5 Steps to a 5: AP Chemistry 2017 (2016)

STEP 4

Review the Knowledge You Need to Score High

CHAPTER 6

Reactions and Periodicity

IN THIS CHAPTER

Summary: Chemistry is the world of chemical reactions. Chemical reactions power our society, our environment, and our bodies. Some chemical species called reactants are converted into different substances called products . During this process, there are energy changes that take place. It takes energy to break old bonds. Energy is released when new bonds are formed. Does it take more energy to break the bonds than is released in the formation of the new bonds? If so, energy will have to be constantly supplied to convert the reactants into products. This type of reaction is said to be endothermic , absorbing energy. If more energy is released than is needed to break the old bonds, then the reaction is said to be exothermic , releasing energy. The chemical reactions that provide the energy for our world are exothermic reactions. In Chapter 9 , Thermodynamics, you can read in more depth about the energy changes that occur during reactions.

Reactions occur because of collisions. One chemical species collides with another at the right place and transfers enough energy, and a chemical reaction occurs. Such reactions can be very fast or very slow. In Chapter 14 on Kinetics, you can study how reactions occur and the factors that affect the speed of reactions. But in this chapter we will review the balancing of chemical equations, discuss the general types of chemical reactions, and describe why these reactions occur.

Keywords and Equations

There are no keywords or equations listed on the AP exam that are specific to this chapter.

Reactions questions will always appear in the free-response section of the AP exam.

AP Exam Format

Beginning with the 2007 AP exam, the treatment of chemical reactions was changed from previous years. In the past, the free-response questions concerning chemical reactions simply involved the formulas of the reactants and products chosen from a series of reactions. You were not expected to write balanced chemical equations. However, under the current AP Chemistry exam format you no longer are able to choose from a list of reactions. You are expected to write a balanced chemical equation for every reaction given and answer one or more questions about each reaction. If the reaction occurs in aqueous solution, you will have to write the net ionic equation for the process. For the reactions question of the AP exam, you will be expected not only to balance the equation, but also to have an understanding of why the reaction occurs. The reactions and concepts described may also appear in other parts of the AP exam, such as the multiple-choice sections. Again, you will need to have an understanding of why a particular reaction occurs. As you study this chapter, pay particular attention to the explanations that accompany the reactions and equations. You will be expected to demonstrate your understanding on the AP exam.

General Aspects of Chemical Reactions and Equations

Balancing Chemical Equations

The authors hope that, because you are preparing to take the AP exam, you have already been exposed to the balancing of chemical equations. We will quickly review this topic and point out some specific aspects of balancing equations as the different types of chemical reactions are discussed.

A balanced chemical equation provides many types of information. It shows which chemical species are the reactants and which species are the products. It may also indicate in which state of matter the reactants and products exist. Special conditions of temperature, catalysts, etc. may be placed over or under the reaction arrow. And, very important, the coefficients (the integers in front of the chemical species) indicate the number of each reactant that is used and the number of each product that is formed. These coefficients may stand for individual atoms/molecules, or they may represent large numbers of them called moles (see Chapter 7 , Stoichiometry, for a discussion of moles). The basic idea behind the balancing of equations is the Law of Conservation of Matter , which says that in ordinary chemical reactions matter is neither created nor destroyed. The number of each type of reactant atom has to equal the number of each type of product atom. This requires adjusting the reactant and product coefficients—balancing the equation. When finished, the coefficients should be in the lowest possible whole-number ratio.

Most equations are balanced by inspection. This means basically a trial-and-error, methodical approach to adjusting the coefficients. One procedure that works well is to balance the homonuclear (same nucleus) molecule last. Chemical species that fall into this category include the diatomic elements, which you should know: H2 , O2 , N2 , F2 , Cl2 , Br2 , and I2 . This is especially useful when balancing combustion reactions. If a problem states that oxygen gas was used, then knowing that oxygen exists as the diatomic element is absolutely necessary in balancing the equation correctly.

Periodic Relationships

The periodic table can give us many clues as to the type of reaction that is taking place. One general rule, covered in more detail in Chapter 11 , Bonding, is that nonmetals react with other nonmetals to form covalent compounds, and that metals react with nonmetals to form ionic compounds. If the reaction that is producing the ionic compound is occurring in solution, you will be expected to write the net ionic equation for the reaction. Also, because of the wonderful arrangement of the periodic table, the members of a family or group (a vertical grouping) all react essentially in the same fashion. Many times, in reactions involving the loss of electrons (oxidation), as we proceed from top to bottom in a family the reaction rate (speed) increases. Conversely, in reactions involving the gain of electrons (reduction) the reaction rate increases as we move from the bottom of a family to the top. Recall also that the noble gases (VIIIA) undergo very few reactions. Other specific periodic aspects will be discussed in the various reaction sections.

General Properties of Aqueous Solutions

Many of the reactions that you will study occur in aqueous solution. Water is called the universal solvent, because it dissolves so many substances. It readily dissolves ionic compounds as well as polar covalent compounds, because of its polar nature. Ionic compounds that dissolve in water (dissociate) form electrolyte solutions, which conduct electrical current owing to the presence of ions. The ions can attract the polar water molecules and form a bound layer of water molecules around themselves. This process is called solvation . Refer to Chapter 13 , Solutions and Colligative Properties, for an in-depth discussion of solvation.

Even though many ionic compounds dissolve in water, many others do not. If the attraction of the oppositely charged ions in the solid for each other is greater than the attraction of the polar water molecules for the ions, then the salt will not dissolve to an appreciable amount. If solutions containing ions such as these are mixed, precipitation will occur, because the strong attraction of the ions for each other overcomes the weaker attraction for the water molecules.

As mentioned before, certain covalent compounds, like alcohols, readily dissolve in water because they are polar. Since water is polar, and these covalent compounds are also polar, water will act as a solvent for them (general rule of solubility: “Like dissolves like”). Compounds like alcohols are nonelectrolytes —substances that do not conduct an electrical current when dissolved in water. However, certain covalent compounds, like acids, will ionize in water, that is, form ions:

HCl(aq) → H+ (aq) + Cl (aq)

There are several ways of representing reactions that occur in water. Suppose, for example, that we were writing the equation to describe the mixing of a lead(II) nitrate solution with a sodium sulfate solution and showing the resulting formation of solid lead(II) sulfate. One type of equation that can be written is the molecular equation , in which both the reactants and products are shown in the undissociated form:

Pb(NO3 )2 (aq) + Na2 SO4 (aq) → PbSO4 (s) 2NaNO3 (aq)

Molecular equations are quite useful when doing reaction stoichiometry problems (see Chapter 7 ).

Showing the soluble reactants and products in the form of ions yields the ionic equation (sometimes called the total ionic equation):

Pb2+ (aq) + 2NO3  (aq) + 2Na+ (aq) + SO4 2– (aq) → PbSO4 (s) 2Na+ (aq) 2NO3  (aq)

Writing the equation in the ionic form shows clearly which species are really reacting and which are not. In the example above, Na+ and NO3  appear on both sides of the equation. They do not react, but are simply there in order to maintain electrical neutrality of the solution. Ions like this, which are not actually involved in the chemical reaction taking place, are called spectator ions .

The net ionic equation is written by dropping out the spectator ions and showing only those chemical species that are involved in the chemical reaction:

Pb2+ (aq) + SO4 2– (aq) → PbSO4 (s)

This net ionic equation focuses only on the substances that are actually involved in the reaction. It indicates that an aqueous solution containing Pb2+ (any solution, not just Pb(NO3 )2 (aq)) will react with any solution containing the sulfate ion to form insoluble lead(II) sulfate. If this equation form is used, the spectator ions involved will not be known, but in most cases, this is not a particular problem, since the focus is really the general reaction, and not the specific one. You will be expected to write the balanced net ionic equation for many of the reactions on the test.

Precipitation Reactions

Precipitation reactions involve the formation of an insoluble compound, a precipitate , from the mixing of two soluble compounds. Precipitation reactions normally occur in aqueous solution. The example above that was used to illustrate molecular equations, ionic equations, etc., was a precipitation reaction. A solid, lead(II) sulfate, was formed from the mixing of the two aqueous solutions. In order to predict whether or not precipitation will occur if two solutions are mixed, you must:

  1. Learn to write the correct chemical formulas from the names; on the AP exam, names are frequently given instead of formulas in the reaction section.
  2. Be able to write the reactants and products in their ionic form, as in the ionic equation example above. Be sure, however, that you do not try to break apart molecular compounds such as most organic compounds, or insoluble species.
  3. Know and be able to apply the following solubility rules by combining the cation of one reactant with the anion of the other in the correct formula ratio, and determining the solubility of the proposed product. Then do the same thing for the other anion/cation combination.
  4. On the AP exam, you will be expected to explain why a substance is soluble/insoluble. Simply quoting the solubility rule is not sufficient.

Learn the following solubility rule:

All sodium, potassium, ammonium, and nitrate salts are soluble in water.

Although not required for the AP exam, the following solubility rules are very useful:

  • All salts containing acetate (CH3COO ), and perchlorates (ClO4  ) are soluble .
  • All chlorides (Cl), bromides (Br ), and iodides (I ) are soluble, except those of Cu+ , Ag+ , Pb2+ , and Hg2 2+ .
  • All salts containing sulfate (SO42− ) are soluble, except those of Pb2+ , Ca2+ , Sr2+ , and Ba2+ .

Salts containing the following ions are normally insoluble :

  • Most carbonates (CO32− ) and phosphates (PO4 3− ) are insoluble , except those of Group IA and the ammonium ion.
  • Most sulfides (S2−) are insoluble , except those of Group IA and IIA and the ammonium ion.
  • Most hydroxides (OH) are insoluble , except those of Group IA, calcium, and barium.
  • Most oxides (O2−) are insoluble , except for those of Group IA, and Group IIA, which react with water to form the corresponding soluble hydroxides.

Let’s see how one might apply these rules. Suppose a solution of lead(II) nitrate is mixed with a solution of sodium iodide. Predict what will happen.

Write the formulas:

Pb(NO3 )2 (aq) + NaI(aq) →

Convert to the ionic form:

Pb2+ + (aq) + 2NO3  (aq) + Na+ (aq) + I (aq) →

Predict the possible products by combining the cation of one reactant with the anion of the other and vice versa:

PbI2 + NaNO3

Apply the solubility rules to the two possible products:

Complete the chemical equation and balance it:

If both possible products are soluble, then the reaction would be listed as NR (No Reaction). In the reaction question part of the AP exam, there will be a possible reaction for every part of the question. If at least one insoluble product is formed, the reaction is sometimes classified as a double displacement (replacement) or metathesis reaction .

Oxidation–Reduction Reactions

Oxidation–reduction reactions, commonly called redox reactions , are an extremely important category of reaction. Redox reactions include combustion, corrosion, respiration, photosynthesis, and the reactions involved in electrochemical cells (batteries). The driving force involved in redox reactions is the exchange of electrons from a more active species to a less active one. You can predict the relative activities from a table of activities or a half-reaction table. Chapter 16 , Electrochemistry, goes into depth about electrochemistry and redox reactions.

The AP free-response booklet includes a table of half-reactions, which you may use for help during this part of the exam. A similar table can be found in the back of this book. Alternatively, you may wish to memorize the common oxidizing and reducing agents.

Redox is a term that stands for red uction and ox idation. Reduction is the gain of electrons and oxidation is the loss of electrons. For example, suppose a piece of zinc metal is placed in a solution containing the blue Cu2+ cation. Very quickly, a reddish solid forms on the surface of the zinc metal. That substance is copper metal. As the copper metal is deposited, the blue color of the solution begins to fade. At the molecular level, the more active zinc metal is losing electrons to form the Zn2+ cation, and the Cu2+ ion is gaining electrons to form the less active copper metal. These two processes can be shown as:

The electrons that are being lost by the zinc metal are the same electrons that are being gained by the copper(II) ion. The zinc metal is being oxidized and the copper(II) ion is being reduced. Further discussions on why reactions such as these occur can be found in the section on single displacement reactions later in this chapter.

Something must cause the oxidation (taking the electrons), and that substance is called the oxidizing agent (the reactant being reduced). In the example above, the oxidizing agent is the Cu2+ ion. The reactant undergoing oxidation is called the reducing agent because it is furnishing the electrons that are being used in the reduction half-reaction. Zinc metal is the reducing agent above. The two half-reactions, oxidation and reduction, can be added together to give you the overall redox reaction. When doing this, the electrons must cancel—that is, there must be the same number of electrons lost as electrons gained:

In these redox reactions, there is a simultaneous loss and gain of electrons. In the oxidation reaction (commonly called a half-reaction) electrons are being lost, but in the reduction half-reaction those very same electrons are being gained. So, in redox reactions electrons are being exchanged as reactants are being converted into products. This electron exchange may be direct, as when copper metal plates out on a piece of zinc, or it may be indirect, as in an electrochemical cell (battery).

Another way to determine what is being oxidized and what is being reduced is by looking at the change in oxidation numbers of the reactant species. (See Chapter 5 , Basics, for a discussion of oxidation numbers and how to calculate them.) On the AP exam, you may be asked to assign oxidation numbers and/or identify changes in terms of oxidation numbers. Oxidation is indicated by an increase in oxidation number. In the example above, the Zn metal went from an oxidation state of zero to +2. Reduction is indicated by a decrease in oxidation number. Cu2+ went from an oxidation state of +2 to zero. In order to figure out whether a particular reaction is a redox reaction, write the net ionic equation. Then determine the oxidation numbers of each element in the reaction. If one or more elements have changed oxidation number, it is a redox reaction.

There are several types of redox reaction that are given specific names. In the next few pages, we will examine some of these types of redox reaction.

Combination Reactions

Combination reactions are reactions in which two or more reactants (elements or compounds) combine to form one product. Although these reactions may be of a number of different types, some types are definitely redox reactions. These include reactions of metals with nonmetals to form ionic compounds, and the reaction of nonmetals with other nonmetals to form covalent compounds.

In the first reaction, we have the combination of an active metal with an active nonmetal to form a stable ionic compound. The very active oxygen reacts with hydrogen to form the stable compound water. The hydrogen and potassium are undergoing oxidation, while the oxygen and chlorine are undergoing reduction.

Decomposition Reactions

Decomposition reactions are reactions in which a compound breaks down into two or more simpler substances. Although not all decomposition reactions are redox reactions, many are. For example, the thermal decomposition reactions, such as the common laboratory experiment of generating oxygen by heating potassium chlorate, are decomposition reactions:

In this reaction the chlorine is going from the less stable +5 oxidation state to the more stable −1 oxidation state. While this is occurring, oxygen is being oxidized from −2 to 0.

Another example is electrolysis , in which an electrical current is used to decompose a compound into its elements:

The spontaneous reaction would be the opposite one; therefore, we must supply energy (in the form of electricity) in order to force the nonspontaneous reaction to occur.

Single Displacement Reactions

Single displacement (replacement) reactions are reactions in which atoms of an element replace the atoms of another element in a compound. All of these single replacement reactions are redox reactions, since the element (in a zero oxidation state) becomes an ion. Most single displacement reactions can be categorized into one of three types of reaction:

  • A metal displacing a metal ion from solution
  • A metal displacing hydrogen gas (H2) from an acid or from water
  • One halogen replacing another halogen in a compound

Remember: It is an element displacing another atom from a compound. The displaced atom appears as an element on the product side of the equation.

Table 6.1 Activity Series of Metals in Aqueous Solution

Reactions will always appear in the free-response section of the AP Chemistry exam. This may not be true in the multiple-choice part.

For the first two types, a table of metals relating their ease of oxidation to each other is useful in being able to predict what displaces what. Table 6.1 shows the activity series for metals , which lists the metal and its oxidation in order of decreasing ease of oxidation. An alternative to the activity series is a table of half-cell potentials, as discussed in Chapter 16 , Electrochemistry. In general, the more active the metal, the lower its potential.

Elements on this activity series can displace ions of metals lower than themselves on the list. If, for example, one placed a piece of tin metal into a solution containing Cu(NO3 )2 (aq), the Sn would replace the Cu2+ cation:

The second equation is the net ionic form that is often required on the AP exam.

If a piece of copper metal was placed in a solution of Sn(NO3 )2 (aq), there would be no reaction, since copper is lower than tin on the activity series. This table allows us to also predict that if sodium metal is placed in water, it will displace hydrogen, forming hydrogen gas:

The Group IA and IIA elements on the activity table will displace hydrogen from water, but not the other metals shown. All the metals above hydrogen will react with acidic solutions to produce hydrogen gas:

Halogen reactivity decreases as one goes from top to bottom in the periodic table, because of the decreasing electronegativity. Therefore, a separate activity series for the halogens can be developed:

The above series indicates that if chlorine gas were dissolved in a KI(aq) solution, the elemental chlorine would displace the iodide ion:

Combustion Reactions

Combustion reactions are redox reactions in which the chemical species rapidly combines with oxygen and usually emits heat and light. Reactions of this type are extremely important in our society as the sources of heat energy. Complete combustion of carbon yields carbon dioxide, and complete combustion of hydrogen yields water. The complete combustion of hydrocarbons , organic compounds containing only carbon and hydrogen, yields carbon dioxide and water:

2 C2 H6 (g) + 7 O2 (g) → 4CO2 (g) + 6 H2 O(g)

If the compound also contains oxygen, such as in alcohols, ethers, etc., the products are still carbon dioxide and water:

2 CH3 OH(l) + 3 O2 (g) → 2CO2 (g) + 4 H2 O(g)

If the compound contains sulfur, the complete combustion produces sulfur dioxide, SO2 :

2 C2 H6 S(g) + 9 O2 (g) → 4 CO2 (g) + 6 H2 O(g) + 2 SO2 (g)

If nitrogen is present, it will normally form the very stable nitrogen gas, N2 .

In all of these reactions, the driving force is the highly reactive oxygen forming a very stable compound(s). This is shown by the exothermic nature of the reaction.

In balancing any of these combustion reactions, it is helpful to balance the oxygen last.

Coordination Compounds

When a salt is dissolved in water, the metal ions, especially transition metal ions, form a complex ion with water molecules and/or other species. A complex ion is composed of a metal ion bonded to two or more molecules or ions called ligands . These are Lewis acid–base reactions. For example, suppose Cr(NO3 )3 is dissolved in water. The Cr3+ cation attracts water molecules to form the complex ion  In this complex ion, water acts as the ligand. If ammonia is added to this solution, the ammonia can displace the water molecules from the complex:

In reactions involving coordination compounds, the metal acts as the Lewis acid (electron-pair acceptor), while the ligand acts as a Lewis base (electron-pair donor). In the reaction above, the ammonia ligand displaced the water ligand from the chromium complex because nitrogen is a better electron-pair donor (less electronegative) than oxygen.

The nitrogen in the ammonia and the oxygen in the water are the donor atoms. They are the atoms that actually donate the electrons to the Lewis acid. The coordination number is the number of donor atoms that surround the central atom. As seen above, the coordination number for Cr3+ is 6. Coordination numbers are usually 2, 4, or 6, but other values can be possible. Silver (Ag+ ) commonly forms complexes with a coordination number of 2; zinc (Zn2+), copper (Cu2+ ), nickel (Ni2+ ), and platinum (Pt2+ ) commonly form complexes with a coordination number of 4; most other central ions have a coordination number of 6.

Acid–Base Reactions

Acids and bases are extremely common, as are the reactions between acids and bases. The driving force is often the hydronium ion reacting with the hydroxide ion to form water. Chapter 15 on Equilibrium describes the equilibrium reactions of acids and bases, as well as some information concerning acid–base titration. After you finish this section, you may want to review the acid–base part of the Equilibrium chapter.

Properties of Acids, Bases, and Salts

At the macroscopic level, acids taste sour, may be damaging to the skin, and react with bases to yield salts. Bases taste bitter, feel slippery, and react with acids to form salts.

At the microscopic level, acids are defined as proton (H+ ) donors (Brønsted–Lowry theory) or electron-pair acceptors (Lewis theory). Bases are defined as proton (H+ ) acceptors (Brønsted–Lowry theory) or electron-pair donors (Lewis theory). Consider the gas-phase reaction between hydrogen chloride and ammonia:

HCl is the acid, because it is donating an H+ and the H+ will accept an electron pair from ammonia. Ammonia is the base, accepting the H+ and furnishing an electron pair that the H+ will bond with via coordinate covalent bonding. Coordinate covalent bonds are covalent bonds in which one of the atoms furnishes both of the electrons for the bond. After the bond is formed, it is identical to a covalent bond formed by donation of one electron by both of the bonding atoms.

Acids and bases may be strong , dissociating completely, or weak , partially dissociating and forming an equilibrium system.(See Chapter 15 for the details on weak acids and bases.) Strong acids include:

  1. Hydrochloric, HCl
  2. Hydrobromic, HBr
  3. Hydroiodic, HI
  4. Nitric, HNO3
  5. Chloric, HClO3
  6. Perchloric, HClO4
  7. Sulfuric, H2SO4

The strong acids above are all compounds that ionize completely in aqueous solution, yielding hydrogen ions and the anions from the acid.

Strong bases include:

  1. Alkali metal (Group IA) hydroxides (LiOH, NaOH, KOH, RbOH, CsOH)
  2. Ca(OH)2, Sr(OH)2 , and Ba(OH)2

The strong bases listed above are all compounds that dissociate completely, yielding the hydroxide ion (which is really the base, not the compound).

Unless told otherwise, assume that acids and bases not on the lists above are weak and will establish an equilibrium system when placed into water.

Some salts have acid–base properties. For example, ammonium chloride, NH4 Cl, when dissolved in water will dissociate and the ammonium ion will act as a weak acid, donating a proton. We will examine these acid–base properties in more detail in the next section.

Certain oxides can have acidic or basic properties. These properties often become evident when the oxides are dissolved in water. In most case, reactions of this type are not redox reactions.

Many oxides of metals that have a +1 or +2 charge are called basic oxides (basic anhydrides), because they will react with acids.

Many times they react with water to form a basic solution:

Many nonmetal oxides are called acidic oxides (acidic anhydrides), because they react with water to form an acidic solution:

CO2 (g) + H2 O(l) → H2 CO3 (aq)

H2 CO3 (aq) is named carbonic acid and is the reason that most carbonated beverages are slightly acidic. It is also the reason that soft drinks have fizz, because carbonic acid will decompose to form carbon dioxide and water.

Acid–Base Reactions

In general, acids react with bases to form a salt and, usually, water. The salt will depend upon which acid and base are used:

Reactions of this type are called neutralization reactions .

The first two neutralization equations are represented by the same net ionic equation:

H+ (aq) + OH (aq) → H2 O(l)

In the third case, the net ionic equation is different:

H+ (aq) + NH3 (aq) → NH4 + (aq)

As mentioned previously, certain salts have acid–base properties. In general, salts containing cations of strong bases and anions of strong acids are neither acidic nor basic. They are neutral, reacting with neither acids nor bases. An example would be potassium nitrate, KNO3 . The potassium comes from the strong base KOH and the nitrate from the strong acid HNO3 .

Salts containing cations not of strong bases but with anions of strong acids behave as acidic salts. An example would be ammonium chloride, NH4 Cl.

Cations of strong bases and anions not of strong acids are basic salts. An example would be sodium carbonate, Na2 CO3 . It reacts with an acid to form carbonic acid, which would then decompose to carbon dioxide and water:

The same type of reaction would be true for acid carbonates, such as sodium bicarbonate, NaHCO3 .

Another group of compounds that have acid–base properties are the hydrides of the alkali metals and of calcium, strontium, and barium. These hydrides will react with water to form the hydroxide ion and hydrogen gas:

Note that in this case, water is behaving as H+ OH .

Acid–Base Titrations

A common laboratory application of acid–base reactions is a titration. A titration is a laboratory procedure in which a solution of known concentration is used to determine the concentration of an unknown solution. For strong acid/strong base titration systems, the net ionic equation is:

H+ (aq) + OH (aq) → H2 O(l)

For example, suppose you wanted to determine the molarity of an HCl solution. You would pipet a known volume of the acid into a flask and add a couple of drops of a suitable acid–base indicator. An indicator that is commonly used is phenolphthalein, which is colorless in an acidic solution and pink in a basic solution. You would then fill a buret with a strong base solution (NaOH is commonly used) of known concentration. The buret allows you to add small amounts of the base solution to the acid solution in the flask. The course of the titration can also be followed by the use of a pH meter. Initially the pH of the solution will be low, since it is an acid solution. As the base is added and neutralization of the acid takes place, the pH will slowly rise. Small amounts of the base are added until one reaches the equivalence point. The equivalence point is that point in the titration where the number of moles of H+ in the acid solution has been exactly neutralized with the same number of moles of OH :

moles H+ = moles OH at the equivalence point

For the titration of a strong acid with a strong base, the pH rapidly rises in the vicinity of the equivalence point. Then, as the tiniest amount of base is added in excess, the indicator turns pink. This is called the endpoint of the titration. In an accurate titration, the endpoint will be as close to the equivalence point as possible. For simple titrations that do not use a pH meter, it is assumed that the endpoint and the equivalence point are the same, so that:

moles H+ = moles OH at the endpoint

After the equivalence point has been passed, the pH is greater than 7 (basic solution) and begins to level out somewhat. Figure 6.1 shows the shape of the curve for this titration.

Reaction stoichiometry can then be used to solve for the molarity of the acid solution. See Chapter 7 , Stoichiometry, for a discussion of solution stoichiometry.

An unknown base can be titrated with an acid solution of known concentration. One major difference is that the pH will be greater than 7 initially and will decrease as the titration proceeds. The other major difference is that the indicator will start pink, and the color will vanish at the endpoint.

Figure 6.1   Titration of a strong acid with a strong base.

Experiments

Laboratory experiments involving reactions are usually concerned with both the reaction and the stoichiometry. You need some idea of the balanced chemical equation. In the case of an acid–base reaction, an acid reacts with a base. The acid supplies H+ and the base accepts the H+ . If the acid is diprotic, such as H2 SO4 , it can donate two H+ .

The key to any reaction experiment is moles. The numbers of moles may be calculated from various measurements. A sample may be weighed on a balance to give the mass, and the moles calculated with the formula weight. Or the mass of a substance may be determined using a volume measurement combined with the density. The volume of a solution may be measured with a pipet, or calculated from the final and initial readings from a buret. This volume, along with the molarity, can be used to calculate the moles present. The volume, temperature, and pressure of a gas can be measured and used to calculate the moles of a gas. You must be extremely careful on the AP exam to distinguish between those values that you measure and those that you calculate.

The moles of any substance in a reaction may be converted to the moles of any other substance through a calculation using the balanced chemical equation. Other calculations are presented in Chapter 7 , Stoichiometry.

Common Mistakes to Avoid

  1. In balancing chemical equationsdon’t change the subscripts in the chemical formula, just the coefficients.
  2. Molecular compounds ionize, ionic compounds dissociate.
  3. In writing ionic and net ionic equations, show the chemical species as they actually exist in solution (i.e., strong electrolytes as ions, etc.).
  4. In writing ionic and net ionic equations, don’t break apart covalently bonded compounds unless they are strong acids that are ionizing.
  5. Know the solubility rules as guidelines, not explanations.
  6. Oxidizing and reducing agents are reactants, not products.
  7. The products of the complete combustion of a hydrocarbon are carbon dioxide and water. This is also true if oxygen is present as well; but if some other element, like sulfur, is present you will also have something else in addition to carbon dioxide and water.
  8. If a substance that does not contain carbon, like elemental sulfur, undergoes complete combustion, no carbon dioxide can be formed.
  9. If an alcohol like methanol, CH3OH, is dissolved in water, no hydroxide ion, OH , will be formed.
  10. Know the strong acids and bases.
  11. HF is not a strong acid.
  12. In titration calculations, you must consider the reaction stoichiometry.
  13. Be sure to indicate the charges on ions correctly.
  14. The common coordination numbers of complex ions are 2, 4, and 6.
  15. Do not confuse measured values and calculated values.

 Review Questions

Here are questions you can use to review the content of this chapter and practice for the AP Chemistry exam. First are 16 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry exam. Following those is a single-part free-response question like ones in Section II of the exam. To make these review questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

Multiple-Choice Questions

Answer the following questions in 20 minutes. You may not use a calculator. You may use the periodic table and the equation sheet at the back of this book.

1 . ___ Fe(OH)2 (s) + ___ H3 PO4 (aq) → ___ Fe3 (PO4 )2 (s) + ___ H2 O(l)

After the above chemical equation is balanced, the lowest whole-number coefficient for water is:

(A) 3

(B) 1

(C) 9

(D) 6

2 . This ion will generate gas bubbles upon the addition of hydrochloric acid.

(A) Cu2+

(B) CO3 2–

(C) Fe3+

(D) Al3+

3 . Aqueous solutions of this ion are blue.

(A) Cu2+

(B) CO3 2–

(C) Fe3+

(D) Al3+

4 . Which of the following best represents the balanced net ionic equation for the reaction of lead(II) carbonate with concentrated hydrochloric acid? In this reaction, all lead compounds are insoluble.

(A) Pb2 CO3 + 2 H+ + Cl → Pb2 Cl + CO2 + H2 O

(B) PbCO3 + 2 H+ + 2 Cl → PbCl2 + CO2 + H2 O

(C) PbCO3 + 2 H+ → Pb2+ + CO2 + H2 O

(D) PbCO3 + 2 Cl → PbCl2 + CO3 2–

5 . A sample of copper metal is reacted with concentrated nitric acid in the absence of air. After the reaction, which of these final products are present?

(A) CuNO3 and H2 O

(B) Cu(NO3 )3 , NO, and H2 O

(C) Cu(NO3 )2 , NO, and H2 O

(D) CuNO3 , H2 O, and H2

6 . Which of the following is the correct net ionic equation for the reaction of acetic acid with potassium hydroxide?

(A) HC2 H3 O2 + OH → C2 H3 O2  + H2 O

(B) HC2 H3 O2 + K+ → KC2 H3 O2 + H+

(C) HC2 H3 O2 + KOH → KC2 H3 O2 + H2 O

(D) H+ + OH → H2 O(l)

7 . Which of the following is the correct net ionic equation for the addition of aqueous ammonia to a precipitate of silver chloride?

(A) AgCl + 2 NH3 → [Ag(NH3 )2 ]+ + Cl

(B) AgCl + 2 NH4 + → [Ag(NH4 )2 ]3+ + Cl

(C) AgCl + NH4 + → Ag+ + NH4 Cl

(D) AgCl + NH3 → Ag+ + NH3 Cl

8 . Potassium metal will react with water to release a gas and form a potassium compound. Which of the following is true?

(A) The final solution is basic.

(B) The gas is oxygen.

(C) The potassium compound precipitates.

(D) The potassium compound will react with strong bases.

9 . A sample is tested for the presence of the  ion. This ion, along with others, may be precipitated with chloride ion. If  is present in the chloride precipitate, a black color will form upon treatment with aqueous ammonia. The balanced net ionic equation for the formation of this black color is:

10 . How many moles of Pb(NO3 )2 must be added to 0.10 L of a solution that is 1.0 M in MgCl2 and 1.0 M in KCl to precipitate all of the chloride ion? The compound PbCl2 precipitates.

(A) 1.0 mol

(B) 0.20 mol

(C) 0.50 mol

(D) 0.15 mol

11 . When 50.0 mL of 1.0 M AgNO3 is added to 50.0 mL of 0.50 M HCl, a precipitate of AgCl forms. After the reaction is complete, what is the concentration of silver ions in the solution?

(A) 0.50 M

(B) 0.0 M

(C) 1.0 M

(D) 0.25 M

12 . A student mixes 50.0 mL of 0.10 M Pb(NO3 )2 solution with 50.0 mL of 0.10 M KCl. A white precipitate forms, and the concentration of the chloride ion becomes very small. Which of the following correctly places the concentrations of the remaining ions in order of decreasing concentration?

(A) [NO3  ] > [Pb2+ ] > [K+ ]

(B) [NO3  ] > [K+ ] > [Pb2+ ]

(C) [K+ ] > [NO3  ] > [Pb2+ ]

(D) [Pb2+ ] > [NO3  ] > [K+ ]

13 . A solution is prepared for qualitative analysis. The solution contains the following ions: Co2+ , Pb2+ , and Al3+ . Which of the following will cause no observable reaction?

(A) Dilute NH3 (aq) is added.

(B) Dilute K2 CrO4 (aq) is added.

(C) Dilute HNO3 (aq) is added.

(D) Dilute K2 S(aq) is added.

14 . Chlorine gas is bubbled through a colorless solution and the solution turns reddish. Adding a little methylene chloride to the solution extracts the color into the methylene chloride layer. Which of the following ions may be present in the original solution?

(A) Cl

(B) I

(C) 

(D) Br

15 . The addition of excess concentrated NaOH(aq) to a 1.0 M (NH4 )2 SO4 solution will result in which of the following observations?

(A) The solution becomes neutral.

(B) The formation of a brown precipitate takes place.

(C) Nothing happens because the two solutions are immiscible.

(D) The odor of ammonia will be detected.

16 . ____C4 H11 N(l) + ____O2 (g) → ____CO2 (g) + ____H2 O(l) + ____N2 (g)

When the above equation is balanced, the lowest whole number coefficient for CO2 is:

(A) 4

(B) 16

(C) 27

(D) 22

Answers and Explanations for the Multiple-Choice Questions

1 . D —The balanced equation is:

2 . B —Carbonates produce carbon dioxide gas in the presence of an acid. None of the other ions will react with hydrochloric acid to produce a gas.

3 . A —Aqueous solutions of Cu2+ are normally blue. Iron ions give a variety of colors but are normally colorless, or nearly so, in the absence of complexing agents. The other ions are colorless.

4 . B —Lead(II) carbonate is insoluble, so its formula should be written as PbCO3 . Hydrochloric acid is a strong acid so it should be written as separate H+ and Cl ions. Lead(II) chloride, PbCl2 , is insoluble, and carbonic acid, H2CO3 , quickly decomposes to CO2 and H2 O. Also notice that A cannot be correct because the charges do not balance.

5 . C —The balanced chemical equation is:

3 Cu(s) + 8 HNO3 (aq) → 3 Cu(NO3 )2 (aq) + 2 NO(g) + 4 H2 O(l)

The copper is below hydrogen on the activity series, so H2 cannot form by this acid-metal reaction. Nitric acid causes oxidation, which will oxidize copper to Cu2+ giving Cu(NO3 )2 . Some of the nitric acid reduces to NO. An oxidation and a reduction must ALWAYS be together.

6 . A —Acetic acid is a weak acid; as such, it should be written as HC2 H3 O2 . Potassium hydroxide is a strong base so it will separate into K+ and OH ions. Any potassium compound that might form is soluble and will yield K+ions. The potassium ions are spectator ions and are left out of the net ionic equation.

7 . A —Aqueous ammonia contains primarily NH3 , which eliminates choices B and C. NH3 Cl does not exist, which eliminates choice D. The reaction produces the silver-ammonia complex, [Ag(NH3 )2 ]+ .

8 . A —The reaction of potassium to produce a potassium compound is an oxidation; therefore, there must be a reduction, and the only species available for reduction is hydrogen. The reaction is:

2 K(s) + 2 H2 O(l) → 2 KOH(aq) + H2 (g)

KOH is a water-soluble strong base, which will not react with other strong bases.

9 . B —The black color is due to the formation of metallic mercury. Aqueous ammonia contains primarily NH3 , which eliminates choices C and D. The total charges on each side of the reaction arrow must be equal, which eliminates choice A.

10 . D —The magnesium chloride gives 0.20 moles of chloride ion, and the potassium chloride gives 0.10 moles of chloride ion. A total of 0.30 moles of chloride will react with 0.15 moles of lead, because two Cl require one Pb2+ .

11 . D —The HCl is the limiting reagent. The HCl will react with one-half the silver to halve the concentration. The doubling of the volume halves the concentration a second time.

12 . B —All the potassium and nitrate ions remain in solution, leaving PbCl2 as the only possible precipitate. Equal volumes of equal concentrations give the same number of moles of reactants; however, two nitrate ions are produced per solute formula as opposed to only one potassium ion. Initially, the lead and potassium would be equal, but some of the lead is precipitated as PbCl2 .

13 . C —Ammonia, as a base, will precipitate the metal hydroxides since the only soluble hydroxides are the strong bases. Chromate, sulfide, and chloride ions might precipitate one or more of the ions. Nitrates, from nitric acid, are soluble; therefore, this is the solution that is least likely to cause an observable change.

14 . D —Chlorine causes oxidation. It is capable of oxidizing both B and D. Answer B gives I2 , which is brownish in water and purplish in methylene chloride. Bromine solutions are reddish in both.

15 . D —Excess strong base will ensure the solution to be basic and not neutral. Both ammonium and sodium salts are soluble; therefore, no precipitate can form. One aqueous solution will mix with another aqueous solution. The following acid–base reaction occurs to release ammonia gas: NH4 + (aq) + OH (aq) → NH3 (g) + H2 O(l).

16 . B —4 C4 H11 N(l) + 27 O2 (g) → 16 CO2 (g) + 22 H2 O(l) + 2 N2 (g)

Free-Response Question

On the new AP exams, there will be both long-answer and short-answer free-response questions. The following is an example of a short-answer question.

You have 5 minutes to do the following question. You may use a calculator and the tables in the back of the book.

Question 1

A lead(II) nitrate, Pb(NO3 )2 , solution is mixed with an ammonium sulfate, (NH4 )2 SO4 , solution and a precipitate forms. What is the precipitate, and which ions, if any, are spectator ions in this reaction? Explain how you arrived at your answers.

Answer and Explanation for the Free-Response Question

There are four ions present. These ions are  , and  . The starting materials are in solution; therefore, they are soluble. The compounds that might form in the reaction are NH4 NO3 and PbSO4 . Since ammonium (and nitrate) salts are normally soluble, the precipitate must be PbSO4 . The spectator ions are the nitrate ions ( ) and the ammonium ions ( ), the ions not in the precipitate.

You get 1 point for correctly identifying the precipitate. You get an additional point for the identification of the spectator ions and 2 points for the explanation. There is a maximum of 4 points possible.

 Rapid Review

  • Reaction questions will always appear in the free-response section of the AP exam. This may not be true in the multiple-choice part.
  • Energy may be released in a reaction (exothermic) or absorbed (endothermic).
  • Chemical equations are balanced by adding coefficients in front of the chemical species until the number of each type of atom is the same on both the right and left sides of the arrow.
  • The coefficients in the balanced equation must be in the lowest whole-number ratio.
  • Water is the universal solvent, dissolving a wide variety of both ionic and polar substances.
  • Electrolytes are substances that conduct an electrical current when dissolved in water; nonelectrolytes do not.
  • Most ions in solution attract and bind a layer of water molecules in a process called solvation.
  • Some molecular compounds, like acids, ionize in water, forming ions.
  • In the molecular equation, the reactants and products are shown in their undissociated/un-ionized form; the ionic equation shows the strong electrolytes in the form of ions; the net ionic equation drops out all spectator ions and shows only those species that are undergoing chemical change.
  • Precipitation reactions form an insoluble compound, a precipitate, from the mixing of two soluble compounds.
  • Learn and be able to apply the solubility rules.
  • Redox reactions are reactions where oxidation and reduction take place simultaneously.
  • Oxidation is the loss of electrons, and reduction is the gain of electrons.
  • Combination reactions are usually redox reactions in which two or more reactants (elements or compounds) combine to form one product.
  • Decomposition reactions are usually redox reactions in which a compound breaks down into two or more simpler substances.
  • Single displacement reactions are redox reactions in which atoms of an element replace the atoms of another element in a compound.
  • Know how to use the activity series to predict whether or not an element will displace another element.
  • Combustion reactions are redox reactions in which the chemical species rapidly combine with diatomic oxygen gas, emitting heat and light. The products of the complete combustion of a hydrocarbon are carbon dioxide and water.
  • Indicators are substances that exhibit different colors under acidic or basic conditions.
  • Acids are proton donors (electron-pair acceptors).
  • Bases are proton acceptors (electron-pair donors).
  • Coordinate covalent bonds are covalent bonds in which one atom furnishes both of the electrons for the bond.
  • Strong acids and bases completely ionize/dissociate, and weak acids and bases only partially ionize/dissociate.
  • Know the strong acids and bases.
  • Acids react with bases to form a salt and usually water in a neutralization reaction.
  • Many hydrides react with water to form the hydroxide ion and hydrogen gas.
  • A titration is a laboratory procedure for determining the concentration of an unknown solution using a solution of known concentration.
  • The equivalence point of an acid–base titration is the point at which the moles of H+from the acid equals the moles of OH from the base. The endpoint is the point at which the indicator changes color, indicating the equivalence point.
  • A complex ion is composed of a metal ion covalently bonded to two or more molecules or anions called ligands.
  • The coordination number (usually 2, 4, or 6) is the number of donor atoms that can surround a metal ion in a complex.