step correctly by numbering the carbon chain from the opposite end as well. The correct numbering sequence is the one in which the substituent branches extend from the lowest-numbered carbons. For example, as it is drawn and numbered in Figure 25-3, the alkane has substituent groups branching off of its third, fourth, and fifth carbons. If the carbon chain had been numbered backward, these would be the fourth, fifth, and sixth carbons in the chain. Because the first set of numbers is lower, the chain is numbered properly. The longest chain in a branched alkane is called the Parent chain.

3. Count the number of carbons in each branch. These groups are called Alkyl groups And are named by adding the suffix -yl To the appropriate alkane prefix (Table 25-1 awaits your visit). The three most common alkyl groups are the methyl (one carbon), ethyl (two carbons), and propyl (three carbons) groups. Figure 25-3 has two methyl groups, one ethyl group, and no propyl groups.

Be careful when you find yourself dealing with alkyl groups made up of more than just a few carbons. A tricky drawing may cause you to misnumber the parent chain!

4. Attach the number of the carbon from which each substituent branches to the front of the alkyl group name. For example, if a group of two carbons is attached to the third carbon in a chain, like it is in Figure 25-3, the group is called 3-ethyl.

5. Check for repeated alkyl groups. If multiple groups with the same number of carbons branch off the parent chain, do not repeat the name. Rather, include multiple numbers, separated by commas, before the alkyl group name. Also, specify the number of instances of the alkyl group by using the prefixes di-, tri-, tetra-, and so on. For example, if one-carbon groups (in other words, methyl groups) branch off carbons four and five of the parent chain, the two methyl groups appear together as "4,5-dimethyl."

6. Place the names of the substituent groups in front of the name of the parent chain In alphabetical order. Prefixes like di-, tri-, and tetra – do not figure into the alphabetizing. So, the proper name of the organic molecule in Figure 25-3 is 3-ethyl-4,5-dimethyloctane.

Note that hyphens are used to connect all the naming elements except for the last connection to the parent chain, which includes neither a hyphen nor a space (. . . dimethyl-octane is wrong). We recommend practicing alkanes till you get them in your sleep before moving on to other organic compounds!

Getting unsaturated: Alkenes and alkynes

Carbons can do more than engage in four single bonds. There’s more to organic molecules than substituent-for-hydrogen swaps:

When carbons in an organic compound fill their valence shells entirely with single bonds, we say the compound is Saturated.

When hydrocarbons contain carbons that bond to each other more than once, creating double or triple covalent bonds, we say these hydrocarbons are Unsaturated Because they have fewer than the maximum possible number of hydrogens or substituents.

For every additional carbon-carbon bond formed in a hydrocarbon, two fewer covalent bonds to hydrogen are formed.

When neighboring carbons share four valence electrons to form a double bond, the resulting hydrocarbon is called an Alkene. Alkenes are characterized by these chemically interesting double bonds, which are more reactive than single carbon-carbon bonds (see Chapter 7 for a review of sigma and pi bonding). Double bonds also change the shape of a hydrocarbon, because the Sp2 Hybridized valence orbitals assume a trigonal planar geometry, as shown by the carbons of ethene in Figure 25-4. Saturated carbon is Sp3 Hybridized and has tetrahedral geometry (again, see Chapter 7 to review hybridization).

H

\

Figure 25-4′

Ethene carbons.

H

H

H

Naming alkenes is slightly more complicated than naming alkanes. In addition to the number of carbons in the main chain and any branching substituents, you must also note the location of the double bonds in an alkene and incorporate that information into the name. Nevertheless, the essential naming strategy for alkenes is quite similar to that for alkanes:

1. Locate the longest carbon chain that includes the double bond, and number it, Starting at the end closest to the double bond. In other words, double bonds trump sub-stituents when it comes to numbering the parent chain. Build the name of the parent chain by using the same prefixes as used for alkanes (refer to Table 25-1), but match the prefix with the suffix -ene. A three-carbon chain with a double bond, for example,

Is called "propene."

2. Number and name substituents that branch off the alkene In the same way as done for alkanes (see the section earlier in this chapter, "Starting with straight-chain alkanes — linear hydrocarbons" for more on naming alkanes). List the number of the substituted carbon, followed by the name of the substituent. Separate the sub-stituent number and name with a hyphen.

3. Identify the lowest-numbered carbon that participates in the double bond, and put that number Between the substituent names and the parent chain name (sandwiched by hyphens), but after all the substituent names. For example, if the second and third carbons of a five-carbon alkene engage in a double bond, then the molecule is called 2-pentene, not 3-pentene. If that same molecule has a methyl substituent at the fourth carbon, then the molecule is called 4-methyl-2-pentene.

Alternately, and especially when there are substituents present, the position of an unsaturation is indicated between the prefix and suffix of the parent chain name. So, 4-methyl-2-pentene may also be written 4-methylpent-2-ene.

Alkynes Are hydrocarbons in which neighboring carbons share six electrons to engage in triple covalent bonds. The naming strategy for alkynes is the same as that for alkenes, except that the alkyne parent chain is named by matching the prefix with the suffix -yne.

An important consequence of the presence of multiple bonds in alkenes and alkynes is their marked increase in reactivity over alkanes. Generally speaking, the more multiple bonds a compound has, the more reactive it tends to be. This is because of the propensity of multiple bonded carbons to undergo addition reactions, which are discussed at the end of this chapter.

Rounding ‘em Up: Circling Carbons with Cyclics and Aromatics

The compounds we discuss earlier in this chapter are linear or branched. However, hydrocarbons can be circular, or Cyclical, Hydrocarbons There are two important categories of cyclical hydrocarbons:

Cyclic aliphatic Hydrocarbons Aromatic Hydrocarbons

Chemists sometimes divide hydrocarbons into aliphatic and aromatic categories to highlight important differences in structure and reactivity. Without going into more technical detail than is useful here, aliphatic molecules and aromatic molecules have significantly different electronic configurations (which electrons go into which orbitals). As a result, the two types of hydrocarbons typically undergo different kinds of reactions. In particular, they tend to undergo different kinds of substitution reactions, ones in which some atom or group substitutes for hydrogen.

Shutting the circle with cyclic aliphatic hydrocarbons

Cyclic aliphatic hydrocarbons are like the hydrocarbons that we discuss earlier in this chapter, except that they form a closed ring. The rules for naming these compounds build on the rules for naming alkanes, alkenes, and alkynes. For example, a cyclical six-carbon alkane includes the name "hexane," but is preceded by the prefix Cyclo-, Making the final name "cyclohexane."

A single substituent or unsaturation on a cyclic aliphatic hydrocarbon does not require a number. So, a single methyl-for-hydrogen substitution on cyclohexane yields a compound with the name methylcyclohexane. Likewise, a lone double bond unsaturation on cyclo-hexane yields a compound with the name cyclohexene.

Multiple substitutions or unsaturations require numbering. In these cases, the same rules apply for deciding the rank of substituents as applied to alkenes and alkynes. Triple bonds

Outrank double bonds. Double bonds outrank other substitutions. So, number the carbons in the way that respects these rankings and produces the lowest overall numbers. A cyclo-hexane molecule with two methyl substituents on neighboring carbons, for example, is called 1,2-dimethylcyclohexane.

Analyzing aromatic hydrocarbons

Aromatic hydrocarbons have special properties because of their electronic structure. Aromatics are both cyclic and conjugated:

I Cyclic: The carbons form a ringlike structure.

Conjugated: Conjugation results from an alternation of double or triple bonds with single bonds.

Aromatic molecules have clouds of Delocalized pi electrons, Electrons that move freely through a set of overlapping P Orbitals. The model aromatic compound is benzene. Because of its cyclical, conjugated bonding, pi electrons delocalize evenly into rings above and below the plane of the flat benzene molecule. Aromatic compounds are very stable compared to their aliphatic counterparts.

Numbering substituents on aromatics follows the same basic pattern as followed for cyclic aliphatic compounds. A single substituent requires no numbering, as in methylbenzene. Multiple substituents are numbered by rank, with the highest-ranked substituent placed on carbon number one, and proceeding in a way that results in the lowest overall numbers. A benzene ring with ethyl and methyl substituents situated two carbons away from one another, for example, would be called 1-ethyl-3-methylbenzene.

Decorating Skeletons: Organic Functional Groups

Surely you’ve noticed that in this chapter we have thus far limited our study of organic compounds to those made entirely of hydrogen and carbon. When studying the complex rules governing hydrocarbons, it is easy to forget that the periodic table contains more than 100 other elements. Although organic chemistry often concentrates on just a few of those elements, you’ll need to be familiar with many other compound types, and this time they’ll be made of more than just hydrogen and carbon. All of these exotic new compounds contain plain vanilla carbon chains, but each has a distinguishing feature composed of another, more exotic element. These embellishments are called Functional groups. These groups include noncarbon elements, and give organic compounds a dizzying array of different properties.

Because hydrocarbons are old news, we sometimes refer to them simply using the letter R, representing any hydrocarbon group: branched or unbranched, saturated or unsaturated.

Table 25-2 summarizes the other important organic compounds, their distinguishing features, and their endings. Some molecules contain more than one of these distinguishing features. Locations where one of these groups or a multiple bond appear are prime sites for reactions to occur. These sites are therefore called Functional groups. Carefully study the functional group column in Table 25-2 so you can recognize them quickly. These are the structures you should keep an eye out for later in this chapter.

Alcohols: Hosting a hydroxide

Alcohols Are hydrocarbons with a hydroxide group attached to them. Their general form is R-OH. To name an alcohol, you simply count the longest number of consecutive carbon atoms in the chain, find its prefix Table 25-2, and then attach the ending -ol To that prefix. For example, a one-carbon chain with an OH group attached is methanol, two is ethanol, three is propanol, and so on. Substituents (groups branching off the main chain) must, as always, be accounted for, and you must specify the number of the carbon atom in the chain to which the hydroxyl (OH) group is attached. Begin your numbering at the end of the chain closest to the OH, and attach the number before the prefix + -ol. For example, the compound shown in Figure 25-5 contains a six-carbon chain with a hydroxyl group. Its base name is therefore

Hexanol. It has methyl groups (with one carbon each) on the second and fourth carbons and its OH group lies on the third carbon. Its proper name is therefore 2,4-dimethyl-3-hexanol.

Some alcohols are Polyhydroxyl, Or contain multiple hydroxyl groups. In order to avoid confusion with multiples of any substituent groups, the prefixes indicating numbers of hydroxyl groups in an alcohol are attached to the suffix Ol, Making them diols, triols, etc. For example, a four carbon alcohol containing hydroxyl groups on the first and third carbons is 1,3-butane-diol. The numbers indicating the locations of the hydroxyl groups must come before the base name but after any substituent groups.

_ HH OH H H H

Figure 25-5:

An example H-C-C-C-C-C-C-H

Of an alcohol.

_ H CH, H CH, H H

Ethers: Invaded by oxygen

Perhaps most commonly known for their use as early anesthetics, Ethers Are carbon chains that have been infiltrated by an oxygen atom. These oxygen atoms lie conspicuously in the middle of a carbon chain like badly disguised spies in an enemy camp and have the general formula R-O-R. Ethers are named by naming the alkyl groups on either side of the lone oxygen as substituent groups individually (adding the ending -yl To their prefixes), and then attaching the word Ether To the end. For example, the compound shown in Figure 25-6 is an ether with a methyl group (one carbon) on one side of the oxygen and an ethyl (two carbons) on the other. Placing the substituents in alphabetical order, the compound’s proper name is ethyl methyl ether.

Figure 25-6:

An example of an ether.

3

CH

Oxygen atoms in ethers are often surrounded by two identical alkyl groups, in which case the prefix Di – Must be attached to the name of that substituent. For example, an oxygen surrounded by two propyl groups (with three carbons each) is called dipropyl ether.

Carboxylic acids: R-COOH brings up the rear

Carboxylic acids Appear to be ordinary hydrocarbons until you reach the very last carbon in their chain, whose three ordinary hydrogens have been usurped by a double-bonded oxygen and a hydroxyl group (an OH group). This gives a carboxylic acid the general form R-COOH. These compounds are named by attaching the suffix -oic acid To the end of the prefix. For example, the compound shown in Figure 25-7 has four carbons, the last in the chain attached to a double-bonded oxygen and a hydroxyl group. The compound is therefore called butanoic acid. The hydrogen of the COOH group can pop off as H+ leaving an R-COO – ion, which is why these compounds are called acids. Carboxylic acids are a favorite compound type for AP exam writers because they are both organics and acids. It is important to note, however, that all car-boxylic acids are weak acids, so learn to spot them among acid lists and keep their relative weakness in mind.

Figure 25-7:

An example of a carboxylic acid.

H3C’

3

CH,

CH,

OH

O

C

Esters: Creating two carbon chains

What if the same double-bonded oxygen/hydroxyl pair from a carboxylic acid has infiltrated deeper into the hydrocarbon chain and is not on an end, but rather in the middle? In order to accomplish this, the COOH group must lose the hydrogen of its hydroxyl group (which, as an acidic hydrogen, is no problem) to open up a bond to which a second hydrocarbon can attach. A compound of this nature, with the general formula R-COOR, is called an Ester. Lots of nice, smelly coumpounds are esters, including the compounds that give pears, apples, and bananas their pleasant aromas.

Because the carbon chain in an ester is also broken by an oxygen, you’ll need to choose one carbon chain as a lowly substituent group and the other to bear the proud suffix of -oate. This high-priority group is always the carbon chain that includes the carbon double-bonded to one oxygen and single bonded to the other. The group on the other side of the single-bonded oxygen is named as an ordinary substituent. For example, the compound shown in Figure 25-8 is phenyl ethanoate, because the high priority group has two carbons and the low priority group is a benzene ring.

H3C

3

C

C

/

Figure 25-8:

An example of an ester.

—O

\

C HC

HCCH

HCCH

\ _ /

CH CH

That’s right! The low-priority group on the other side of the oxygen doesn’t have to be a hydrocarbon chain! It can be a ring or a metal as well. In Figure 25-8, this group is a benzene ring.

Aldehydes: Holding tight to one oxygen

Aldehydes Are much like carboxylic acids except that they lack the second oxygen in the COOH group. Their final carbon shares a single bond with its neighboring carbon, a double bond with an oxygen, and a single bond with hydrogen. Aldehydes are named with the suffix -al (not to be confused with the -ol Of alcohols) and have the general formula R-CHO. For example, the compound in Figure 25-9 is pentanal, a five-carbon chain with a double-bonded oxygen taking the place of two of the hydrogen bonds on the final carbon.

Figure 25-9:

An example of an aldehyde.

. CH,

-CH,

3

*CH,

CH

O

Ketones: Lone oxygen sneaks up the chain

Much like esters are to carboxylic acids, Ketones Share the same basic structure as aldehydes, except that their double-bonded oxygen can be found hiding out in the midst of the carbon chain, giving them the general form R-COR. Ketones are named by adding the suffix -one To the prefix generated by counting up the longest carbon chain. Unlike esters, however, the carbon chain of a ketone is not broken by the double-bonded oxygen group, making naming them much simpler. The compound shown in Figure 25-10, for example, is called simply 2-butanone; the number before the name specifies the number of the carbon to which the oxygen is attached.

Figure 25-10:

An example of a ketone.

H3C

CH

CH

O

Halocarbons: Hello, halogen!

Halocarbons Are simply hydrocarbons with a halogen or more tacked on to a carbon in place of a hydrogen atom. (A Halogen Is a group VIIA element.) Halogens in a halocarbon are always named as substituent groups — Fluoro, chloro, bromo, Or Iodo, One for each of the four halogens that form halocarbons (F, Cl, Br or I). The shorthand for a halocarbon is R-X, where X is the halogen. The compound shown in Figure 25-11 has two bromo groups attached, one to the second carbon and one to the third in a five-carbon alkane. Its official name is, therefore, 2,3-dibromopentane.

Br

CH

CH

Figure 25-11:

An example of a

Halocarbon.

H3C

3

CH

CH

Br

Amines: Hobnobbing with nitrogen

Amines Are as conspicuous as Waldo on the very first page of a Where’s Waldo Book. No tricksters wearing red-striped shirts are waiting to fool you among these organic molecules. Amines and their derivatives are the only compounds that you’ll encounter in basic organic chemistry that contain nitrogen atoms. Amines have the basic form R-NH2, and they’re named by naming the carbon chain as a substituent (with the ending -yl), And then adding the suffix -amine. The structure in Figure 25-12, for example, is called ethylamine because it’s a two-carbon ethyl chain with an amine group.

Figure 25-12:

An example of an amine.

-CH,

H, C

3

‘ NH,

Rearranging Carbons: Isomerism

How about this: Two organic molecules have identical chemical formulas. Each atom in one molecule is bonded to the same groups as in the other. They’re identical molecules, right? Wrong! (Mischievous chemistry gods point and snicker.) Many organic molecules are Iso-mers, Compounds with the same formula and types of bonds, but with different structural or spatial arrangements. Who cares about such subtle differences? Well, you might. Consider thalidomide, a small organic molecule widely prescribed to pregnant women in the late 1950s and early 1960s as a treatment for morning sickness. Thalidomide exists in two isomeric forms that rapidly switch from one to the other in the body. One isomer is very effective at combating morning sickness. The other isomer causes serious birth defects. Isomers matter.

Isomers can be confusing. They fall into different categories and subcategories. So, before committing your brain to a game of isomeric Twister, peruse the following breakdown: Keep in mind, however that all isomers, no matter what their specific classification, have both the same number and type of atoms.

Structural isomers Have identical molecular formulas but differ in the arrangement of bonds.

Stereoisomers Have identical connectivities — all atoms are bonded to the same types of other atoms — but differ in the arrangement of atoms in space.

• Diastereomers Are stereoisomers that are Not Non-superimposable mirror images of each other. Two types of diastereomers exist: Geometric isomers (or Cis-trans isomers) Are diastereomers that differ in the arrangement of groups around a double bond, or the plane of a ring. Conformers And Rotamers Are diastereomers that differ because of rotations about individual bonds (we don’t cover them in this book because they are beyond the scope of general chemistry).

• Enantiomers Are stereoisomers that Are Non-superimposable mirror images of each other.

You already know how to recognize and name stereoisomers appropriately by carefully studying the structure of the main chain and branching substituents. This section focuses on the trickier category: stereoisomers.

Picking sides with geometric isomers

Geometric isomers Or Cis-trans isomers Are a good place to start in the world of stereoisomers because they’re the easiest of the stereoisomers to understand. In the following sections, we explain how isomers relate to alkenes, alkanes that aren’t straight-chain, and alkynes.

Straight-chain alkanes are immune from geometric isomerism because their carbon-carbon single bonds can rotate freely. Unsaturate (or add another bond to) one of those bonds, however, and you’ve got a different story. Alkenes have double bonds that resist rotation. Furthermore, the Sp2 Hybridization of double-bonded carbons gives them trigonal planar bonding geometry (see Chapter 7 for an introduction to hybridization). The result is that groups attached to these carbons are locked on one side or the other of the double bond. Convince yourself of this by examining Figure 25-13.

Figure 25-13:

Cis And Trans Isomers ofan alkene.

H3C

3 ‘

H

\ / 3

H

H3C

3

H

\ /

H CH

A)

B)

In the left-hand structure of Figure 25-13, the carbon chain continues along the same side of the carbon-carbon double bond. Both methyl (-CH3) groups lie on the same side of the unsat-uration. This is called Cis Configuration. In the right-hand structure, the carbon chain swaps sides as it proceeds across the double bond. The methyl groups lie on opposite sides of the unsaturation. This is called a Trans Configuration.

Naming Cis-trans Isomers is simple. Attach the appropriate Cis – Or Trans – Prefix before the number referring to the carbon of the double bond. For example, the left-hand structure in Figure 22-1 is Cis-2-butene, while the structure on the right is Trans-2-butene.

Although straight-chain alkanes happily avoid isomerism by rotating merrily about their single bonds, the four bonds of sp3-hybridized carbons assume tetrahedral geometry. Detailed representations like the one shown for methane in Figure 25-14 reveal this geometry. In the structure of methane, the bonds depicted as straight lines run in the plane of the page. The bond drawn as a filled wedge projects outward from the page. The bond drawn as a dashed wedge projects behind the page. These filled and dashed wedge symbols are known as Stereo bonds Because they are helpful in identifying stereoisomers.

Figure 25-14:

Stereo bonds in methane.

C

HH

H

When alkanes close into rings, they can no longer freely rotate about their single bonds, and the tetrahedral geometry of sp3-hybridized carbons creates Cis-trans Isomers. Groups bonded to ring carbons are locked above or below the plane of the ring, as shown in Figure 25-15.

The figure depicts two different versions of trans-1,2-dimethylcyclohexane. In both versions, the adjacent methyl substituents are locked in Trans Positions, on opposite sides of the ring.

The upper set of structures shows the plane of the ring as seen from above, and highlights the Trans – Configuration of the methyl groups with stereo bonds.

The lower set of structures shows the same rings rotated 90 degrees downward and toward you.

H, C,

CH,

____■ ,

CH CH,

H3CV CH2

,

CH CH,

Figure 25-15:

Two isomers of Trans-1, ,-dimethyl-cyclo-hexane.

H3C

3

CH CH,

H3C

3

CH CH,

CH

The Trans Configuration of the methyl groups is most clear in the lower structures. The front-most methyl group is highlighted with explicit hydrogen atoms to emphasize its position above or below the plane of the ring.

Alkynes also contain carbon-carbon bonds that cannot rotate freely. However, the Sp Hybridization of the carbons in these bonds leads to linear bonding geometry. The two-carbon alkyne, ethyne, is shown in Figure 25-16. Each carbon locks three of its valence electrons into the axis of the triple bond. Each has only one valence electron remaining with which to bond to hydrogen. No Cis-trans Isomerism is possible in this scenario.

Figure 25-16:

No iso-merism is possible at the triple bonds of alkynes, such as ethyne.

■ CC-H

H

Changing Names: Some Common Organic Reactions

The carbon-based organic molecules presented in this chapter can morph into one another through chemical reactions. Such reactions are rather common, and the two most common types are substitution and addition reactions.

Alcohols are particularly prone to a chemical process called Substitution In which their OH group is replaced by a different atom. For example, the OH group on 2-pentanol can be replaced by the halogen fluorine, turning the alcohol into a halocarbon called 2-fluoro-pentane (see Figure 25-17).

OH

CH3— CH2— CH2— CH — CH3 + F2

3 2 2 3 2

Figure 25-17:

The

Process of F substitution.

CH3 CH2 CH2 CH CH3

3 2 2 3

The double bonds of alkenes also make them particularly likely to react with other compounds through a process called Addition. If the double bond between two carbon molecules is broken, it allows each of those two carbon atoms to form a bond with another atom or molecule. The reaction shown in Figure 25-18, for example, shows water being added across the double bond of 1-hexene. The water molecule itself is split into two pieces — simple hydrogen and a hydroxide — and each of them is added to one of the two carbons that used to share a double bond, forming either 1-hexanol or 2-hexanol.

Figure 25-18: HOH

The process

Of addition. H2C CH — CH2— CH2— CH2— CH3

2 2 2 2 3

H3C

3

OH

CH CH

CH

CH

CH

In This Chapter

► Keeping up with kinetic theory

► Going through the Gas Laws

► Digging in Dalton’s Law of Partial Pressures

► Checking out diffusion and Graham’s Law

► Encountering ideal and nonideal gases

T first pass, gases may seem to be the most mysterious of the states of matter. Nebulous and wispy, gases easily slip through our fingers. For all their diffuse fluidity, however, gases are actually the best understood of the states. The key to understanding gases is that they all tend to behave in the same ways — physically, if not chemically. For example, gases expand to fill the entire volume of any container in which you put them. Also, gases are easily compressed to fill smaller volumes.

Before taking the AP chemistry exam, you will need to become familiar with a body of equations called the ideal gas laws and should be able to choose the equation appropriate to a particular situation and manipulate it. You will also need to have a solid understanding of the underlying theory that describes the behavior of gases, including kinetic molecular theory and the differences between real and ideal gases. This chapter will familiarize you with all of these concepts, after that you should tackle Chapter 10, which includes questions on gases similar to those which will appear on the actual exam.

Moving and Bouncing: Kinetic Molecular Theory of Gases

Imagine two billiard balls, each glued to either end of a spring. How many different kinds of motion could this contraption undergo? You could twist along the axis of the spring. You could bend the spring or stretch it. You could twirl the whole thing around, or you could throw it through the air. Diatomic molecules have just such a structure and can undergo these same kinds of motions when you supply them with energy. As collections of molecules undergo changes in energy, those collections move through the states of matter — solid, liquid, and gas. Transitions between phases of matter will be described in detail in Chapter 11, while this chapter will lay out the basics of kinetic theory.

*JJIBE*

The basics of kinetic theory

Kinetic theory, the body of ideas that explains the internal workings of gases and the phenomena caused by them, first made a name for itself when scientists attempted to explain and predict the properties of gases. Kinetic theory also explains the behavior of solids and liquids, and Chapter 11 will cover how it applies to each of those phases. Kinetic theory as applied to gases is particularly concerned with how the properties of a gas change with varying temperature and pressure. The underlying idea behind kinetic theory is the concept of kinetic energy, or the energy of motion. The particles of matter within a gas (atoms or molecules) undergo vigorous motion as a result of the kinetic energy within them.

Gas particles have a lot of kinetic energy and constantly zip about, colliding with one another or with other objects. This is a complicated picture, but scientists simplify things by imagining an ideal gas where

Particles move Randomly.

The only type of motion is Translation (moving from place to place, as opposed to twisting, vibrating, and spinning).

When particles collide, the collisions are Elastic (perfectly bouncy, with no loss of energy).

Gas particles Neither attract nor repel One another. Gases that actually behave in this simplified way are called Ideal.

The model of ideal gases explains why gas pressure increases with increased temperature. By heating a gas, you add kinetic energy to the particles. As a result, the particles collide with greater force upon other objects, so those objects experience greater pressure. In other words, as temperature increases, the average kinetic energy of gas particles increases and pressure increases proportionally as well.

Statistics in kinetic theory

The study of all matter, and gases in particular, is a study of statistics. Gas properties are measured in averages rather than absolutes. Note that above we spoke of temperature as being proportional to the average kinetic energy of the gas particles, not the kinetic energy of an individual gas particle. You could spend your whole life searching for a collection of gas molecules in which each individual particle is moving at the same speed as the next. Because gas particles collide and transfer energy between one another at random, neighboring gas particles generally experience different numbers of collisions and end up with different kinetic energies. If you were to measure the velocities of a whole slew of gas molecules, you would find that although they will have a distribution of varying velocities, they will be clustered around an average value. Remember that a liter of gas at STP contains 6.02 x 1023 gas particles, which is a lot to average!

Examine the two velocity distributions in Figure 9-1. Both show a similar bell-shaped curve (called a Maxwell-Boltzman distribution). Note that the probability of finding a gas particle at a certain velocity drops off precipitously as that velocity gets farther from the average velocity peak. Although both of the distributions in Figure 9-1 are Gaussians, there is one key difference between them: The figure on the left represents the velocity distribution of gas particles at a low temperature, and the figure on the right shows a similar distribution at a higher temperature.

Note the features of the low-temperature distribution: Ji* The peak of the velocity distribution is at a lower velocity.

Ji* The velocity distribution shows a pronounced peak with a relatively narrow range of velocities surrounding it.

Note the features of the high-temperature distribution:

Ji* The peak of the velocity distribution is at a higher velocity.

Ji* The velocity distribution shows a wider, flattened curve with a large range of velocities.

Figure 9-1:

Velocity distributions for gases at high and low temperatures.

Low Temperature

-.^ High Temperature

Speed

Particle velocities translate into average kinetic energies through the equation: KE = 34 mv2

This is the kinetic energy of an individual gas molecule, where m is the mass of the molecule and v is the velocity. To calculate an average kinetic energy, we therefore need an expression for the average velocity.

/3RT

U = .1—R-r-

V M

In this expression U Is the average velocity of a collection of particles, and is alternately written as urms (the root-mean-square velocity), R Is the ideal gas constant 8.314 J K"1 mol"1, T Is temperature, and M Is the molar mass. Although we will not derive this expression here, take a minute to look at it and assure yourself that it makes sense. The expression has temperature in the numerator of the velocity calculation, which means that as temperature increases, average velocity should as well. This is consistent with kinetic theory. Molar mass, however, appears in the denominator of the equation, which means that as molar mass increases, average velocity decreases. This too makes sense; kinetic energy depends on both mass and velocity, so heavier particles move more slowly than lighter particles at the same kinetic energy. You will also often see this expression written as the following, which is simply a variation of the equation above and will yield the same result:

/3kT U = A-

V M

Here, Boltzmann’s constant K (1.38 x 10"23 J K"1) replaces the ideal gas constant R, And M, The mass of an individual gas particle, replaces molar mass M.

The final equation that you may consider useful when calculating the kinetic energy of a system of molecules is the following expression for the kinetic energy Per mole Of a gas, which can be derived by plugging the first expression you were given for the average velocity of gas particles into the general equation for kinetic energy and solving for kinetic energy per mole, using this:

KE = KRT

Remember to always use R= 8.314 J K"1 mol"1 for these kinetic energy problems. If you use the wrong units for the gas constant, you will lose points on the AP exam.

Aspiring to Gassy Perfection: Ideal Gas Behavior

Like all ideals (the ideal job, the ideal mate, and so on), ideal gases are entirely fictional. All gas particles occupy some volume. All gas particles have some degree of interparticle attraction or repulsion. No collision of gas particles is perfectly elastic. But lack of perfection is no reason to remain unemployed or lonely. Neither is it a reason to abandon kinetic molecular theory of ideal gases. In this chapter, you will be introduced to a wide variety of applications of kinetic theory, which come in the form of the so-called "gas laws."

Relationships between four factors: pressure, volume, temperature, and number of particles are the domain of the gas laws. We take a look at the gas laws in the sections that follow.

Boyle’s Law

The first of these relationships to have been formulated into a law is that between pressure and volume. Robert Boyle, an Irish gentleman regarded by some as the first chemist (or "chymist," as his friends might have said), is typically given credit for noticing that gas pressure and volume have an inverse relationship:

Volume = Constant x (1/Pressure)

This statement is true when the other two factors, temperature and number of particles, are fixed. Another way to express the same idea is to say that although pressure and volume may change, they do so in such a way that their product remains constant. So, as a gas undergoes change in pressure (P) And volume (V) Between two states, the following is true:

P1 x V1 = P2 x V2

The relationship makes good sense in light of kinetic molecular theory. At a given temperature and number of particles, more collisions will occur at smaller volumes. These increased collisions produce greater pressure. And vice versa. Boyle had some dubious ideas about alchemy, among other things, but he really struck gold with the pressure-volume relationship in gases.

Charles’s Law

*JJIBE*

Lest the Irish have all the gassy fun, the French contributed a gas law of their own. History attributes this law to French chemist Jacques Charles. Charles discovered a direct, linear relationship between the volume and the temperature of a gas:

Volume = Constant x Temperature

This statement is true when the other two factors, pressure and number of particles, are fixed. Another way to express the same idea is to say that although temperature and volume may change, they do so in such a way that their ratio remains constant. So, as a gas undergoes change in temperature (T) And volume (V) Between two states, the following is true:

Not to be outdone by the French, another Irish scientist took Charles’s observations and ran with them. William Thomson, eventually to be known as Lord Kelvin, took stock of all the data available in his mid-nineteenth century heyday and noticed a couple of things:

First, plotting the volume of a gas versus its temperature always produced a straight line.

Second, extending these various lines caused them all to converge at a single point, corresponding to a single temperature at zero volume. This temperature — though not directly accessible in experiments — was about "273 degrees Celsius. Kelvin took the opportunity to enshrine himself in the annals of scientific history by declaring that temperature as Absolute zero, The lowest temperature possible.

This declaration had at least two immediate benefits. First, it happened to be correct. Second, it allowed Kelvin to create the Kelvin temperature scale, with absolute zero as the Official Zero. Using the Kelvin scale (where °C = K + 273), everything makes a whole lot more sense. For example, doubling the Kelvin temperature of a gas doubles the volume of that gas. When you work with any gas law involving temperatures, converting Celsius temperatures to Kelvin is crucial.

The Combined and Ideal Gas Laws

Boyle’s and Charles’s laws are convenient if you happen to find yourself in situations where only two factors change at a time. However, the universe is rarely so well behaved. What if pressure, temperature, and volume all change at the same time? Is aspirin and a nap the only solution? No. Enter the Combined Gas Law:

P1 x V = P2 x V2 T T2

1 2

Of course, the real universe can fight back by changing another variable. In the real universe, for example, tires spring leaks. In such a situation, gas particles escape the confines of the tire. This escape decreases the number of particles, N, Within the tire. Cranky, tire-iron wielding motorists on the side of the road will attest that decreasing N Decreases volume. This relationship is sometimes expressed as Avogadro’s Law:

Volume = Constant x Number of particles

Combining Avogadro’s Law with the Combined Gas Law produces the wonderfully comprehensive relationship:

P1xV1 P2xV2

N1 x T N2 x T2

The final word on ideal gas behavior summarizes all four variables (pressure, temperature, volume, and number of particles) in one easy-to-use equation called the Ideal Gas Law:

PV = nRT

Here, R Is the gas constant, the one quantity of the equation that can’t change. Of course, the exact identity of this constant depends on the units you are using for pressure, temperature, and volume. A very common form of the gas constant as used by chemists is R = 0.08206L atm K"1 mol"1 (which can also be expressed as 62.4 L torr mol"1 K"1). Alternately, you may encounter R = 8.314L kPa K"1 mol"1.

Dalton’s Law of Partial Pressures

Gases mix. They do so better than liquids and infinitely better than solids. So, what’s the relationship between the total pressure of a gaseous mixture and the pressure contributions of the individual gases? Here is a satisfyingly simple answer: Each individual gas within the mixture contributes a partial pressure, and adding the partial pressures yields the total pressure. Fortunately, because the gases are ideal, the mass of the gas particles does not influence partial pressure. The number of molecules is the only factor that affects partial pressure. This relationship is summarized by Dalton’s Law of Partial Pressures, For a mixture of a number of individual gases:

PTotal = P1 + P2 + P3 + . . . + PN

This relationship makes sense if you think about pressure in terms of kinetic molecular theory. Adding a gaseous sample into a volume that already contains other gases increases the number of particles in that volume. Because pressure depends on the number of particles colliding with the container walls, increasing the number of particles increases the pressure proportionally. Remember, this assumes that the individual gas molecules are behaving ideally.

You can also calculate the partial pressure of an individual gas A in a mixture of gases using the equation:

PA = Ptotal x XA

Here XA represents the mole fraction of the gas in question, or the number of moles of that gas divided by the total number of gas moles in the mixture.

Graham’s Law and diffusion

"Wake up and smell the coffee." This command is usually issued in a scornful tone, but most people who have awakened to the smell of coffee remember the event fondly. The morning gift of coffee aroma is made possible by a phenomenon called Diffusion. Diffusion is the movement of a substance from an area of higher concentration to an area of lower concentration. Diffusion occurs spontaneously, on its own. Diffusion leads to mixing, eventually producing a homogenous mixture in which the concentration of any gaseous component is equal throughout an entire volume. Of course, that state of complete diffusion is an equilibrium state; achieving equilibrium can take time.

4

Different gases diffuse at different rates, depending on their molar masses (see Chapter 7 for details on molar masses). The rates at which two gases diffuse can be compared using Graham’s Law. Graham’s Law also applies to Effusion, The process in which gas molecules flow through a small hole in a container. Whether gases diffuse or effuse, they do so at a rate inversely proportional to the square root of their molar mass. In other words, more massive gas molecules diffuse and effuse more slowly than less massive gas molecules. So, for gases A and B

RateA RateB

.y/molar massB molar mass A

Getting Real: Nonideal Gas Behavior

«$JAB^ Ideal gas equations are idealizations based on the oversimplified assumptions of ideal gas theory. Real gases, however, do not obey the ideal gas laws. At certain temperatures and pressures, they provide good approximations, but there are situations when a more sophisticated (and therefore more complicated) equation is needed.

Chemists use a rearranged version of the ideal gas law to determine whether or not they can use it.

PV n RT

For an ideal gas, N Should equal one, so a gas for which the ideal gas law is a good approximation should have an N Value close to one. At high pressures and at low temperatures, this becomes increasingly unlikely and a new equation is needed.

Assuming that a gas is ideal means assuming that its particles occupy no space and do not attract one another, but real gas particles defy both of those postulates: They do occupy space and they do attract one another. At high pressures, this fact becomes increasingly difficult to ignore because the space between gas particles becomes smaller and smaller and the assumption is only valid when the space between gas molecules is much greater than the size of an individual gas molecule. At high temperatures, the internal energy of a gas molecule is great enough to vastly outweigh the attraction to neighboring molecules, but at low temperatures, hence low kinetic energy, the effect of molecular attraction becomes increasingly significant. In these situations it is necessary to add in a correction for the volume of molecules and another for intermolecular attractions. These corrections yield a modified version of the ideal gas law called the van der Waals equation, which is given on the AP exam, so there is no need to memorize it!

This equation should make it clear why chemists prefer to use the ideal gas law to minimize necessary computations. Two new constants, called the van der Waals constants, appear in this equation. The constant A Is a measurement of how strongly the gas molecules attract one another and has units of L2atm mol2. The constant B Is a measurement of the volume occupied by a mole of gas molecules and has units of L mol. These constants vary as the identity of the gas in question varies. Larger and more complex molecules generally have larger A And B Values because of their greater size and greater propensity to experience inter-molecular forces, or interactions between molecules.

In This Chapter

^ Moving from molecular formulas to Lewis structures

^ Overlapping orbitals to make molecular shapes

^ Understanding how shape contributes to polarity and isomers

Olecular shapes help determine how molecules interact with each other and within themselves. For example, molecules that stack nicely on one another are more likely to form solids. And two molecules that can fit together so their reactive bits lie closer together in space are more likely to react with one another. In this chapter, we describe how to get a sense of the shape of a molecule, starting with only its molecular formula. Then, we explore a few ways that molecular shape can impact molecular properties.

Drawing Dots and Lines with Lewis

Valence electrons are critical for bonding. If that sentence confuses you, then you need to amble through Chapter 5 before moving further into this chapter. Because valence electrons are the ones important for bonding, chemists use Lewis electron dot structures: Symbols that represent valence electrons as dots surrounding an atom’s chemical symbol.

You should be able to draw and interpret electron dot structures for atoms so that you understand how the dots relate to the valence shell electrons possessed by an atom. By keeping track of how filled or unfilled an atom’s valence shell is, you can predict whether it is likely to bond with other atoms in an attempt to fill its valence shell.

Drawing electron dot structures

Figure 7-1 shows the electron dot structures for elements in the periodic table’s first two rows. Electrons in an atom’s valence shell are represented by dots surrounding the symbol for the element. Completely filled shells are surrounded by eight (8) dots (or two dots, in the lone case of helium). Valence shells progressively fill moving from left to right in the table. This pattern repeats itself with each successive row of the periodic table, as each row corresponds to adding a new outermost valence shell. Because atoms are most stable (think "happiest") when their valence shells are full, they tend to seek full-shell states in one of two ways. Atoms may gain or lose electrons (forming anions or cations, respectively) to end up with filled shells. Or, atoms may covalently bond with each other so that each atom can lay claim to the electrons within the bond.

To draw the electron dot structure of any element

1. Write the element’s name.

2. Count the number of electrons in that element’s valence shell.

3. Draw that number of dots around the chemical symbol for the element. As a general rule, space the dots evenly around the element’s symbol.

Figure 7-1:

Electron dot structures

For elements in the first

Two rows of

The periodic table.

IA

IIA

•Be*

IIIA IVA VA VIA VIIA VIIIA

He!

• • • •

•ES« ‘OIIM! 101 IF* INeL

H

Li

Constructing Lewis structures

Just as electron dot structures show the number of valence electrons that surround individual atoms, Lewis structures use dots and lines to show the distribution of electrons around all the atoms within a compound. Lewis structures are great tools for figuring out which molecules are reasonable (filling up all the atomic valence shells) and which molecules aren’t so reasonable (leaving some valence shells unhappily unfilled).

If you know the molecule’s formula you can figure out the correct Lewis structure for that molecule. The following example gives you the steps you need to work out a Lewis structure. This example uses formaldehyde, CH2O. You can follow along with Figure 7-2:

1. Add up all the valence electrons for all the atoms in the molecule.

These are the electrons you can use to build the structure. Account for any extra or missing electrons in the case of ions. For example, if you know your molecule has +2 charge, remember to subtract two from the total number of valence electrons. In the case of formaldehyde, C has four valence electrons, each H has one valence electron, and O has six valence electrons. The total number of valence electrons is 12.

2. Pick a "central" atom to serve as the anchor of your Lewis structure.

The central atom is usually one that can form the most bonds, which is often the atom with the most empty valence orbital slots to fill. In larger molecules, some trial-and-error may be involved in this step, but in smaller molecules, some choices are obviously better than others. For example, carbon is a better choice than hydrogen to be the central atom because carbon tends to form four bonds, whereas hydrogen tends to form only one bond. In the case of formaldehyde, carbon is the obvious first choice because it can form four bonds, while oxygen can form only two, and each hydrogen can form only one.

3. Connect the other, "outer" atoms to your central atom using single bonds only.

Each single bond counts for two electrons. In the case of formaldehyde, attach the single oxygen and each of the two hydrogen atoms to the central carbon atom.

4. Fill the valence shells of your outer atoms. Then put any remaining electrons on the central atom.

In our example, carbon and oxygen should each have eight electrons in their valence shells; each hydrogen atom should have two. However, by the time we fill the valence shells of our outer atoms (oxygen and the two hydrogens), we have used up our allotment of 12 electrons.

5. Check whether the central atom now has a full valence shell.

If the central atom has a full valence shell, then your Lewis structure is drawn properly — it’s formally correct even though it may not correspond to a real structure. If the central atom still has an incompletely filled valence shell, then use electron dots (nonbonding electrons) from outer atoms to create double and/or triple bonds to the central atom until the central atom’s valence shell is filled. Remember, each added bond requires two electrons. In the case of our formaldehyde molecule, we must create a double bond between carbon and one of the outer atoms. Oxygen is the only choice for a double-bond partner, because each hydrogen can accommodate only two electrons in its shell. So, we use two of the electrons assigned to oxygen to create a second bond with carbon.

1. C(4 e-) + H(1 e-) + H(1 e-) + 0(6 e-) = 12e -

2. Carbon is the central atom; it can form more bonds (4) than 0, H.

Figure 7-2:

Putting together a Lewis structure.

HCH

C

HH

Getting bonds straight with line structures

Atoms involved in covalent bonds share electrons such that each atom ends up with a completely filled valence shell, as described more fully in Chapter 5. The simplest and best studied covalent bond is the one formed between two hydrogen atoms (dihydrogen). Separately, each atom has only one electron with which to fill its 1s orbital. By forming a covalent bond, each atom lays claim to two electrons within the molecule of dihydrogen. Covalent bonds can be represented in different ways, as shown for dihydrogen in Figure 7-3.

Figure 7-3:

Three representations of the formation of a covalent bond in dihydrogen.

3.

0

4.

Lewis structures for compounds more complicated than dihydrogen can get pretty busy-looking. Most Lewis structures get rid of the dizzying effect of swarms of electron dots by representing shared electron pairs —that is, covalent bonds — as lines. Each two-dot electron pair that is shared between two atoms is rewritten as a nice, clean line connecting the atom symbols. Only nonbonding electrons (lone pairs) are left as dots. Whether a Lewis structure for a compound uses electron dot pairs or lines to represent bonds, those dots or lines refer to preceisely the same entities: covalent bonds.

Atoms can share more than a single pair of electrons. When atoms share two pairs of electrons, they are said to form a Double bond, And when they share three pairs of electrons they are said to form a Triple bond. Examples of double and triple bonds are shown with Lewis structures using both electron dots and lines in Figure 7-4.

Figure 7-4:

The formation of double bonds in carbon dioxide and triple bonds in dinitrogen.

Rummaging through resonance structures

Sometimes a given set of atoms can covalently bond with each other in multiple ways to form a compound. In these cases, you can draw several possible valid Lewis structures for the compound. This situation leads to something called Resonance. Each of the possible Lewis structures is called a Resonance structure. The actual structure of the compound is a Resonance hybrid, A sort of average of all the resonance structures. Resonance is particularly common in molecules where much bonding occurs in a side-by-side manner with P-orbital electrons from adjacent atoms. These kinds of bonds are called n bonds (see "Overlapping Orbitals to Form Valence Bonds " for more on that type of bond).

When resonance structures involve n bonds, electrons in the bonds can become "delocal-ized," forming an extended orbital that bonds more than two atoms. The benzene molecule, shown in Figure 7-5, is a classic example of resonance and electron delocalization.

A)

H i

C

HCH H

H i

HCH

C

CC HCH

H

Figure 7-5:

The resonance structures (a)resonance hybrid (b) and electron delocaliza-tion of benzene(c).

B)

C)

Adjacent P Orbitals

H i

I: !l

CC HCH

H

Delocalization among n bonds

I—I

I—I

I—I

I—I

Overlapping Orbitals to Form Valence Bonds

When we say that covalent bonds "share" valence electrons (if you’re asking yourself, "What in the world are covalent bonds?" see Chapter 5 for more info), what do we really mean? We mean that shared electron pairs now flit about within overlapping atomic orbitals (if your head is orbiting right now trying to figure out what atomic orbitals are, check out Chapter 3).

From a molecular formula you can figure out a Lewis structure. From the Lewis structure and by using valence bond theory and VSEPR theory you can often get a pretty good idea of the shape of a molecule, which is a major factor in determining other properties, like polarity, phase behavior, and the tendency to interact or react with other molecules. Going from a Lewis structure to a molecular shape requires a little know-how, however. In the next few sections, we give you the skills you need.

We realize the word "theory" sounds stuffy and perhaps intimidating, but, hey, we didn’t pick it. However, we take the stuffiness out of the valence bond theory as well as the VSEPR theory in the following sections so that you can predict the shape of molecules.

Getting a grasp on valence bond theory

According to valence bond theory, two atoms approach each other as they form a covalent bond, overlapping their electron shells, until they reach a minimum energy (see Figure 7-6). The positively charged nuclei draw closer together as each is attracted to the negative charge of the electrons between them. After a point, drawing any closer together would crowd their positively charged nuclei too closely for comfort. The positive charges begin to repel at such short distances. So, the length of the covalent bond is the result of a balancing act between attractive and repulsive forces.

Figure 7-6:

Two hydrogen atoms overlapping to form a covalent bond.

HH (too c lose)

(\\ …….. (too far)

Energy minimum at 74 x 10-12m

Distance between hydrogen nuclei

Different kinds of bonds result from the different ways the orbitals can overlap in space. The kind of symmetry the resulting bond has with the Bond axis, (an imaginary line that connects the centers of the bonded atoms) determines what kind of bond is formed:

Sigma bond: When orbitals overlap in a way that is Completely Symmetrical with the bond axis, a o bond (sigma bond) is formed. Sigma bonds form when S Or P Orbitals overlap in a head-on manner. Single bonds are usually sigma bonds.

I Pi bond: When orbitals overlap in a way that is symmetrical with the bond axis In only one plane, A n bond (pi bond) is formed. Pi bonds form when adjacent P Orbitals overlap above and below the bond axis.

Sigma bonds are stronger than pi bonds because the electrons within sigma bonds lie entirely between the two atomic nuclei, simultaneously attracted to both. A double bond is one sigma bond and one pi bond, and a triple bond is one sigma bond and two pi bonds.

+

0

Figure 7-7 shows you an example of a sigma bond forming from two S Orbitals and the formation of a pi bond from two adjacent P Orbitals.

Figure 7-7:

Formation of sigma and pi bonds.

Ss

+

Pp

S:

Shaping up with the VSEPR theory

Sigma and pi bonds form by overlapping the valence orbitals of atoms (see "Getting a grasp on valence bond theory" for more on sigma and pi bonds), so the overall shape of a molecule depends largely on the geometric arrangement of valence orbitals around each atom. That’s where VSEPR theory comes into play. Now don’t worry: We’ll start with the hard part: VSEPR stands for Valence shell electron pair repulsion. Okay, now it gets easier. VSEPR is simply a model that helps predict and explain why molecules have the shapes they do.

Only use VSEPR theory on the "p-block" elements. These elements are the ones in Groups IIIA, IVA, VA, VIA, VIIA, and VIIIA (except helium) on the right side of the periodic table. For the most part, this restriction means that you use VSEPR theory to predict geometry around nonmetal atoms.

In the following sections, we break down the basics of VSEPR theory so you can understand and predict the shape of molecules.

Predicting shapes with VSEPR theory

Simply by combining the principle that valence electron pairs (lone or bonds) want to get as far apart as they can (thus "repulsion" in the acronym VSEPR) with the fact that lone pairs repel more strongly than do bonding pairs, you can predict an impressive array of molecular shapes. The total number of valence electron pairs determines the overall geometry around a central atom. The distribution of these pairs between bonding and nonbonding orbitals adjusts that geometry. So VSEPR theory can predict several shapes that appear over and over in real-life molecules. These shapes, shown in Figure 7-8, resemble those observed in real-life molecules.

—

K

# of e pairs

E pair geometry

Bonding pairs

Lone pairs

Molecular

Shape

2 pairs

Linear

B A B Linear

4 pairs

Tetrahedral

40 A

B

B B Tetrahedral

Figure 7-8:

Molecular

Shapes predicted by VSEPR theory.

31 A

Trigonal pyramidal

B *

Bent

2

0

Making molecular shapes with hybridization

VSEPR theory is pretty good at making predictions about what shapes will emerge from a set of mostly equivalent valence shell electron pairs. But wait — what about the difference between valence electrons in S Orbitals versus those in P Orbitals? These electrons don’t seem equivalent. That’s when hybridization jumps in. Hybridization Refers to the mixing of atomic orbitals into new, hybrid orbitals. Valence electron pairs occupy equivalent hybrid orbitals.

Check out the electron configuration of carbon in Figure 7-9. Carbon contains a filled 1s orbital, but this is an inner-shell orbital, so it doesn’t impact the geometry of bonding. However, the valence shell of carbon contains one filled 2s orbital, two half-filled 2p orbitals, and one empty P Orbital. Not the picture of equality, so the valence orbitals must form new shapes or hybridize.

.fcjABEfl Different combinations of orbitals produce different hybrids. One S Orbital mixes with two P Orbitals to create three identical Sp2 Hybrids. One S Orbital mixes with one P Orbital to create HJUl ) two identical Sp Hybrids. Centers with sp3, sp2, and Sp Hybridization tend to possess tetra-hedral, trigonal planar, and linear shapes, respectively.

Figure 7-9:

The electron configuration of carbon.

Tl tl t t

1s

2s

2p

C

Doing the math to predict molecular shapes

The shapes of real molecules emerge from the geometry of valence orbitals — the orbitals that bond to other atoms. Here’s how to predict this geometry:

1. Count the number of lone pairs and bonding partners an atom actually has within a molecule. You can do this by looking at the Lewis structure.

In formaldehyde (H2C = O), for example, carbon bonds with two hydrogen atoms and double bonds with one oxygen atom. So, carbon effectively has three valence orbitals.

2. Next, inspect the electron configuration, looking for the mixture of orbital types (like s and P) That the valence electrons occupy.

Carbon has four valence electrons in 2s22p2 configuration. Two valence electrons occupy an S Orbital, and one electron occupies each of two identical P Orbitals. The S Orbital isn’t equivalent to the P Orbitals. So, we mix the S Orbital with the two P Orbitals to create three identical Sp2 Hybrid orbitals, as shown in Figure 7-11.

Note that the total number of orbitals doesn’t change; in the example, formaldehyde has three valence orbitals before mixing and still has three valence orbitals after mixing. VSEPR theory predicts that electrons in three identical orbitals mutually repel to create a trigonal planar geometry like the one shown in Figure 7-8, which is the shape of the formaldehyde molecule.

Elements in periods 3 and below of the periodic table can engage in Hypervalency, Meaning that they can possess more than eight electrons in their valence shell. Hypervalency complicates things a bit, but the basic principles of VSEPR theory still apply — electron pairs distribute themselves as far apart as possible around the hypervaent atom. Depending on the distribution of bonding and nonbonding electron pairs, hypervalent atoms participate in "expanded octet" geometries like octahedral, square planar and T-shaped, as shown in Figure 7-10.

Figure 7-10:

"Expanded octet" geometries.

Xe

90°

■ CI-F

87.5°

Octahedral

Square planar

F

T-shaped

Figure 7-11:

Formation ofthree ^

Equivalent

Sp2 Orbitals from one SOrbital and two

POrbitals.

O

/…

Pp

Hybridization

§-\ /o"\ /°i

Polarity and Isomers

You can often get a pretty good idea of the shape of a molecule (if you can’t, see the section earlier in this chapter, "Overlapping Orbitals to Form Valence Bonds"). Knowing the shape of a molecule helps you to understand the molecule’s other properties, like polarity, phase behavior, and the tendency to interact or react with other molecules. These are exactly the kinds of things that chemists and the rest of us actually Care About, whether we know it or not. For example, life on Earth depends largely upon the unique polarity and shape of the water molecule. The tasty combination of oil and vinegar in a raspberry vinaigrette depends on the phase behavior of the oil and vinegar components.

F

F

F

S

F

F

F

F

+

+

S

Polarity

The shape of a molecule and its polarity have a fairly simple relationship; the polarity of a molecule emerges from the polarities of the bonds within that molecule. As described in Chapter 5, differences in the electronegativity of atoms create polarity in bonds (which causes those bonds to be polar bonds) between those atoms. Each polar bond possesses a Bond dipole, Which is represented by this symbol: |i. Depending of the shape of the molecule, individual bond dipoles may or may not lead to an overall molecular dipole, a polarity in the molecule as a whole.

You can calculate a bond dipole by multiplying the amount of charge separated along that bond (Q) by the distance of the separation, d. Because the SI units of charge and distance are Coulombs (C) and meters (m), this calculation yields units of Coulomb-meters. A Coulomb-meter is a ridiculously big unit compared to the size of bond dipole, so bond dipoles are typically reported in Debyes (D), where 1 D = 3.336 x 10-30 Cm.

You won’t be asked to calculate a bond dipole on the AP exam, but knowing how bond dipoles are calculated helps you to understand how they add up within a molecule.

You won’t be asked to calculate a bond dipole on the AP exam, but knowing how bond dipoles are calculated helps you to understand how they add up within a molecule.

The precise way in which individual bond dipoles contribute to the overall Molecular dipole Depends on the shape of the molecule. Dipoles add as Vectors. Basically, this means

If two equivalent bond dipoles point in opposite directions, they cancel each other out. IU If the two bond dipoles point in the same direction, they add.

I If two bond dipoles point at a diagonal to one another, the horizontal and vertical components of the dipoles add (or cancel) separately.

The molecular dipoles of carbon dioxide and water demonstrate vector addition of bond dipoles nicely, as shown in Figure 7-12:

I Each molecule consists of a central atom bonded to two identical partners.

• In the case of carbon dioxide, carbon double-bonds to two oxygens to produce a linear molecule (thanks to Sp Hybridization!). The two C-O bond dipoles of carbon dioxide cancel each other out, so the molecular dipole is zero.

• In the case of water, a single oxygen atom single-bonds to two hydrogens that leaves the oxygen with two lone pairs.

I The bond pairs and the lone pairs (which are Sp3 Hybridized) lead to a nearly tetra-hedral orbital geometry and result in a bent shape of the molecule as you trace from hydrogen to oxygen to hydrogen.

I The two C-O bond dipoles of carbon dioxide cancel each other out, so the molecular dipole is zero.

I Only the horizontal components of the two H-O bond dipoles of water cancel each other out.

I The vertical components of those bond dipoles add, making the molecular dipole of water the sum of those components.

Because carbon dioxide has no molecular dipole, it interacts very weakly with itself and therefore is a gas at room temperature. Because water has a significant molecular dipole, it acts as a good solvent for other polar compounds.

+ +

A)

U(CO,)= A, + u, = 0

B)

Figure 7-12:

Vector addition of bond dipoles to produce molecular dipoles.

C)

Isomers

Molecular shapes can contribute to chemical diversity (an abundance of compounds with differing properties) through the creation of Isomers. Isomers are compounds made up of the same atoms but put together in different arrangements. Isomers with the same molecular formula may freeze or boil at different temperatures, may undergo different kinds of reactions, and may have important differences in situations where exact shape is critically important. Distinguishing between isomers is especially important in organic chemistry and biochemistry (see Chapter 25 for more detail on these) because these branches of chemistry involve many large molecules built with many individual bonds. With many bonds, there are many opportunities for isomerism. It turns out that big organic molecules can be very selective, reacting specifically with one isomer but not with others.

The following list gives you an overview of the different kinds of isomers, and how each is related to the others:

Structural isomers Have the same set of atoms, but the atoms are connected differently.

Stereoisomers Have the same bonds, but different three-dimensional arrangements of atoms. Among stereoisomers, you have

• Geometric isomers: Geometric isomers can occur when atoms are restricted from rotating freely about their bonds, such as in the case of a double bond or when bulky groups of atoms bump into each other. In these cases, atoms or groups of atoms can get trapped on one side or the other of the bond, creating geometrically distinct versions of the molecule.

• Optical isomers: Optical isomers (also called Enantiomers) Are mirror images of one another, but ones you can’t superimpose — in the same way you can’t match your left hand with your right hand when both palms face in the same direction. Molecules that have this handedness are called Chiral.

In This Chapter

^ Making sure the exam is right for you

^ Breaking down the exam into all its parts

^ Ticking through the topics covered on the exam

^ Acing the questions

^ Understanding your score

Th

‘ust the words "Advanced Placement" can bring about trembling fits of fear in even ■ the best of Chemistry students. The good ‘ol folks at the Educational Testing Service (ETS) — the company hired by the College Board to write and grade the AP exams — have been disseminating this fear for over 50 years, and they’ve had the time to get quite good at it. Consider this chapter your best bet at releasing your fear, getting some early college credit, and planning what to do with all the free time you’ll have in college by not having to take an intro chemistry course.

As you get intimate with the AP test, you will become familiar with why you should (or shouldn’t) take this test, the test’s structure, how you will be scored, and how to plan your time effectively. In this chapter we uncover the often-overlooked strategies for taking the exam itself.

You already know a whole bunch about chemistry. After all, you’ve probably spent the best part of this year listening eagerly to your high school chemistry teacher (in between texting your friend and downloading the latest Modest Mouse tune). Of course, nothing takes the place of full comprehension of the material, but knowing chemistry is only part of the puzzle (albeit a really big part). We can show you some other things that you can do to help raise your score that have little to do with chemistry. So sit back, relax, and let us show you the way.

Seeing If You Have the Right Stuff

Taking the AP exam sends out a beacon to the folks at any college that you believe you can think and perform on the college level, but can you? Although taking an AP chemistry course is not a prerequisite to taking the AP chemistry exam, we highly recommend that you do. Plus, you must be honest with yourself. If you received poor grades in chemistry in high school and could care less about stoichiometry, or if you accidentally blew up your high school chem lab by combining various chemicals, you might want to consider spending your time in an area you truly enjoy (perhaps a money-making vocation where you can pay for that lab to be rebuilt).

However, check out what your high score on the AP exam (see "Getting the Skinny on the Scoring," later in this chapter) can get you:

You prove to the college that you understand advanced material and already possess what it takes to be successful in college. Well, go you!

You have the opportunity to receive credit or advanced standings at most universities around the country. In plain English, this means you can sleep in a bit, because you won’t have to take an intro chemistry class in college because you already proved that you know this stuff! Even better, if you aren’t planning on studying science in college, you probably won’t ever have to take another science class again! Yippee!

Those smart AP creators have AP tests in a variety of different subjects, so if you’re not sure about your dedication to this particular test, you might want to visit their Web site and glance at the other subjects they offer. If you’re just taking the AP chemistry class and don’t plan on taking the exam at the end, then that’s cool too. Skip this chapter and use the rest of the book for succeeding in the AP chemistry class.

Knowing the Breakdown — So You Don’t Have One!

Considering that the AP folks don’t provide anti-anxiety salt licks as you walk into the exam, knowing the breakdown of the exam ahead of time prepares you for what you’re up against and makes you more confident so that you’re better relaxed to take the exam.

The AP Chemistry exam takes three hours and includes two sections; a 90-minute, 75-question, multiple-choice section and a 95-minute, 6-question, free-response section. We explain these sections further in the sections that follow.

Making your Way through multiple-choice questions: Section 1

You can’t use a calculator on the multiple-choice section. With the recent capabilities of graphing calculators, it could probably take the test for you! The AP higher-ups figured this one out a long time ago.

You have to answer (or try to answer) 75 questions on the multiple-choice section. Each question has five possible answer choices. We have a feeling that the AP creators couldn’t agree on what to put on the exam, so they just tossed everything into it. The questions you will encounter cover a large amount of material. Not all students will be exposed to all the material that the test may cover, so don’t be surprised to see topics that may be unfamiliar to you. You can expect basic factual questions as well as heavy-duty, thought-provoking problems, including 10 to 12 math problems to be done without a calculator. This section is worth 50 percent of your score.

You will only have about 1.2 minutes to answer each question. If you don’t know the immediate answer, skip the problem and move on.

Taking on free-response questions: Section 2

The second section of the exam consists of free-response questions. You will find a total of six questions:

Three quantitative problems (one on chemical equilibrium) lasting 55 minutes.

One of the questions will be on chemical reactions, which will require you to write balanced net equations for chemical reactions.

^ Reaction question and two essay questions lasting 40 minutes.

I One of the six questions will be a laboratory-based question that could be located in either the quantitative section or the essay section, so pay close attention when we cover labs throughout this book because the lab question could be on anything.

The free-response questions are long, make no bones about it. They will require you to demonstrate your problem-solving skills, show knowledge of chemical reactions, and question your ability to reason and explain ideas logically and clearly. They are broken down into multiple parts. Basically, the test will ask you multiple questions about the concept and require you to show concept mastery.

You won’t be given an incomprehensible topic for any of the free-response questions. You will be asked to solve a fairly basic chemistry concept, but they will want you to go into detail and ask you several different questions about it. There will be formulas to use and numbers to work with. Although the free-response questions are long and daunting, they’re not there to trick you. The multiple parts often help to lead you to the right way to tackle the problem. The AP people want to see your true chemistry acumen, and this is the best way to assess that. We will discuss how to tackle free-response problems later in this chapter.

You can use certain tools during this portion of the test. The following list shows you what you can bring in and what the AP folks provide, as well as the restrictions on these tools:

^ Calculators: You can use your calculator for the first 55 minutes of the free-response section, but you can’t share your calculator with another student. The AP folks also restrict which calculators you can use on the AP test. Because technology changes so rapidly, you should check the College Board Web site (Www. collegeboard. com) For an up-to-the-minute list of acceptable calculators.

^ Tables containing commonly used equations: You can use these tables only during the free-response section. You can’t use them on the multiple-choice section. The free-response section requires you to solve in-depth problems and to write essays where the knowledge of the concepts and how to apply the principles are the most important parts of solving the problem. The College Board gives everybody the table of equations to make it fair for the students who may not have the equations stored in their calculators. The AP folks have some heart, after all! But remember, because the test gives you the equations, you will receive no credit for answers simply written down in equation form without supported explanations or some type of logical development.

Tackling the Topics Covered

Before the AP exam was written, a few geeky chemists studied the chemistry curricula of many of the nation’s best colleges. They combined their reports and came away with a clear idea of the stuff being taught to chemistry college students around the country. The end result culminated in the AP chemistry exam covering five key areas, and because we’re

Obsessive about making sure you’re the most informed you can be (and because we’re control freaks), we’ve outlined each area for you below:

IU Structure of matter (20 percent of test; see Chapters 3 through 8):

• Atomic theory and atomic structure: Evidence for the atomic theory; atomic masses; atomic number and mass numbers; electron energy levels; periodic relationships

• Chemical bonding: Binding forces; molecular models, geometry of molecules and ions, structural isomerism of simple organic molecules and coordination complexes

• Nuclear chemistry: nuclear equations, half-lives and radioactivity I States of matter (20 percent of test; see Chapters 9 through 12):

• Gases: Laws of ideal gases; kinetic molecular theory

• Liquids and solids: Liquids and solids from the kinetic-molecular viewpoint; phase diagrams of one component systems; changes of state, including critical points and triple points; structure of solids

• Solutions: Types of solutions and factors affecting solubility; methods of expressing concentration; Raoult’s law and colligative properties; nonideal behavior

I Reactions (35 to 40 percent of test; see Chapters 13 through 22):

• Reaction types: Acid-base reactions; precipitation reactions; oxidation-reduction reactions

• Stoichiometry: Ionic and molecular species present in chemical systems; balancing of equations including those for redox reactions; mass and volume relations

• Equilibrium: Concept of dynamic equilibrium, physical and chemical; quantitative treatment

• Kinetics: Concept of rate of reaction; use of experimental data and graphical analysis to determine reactant order, rate constants, and reaction rate laws; effect of temperature change on rates; energy of activation; relationship between the rate-determining step and a mechanism

• Thermodynamics: State functions; first law; second law; relationship of change in free energy to equilibrium constants and electrode potentials

I Descriptive chemistry (10 to 15 percent of test; see Chapters 23 to 27): Relationships in the periodic table (periodicity); chemical reactivity and products of chemical reactions; introduction to organic chemistry

I Laboratory (5 to 10 percent of test; see Chapters 27 and 28): Making observations of chemical reactions and substances; recording data; interpreting results; communicating the results

Understanding the AP Test Questions

The AP chemistry test offers no surprises. Every question covers one or more basic fundamental chemistry concept, tests you on your knowledge of the five areas described in the section above, "Tackling the Topics Covered." Following the basic guidelines in this section helps you tackle the multiple-choice and free-response sections of the exam.

Making sense of multiple choice

Of the two parts of this test, the multiple-choice part is easiest because the correct answer is staring you right in your face; you just need to find it. In the following sections, we describe the different setups of multiple-choice questions, and we’ve also included a section that shows you some educated-guessing techniques to help you whittle down the obvious wrong answers.

Clearing out the "crowded" questions

Basically, there are three types of multiple-choice questions. The first type of questions you encounter we call "crowded together" questions. These are the easiest because they take very limited time to complete. With crowded together questions, you get some information upfront lettered A through E. After this information, you get two to three questions pertaining to the initial A through E information. You just pick the correct letter choice to solve the questions.

Standing out from the crowd: Dealing with "loner" questions

Another type of question that makes up much of the multiple-choice questions we like to call "loner questions." Each loner question covers only one lonely topic at a time. You’re presented with a question providing you with the information you need to complete the problem, usually five possible choices, again lettered A through E.

Putting a spin on interpretative questions

The interpretation questions do not occur very often, but they’re important to know about. Basically, the interpretation questions take two somewhat related loner questions and stuff them together and then ask you two or more questions about the information. You are given a graph, diagram, or data table, and are then asked two questions about the presented visual.

BEll

Getting through the questions… With as many right answers as possible

The AP test creators throw everything at you at once, so don’t think that the test starts easy and gets more difficult — it doesn’t. Manage your time on the text using these tips:

I Short questions: Answer the questions that are the shortest first, leaving more time for you to put a little more time into the longer, more intense questions.

I Easy questions: Answer the questions that make you want to smile first (the concepts you know the most about), and leave the questions that make you want to vomit for last.

I Educated guessing: Educated guessing is a good thing. Eliminating even one possible answer increases your choices of hitting the right one. You don’t get penalized for leaving an answer blank, but you do get a fraction (>4) removed from your score for a wrong answer. On the other hand, guessing correctly earns you a full point. In other words, completely random guessing neither hurts nor harms you. Educated guessing helps you. If you don’t know how to solve the problem but want to make an educated guess, keep these ideas in mind to make the best possible educated guess:

• Take a quick glance through the possible answers.

• The AP folks are not trying to trick you, so if something looks blatantly wrong, it probably is. So you’re probably safe to eliminate it.

• Eliminate the answers that don’t seem to match up to the question.

• Eliminate answers that appear too close to the actual question.

Finagling the free-response questions

The free-response section takes 95 minutes and has six questions divided into two parts:

IU Part A: Part A takes 55 minutes, in which you answer three questions for which you’re allowed to use your calculator. Part A covers one equilibrium problem and two other problems.

IU Part B: Part B takes 40 minutes and covers the last three questions, one being a reactions question and the other two being essay questions. A lab question is thrown in the mix somewhere. Calculators are Not Permitted in Part B.

Part A counts for 60 percent of your score in the free-response section with each question counting 20 percent. Spending a little extra time in Part A is worth it.

The test giVeth…

Before you begin the free-response section, you are given a plethora of information:

IU You will be bombarded with four, highly coveted pages of chemistry-related material, consisting of a whole bunch of equations and constants covering atomic structure, equilibrium, thermochemistry, liquids, gases, solutions, oxidation-reduction, and electrochemistry.

IU You will also receive a limited periodic table and a reduction potential table.

Right now, you might be thinking, "Cool, I’ll be getting all the info I need to solve the problems." But, when you get to the test site and look at the questions, you’ll soon realize that the good AP pros didn’t just give you Only The info you need, but they also gave you a ton more than you’ll ever use on the test. So, even though you have information to take from, you still need to be chem smart to know when and where to use the info given to you.

Breathing easy by knowing what to expect

On the free-response portion of the exam, you need to answer the questions in your own logical words and commit to your answers. The topics on the AP exam refer to general chemistry topics, so rest assured that the AP pros won’t toss some rarely taught concept your way to see if they can throw you a curve ball and screw up your game.

Most questions have multiple parts. You’ll first encounter some type of initial chemistry information (it could be a figure, a graph or a concept) and then see questions labeled a, b, c, and so on. Each of these subquestions requires you to write an answer of at least a sentence and sometimes a paragraph or two or to give a multistep equation as your answer.

You have approximately 18 minutes (three problems in 55 minutes) and 13 minutes (three problems in 40 minutes) to complete each free-response problem.

Putting partial answers into practice

Don’t be stingy with your answers! Partial credit is given for saying at least Something Right in the answer. You receive points for writing certain correct portions of the answer, so take your time, write all you know, and do not rush. However, be sure what you write is relevant to the question. No points are given for writing about chemistry in general when the subject is sodium chloride.

If you come across a problem that freaks you out, remind yourself that there must be some part of the subquestion that you’re more familiar with. Read the entire problem before starting it. Do not skip a problem simply because the vomit feeling is welling up inside of you. Take a deep breath, scan for any part of the question that you might be able to write about, and begin writing. You can still get a fairly high score even when you receive only partial credit for certain subquestions.

More tips directly from the source

The College Board is nice enough to offer some tips of its own. In the following list, we offer you a condensed version of these tips to make your life easier:

I Show all your work: Partial credit is given for partial solutions.

I If you do work that is incorrect, simply put an X through it: Don’t take the time to erase.

I Organize your answers clearly: If the scorers cannot follow your reasoning, they might not give you credit.

IU You do not need to simplify, but if you’re asked to calculate, you must simplify all numerical expressions or carry out all numerical calculations to get credit.

IU Do not use what they call a "scattershot" approach: Avoid writing a whole bunch of equations or nonsense, hoping that that one among them will be correct.

Getting the Skinny on the Scoring

While you wait patiently to find out if you’re a genius, the secret scorers are busy at work squinting their eyes, rubbing their chins, and contemplating your entire chemistry future. You ultimately receive a score between 1 and 5. We know, after all the hard work you did, the hours of studying, and the sweating at the exam, the highest score you can receive is a measly 5! The nerve of those people!

While those AP pros might not be the most creative bunch of folk, they did do a lot of researching to figure out how to score you. They periodically compare the performance of AP students with that of college students tested on the same material. It comes down to this:

IU A score of 5 On the AP test is comparable to a college student’s earning an A in his college-level chemistry course.

IU A score of 4 Equates to B in a college course, and so on.

IU A score of 3 or higher Equates to a C in college and could still qualify you for college credit. Anything less than a 3 won’t qualify you.

But the College Board likes the word "qualified" so here’s how it puts it:

5 — Extremely well qualified

4 — Well qualified

3 — Qualified

2 — Possibly qualified

1 — No recommendation

You might be wondering how they come up with the 1 to 5 score. If you weren’t wondering, well, we were, and we figured while we’re at it that we might as well make the information available to you.

The multiple-choice section is scored by a computer. The answer sheets are scanned and the computer adds the number of correct responses and subtracts a fraction for each wrong answer. You don’t get penalized for answers left blank, but you subtract a fraction from your score for each wrong answer. Thus, make only educated guesses (see the section, " Getting through the questions. . . with as many right answers as possible" for more on making educated guesses.)

The free-response section is scored by real, live people during the first half of June. Really, they’re living, breathing people. Basically, major special college professors and veteran super-duper AP teachers come together in the summer and have an AP scoring party. They all gather, distribute their pocket protectors, and get down to the fun of reading your responses.

The scores from the multiple-choice section and free-response section are combined to give you a composite score.

In This Chapter

^ Defining Arrhenius, Br0nsted-Lowry and Lewis acids and bases ^ Titrating

^ Using buffered solutions

^ Defining the solubility product constant

^ Working with acid and base dissociation constants

/f you’ve read any comic books, watched any superhero flicks, or even tuned in to one of those fictional solve-the-crime-in-50-minutes shows on TV, you’ve likely come across a reference to acids being dangerous substances. Acids are generally thought of as something that evil villains intend to spray in the face of a hero or heroine, but somehow usually manage to spill on themselves. However, you encounter and even ingest a wide variety of fairly innocuous acids in everyday life. Citric acid, present in citrus fruits, such as lemons and oranges, is very ingestible, as is acetic acid, a. k.a. vinegar.

Strong acids can indeed burn the skin and must be handled with care in the laboratory. However, strong bases can burn skin as well. Chemists must have a more sophisticated understanding of the difference between an acid and a base and their relative strengths than simply their propensity to burn. This chapter focuses on how you can identify acids and bases, as well as several ways to determine their strengths, all of which are fair game on the AP chemistry exam.

Thinking Acidly: Three Different Ways to Define Acid-Base Behavior

As chemists came to understand acids and bases as more than just "stuff that burns," their understanding of how to define them evolved as well. It’s often said that acids taste sour, while bases taste bitter, but we do Not Recommend that you go around tasting chemicals in the laboratory to identify them as acids or bases. In the following sections, we explain three much safer methods you can use to tell the difference between the two.

Method 1: Arrhenius sticks to the basics

Svante Arrhenius was a Swedish chemist who is credited not only with the acid-base determination method that is named after him, but with an even more fundamental chemical concept: that of Dissociation. Arrhenius proposed an explanation in his PhD thesis for a phenomenon that, at the time, had chemists all over the world scratching their heads. What had them perplexed was this: Although neither pure salt nor pure water are good conductors of electricity, solutions in which salts are dissolved in water tend to be excellent conductors of electricity.

Arrhenius proposed that aqueous solutions of salts conducted electricity because the bonds between atoms in the salts had been broken simply by mixing them into the water, forming ions. Although the ion had been defined several decades earlier by Michael Faraday, chemists generally believed at the time that ions could only form through Electrolysis, Or the breaking of chemical bonds using electric currents, so Arrhenius’s theory was met with some skepticism. Ironically, although his thesis committee wasn’t overly impressed and gave him a grade just barely sufficient to pass, Arrhenius eventually managed to win over the scientific community with his research. He was awarded the Nobel Prize in Chemistry in 1903 for the same ideas that nearly cost him his doctorate.

Arrhenius subsequently expanded his theories to form one of the most widely used and straightforward definitions of acids and bases. Arrhenius said that acids are substances that form hydrogen (H+) ions when they dissociate in water, while bases are substances that form hydroxide (OH-) ions when they dissociate in water.

Peruse Table 17-1 for a list of common acids and bases and note that all of the acids in the list contain a hydrogen and most of the bases contain a hydroxide. The Arrhenius definition of acids and bases is very straightforward and works for many common acids and bases, but it’s limited by its narrow definition of bases.

Method 2: Brpnsted-LoWry tackles bases without a hydroxide ion

You no doubt noticed that some of the bases in Table 17-1 do not contain a hydroxide ion, which means that the Arrhenius definition of acids and bases can’t apply. When chemists realized that several substances without a hydroxide ion still behaved like bases, they reluctantly

Acknowledged that another determination method was needed. Independently proposed by Johannes Br0nsted and Thomas Lowry in 1923 and therefore named after both of them, the Br0nsted-Lowry method for determining acids and bases accounts for those pesky non-hydroxide containing bases.

Under the Br0nsted-Lowry definition, an acid is a substance that donates a hydrogen-ion (H+) in an acid-base reaction, while a base is a substance that accepts that hydrogen ion from the acid. When ionized to form a hydrogen cation (an atom with a positive charge), hydrogen loses its one and only electron and is left with only a single proton. For this reason, Br0nsted-Lowry acids are often called Proton donors And Br0nsted-Lowry bases are called Proton acceptors.

The best way to spot Br0nsted-Lowry acids and bases is to keep careful track of hydrogen ions in a chemical equation. Consider, for example, the dissociation of the base sodium carbonate in water. Note that although sodium carbonate is a base, it doesn’t contain a hydroxide ion.

NaCO3 + H2O — H2CO3 + NaOH

This is a simple double replacement reaction (see Chapter 13 for an introduction to these types of reactions). A hydrogen ion from water switches places with the sodium of sodium carbonate to form the products: carbonic acid and sodium hydroxide. By the Br0nsted-Lowry definition, water is the acid because it gave up its hydrogen to NaCO3. This makes NaCO3 the base because it accepted the hydrogen ion from H2O.

On the AP exam, if you ever see carbonic acid formed as a product, you will be expected to know that it breaks down into CO2 and H2O, making this reaction NaCO3 + H2O — H2O + CO2 + NaOH.

Almost (but not quite) All The acids and bases you will encounter on the AP test will be Bronsted-Lowry type so master these first.

What about the substances on the righthand side of the equation? Br0nsted-Lowry theory calls the products of an acid-base reaction the Conjugate acid And Conjugate base. The conjugate acid is produced when the base accepts a proton (in this case, H2CO3), while the conjugate base is formed when the acid loses its hydrogen (in this case, NaOH). This reaction also brings up a very important point about the strength of each of these acids and bases. Although sodium carbonate is a very strong base, its conjugate acid, carbonic acid, is a very weak acid. Weak acids always form strong conjugate bases and vice versa. The same is true of strong acids.

Method 3: Lewis relies on electron pairs

In the same year that Br0nsted and Lowry proposed their definition of acids and bases, an American chemist named Gilbert Lewis proposed an alternate definition that not only encompassed the Br0nsted-Lowry theory, but also accounted for acid-base reactions in which a hydrogen ion is not exchanged. Lewis’s definition relies on tracking lone pairs of electrons. Under his theory, a base is any substance that donates a pair of electrons to form a coordinate covalent bond with another substance, while an acid is a substance that accepts that electron pair in such a reaction. Recall that a coordinate covalent bond is a covalent bond in which both of the bonding electrons are donated by one of the atoms forming the bond.

All Br0nsted-Lowry acids are Lewis acids, but in practice, the term Lewis acid Is generally reserved for Lewis acids that don’t also fit the Br0nsted-Lowry definition. The best way to spot a Lewis acid/base pair is to draw a Lewis dot structure of the reacting substances,

Noting the presence of lone pairs of electrons. (We introduce Lewis structures in Chapter 7.) For example, consider the reaction between ammonia (NH3) and boron trifluoride (BF3):

NH3 + BF3 — NH3BF3

At first glance, neither the reactants nor the product appear to be acids or bases, but the reactants are revealed as a Lewis acid-base pair when drawn as Lewis dot structures as in Figure 17-1. Ammonia donates its lone pair of electrons to the bond with boron trifluoride, making ammonia the Lewis base and boron trifluoride the Lewis acid.

Figure 17-1:

The Lewis H F H F

Dot struc- || ||

Tures of H — N:+B — F —*~ H —N— B—F

Ammonia | | | |

And boron H F H F trifluoride.

Sometimes you can identify the Lewis acid and base in a compound without drawing the Lewis dot structure. You can do this by identifying reactants that are electron rich (bases) or electron poor (acids). Metal cations, for example, are electron poor and tend to act as a Lewis acid in a reaction, accepting a pair of electrons.

In practice, it’s much simpler to use the Arrhenius or Bronsted-Lowry definitions of acid and base, but you’ll need to use the Lewis definition when hydrogen ions aren’t being exchanged. You can pick and choose among the definitions when you’re asked to identify acid and base in a reaction.

Describing Acid-Base Equilibria with Constants

Arrhenius’s concept of dissociation gives us another convenient way of measuring the strength of an acid or base. Although water tends to dissociate all acids and bases, the degree to which they dissociate depends on their strength. Strong acids such as HCl, HNO3, and H2SO4 dissociate completely in water, while weak acids dissociate only partially. Practically speaking, a weak acid is any acid that doesn’t dissociate completely in water. Note that while acid-base reactions do occur in other solvents, the AP exam will only deal with aqueous (water) solutions.

To measure the amount of dissociation occurring when a weak acid is in aqueous solution, chemists use a constant called the Acid dissociation constant (Ka). Ka Is a special variety of the equilibrium constant introduced in Chapter 15. As we explain in Chapter 15, the equilibrium constant of a chemical reaction is the concentration of products divided by concentration of reactants and indicates the balance between products and reactants in a reaction. The acid dissociation constant is simply the equilibrium constant of a reaction in which an acid is mixed with water and from which the water concentration has been removed. This is done because the concentration of water is a constant in dilute solutions, and a better indicator of acidity is the concentration of the dissociated products divided by the concentration of the acid reactant. The general form of the acid dissociation constant is therefore

K [H3O +]X[ A-]

K =-1-1-

A [HA]

Where [HA] is the concentration of the acid before it loses its hydrogen, and [A] is the concentration of its conjugate base. Notice that the concentration of the hydronium ion (H3O+) is used in place of the concentration of H+, which we use to describe acids earlier in this chapter. In truth, they are one and the same and this is just shorthand for this species, since "bare" protons never exist in water. Generally speaking, H+ ions in aqueous solution will be caught up by atoms of water in solution, making H3O+ ions.

A similar situation exists for bases. Strong bases such as KOH, NaOH, and Ca(OH)2 dissociate completely in water. Weak bases don’t dissociate completely in water, and their strength is measured by the Base dissociation constant, Or Kb.

Kb =

[OH -]x[B+] [ BOH ]

Here, BOH is the base, and B+ is its conjugate acid. This can also be written in terms of the acid and base:

Kb

[OH-]x[HA] [A ]

In problems where you are asked to calculate Ka Or Kb, You’ll generally be given the concentration, or molarity, of the original acid or base and the concentration of its conjugate Or Hydronium/hydroxide, but rarely both. Dissociation of a single molecule of acid involves the splitting of that acid into one molecule of its conjugate base and one hydrogen ion, while dissociation of a base always involves the splitting of that base into one molecule of the conjugate acid and one hydroxide ion. For this reason, the concentration of the conjugate and the concentration of the hydronium or hydroxide ion in any dissociation are equal, so you only need one to know the other. You may also be given the pH and be asked to figure out the [H+] (equivalent to [H3O+]) or [OH-]. Ka And Kb Are also constants at constant temperature. Remember the above only works if you can assume that Only The acid (or base) + water is ini-taially present. What solutions can be acidic or basic without an acid or base present? Salt solutions!

In chemistry, a "salt" is not necessarily the substance you sprinkle on French fries, but has a much broader definition. Rather, a Salt Is any substance that is a combination of an anion (an atom with a negative charge) and a cation (an atom with a positive charge) and is created in a neutralization reaction. Salts, therefore, have a tendency to dissociate in water. The degree of dissociation possible — in other words, the solubility of the salt — varies greatly from one salt to another.

Chemists use a quantity called the Solubility product constant Or Ksp To compare the solubilities of salts. Ksp Is calculated in much the same way as an equilibrium constant (Ke,). The product concentrations are multiplied together, each raised to the power of its coefficient in the balanced dissociation equation. There is one key difference, however, between a KSp And a Ke,. Ksp Is a quantity specific to a Saturated Solution of salt, so the concentration of the undissociated salt reactant has absolutely no bearing on its value. If the solution is saturated, then the amount of possible dissociation is at its maximum, and any additional solute added merely settles on the bottom.

_

Measuring Acidity and Basicity with pH

A substance’s identity as an acid or a base is only one of many things that a chemist may need to know about it. Sulfuric acid and water, for example, can both act as acids, but using sulfuric acid to wash your face in the morning would be a grave error indeed. Sulfuric acid

And water differ greatly in Acidity, A measurement of an acid’s strength. A similar quantity, called Basicity, Measures a base’s strength.

Acidity and basicity are measured in terms of quantities called PH And pOH, respectively. Both are simple scales ranging from 0 to 14, with low numbers on the pH scale representing a higher acidity and therefore a stronger acid. On both scales, a measurement of 7 indicates a Neutral solution. On the pH scale, any number lower than 7 indicates that the solution is acidic, with acidity increasing as pH decreases, while any number higher than 7 indicates a basic solution, with basicity increasing as pH increases. In other words, the farther the pH gets away from 7, the more acidic or basic a substance gets. pOH shows exactly the same relationship between distance from 7 and acidity or basicity, only this time low numbers indicate very basic solutions, while high numbers indicate very acidic solutions. Both the pH and the pOH scales are logarithmic, meaning that every time you increase them by 1, the concentration of H+ or OH – in the solution increases by 10. An acid with a pH of 1 has 100 times more H+ ions than and acid with a pH of 3.

PH is calculated using the formula pH = – log[H+], where the brackets around H+ indicate that it’s a measurement of the concentration of hydrogen ions in moles per liter (or molarity; see Chapter 11). pOH is calculated using a similar formula, with OH – concentration replacing the H+ concentration: pOH = – log[OH-]. (The word Log In each formula stands for base 10 Logarithm.)

Because a substance with high acidity must have low basicity, a low pH (high acidity) indicates a high pOH for a substance and vice versa. In fact, a very convenient relationship between pH and pOH allows you to solve for one when you have the other: pH + pOH = 14.

You’ll often be given a pH or pOH and be asked to solve for the H+ or OH- concentrations instead of the other way around. The logarithms in the pH and pOH equations make it tricky to solve for [H+] or [OH-], but if you remember that a log is inversed by raising 10 to the power of each side of an equation, you will quickly arrive at a convenient formula for [H+], namely [H+] = 10-pH. Similarly, [OH-] can be calculated using the formula [OH-] = 10-pOH. As with pH and pOH, a convenient relationship exists between [H+] and [OH-], which multiply together to equal a constant. This constant, called the Ion product constant for water, Or KW, Is calculated as follows: KW = [H+] x [OH] = 1 x 10-14.

Taking It One Proton at a Time: Titration

So far we have treated acids and bases separately, but chemists are awfully fond of mixing things. In the real world of chemistry, acids and bases often meet in solution and when they do, they’re drawn to one another. These unions of acid and base are called Neutralization reactions Because the low pH of the acid and the high pH of the base neutralize one another. This does not, however, mean that the final solution will be neutral (pH=7) because not all acids and bases have equal abilities to neutralize nor do they have to be mixed in equal amounts or at equal concentrations to create a neutralization reaction.

Although strong acids and bases have their uses, the prolonged presence of a strong acid or base in an environment not equipped to handle it can be very damaging. In the laboratory, for example, you need to handle strong acids and bases carefully, and deliberately perform neutralization reactions when appropriate. A lazy chemistry student will get a brow-beating from her hawk-eyed teacher if she attempts to dump a concentrated acid or base down the laboratory sink. Doing so can damage the pipes and is generally unsafe. Instead, a responsible chemist will neutralize acidic laboratory waste with a base such as baking soda, and will neutralize basic waste with an acid. Doing so makes most solutions perfectly safe to dump down the drain and often results in the creation of a satisfyingly sizzly solution while the reaction is occurring.

Neutralizing equivalents

At heart, neutralization reactions in which the base contains a hydroxide ion are simple double replacement reactions of the form HA + BOH — BA + H2O (or in other words, an acid reacts with a base to form a salt and water). You’re asked to write a number of such reactions in this chapter, so be sure to review double replacement reactions and balancing equations in Chapter 13 before you delve into the new and exciting world of neutralization.

Unlike people, not all acids and bases are created equally. Some have an innate ability to neutralize more effectively than others. Consider hydrochloric acid (HCl) and sulfuric acid (H2SO4), for example. If you mixed 1M sodium hydroxide (NaOH) together with 1M hydrochloric acid, you’d need to add equal amounts of each to create a neutral solution. If you mixed sodium hydroxide with sulfuric acid, however, you’d need to add twice as much sodium hydroxide as sulfuric acid. Why this blatant inequality of acids? The answer lies in the balanced neutralization reactions for both acid/base pairs.

HCl + NaOH — NaCl + H2O H2SO4 + 2NaOH — Na2SO4 + 2H2O

The coefficients in the balanced equations are the key to understanding this inequality. To balance the equation, the coefficient 2 needs to be added to sodium hydroxide, indicating that two moles of it must be present to neutralize one mole of sulfuric acid. On a molecular level, this happens because sulfuric acid has two acidic hydrogen atoms to give up, and the single hydroxide in a molecule of sodium hydroxide can only neutralize one of those two acidic hydrogens to form water. Therefore, two moles of sodium hydroxide are needed for every one mole of sulfuric acid. Hydrochloric acid, on the other hand, has only one acidic hydrogen to contribute, so it can be neutralized by an equal amount of sodium hydroxide, which has only one hydroxide to contribute to neutralization.

The number of moles of an acid or base multiplied by the number of hydrogens or hydroxides that a molecule has to contribute in a neutralization reaction is called the number of Equivalents Of that substance. Basically, the number of Effective Neutralizing moles available determines the ratio of acid to base in a neutralization reaction.

Titrating with equivalents

Imagine you’re a newly hired laboratory assistant who’s been asked to alphabetize the chemicals on the shelves of a chemistry laboratory during a lull in experimenting. As you reach for the bottle of sulfuric acid, your first-day jitters get the better of you, and you knock over the bottle. Some careless chemist failed to screw the cap on tightly! You quickly neutralize the acid with a splash of baking soda and wipe up the now-nicely neutral solution. As you pick up the bottle, however, you notice that the spilled acid burned away most of the label! You know it’s sulfuric acid, but there are several different concentrations of sulfuric acid on the shelves, and you don’t know the molarity of the solution in this bottle. Knowing that your boss will surely blame you if she sees the damaged bottle, and not wanting to get sacked on your very first day, you quickly come up with a way to determine the molarity of the solution and save your job.

You know that the bottle contains sulfuric acid of a mystery concentration, and you notice bottles of 1M sodium hydroxide, a strong base, and phenolphthalein, a pH indicator, among the chemicals on the shelves. You measure a small amount of the mystery acid into a beaker and add a little phenolphthalein. You reason that if you drop small amounts of sodium hydroxide into the solution until the phenolphthalein indicates that the solution is neutral by turning the appropriate color, you’ll be able to figure out the acid’s concentration.

You can do this by making a simple calculation of the number of moles of sodium hydroxide you have added, and then reasoning that the mystery acid must have an equal number of equivalents to have been neutralized. This then leads to the number of moles of acid, and that, in turn, can be divided by the volume of acid you added to the beaker to get the molar-ity. Whew! You relabel the bottle and rejoice in the fact that you can come in and do menial labor in the lab again tomorrow.

This process is called a Titration, And it’s often used by chemists to determine the molarity of acids and bases. In a titration calculation, you generally know the identity of an acid or base of unknown concentration, and the identity and molarity of the acid or base which you’re going to use to neutralize it. Given this information, you then follow six simple steps:

1. Measure out a small volume of the mystery acid or base.

2. Add a pH indicator such as phenolphthalein.

Be sure to take note of the color that will indicate that the solution has reached neutrality. Again, this does not necessarily mean pH=7. Rather, the solution will reach neutrality when the reaction between the acid and base is complete, which may leave the solution with a pH other than 7.

3. Neutralize.

Drop the acid or base of known concentration into the solution until the indicator shows that it’s neutral, keeping careful track of the volume added.

4. Calculate the number of moles added.

Multiply the number of liters of acid or base added by the molarity of that acid or base to get the number of moles added.

5. Account for equivalents.

Determine how many moles of the mystery substance being neutralized are present using equivalents. How many moles of your acid do you need to neutralize one mole of base or vice versa?

6. Solve for molarity.

Divide the number of moles of the mystery acid or base by the number of liters measured out in Step 1, giving you the molarity.

The titration process is often visualized using a graph showing concentration of base on one axis and concentration of acid on the other as in Figure 17-2. The interaction of the two traces out a Titration curve, Which has a characteristic S Shape.

Keeping Steady: Buffers

You may have noticed that the titration curve shown in Figure 17-2 has a flattened area in the middle where pH doesn’t change significantly, even when you add a conspicuous amount of base. This region is called a Buffer region.

Certain solutions, called Buffered solutions, Resist changes in pH like a stubborn child resists eating her Brussels sprouts: steadfastly at first, but choking them down reluctantly if enough pressure is applied (such as the threat of no dessert). Although buffered solutions maintain their pH very well when small amounts of acid or base are added to them or the solution is diluted, they can only withstand the addition of a certain amount of acid or base before becoming overwhelmed.

Buffers are most often made up of a weak acid and its conjugate base, though they can also be made of a weak base and its conjugate acid. A weak acid in aqueous solution will be partially dissociated, the amount of dissociation depending on its PKa Value (the negative logarithm of its acid dissociation constant). The dissociation will be of the form HA + H2O — H3O+ + A-, where A – is the conjugate base of the acid HA. The acidic proton is taken up by a water molecule, forming hydronium. If HA were a strong acid, 100 percent of the acid would become H3O+ and A-, but because it’s a weak acid, only a fraction of the HA dissociates, and the rest remains HA.

BE*

4^ II*

Solving the earlier expression for Ka For the [H3O+] concentration allows you to devise a relationship between the [H3O+] and the Ka Of a buffer.

[ H3O +] = K A X

Taking the negative logarithm of both sides of the equation and manipulating logarithm rules yields an equation called the Henderson-Hasselbalch equation That relates the pH and the pKa.

PH = pK a + log

[ A ]

[HA]

[A ]

This equation can be manipulated using logarithm rules to get ■ more useful in certain situations. [[HA]

10p

Ka, which may be

If you prefer to (or are asked to) calculate the pOH instead, you can use the equation

POH =pKb +log

[ B ]

The very best buffers and those best able to withstand the addition of both acid and base are those for which [HA] and [A-] are approximately equal. Because the [A-]/[HA] = 1 and the log of 1 is 0, the logarithmic term in the Henderson-Hasselbalch equation disappears, and the equation becomes pH = pKa. When creating a buffered solution, chemists therefore choose an acid that has a pKa close to the desired pH.

If you add a strong base such as sodium hydroxide (NaOH) to this mixture of dissociated and undissociated acid, its hydroxide is absorbed by the acidic proton, replacing the exceptionally strong base OH – with a relatively weak base A-, and minimizing the change in pH.

HA + OH —H2O + A

This causes a slight excess of base in the reaction, but doesn’t affect pH significantly. You can think of the undissociated acid as a reservoir of protons that are available to neutralize any strong base that may be introduced to the solution. As we explain in Chapter 15, when a product is added to a reaction, the equilibrium in the reaction changes to favor the reactants or to "undo" the change in conditions. Because this reaction generates A-, the acid dissociation reaction happens less frequently as a result, further stabilizing the pH.

When a strong acid, such as hydrochloric acid (HCl), is added to the mixture, its acidic proton is taken up by the base A-, forming HA.

H+ + A- — HA

This causes a slight excess of acid in the reaction but doesn’t affect pH significantly. It also shifts the balance in the acid dissociation reaction in favor of the products, causing it to happen more frequently and re-creating the base A-.

The addition of acid and base and their effect on the ratio of products and reactants is summarized in Figure 17-3.

Buffer after Buffer with equal Buffer after

Addition of OH- concentrations of addition of H+

Weak acid and its

Conjugate base _

Figure 17-3:

The effect of adding acid and base on the ratio of products and reactants.

HCl

Cl-

I OH-

HCl

Cl-

HCl

Cl-

Cl- + H, O

HCl + OH-

H++ Cl-

HCl

>

Cl-

Buffers have their limits, however. The acid’s proton reservoir, for example, can only compensate for the addition of a certain amount of base before it runs out of protons that can neutralize free hydroxide. At this point, a buffer has done all it can do, and the titration curve resumes its steep, upward slope.

In This Chapter

^ Surveying the range of College Board-recommended AP chemistry experiments ^ Highlighting key concepts and calculations from the labs

Oggles are the geeky, timeless emblems of a chemistry lab. But before you laugh off this chapter, remember that the laboratory is where chemistry really takes place. That’s why the College Board has put forth an extensive list of recommended AP chemistry labs, many of them interrelated. Any actual AP chemistry course will not include the full range of these labs — but you’ll probably have done a significant fraction of them. To expose yourself to the pith of any labs your class skipped, and to shore up your understanding of those you did perform, review the following summary of the recommended labs, 22 things you might have done behind goggles. Examples offered for individual experiments may or may not exactly describe what you have done in your own lab classes but reflect the essential principles of the labs, and data like those you might encounter in lab are included in the examples. As you go through the steps of an experiment, keep alert to possible sources of error. Each measurement you make (massing a sample, measuring a temperature, etc.) carries some error with it. That error propagates through every subsequent calculation that you make using those measurements, and results in error (uncertainty) in any final experimental values.

Determination of the Formula of a Compound

By combusting a metal sample of known mass, you form a metal oxide with an unknown formula. Taking the mass of the oxide allows you to calculate the ratio of oxygen to metal within the oxide, and thereby determine the formula.

Example:

3.65g Mg is combusted within a 16.26g crucible. After combustion completes, the mass of the sample-containing crucible is 22.38g. What is the formula of the oxide?

Follow these steps to find the formula of this metal oxide:

1. Mass of crucible + combusted sample = 22.38g

2. Mass of combusted sample = 22.38g – 16.26g = 6.12g

3. Increase in mass of sample upon combustion = 6.12g – 3.65g = 2.47g

Because combustion is essentially a combination reaction with oxygen, any added mass derives from oxygen. The final sample therefore contains the original amount of Mg:

4. Moles Mg = 3.65g Mg x (1mol Mg / 24.3g Mg) = 0.150mol Mg

5. Moles O added = 2.47g O x (1mol O / 16.0g ) = 0.154mol O

6. Moles Mg : moles O ~ 1:1

The formula of the oxide is MgO.

Determination of the Percentage of Water in a Hydrate

By heating a hydrated ionic compound of known mass, you drive off hydrating water molecules. Often, this process can be followed visually by observing the colored hydrate compound lose its color. Taking the mass of the dehydrated compound allows you to calculate the percent of water in the original hydrated compound.

Example:

5.91g of hydrated copper (II) sulfate, CuSO4nH2O, is heated within a 15.73g crucible until its original blue color changes entirely to white. The sample should be cooled and massed, and then heated and cooled again until the mass remains constant indicating all the water has been removed. The mass of the sample-containing crucible after dehydration is 19.48g. What is the apparent percentage of water in the hydrated compound?

Here are the steps to determine the apparent percent water in the original compound:

1. Mass of sample + crucible = 5.91g + 15.73g = 21.64g

2. Mass of dehydrated sample + crucible = 19.48g

3. Mass of dehydrated water = 21.64 – 19.48g = 2.16g

4. Moles of dehydrated water = 2.16g H2O x (1mol H2O / 18.02 g H2O) = 0.120mol H2O

Because the mass that remains in the crucible after dehydration is CuSO4 (not the hydrate), you can calculate the original moles of hydrated compound:

5. Moles CuSO4 • nH2O = moles CuSO4

6. Mass remaining in crucible = 19.48g – 15.73g = 3.75g

7. Moles CuSO4 = 3.75 g CuSO4 x (1mol CuSO4 / 159.6g CuSO4) = 0.0235mol CuSO4

8. Moles dehydrated water : mole CuSO4 = 0.1120 : 0.0235 = 5.1: 1 ~ 5:1

So, given a plausible amount of experimental error, the original hydrate appeared to have the formula CuSO4 • 5H2O.

To calculate the mass percent of water in the original compound, you simply divide the mass of the hydrating waters (5.1 x 18.02g) by the mass of the whole hydrated compound and multiply by 100%:

(91.9g / 249.7g) x 100% = 36.8%

Determination of the Molar Mass by Vapor Density

By evaporating a volatile liquid such that it fills a flask of known volume, and then measuring the mass of recondensed liquid, you determine the vapor density (mass of compound / volume of flask) of the compound. Then, you use the ideal gas law to determine the molar mass of the compound.

Example:

4.00g of an unknown volatile liquid is loaded into a flask of 312mL interior volume. The flask is capped with a single-hole stopper to allow controlled escape of evaporating material. The stoppered flask is immersed in a 373K water bath, causing the volatile liquid to begin to evaporate. Once the last visible trace of liquid evaporates, the flask is plunged into an ice bath, recondensing any volatile liquid that remains as vapor within the flask. By comparing the mass of the flask before and after the heating/cooling procedure, you calculate that 0.854g of recondensed liquid remains. What is the apparent molar mass of the volatile liquid?

Follow these steps to find the apparent molar mass:

1. The experiment takes place in atmospheric conditions, so the pressure ~ 1.0atm.

2. At the moment the flask was immersed in an ice bath to recondense the volatile liquid, the moles of the sample filled the 312mL volume (0.312L). You know that the temperature of the original water bath was 373K and the pressure ~ 1.0atm. Using the ideal gas law, you can calculate the moles of gas in the flask under those conditions:

PV= NRT

N = (PV) / (RT) = (1.0atm x 0.312L) / (0.0821 Latm mol-1 K-1 x 373K) = 0.010mol

3. Because the recondensed liquid had mass 0.854g, we calculate the apparent molar mass as follows:

Molar mass ~ 0.854g / 0.010mol = 85 g mol-1

Determination of the Molar Mass by Freezing Point Depression

You dissolve a known mass of an unknown, nonelectrolyte solute into a known mass of solvent. Then you measure the freezing point of the resulting solution and compare it to the freezing point of the pure solvent. By calculating the change in the freezing point, you determine the molar mass of the unknown solute based on the colligative property of freezing point depression.

Constraining the unknown solute as an nonelectrolyte ensures that each mole of dissolved compound contributes one mole of dissolved particles, as opposed to the more complicated cases of acids, bases, and salts.

Example:

You dissolve 1.32g of an unknown, nonelectrolyte compound into 50.0g of benzene solvent. You measure the freezing point of the solution to be 2.6°C. What is the apparent molar mass of the solute?

Here are the steps to determine the apparent molar mass of this solute:

1. Freezing point depression depends on the molality (m) of the solution and on the solvent according to the freezing point depression equation, ATf = Kf x m.

The parameter Kf is the freezing point depression constant and varies from solvent to solvent. For benzene, Kf = 5.12 °C M-1.

The parameter ATF is the change in freezing point compared to that of the pure solvent, not the freezing point temperature itself. The normal freezing point of pure benzene is 5.5 °C.

Substituting in known values, you get (5.5 °C – 2.6 °C) = (5.12 °C m"1) x M

2. Solving for M Gives you 0.57 molality, or 0.57 moles solute per kg solvent. Because you dissolved the unknown solute into 50g = 0.050kg solvent, you can calculate the apparent moles of solute:

0.57mol kg-1 = N / 0.050 kg

3. Solving gives you 0.029 moles. Those moles had a mass of 1.32g, so the apparent molar mass is 1.32g / 0.029mol = 46 gmol-1.

Determination of the Molar Volume of a Gas

You react a known mass of a known metal with acid within a flask such that the reaction evolves hydrogen gas H2. The volume of gas is measured by its ability to displace water within a connected eudiometer, a graduated column partially filled with water. By measuring the volume of gas evolved, you can use the ideal gas law to calculate the apparent molar volume of hydrogen gas. In so doing, you must account for the contribution of water vapor within the eudiometer to the total pressure.

Example:

You react 0.027g Mg with concentrated HCl within a stoppered flask that is connected by a tube to a eudiometer. As the reaction proceeds, hydrogen gas bubbles from the mixture into the eudiometer. By observing the water displacement, you estimate that the reaction evolved 26.15mL of H2. The temperature of the lab is 22°C, a temperature at which the vapor pressure of water is 0.030atm. What is the apparent molar volume of the hydrogen gas? How do the moles of H2 estimated by this method compare to the number of moles expected from the reaction?

To compare the experimentally estimated number of moles to the expected number, follow these steps:

1. Set yourself on solid ground by writing a balanced equation for the reaction:

Mg(s) + 2HCl(aq) — Mg2+(aq) + 2Cl-(aq) + H2(g)

2. Given the stoichiometry of the balanced reaction equation, you can calculate a theoretical yield of H2 gas from the starting mass of Mg:

0.027g Mg x (1mol Mg / 24.3g Mg) x (1mol H2 / 1 mol Mg) = 1.1 x 10-3 mol H2

Potentially, 1.1 x 10-3 mol H2 gas could coexist with water vapor within the eudiometer tube.

3. You can estimate the moles of water in the displaced volume within the tube (26.15mL = 0.02615L) by using the ideal gas law and the given vapor pressure for water at 22 °C, 0.030atm:

PV = NRT

N = (PV) / (RT) = (0.030atm x 0.02615L) / (0.0821 Latm mol-1 K-1 x 295K) N = 3.2 x 10-5 mol H2O

4. Raoult’s law states that the partial pressure of a substance in the vapor pressure over a sample is proportional to the mole fraction of that substance within the sample. So, you can estimate the fractional pressure exerted by water vapor trapped in the eudiometer tube:

Fractional pressure ~ (3.2 x 10-5 mol H2O) / (3.2 x 10-5 mol H2O + 1.1 x 10-3 mol H2) Fractional pressure ~ 0.028

5. So, the pressure that caused the displacement within the eudiometer derived not only from hydrogen gas but also from about 3 mole percent water vapor. Using this information, you can recalculate the moles of hydrogen within the eudiometer, including a correction for the partial pressure contributed by water vapor:

N = [(1.0atm - 0.028atm) x (0.02615L) / (0.0821 Latm mol-1 K-1 x 295K)

N = 1.0 x 10-3 mol H2

6. The volume occupied by this amount of hydrogen ~ 0.02615L x (1 - 0.028) ~ 0.025L

7. So, the apparent molar volume of the hydrogen ~ 0.025L / 1.0 x 10-3 mol = 25 L mol-1

8. Finally, the percent error between the actual moles H2 produced versus those estimated from the balanced equation calculates as:

[(1.1 x 10-3mol - 1.0 x 10-3mol) / 1.1 x 10-3mol] x 100% = 9%

Standardization of a Solution Using a Primary Standard

To reliably prepare working solutions from stock solutions, it helps to be able to measure the concentration of stock solutions by reacting them with "primary standards" of very certain concentration. This process is called Standardization. To standardize solutions in this way, you must know the balanced reaction equation for the reaction undergone between the primary standard and sample solutions, and you must have some means to assess the extent of the reaction, such as a color change.

Analysis of the data from a standardization titration is essentially identical to the analysis used for acid-base titrations. Both experiments employ titration. Both require you to make use of a balanced reaction equation. Both often use a color change to indicate the endpoint of a titration. A common standard substance is potassium acid phthallate (or KHP for short), which can be prepared very pure, is solid at room temperature, and so easily weighed to obtain an exact mass for reaction. For an example of the types of calculations used in a standardization experiment, see "Determination of the Equilibrium Constant for a Chemical Reaction," later in this chapter.

Determination of Concentration by Acid-Base Titration, Including a Weak Acid or Base

To reliably determine the concentration of acid or base solutions, you can use the solution of unknown concentration in a neutralization reaction against an acid or base of highly certain concentration. To use this method, you must know the balanced reaction equation of the neutralization reaction and you must have a means to assess the extent of the reaction, such as the change in color of an indicator. If you know the number of acid or base equivalents contributed by your sample compound (HCl contributes one acid equivalent, H2SO4 contributes two acid equivalents, for example), you can determine the molar concentration of the compound. If you do not know the number of equivalents per mole of your compounds, you can still determine the normality, N, of your sample solution. (You may or may not have worked with "normality" in your AP class, but the AP exam does not include it.)

Example:

You wish to determine the concentration of a solution of ascorbic acid, C6H8O6, a weak acid. You know that ascorbic acid is monoprotic, contributing one mole of acid equivalent per mole of compound. To determine the concentration of your solution, you titrate 50.0mL of ascorbic acid solution with a 0.10M NaOH standard solution. To follow the progress of the neutralization reaction, you use a colorimetric indicator, phenolph-thalein. The phenolphthalein changes from colorless to pink when you have added 19.6mL of standard to the ascorbic acid sample solution. What is the apparent concentration of the sample solution?

To determine the apparent concentration of the original ascorbic acid solution, follow these steps:

1. To begin the analysis, you must know the stoichiometry of the neutralization reaction, as expressed in a balanced reaction equation:

C6H8O6(aq) + NaOH(aq) — C6H7O6Na(aq) + H2O(l)

Expressed as a net ionic equation,

C6H8O6(aq) + OH-(aq) — C6H7O6- (aq) + H2O(l)

In other words, the soichiometry of the interaction between NaOH and ascorbic acid is 1:1.

2. Phenolphthalein, the colorimetric indicator added to the titration solution, changed from colorless to pink at the equivalence point, indicating that the moles of acid in solution equal the moles of base. The color changed at 19.6mL (0.0196L) added 0.10M NaOH.

3. The moles of base added at the equivalence point is simply the concentration of the standard NaOH solution multiplied by the volume added:

Moles base = 0.10 mol L-1 x 0.0196L = 2.0 x 10-3 mol

4. Because moles base = moles acid at the equivalence point, the original 50.0mL (0.0500L) ascorbic acid solution apparently contained 2.0 x 10-3 mol ascorbic acid. So, the molar concentration of the ascorbic acid solution is

Molarity = (2.0 x 10-3 mol) / (0.0500L) = 4.0 x 10-2 M

Determination of a Concentration

By Oxidation-Reduction Titration

To determine the concentration of a solution of an oxidizing agent or reducing agent, you can use the sample solution in an oxidation-reduction reaction against a reducing agent or oxidizing agent of highly certain concentration. To use this method, you must know the balanced reaction equation of the redox reaction and you must have a means to assess the extent of the reaction, such as a color change.

For example, here is a balanced equation for a redox reaction between hydrogen peroxide and permanganate:

5H2O2(aq) + 6H+(aq) + 2MnO4-(aq) — 5O2(g) + 2Mn2+(aq) + 8H2O(l)

In this reaction, the permanganate ion, MnO4-, itself acts as the indicator because it has an intense purple color. So, as the reaction proceeds and permanganate is reduced, the purple color disappears. When the reaction is complete, the solution entirely loses purple color, becoming colorless.

Analysis of the data from a oxidation-reduction titration is essentially identical to the analysis used for standardization and acid-base titrations. The key elements are the use of a color change to indicate a reaction endpoint and the use of a balanced reaction equation to make correct stoichiometric calculations. The fact that the reaction in question is an oxidation-reduction reaction doesn’t really change the method of analysis. So, to see an example of the types of calculations used in a redox titration experiment, see determination of concentration by acid-base titration, earlier in this chapter.

Determination of Mass and Mole Relationships in a Chemical Reaction

This experiment is not typically performed on its own but rather as a necessary part of other experiments. The basic principle shows up in two main ways. First, you use a measured mass to calculate the apparent number of moles in a sample. Second, you use a measured number of moles to calculate the apparent mass of a sample. The measured quantities may be determined directly or indirectly, depending on the experiment. In either case, the apparent ratio of the mass of a substance to the corresponding number of moles represents the experimental molar mass (g mol-1) of the substance.

Determination of the Equilibrium Constant for a Chemical Reaction

The equilibrium constant, Keq, For a reaction is a ratio of the concentrations of products to reac-tants when the reaction has achieved equilibrium. So, you can determine the equilibrium constant by measuring the concentrations of reactants and products in an equilibrium solution. To use this method, you must know the balanced reaction equation for the reaction in question, and you must have a reliable means to measure the reactant and product concentrations.

Example:

You wish to determine the equilibrium constant for the ionization reaction of ascorbic acid, C6H8O6, a weak acid. You know that ascorbic acid is monoprotic, contributing one mole of acid equivalent per mole of compound. Because one product of the ionization reaction is H+, you choose to measure the concentration of reactants and products by measuring the pH of a standardized, 1.00 x 10-2 M, Solution of ascorbic acid. You observe the pH to be 3.08. What is the apparent Keq of ascorbic acid?

To find the apparent Keq of ascorbic acid by this method, do the following:

1. To calculate the Keq from a set of measured concentrations, you must have access to a balanced reaction equation. For the ionization reaction of ascorbic acid, the equation is

C6H8O6(aq) — C6H7O6-(aq) + H+(aq)

Or to be more rigorously correct,

C6H8O6(aq) + H2O(l) — C6H7O6-(aq) + H3O+(aq)

2. For simplicity, we’ll use the first equation to set up an expression for Keq:

Keq = products / reactants = [C6H7O6-]x[ H+] / [C6H8O6]

So, if you can measure the concentrations of C6H8O6, C6H7O6-, and H+ in a solution of ascorbic acid at equilibrium, you can use those values to solve for Keq.

3. Because the reaction in question is an acid ionization reaction, measuring the pH of an equilibrium solution is a clever way to obtain the needed data. Prior to ionizing, the ascorbic acid in solution exists at 1.00 x 10-2 M Concentration. Each mole of acid that ionizes depletes that initial concentration and adds one mole of ascorbate (the conjugate base) and one mole of H+ to solution. In other words, the equilibrium mixture contains the following concentrations:

[C6H8O6] = (1.00 x 10-2 – X) M

[C6H7O6-]= XM

[H+]= XM

All concentrations are expressed in terms of x, which means that if you can measure x, then you know all concentrations and can calculate the equilibrium constant.

4. The concentration of H+ can be calculated from the measured pH:

PH = -log[H+] therefore 10-pH = [H+] The pH of the solution is measured to be 3.08, so [H+] = 10-3.08 = 8.3 x 10-4M.

5. So, X = 8.3 x 10-4M. Knowing this, you can substitute into the expression for Keq and solve:

Keq = (8.3 x 10-4 x 8.3 x 10-4) / (1.00 x 10-2 – 8.3 x 10-4) = 7.5 x 10-5

Determination of Appropriate Indicators for Various Acid-Base Titrations and Determining pH

This experiment is often performed as part of a larger acid-base titration lab. The core concept is that common colorimetric pH indicators work best within limited pH ranges. Different indicators work best in different pH ranges. The reason for this is that indicator dyes are simply molecules that can exist in protonated or deprotonated form, with each form exhibiting a

Different color. The affinity of a given dye molecule for a proton, H+, is expressed by either of two related quantities: the Ka Or the pKa of the molecule.

The Ka is the acid dissociation constant, which is simply the equilibrium constant for the ionizing dissociation reaction of the molecule, HA H+ + A-. The pKa is simply the negative logarithm of the KA, pKA = – log KA.

The pKA is a convenient quantity to use for thinking about the effective pH range of an indicator dye. The pKA represents the pH at which one half the indicator molecules exist in proto-nated form and one half exist in deprotonated form. Beyond about one pH unit above or below its characteristic pKa, an indicator dye is outside of its useful range — no discernible changes in color can be detected this far from the pKA. So, it is best to use indicator dyes with pKa values as close as possible to the expected pH range of an experiment. If no reasonable pH range can be estimated prior to the experiment, it may be necessary to conduct the experiment multiple times, in the presence of indicators covering a range of pKA values.

Here are the most common indicator dyes alongside their respective pKa values:

Determination of the Rate of a Reaction and Its Order

You determine the rate at which reactant is converted into product and, by varying the initial concentrations of reactants across a series of trials, you gather data that allow you to determine the reaction order. The overall reaction order is the sum of the orders of the individual reactants. Reaction order is a measure of the sensitivity of the reaction rate to changes in concentration.

Example:

You measure the initial rate of the following (unbalanced) reaction — the acid-catalyzed iodination (addition of iodine) of acetone — under four different sets of reactant concentrations.

C3H6O + I2 — C3H6IO + H+ + I-

You observe the following results.

What is the reaction order of each species (C3H6O, I2 and H+) and what is the overall reaction order? What are the rate law and the apparent rate constant? To determine the rate law for each reactant, observe the following:

1. To determine rates like those listed in the previous table, you need a means of detecting the progress of the reaction, which typically means having some way to measure the appearance of product or the disappearance of reactant. Often, the simplest and most precise way to measure reaction progress is spectrophotometry, as described in the determination of electrochemical series, later in this chapter.

Because rates are often so dependent on reactant concentrations, it is usually advisable to measure "initial rates," which are fast, linear portions of the early reaction, before depletion of reactants appreciably slows the observed rate.

2. Once you have a set of initial rates at varying concentrations of reactants and/or products, you can seek patterns in the variation of rate with concentration. In the example data, it is clear that doubling the concentration of I2 had no effect on rate when other concentrations remained constant, suggesting that the reaction is zero order in I2.

3. Determining that I2 concentrations do not impact rate allows you to focus exclusively on C3H6O and H+ concentrations. Doubling the concentration of H+ doubles the observed rate suggesting that the reaction is first order in H+.

4. Maintaining the increased concentration of H+ while doubling the concentration of C3H6O causes a further doubling of the rate, suggesting that the reaction is first order in C3H6O. You can convert the individual reaction orders to an overall reaction order and a rate law with a known rate constant, K, As follows:

1. Because the reaction is first order in two reactants, the overall reaction order is 2.

2. The reaction order is reflected in the rate law for the reaction, Rate = K[C3H6O][H+].

3. To determine the value of the rate constant, K, Simply substitute in any set of data from any of the individual rate experiments:

Rate = K[C3H6O][H+]

K = Rate / ([C3H6O][H+]) = (4.8 x 10-7 MS-1) / (0.100Mx 0.100M) = 4.8 x 10-5 M~l S-1

Determination of the Enthalpy Change Associated with a Reaction

This lab involves Calorimetry, The measurement of heat transfer associated with a process (such as a chemical reaction). One common reaction employed in this lab is the dissolution reaction of a solute into water, although many other types of reaction can be used. At constant pressure, the absorption or release of heat that accompanies a chemical reaction is equivalent to the reaction enthalpy, AHRxn . Reactions that release heat are exothermic and have negative reaction enthalpy. Reactions that absorb heat are endothermic and have positive reaction enthalpy. By measuring the change in temperature of a reaction system of known mass and composition, you can calculate the molar reaction enthalpy, the amount of heat released or absorbed per mole of reaction.

Example:

You add 100.mL of water to a well-insulated (probably polystyrene foam) cup. You observe that the water temperature is 22.0°C. You then add 2.70g solid NaOH to the water, cap the cup with an insulated lid, and swirl gently to promote dissolution. While the NaOH dissolves, you monitor the temperature of the solution via a thermometer

Extending through a perforation in the lid. When the temperature levels off for approximately one half minute, indicating that the dissolution reaction has completed, you record the final temperature as 28.5°C. What is the apparent heat of solution, AHsoln, of sodium hydroxide?

Because the reaction performed in this lab is a dissolution reaction, the reaction enthalpy is also known as the "heat of solution." In other words, AHSoln = AHRxn. The dissolution reaction is

NaOH(s) + H2O(l) — Na+(aq) + OH-(aq) AHsoln = ?

The units of the heat of solution are joules per mole, J mol-1. So, if you can measure the heat flow that accompanies the dissolution reaction, you can divide that heat by the moles of NaOH dissolved to find AHSoln.

The key experimental trick is realizing that as NaOH dissolves, heat flows to or from the water in which it is dissolving. We can neglect any heat flow to or through the polystyrene cup as it is such a good insulator. So, qNaOH = – qwater. You can measure the change in the heat content of the water by measuring a change in the temperature of the solution (assuming that the mass of the solute is much smaller than the mass of the solvent — a fair assumption in dilute solutions).

To find the apparent heat of solution of sodium hydroxide from the data in this experiment, do the following:

1. To translate between temperature and heat flow, you need the constant-pressure calorimetry equation,

Q = m x Cp xAT = m x Cp x (Tfinal – Tintitial)

Here, m is the mass of the water, and CP is the specific heat capacity of the water.

2. Because the density of water is 1.00 g mL-1, you can calculate the mass of water in the experiment:

Mass of water = 100mL H2O x (1.00g / mL H2O) = 100.g

3. The specific heat capacity of water is 4.186 J g-1 °C-1. So, you can calculate the heat transfer during dissolution as

Q = (100.g) x (4.186 J g-1 °C-1) x (28.5 °C – 22.0 °C) = 2.72 x 103 J

4. So, as the NaOH dissolved, it released 2.72 x 103 J of heat energy into the water. In other words, QNaOH = -2.72 x 103 J. This amount of heat was released by 2.70g NaOH. You can convert to moles NaOH by using the molar mass:

Moles NaOH = 2.70g NaOH x (1mol NaOH / 40.0g NaOH) = 0.0675mol NaOH

5. So, the apparent heat of solution of NaOH is:

AHsoln = qNaOH / mol NaOH = -2.72 x 103 J / 0.0675mol = -4.03 x 104 J mol-1.

Separation and Qualitative Analysis of Anions and Cations

This lab is essentially about the descriptive chemistry of ionic compounds and their aqueous solutions. Qualitative analysis involves testing for the presence or absence of a compound, as opposed to measuring the precise concentration of that compound. Compounds can often be identified based on the formation of insoluble precipitates or colored compounds. Sometimes it can be difficult to test for the presence of one particular "target" ion within a

Solution of many ions, because the "nontarget" ions can interfere with intended reactions. For example, Cd2+ can be detected in aqueous solution by the formation of a yellow precipitate upon addition of S2-. However, if Pb2+ or Cu2+ happen to be in the sample solution, a black precipitate forms upon addition of S2-, which makes analysis for Cd2+ impossible. In cases like these, it is desirable to be able to selectively separate interfering ions from a solution while preserving other ions targeted for analysis. In other cases, the goal is simply to analyze a mixture of ions for the total ion composition of the solution. Separation schemes typically depend on the differential solubility of different ionic compounds.

Example:

You are given a solution that may contain Cl – anion, I – anion, both, or neither. To determine which of the four possibilities is true, you may use the following reagents: water, nitric acid (HNO3), silver nitrate (AgNO3), and ammonia (NH3). In addition, you have a standard array of glassware and filtration devices.

To separate the anions in the sample solution with the materials provided, do the following:

1. First, acidify the unknown chloride/iodide solution by dropwise addition of HNO3 until the solution is mildly acidic. Because Cl – and I – are weakly basic, addition of acid reduces their solubility and makes them easier to separate.

2. Next, slowly add AgNO3 until any precipitation is complete. Although silver nitrate is soluble, other silver salts, like AgCl and AgI, are insoluble. Filter and wash the precipitate with water. Examine the precipitate for color. AgCl is white, and AgI is yellow.

3. Next, resuspend the precipitate in a small amount of water and add NH3 and AgNO3. Mix well. The added silver nitrate promotes precipitation of silver salts due to the common ion effect. At this point, any iodide anion present in solution should precipitate as yellow silver iodide, AgI. Separate the supernatant from any AgI precipitate.

4. The ammonia that remains in the supernatant forms a cationic complex with silver, Ag(NH3)2+. After dropwise addition of HNO3 to reduce anion solubility, any Cl- in solution precipitates as white AgCl salt.

Synthesis of a Coordination Compound and Its Chemical Analysis

"Coordination compounds" or "coordination complexes" are structures in which a central metal atom is surrounded by a host of nonmetallic groups called Ligands. The ligands are bound to the metal by "coordinate covalent" bonds. These kinds of bonds are particularly common between transition metals and partners that possess lone pairs of electrons. The lone pair containing atom acts as an electron donor (a Lewis base), giving both electrons to a bond with the metal, which acts in turn as an electron acceptor (a Lewis acid). Coordination compounds are often intensely colored and can have properties that are quite different than those of the free metal.

This lab, in which you synthesize a coordination compound, is often combined with the two labs that follow it on this list: gravimetric determination and colorimetric or spectrophoto-metric determination. In keeping with that practice, we’ll stick with one "example" system through these three labs.

Example:

You synthesize a copper-centered coordination compound by the addition of ammonia to a blue, aqueous solution of copper (II) sulfate. When ammonia is added to the copper (II) sulfate, the blue solution turns a deep purple color, indicating that you have successfully

Formed a coordination complex. To separate the complexes from the surrounding solution, you add ethanol to selectively precipitate the complex. This addition results in the formation of crystals of your coordination complex. Using a vacuum filtration apparatus, you repeatedly wash your crystals, rinsing them with an ammonia-ethanol mixture. You allow the crystals to dry overnight. Presumably, you have formed a complex that contains ammonia and/or sulfate ligands surrounding a copper cation center. In the following two labs, you’ll analyze the compound gravimetrically and spectrophotometrically.

Analytical Gravimetric Determination

Gravimetric analysis entails measurements of mass to determine the composition of a compound. Here, we extend the copper coordination complex example from the synthesis of a coordination compound in the previous experiment.

Example:

Based on the chemicals used in the synthesis, the copper coordination complex likely includes ammonia and/or sulfate ligands, although the number of each type of ligand around the copper cation center is unknown. So, at this point, the formula of the complex is

Cua(NH3)b(SO4)c • nH2O

In this formula, a, b and c represent the stoichiometric amounts of copper, ammonia, and sulfate, respectively. The "• NH2O" indicates that some number N Of water molecules may hydrate the complex.

To analyze the composition of your complex, do the following steps:

1. Dissolve 1.00g of your purified crystals in nitric acid, HNO3, which causes the following reaction:

Cua(NH3)b(SO4)c • nH2O + xH+ — aCu2+ + Ј>NH4+ + cSO42- + nH2O

2. To this solution, add lead (II) acetate, Pb(C2H3O2)2, to selectively precipitate sulfate anion as lead sulfate:

Pb2+(aq) + SO42-(aq) — PbSO4(s)

3. Carefully filter the white PbSO4 precipitate and allow it to dry overnight before measuring its mass. From the mass of PbSO4 you can determine the number of moles of sulfate recovered. Within error, this number should be equivalent to the number of moles of sulfate in the original 1.00g of crystals.

4. To determine the amount of ammonia in your compound, you perform an acid-base titration (covered in more detail in previously in the chapter). Briefly, you dissolve 1.00g of your crystals in 50.0mL water, add a pH indicator, and titrate with acid. By doing this, you can indirectly determine the amount of ammonia in your compound. By observing the endpoint of the titration, you can calculate the moles of ammonia that must have been present in the original sample, because the color change of the titra-tion reflects the progress of this reaction:

BNH3(aq) + XH+(aq) — BNH4+(aq)

Within error, the number of moles of ammonia detected by your titration should be equivalent to the number of moles of ammonia in the original 1.00g of crystals.

The analysis for copper content of the coordination complex is done spectrophotomet-rically, as described in determination of electrochemical series, found below.

Colorimetric or Spectrophotometric Analysis

Spectrophotometric analysis measures the amount of a component within a mixture by exploiting the ability of the component to absorb electromagnetic radiation of a specific wavelength. Colorimetric analysis is simply spectrophotometric analysis using wavelength in the range of visible light. In practice, these analyses most often make use of Beer’s law, a mathematical relationship between the concentration of a substance and the absorption of light of a certain wavelength by that substance:

A = Abc

You may have encountered the equivalent equation A = e/c. This equation is absolutely identical to the one given above, but simply uses the variable e for A And the variable / For b.

To properly use Beer’s Law, you must clearly understand the meaning of each variable.

The variable A Is the absorbance of a specific wavelength of light. The wavelength used is typically one that is strongly absorbed by the compound of interest but not well absorbed by other components in the sample.

The variable A Is the molar absorptivity, an experimentally determined constant that depends on the compound of interest and on the wavelength of light used for analysis. Typically, the units of A Are M-1 Cm-1, but other units can be used. The key criterion is that the units used in A Match those used in the variables B And C.

The variable B Is the pathlength, the distance of the path of light as it travels through the sample. Most often, the pathlength is 1cm, but this value can change in the case of strongly or weakly absorbing samples. Whatever the pathlength, it must match the units used in the variable A.

The variable C Is the concentration of the compound of interest. Typically, the units of C Are molar (that is, M, mol L-1), but that can change in special cases. Whatever the concentration units, they must math those used in the variable A.

Example:

To continue the analysis of the copper coordination complex synthesized and gravimet-rically analyzed in the previous two experiments, you subject the compound to spec-trophotometric analysis to determine its copper content.

To subject the coordination compound to spectrophotometric analysis, you must first estimate the molar absorptivity of your copper compound:

1. Calibrate a spectrometer at 645nm wavelength by using a series of Cu2+ standard solutions, diluted as necessary in 1 MHNO3.

2. Dissolve 0.50g of coordination compound crystals in 20.0mL 1M HNO3 and transfer a sample of the solution to a 1cm-pathlength quartz cell.

If the molar absorptivity of your compound, estimated by the Cu2+ calibration, is 5.2 M-1 Cm-1, and the absorbance reading of your dissolved sample is 0.47, what is the apparent copper concentration in the dissolved sample? What is the apparent molar mass of the compound in your crystals?

To determine the apparent copper concentration of your sample, do the following:

1. The apparent concentration of Cu2+ in the sample can be directly determined by using Beer’s Law:

A = Abc Therefore C = A / (ab)

C = 0.47 / (5.2 M~l Cm-1 x 1 cm) = 9.0 x 10-2M Cu2+

2. Assume for the moment that each coordination complex centers on a single Cu2+ ion

(a reasonable working assumption), then the concentration of coordination complex in your sample is also 9.0 x 10-2M.

You can proceed from these results to estimate the molar mass of your copper-centered coordination complex as follows:

1. You prepared the spectrophotometer sample by dissolving 0.50g compound into 20.0mL (0.0200L) 1 MHNO3. Using this information, you can estimate the moles of coordination compound in the sample:

9.0 x 10-2mol L-1 x 0.0200L = 1.8 x 10-3mol

2. So, 1.8 x 10-3 moles of compound appear to have been present in the 0.50g sample you dissolved. Therefore, the apparent molar mass of the compound is

0.50g / 1.8 x 10-3mol ~ 280 g mol-1

Separation by Chromatography

Chromatography is a useful method for separating the components of a mixture. Although many variations of chromatography allow for quantitative analysis of the components of a mixture, the most frequent uses of chromatography in AP chemistry are qualitative, revealing how many components are in a mixture or determining whether a certain component is in the mixture or not.

In a chromatographic analysis, the sample to be analyzed is included in a "mobile phase," a fluid mixture that passes over a "stationary phase," a surface or porous medium that interacts with the mobile phase as it passes. Separation of components within the sample depends on those components interacting differently with the stationary phase. Some components of the sample may be more strongly attracted to the stationary phase than others, and these strongly attracted components take longer to be swept along by the flowing mobile phase. Separations can be based on charge, size, polarity, or other properties.

One common way to characterize the separated components of a mixture is the retention factor, Rf. To calculate an Rf, You must measure two quantities: the distance covered by the fastest-moving part of the mobile phase (often called the "solvent front") and the distance covered by the component in question. The retention factor is simply the ratio of the two quantities.

Rf = (distance of solvent front) / (distance of sample)

Preparation and Properties of Buffer Solutions

Buffer solutions are mixtures of conjugate acid-conjugate base pairs that resist changes in pH. Typically weak acids and their conjugate bases are used as buffers. Essentially, as extra acid or base is added to a buffer solution, the proportion of conjugate acid to conjugate base shifts in response, thereby minimizing the effect of the acid or base addition. The ability of a buffer solution to resist pH change depends on the concentration of the buffer compounds in solution and on the proportion of acid and base conjugate forms (that is, the ratio of HA to A-). A buffer is at maximum buffering capacity when [HA] = [A-]. Under this optimal-buffering condition, the pH of a buffer solution equals the PKa Of the weak conjugate acid. Buffers tend to be useful within about 1 pH unit of their characteristic pKa.

The Henderson-Hasselbach equation describes the relationship between pH, pKa and the relative concentrations of conjugates acid and base in a buffer solution:

PH = pKa + log ([A-]/[HA])

Example:

You are preparing an acetic acid/acetate buffer solution. At room temperature, acetic acid has pKa = 4.76. You have 500.mL of 0.25M acetic acid. You want the final pH of your solution to be 5.00. What mass of sodium acetate salt must you add to achieve the desired pH?

To determine the mass of sodium acetate you must add to achieve a solution with pH 5.00, do the following:

1. Substitute into the Henderson-Hasselbach equation your known values for pH and pKa, and then solve for the desired ratio of conjugate base to conjugate acid:

5.00 = 4.76 + log ([A-]/[HA])

0.24 = log ([A-]/[HA])

100.24 = [A-]/[HA] = 1.74

2. So, for every mole of acetic acid (the conjugate acid, HA) in solution you want to add 1.74 moles acetate (the conjugate base, A-).

Moles acetic acid = (0.25mol L-1) x (0.500L) = 0.13mol

Moles acetate = 1.74 x 0.13mol = 0.23mol

3. So, you want to add 0.23mol sodium acetate, C2H3O2Na. The molar mass of anhydrous (that is, not hydrated) sodium acetate is 82.0 g mol-1.

0.23 mol C2H3O2Na x 82.0 g mol-1 = 19 g C2H3O2Na

Determination of Electrochemical Series

In this lab, you use your knowledge of electrochemical cells to explain and predict observed voltage potentials in cells of differing composition. Key tasks are to identify the anode and cathode of a cell and to use a table of standard reduction potentials to predict the voltage potential of a cell given its composition.

By measuring the voltage potentials of a series of half-cells, all against the same counterpart half-cell, you can construct an electrochemical series, a ranked list of redox reactions in

Order of reduction potential. Tables of standard reduction potentials represent just such a series, one in which the constant counterpart half cell is the standard hydrogen electrode:

2H+(a<7) + 2e – — H2(g) E =0 (by definition)

Example:

A series of half-cell solutions is tested against a constant half-cell counterpart of 1.0M Copper (II) nitrate, Cu(NO3)2, with a solid copper electrode. In all trials, an ammeter indicates that electrons flow toward the copper electrode from the opposing electrode. The following voltage potentials are observed:

1.0M lead (II) nitrate, Pb(NO3)2 0.47 V

1.0M zinc (II) nitrate, Zn(NO3)2 1.10 V

1.0M iron (II) nitrate, Fe(NO3)2 0.78 V

Given that E = +0.34 V for the reduction of copper (II) cation, which divalent cation is the strongest reducing agent: lead (II), zinc (II) or iron (II)?

To determine the strongest reducing agent from the given data, observe the following:

1. The observed voltage potentials are standard cell potentials:

E°cell = E°red(cathode) – E°red(anode)

2. Because electron flow was toward the copper electrode, you know that the copper electrode was the cathode in each case.

3. Because each observed cell potential is positive and larger than +0.34 V, you know that each metal tested has a negative standard reduction potential.

The most negative of the standard reduction potentials must belong to zinc (II) cation. Based on the data given, zinc (II) cation, Zn2+, is the strongest reducing agent.

To calculate the standard reduction potential of the zinc (II) cation, do the following:

1. Substitute into the equation for standard cell potential:

E°cell = E°red(Cu2+) – E°red(Zn2+) 1.10 V = +0.34 V – E°red(Zn2+) E°red(Zn2+) = -0.76 V

2. Because zinc (II) has the most negative standard reduction potential of the three metal ions tested, it is the most difficult to reduce and is therefore the strongest reducing agent.

Measurements Using Electrochemical Cells and Electroplating

This lab requires you to combine your knowledge of electrochemical cells with basic stoi-chiometry and gravimetric analysis. Electroplating occurs at a cathode when electrons reduce metal cations in solution, causing them to deposit onto the cathode. By measuring the mass of a cathode before and after an electrochemical reaction in which electroplating occurs, you can directly determine the mass of metal deposited onto the cathode during the reaction and compare that mass with one expected from calculations.

Example:

Identical carbon electrodes, each with a mass of 5.50g are immersed into a 1.0MSolution of nickel (II) chloride, NiCl2. The poles of the electrodes are connected to a battery, and current flows through the cell for 30. minutes at 1.5 amperes. What is the expected mass of the cathode after 30 minutes?

To determine the expected mass of the cathode, make the following observations and calculations:

1. Deposition of Ni(s) happens at the cathode, where the following reaction takes place:

Ni2+ (aq) + 2e – — Ni(s) So, it takes 2 moles of electrons to electroplate 1 mole of nickel.

2. Calculate the total moles of electrons that flow through the cathode during the experiment. This calculation requires you to use the unit of the Faraday, F. One Faraday is charge associated with one mole of electrons. Each electron has 1.602 x 10-19 Coulombs of charge, so:

F = (6.022 x 1023mol-1)(1.602 x 10-19 C) = 9.649 x 104 C mol-1

3. The experiment lasted for 30 minutes (1.8 x 103 seconds) at 1.5 amperes. One ampere is one Coulomb per second (C s-1), so the total moles of electrons are

Moles e- = (1.8 x 103 s) x (1.5 C s-1) x (1 mol e – / 9.649 x 104 C) = 2.8 x 10-2 mol e-

4. Calculate the mass of nickel that you expect to electroplate as a result of transferring 2.8 x 10-2 moles of electrons into the NiCl2 solution:

Mass Ni(s) = (2.8 x 10-2 mol e-) x (1mol Ni / 2mol e-) x (58.7g / 1mol Ni) = 0.82g

5. Because the initial mass of the carbon electrode was 5.50g, the expected final mass (after electroplating) is 5.50g + 0.82g = 6.32g.

Synthesis, Purification, and Analysis of an Organic Compound

Typically, this lab involves a relatively straightforward synthesis — one with few steps. The synthesis is followed by a purification of the synthesized compound. In organic synthesis, you typically calculate a percent yield following each major step to track the progress of your reactions. Whatever the chemical details of the compound synthesized, the principles of "yield" are the same:

Percent yield = 100% x (actual yield) / (theoretical yield)

Actual yield is simply the amount of synthesized compound actually obtained after purification. Theoretical yield is the (never-actually-obtained) amount of compound that would result if each step of synthesis occurs with perfect efficiency and if there is no loss of compound between synthetic steps or during the purification.