CHAPTER 16 Acid-Base Equilibria and Solubility Equilibriaglencoe.mheducation.com/.../997800/Chang_TM_Ch_16_FINAL_12_18_… · Chang 11e Teacher‘s Manual, Chapter 16 Enduring Understanding

Chang 11e Teacher‘s Manual, Chapter 16

CHAPTER 16 Acid-Base Equilibria and Solubility Equilibria

Chapter Overview This chapter reinforces and refines all of the concepts introduced in the previous chapters

on acids and bases and equilibrium. The key point to be made throughout is that

equilibrium is equilibrium. There are no new equilibrium concepts to be learned, only

specific applications of what has already been discussed. Additionally, the acid-base

refinement is simply a more mathematical analysis of weak acids and bases that were

discussed in the last chapter. While this chapter examines specific examples of

equilibrium conditions, these examples are very important in that they have many

applications in the natural world and industry. Consequently, these concepts are often

found on the AP exam free-response question number one and in the multiple-choice

section. Concepts covered in this chapter fall under the umbrella of Big Idea 1 and Big

Idea 6 and are detailed in the alignment tables below.

The AP curriculum topics included in this chapter are:

II. States of Matter

C.1. Types of solutions and factors affecting solubility

III. Reactions

A.1. Acid-base reactions; concepts of Arrhenius, Brønsted-Lowry and Lewis;

coordination complexes; amphoterism

A.2. Precipitation reactions

C.1. Concept of dynamic equilibrium, physical and chemical; LeChatelier‘s

principle; equilibrium constants

C.2.b. Equilibrium constants for reactions in solution

(1) Constants for acids and bases; pK; pH

(2) Solubility product constants and their application to precipitation and the

dissolution of slightly soluble compounds

(3) Common ion effect; buffers; hydrolysis

ALIGNMENT OF CONTENT WITH CURRICULUM FRAMEWORK

Big Idea 1: The chemical elements are fundamental building materials of matter, and all

matter can be understood in terms of arrangements of atoms. These atoms retain their

identity in chemical reactions.


Enduring Understanding (EU) 1.A: All matter is made of atoms. There are

a limited number of types of atoms; these are the elements.

Essential Knowledge (EK) 1.A.2: Chemical analysis provides a method for

determining the relative number of atoms in a substance, which can be used to identify

the substance or determine its purity.

a. Because compounds are composed of atoms with known masses, there is a

correspondence between the mass percent of the elements in a compound and the

relative number of atoms of each element.

b. An empirical formula is the lowest whole number ratio of atoms in a compound.

Two molecules of the same elements with identical mass percent of their

constituent atoms will have identical empirical formulas.

c. Because pure compounds have a specific mass percent of each element, experimental

measurements of mass percents can be used to verify the purity of compounds.

Aligns with text section 16.11

Enduring Understanding (EU) 1.E: Atoms are conserved in physical and

chemical processes.

Essential Knowledge (EK) 1.E.2: Conservation of atoms makes it possible to

compute the masses of substances involved in physical and chemical processes.

Chemical processes result in the formation of new substances, and the amount of

these depends on the number and the types and masses of elements in the

reactants, as well as the efficiency of the transformation.

a. The number of atoms, molecules, or formula units in a given mass of substance

can be calculated.

b. The subscripts in a chemical formula represent the number of atoms of each type

in a molecule.

c. The coefficients in a balanced chemical equation represent the relative numbers of

particles that are consumed and created when the process occurs.

d. The concept of conservation of atoms plays an important role in the interpretation

and analysis of many chemical processes on the macroscopic scale.

e. In gravimetric analysis, a substance is added to a solution that reacts specifically

with a dissolved analyte (the chemical species that is the target of the analysis) to

form a solid. The mass of solid formed can be used to infer the concentration of

the analyte in the initial sample.

f. Titrations may be used to determine the concentration of an analyte in a solution.

The titrant has a known concentration of a species that reacts specifically

with the analyte. The equivalence of the titration occurs when the analyte

is totally consumed by the reacting species in the titrant. The equivalence point

is often indicated by a change in a property (such as color) that occurs when the

equivalence point is reached. This observable event is called the end point of the

titration.



Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken. These

two processes are in a dynamic competition, sensitive to initial conditions and external

perturbations

Enduring Understanding (EU) 6.A: Chemical equilibrium is a dynamic,

reversible state in which rates of opposing processes are equal.

Essential Knowledge (EK) 6.A.1: in many classes of reactions, it is important to

consider both the forward and reverse reaction.

a. Many readily observable processes are reversible. Examples include evaporating

and condensing water, absorption of a gas, or dissolving and precipitating a salt.

Relevant and interesting contexts include biological examples (binding of oxygen

to hemoglobin and the attachment of molecules to receptor sites in the nose) and

environmental examples (transfer of carbon between atmosphere and biosphere

and transfer of dissolved substances between atmosphere and hydrosphere).

b. Dissolution of a solid, transfer of protons in acid-base reactions, and transfer of

electrons in redox reactions are important examples of reversible reactions.


Essential Knowledge (EK) 6.A.2: the current state of a system undergoing a

reversible reaction can be characterized by the extent to which reactants have been

converted to products. the relative quantities of reaction components are quantitatively

described by the reaction quotient, Q.

a. Given an initial set of reactant and product concentrations, only those sets of

concentrations that are consistent with the reaction stoichiometry can be attained.

ICE (initial, change, equilibrium) tables are useful for determining which sets of

concentration values are possible.

b. The reaction quotient, Q, provides a convenient measure of the current progress of

a reaction. Q does not include substances whose concentrations are independent

of the amount of substance, such as for a solid in contact with a liquid solution or

with a gas, or for a pure solid or liquid in contact with a gas.

c. The value of Q (and so also K) changes when a reaction is reversed. When

reactions are added together through the presence of a common intermediate, Q

(and so also K) of the resulting reaction is a product of the values of Q (or K) for

the original reactions.


Enduring Understanding (EU) 6.B: Systems at equilibrium are responsive to external perturbations, with the response leading to a change in the composition of the system.

Essential Knowledge (EK) 6.B.1: systems at equilibrium respond to disturbances


by partially countering the effect of the disturbance (LeChatelier‘s principle).

a. LeChatelier‘s principle can be used to predict the response of a system to

the following stresses: addition or removal of a chemical species, change in

temperature, change in volume/pressure of a gas phase system, and dilution of a

reaction system with water or other solvent.

b. LeChatelier‘s principle can be used to reason about the effects a stress will have

on experimentally measurable properties, such as pH, temperature, and color of a

solution.


Essential Knowledge (EK) 6.B.2: A disturbance to a system at equilibrium causes Q

to differ from K, thereby taking the system out of the original equilibrium state. the

system responds by bringing Q back into agreement with K, thereby establishing a

new equilibrium state.

a. LeChatelier‘s principle involves qualitative reasoning that is closely connected to

the quantitative approach of 6.A.3.

b. Some stresses, such as changes in concentration, cause a change in Q. A change in

temperature causes a change in K. In either case, the reaction shifts to bring Q and

K back into equality.


Enduring Understanding (EU) 6.C: Chemical equilibrium plays an important

role in acid-base chemistry and in solubility.

Essential Knowledge (EK) 6.C.1: Chemical equilibrium reasoning can be used to describe

the proton-transfer reactions of acid-base chemistry.

a. The concentrations of hydronium ion and hydroxide ion are often reported as pH

and pOH, respectively.

b. Water autoionizes with an equilibrium constant, Kw. For pure water, pH = pOH,

and this condition is called “neutrality,” or a neutral solution. At 25°C, pKw =

14, and thus pH and pOH add to 14. In pure water at 25°C, pH = pOH = 7.

c. Common strong acids include HCl, HBr, HI, HClO4, H2SO4, and HNO3. The

molecules of strong acids completely ionize in solution to produce hydronium

ions. In other words, 100 percent of the molecules of the strong acid are ionized

in a solution (assuming that the concentration is not extremely high). As such, the

concentration of H3O+ in a strong acid solution is equal to the initial concentration

of the strong acid, and thus the pH of the strong acid solution is easily calculated.

d. Common strong bases include group I and II hydroxides. When dissolved in

solution, strong bases completely dissociate to produce hydroxide ions. Note that

some group II hydroxides are slightly soluble in water. However, 100 percent of

the dissolved base is ionized.


e. Weak acid molecules react with water to transfer a proton to the water molecule.

However, weak acid molecules only partially ionize in this way. In other

words, only a small percentage of the molecules of a weak acid are ionized in a

solution (assuming that the initial concentration is not extremely low). Thus, the

concentration of H3O+ does not equal the initial concentration of the molecular

acid, and the vast majority of the acid molecules remain un-ionized. A solution

of a weak acid thus involves equilibrium between an un-ionized acid and its

conjugate base. The equilibrium constant for this reaction is Ka, often reported

as pKa. The pH of a weak acid solution can be determined from the initial acid

concentration and the pKa. The common weak acids include carboxylic acids.

The relative magnitudes of Ka‘s are influenced by structural factors such as bond

strength, solvation, and electronegativity of the atom bonded to the labile proton.

f. The common weak bases include ammonia, amines and pyridines, other

nitrogenous bases, and conjugate bases (defined below in g). Weak base molecules

in aqueous solutions react with water molecules to produce hydroxide ions.

However, only a small percentage of the molecules of a weak base in a solution

ionize in this way (assuming that the initial concentration is not extremely

low). Thus, the concentration of OH- in the solution does not equal the initial

concentration of the molecular base, and the vast majority of the base molecules

remain un-ionized. A solution of a weak base thus involves an equilibrium

between an un-ionized base and its conjugate acid. The equilibrium constant for

this reaction is Kb, often reported as pKb. The pH of a weak base solution can be

determined from the initial base concentration and the pKb.

g. When an acid molecule loses its proton, it becomes a base, since the resultant ion

could react with water as a base. The acid and base are referred to as a conjugate

acid-base pair. The ionization constants for the acid-base pair are related to Kw,

and at 25°C, pKa + pKb = 14. This relation can be used to reason qualitatively

about the relative strengths of conjugate acids and bases. For example, the

conjugate base of a strong acid is a much weaker base than H2O, and therefore

does not react as a base in aqueous solutions.

h. The pH of an acid solution depends on both the strength of the acid and the

concentration of the acid. If we compare solutions of a weak acid and of a strong

acid at the same pH, we find that both solutions have the same concentration of

H3O+ (aq). However, the strong acid is completely dissociated into ions in

solution, whereas the weak acid is only partially dissociated into ions in solution.

Thus, there are vastly more un-ionized acid molecules in the weak acid solution

than in the strong acid solution at the same pH. Thus, to achieve solutions of equal

pH, the weak acid solution must be a much greater concentration than the strong

acid solution. If we compare solutions of a weak acid and of a strong acid of the

same initial concentration, the concentration of H3O+ in the strong acid solution is

much larger (and the pH thus lower) since the strong acid is 100 percent ionized.

i. Reactions of acids and bases are called neutralization reactions, and these

reactions generally have K >1, and thus can be considered to go to completion.

i. For a mixture of a strong acid with a strong base, the neutralization

reaction is H3O+ + OH– → H2O. The K for this reaction is 10

14 at 25°C, so

the reaction goes to completion. This allows the pH of mixtures of strong

acids and bases to be determined from the limiting reactant, either the acid


or the base.

ii. When a strong base is added to a solution of a weak acid, a neutralization

reaction occurs: conjugate acid + OH- → conjugate base + H2O.

iii. When a strong acid is added to a solution of a weak base, a neutralization

reaction occurs: conjugate base + H3O+ → conjugate acid + H2O.

j. For a weak acid solution and a strong acid solution with the same pH, it takes

much more base to neutralize the weak acid solution because the initial acid

concentration is much larger. The weak acid solution contains a large amount of

un-ionized acid molecules. Therefore, a weak acid solution resists changes in pH

for a much greater amount of added base.

k. A titration technique exists for neutralization reactions. At the equivalence

point, the moles of titrant and the moles of titrate are present in stoichiometric

proportions. In the vicinity of the equivalence point, the pH rapidly changes. This

can be used to determine the concentration of the titrant.

l. As base is added to either a strong acid solution or a weak acid solution, the

H3O+ (aq) concentration does not change much. The change in pH is less than ~1.5

for the region where 10 to 90 percent of the base needed to reach the equivalence

point has been added.

m. The pKa of an acid can be determined from the pH at the half equivalence point of

the titration if the equivalence point is known (i.e., the concentration of both the

titrant and analyte are known).

n. For polyprotic acids, the use of titration curves to evaluate the number of labile

protons is important, as well as knowing which species are present in large

concentrations at any region along the curve.

Numerical computation of the concentration of each species present in the

titration curve for polyprotic acids is beyond the scope of this course and the AP

Exam.

o. Halfway to the equivalence point, the contents of a solution, formed by titrating

a weak acid, is different from that formed by titrating a strong acid. For a strong

acid, the main species in a solution halfway to the equivalence point are H3O+(aq),

the anion from the acid (e.g., Cl–, NO3–), and the cation from the base (e.g., Na

+).

The total positive charge is equal to the total negative charge. For a weak acid,

the main species in a solution halfway to the equivalence point are H3O+(aq), the

anion from the acid (e.g., CH3COO–, F–), the cation from the base (e.g., Na+), and

undissociated acid, HA. The total positive charge is equal to the total negative

charge, and [HA] = [A-].

Aligns with text sections 16.3, 16.4 and 16.5

Essential Knowledge (EK) 6.C.2: the ph is an important characteristic of aqueous solutions that can be controlled with buffers. Comparing ph to pKa allows one to

determine the protonation state of a molecule with a labile proton.


The pH of an aqueous solution is determined by the identity and concentration of the

substance that is dissolved in water. The value of the pH is an important feature of the

solution because it characterizes the relative tendency of the solution to accept a proton

from an acid added to the solution, or to donate a proton to a base that is added. For acid-

base systems, pH characterizes the relative availability of protons, much as temperature

characterizes the relative availability of kinetic energy in the environment. It is often

desirable to use a solution as an environment that maintains a relatively constant pH so

that the addition of an acid or base does not change the pH (e.g., amino acids and proteins

in the body — the blood maintains a relatively constant pH).

a. A buffer solution contains a large concentration of both members in a conjugate

acid-base pair. The conjugate acid reacts with added base and the conjugate

base reacts with added acid. The pH of the buffer is related to the pKa and the

concentration ratio of acid and base forms. The buffer capacity is related to

absolute concentrations of the acid and base forms. These relationships can be

used both quantitatively and qualitatively to reason about issues such as the ratio

of acid to base forms in a given buffer, the impact of this on the buffer capacity

for added acid or base, and the choice of an appropriate conjugate acid-base pair

for a desired buffer pH (including polyprotic acids).

Computing the change in pH resulting from the addition of an acid or a base to a buffer is beyond the scope of this course and the AP Exam.

The production of the Henderson-Hasselbalch equation by algebraic manipulation of the relevant equilibrium constant expression is beyond the scope of this course and the AP Exam.

b. If [A–]/[HA] starts as 1, it is not until the ratio changes by a factor of 10 that a 1 pH

unit change occurs; adding small amounts of either acid or base does not change

the ratio much, so the pH changes are much smaller for buffers than unbuffered

solutions.

c. Weak acids and their conjugate bases make good buffers. Strong acids and bases

do not. It takes much more base to change the pH of a weak acid solution because

there is a large reservoir of undissociated weak acid.

d. By comparing the pH of a solution to the pKa of any acid in the solution, the

concentration ratio between the acid and base forms of that acid (the protonation

state) can be determined. For example, if pH < pKa, the acid form has a higher

concentration than the base form. If pH > pKa, the base form has a higher

concentration than the acid form. Applications of this relationship include the

use of acid-base indicators, the protonation state of protein side chains (including

acids or proteins with multiple labile protons), and the pH required for acid-

catalyzed reactions in organic chemistry.

Aligns with text section 16.3 Essential Knowledge (EK) 6.C.3: the solubility of a substance can be understood in terms of chemical equilibrium

a. The dissolution of a substance in a solvent is a reversible reaction, and so has an

associated equilibrium constant. For dissolution of a salt, the reaction quotient,

Q, is referred to as the solubility product, and the equilibrium constant for this

reaction is denoted as Ksp, the solubility-product constant.

b. The solubility of a substance can be calculated from the Ksp for the dissolution


reaction. This relation can also be used to reason qualitatively about the relative

solubility of different substances.

c. The free energy change (ΔG°) for dissolution of a substance reflects both the

breaking of the forces that hold the solid together and the interaction of the

dissolved species with the solvent. In addition, entropic effects must be considered.

Qualitative reasoning regarding solubility requires consideration of all of these

contributions to the free energy.

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

Memorization of other “solubility rules” is beyond the scope of this course and the AP Exam.

e. A salt is less soluble in a solution that has an ion in common with the salt.

This has important consequences for solubility of salts in sea water and other

natural bodies of water. This phenomenon can be understood qualitatively using

LeChatelier‘s principle.

f. The solubility of a salt will be pH sensitive when one of the ions is an acid or base.

Applications include the iron hydroxides of acid-mine drainage and the effects of

acid rain on solubility of carbonates. These effects can be understood qualitatively

with LeChatelier‘s principle.

Computations of solubility as a function of pH are beyond the scope of this course and the AP Exam.

Computations of solubility in such solutions are beyond the scope of this course and the AP Exam. Aligns with text section 16.6, 16.7, 16.8, 16.9 and 16.11

ALIGNMENT OF LEARNING OBJECTIVE WITH CURRICULUM FRAMEWORK

LO 1.20 The student can design, and/or interpret data from, an experiment that uses

titration to determine the concentration of an analyte in a solution.

[See SP 4.2, 5.1] Aligns with text sections 16.4 and 16.5

LO 6.1 The student is able to, given a set of experimental observations regarding physical,

chemical, biological, or environmental processes that are reversible, construct an explanation

that connects the observations to the reversibility of the underlying chemical reactions or

processes. [See SP 6.2] Aligns with text section 16.1

LO 6.2 The student can, given a manipulation of a chemical reaction or set of reactions

(e.g., reversal of reaction or addition of two reactions), determine the effects of that

manipulation on Q or K. [See SP 2.2] Aligns with text section 16.2

LO 6.4 The student can, given a set of initial conditions (concentrations or partial

pressures) and the equilibrium constant, K, use the tendency of Q to approach

K to predict and justify the prediction as to whether the reaction will proceed toward

products or reactants as equilibrium is approached. [See SP 2.2, 6.4] Aligns with text

section 16.2

LO 6.8 The student is able to use LeChatelier‘s principle to predict the direction of the

shift resulting from various possible stresses on a system at chemical equilibrium. [See SP


1.4, 6.4] Aligns with text section 16.2

LO 6.9 The student is able to use LeChatelier‘s principle to design a set of conditions

that will optimize a desired outcome, such as product yield. [See SP 4.2] Aligns

with text section 16.2

LO 6.13 The student can interpret titration data for monoprotic or polyprotic acids

involving titration of a weak or strong acid by a strong base (or a weak or strong base by a

strong acid) to determine the concentration of the titrant and the pKa for a weak acid, or

the pKb for a weak base. [See SP 5.1, connects to 1.E.2] Aligns with text sections 16.13

and 16.5

LO 6.17 The student can, given an arbitrary mixture of weak and strong acids and bases

(including polyprotic systems), determine which species will react strongly with one

another (i.e., with K >1) and what species will be present in large concentrations at

equilibrium. [See SP 6.4] Aligns with text section 16.3

LO 6.18 The student can design a buffer solution with a target pH and buffer capacity by

selecting an appropriate conjugate acid-base pair and estimating the concentrations needed to

achieve the desired capacity. [See SP 2.3, 4.2, 6.4] Aligns with text section 16.3

LO 6.19 The student can relate the predominant form of a chemical species involving a

labile proton (i.e., protonated/deprotonated form of a weak acid) to the pH of a solution

and the pKa associated with the labile proton. [See SP 2.3, 5.1, 6.4] Aligns with text

section 16.3

LO 6.20 The student can identify a solution as being a buffer solution and explain the

buffer mechanism in terms of the reactions that would occur on addition of acid or base.

[See SP 6.4] Aligns with text section 16.3

LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the

relevant Ksp values. [See SP 2.2, 2.3, 6.4] Aligns with text section 16.6

LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. [See SP 2.2, 2.3, 6.4] Aligns with text sections 16.6 and 16.7

LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of

factors (common ions, pH) that influence the solubility. [See SP 5.1] Aligns with text

sections 16.6, 16.7, 16.8, 16.9 and 16.11

LO 6.24 The student can analyze the enthalpic and entropic changes associated with the

dissolution of a salt, using particulate level interactions and representations. [See SP 1.4, 7.1,

connects to 5.E] Aligns with text section 16.6

Key Concepts

1. Common ion effect

2. Buffers

3. Titrations

4. Solubility equilibrium and precipitation reactions

5. Solubility

6. Effects of common ions, pH, and complex ions on solubility


Key Terms

buffer capacity Ksp

buffers molar solubility

central metal cation pKa

common ion precipitation complex ion qualitative analysis

end point saturated

equivalence point solubility

formation constant solubility product

Henderson-Hasselbalch equation supersaturated

indicator titration

ion product (Q) unsaturated

Kf

Pacing Guide This chapter should require 3-4 days depending on the extent to which the basic ideas of

equilibrium have been mastered. Make certain that the students are presented with the

conceptual background as well as a good number of problems illustrating these points.

They will gain a deeper understanding if both aspects are involved during in-class

discussions.

Applying the Key Concepts

1. The common ion effect is a manifestation of Le Châtelier‘s principle. It has a

significant effect on the pH of acid and base solutions and it is the concept underlying

buffers. It is important for students to understand common ions from an equilibrium

standpoint. Use this idea to reinforce equilibrium and Ka values. Give them examples

to work out involving the effects on pH. This will lead to the idea of pKa and its

relationship to pH. Show how the Henderson-Hasselbalch equation is derived from

this equation, making sure they can algebraically manipulate the log portion of an

equation. The Henderson-Hasselbalch equation is on the equation tables given with

the AP exam and problems involving this concept are often used in conjunction with

making and/or analyzing buffer solutions.

A good demonstration that you can use to introduce this concept and also use for the

concept of solubility later on in the chapter is to add concentrated HCl to a saturated

solution (supernatant portion only) of sodium chloride.

2. Introduce buffers by emphasizing their importance to biological as well as chemical

systems. The buffering of blood is very interesting and can be used to begin this

section and when buffer capacity is discussed later on. The textbook includes a nice


summary of this system and much more, including illnesses. Syndromes caused by

unbalanced blood pH can be found on the internet.

Make sure that students understand that buffers resist changes in pH, NOT that they

keep pH at a particular value. You can develop this logic by comparing the pH

changes that accompany the addition of HCl to pure deionized water to the changes

accompanying the addition of HCl to a buffer solution set to 7.0. The two key points

would be that (1) pH does change in the buffer but (2) much more slowly. Another

key point is that buffered solutions resist pH changes if a base or an acid are added to

it. You can bring this point to their attention by asking why a weak acid is not a

buffer since the addition of the base would use the hydronium ions present to form

water, thus removing them from solution. Explain that these ions would then be

replaced by a shift in equilibrium to the right producing more H3O+

eventually back

to a concentration roughly equivalent to the [H3O+] before the base was added. This

should help them understand the necessity of adding the anion of the acid in the form

of salt containing the common ion to resist changes in pH when an acid is added to

the solution.

Give the students practice with determining pH changes in buffered solutions since

this is a common type of problem on the AP exam and it is a good way to review

equilibrium.

Another common question on acid-base equilibrium involves preparing buffered

solutions and, less frequently, discussing buffer capacity. Students should understand

the relationship between pKa and the desired pH as well as the importance of the

[conjugate base]/[weak acid] ratio. Emphasizing the ratio along with the actual

concentrations will help them prepare buffers that have a particular pH with a

reasonable buffer capacity.

3. Titrations are one of the most often used methods for analyzing concentrations. Be

sure to review the uses of titration for redox reactions as well as acid-base reactions

that were first introduced in chapter 4. They are expected to be able to explain and/or

draw titration curves for strong acid-strong base, strong acid-weak base, and weak

acid-strong base titrations as well as diprotic acids. All these types of curves can be

found on the AP exam at one time or another. Students must be absolutely clear what

is meant by the terms ‗end-point‘ and ‗equivalence point‘ and not mistake one for the

other. Relate the different curves to the last chapter and the hydrolysis of water due to

the conjugate ions of weak acids and bases. Show the rationale behind the fact that

the pH=pKa at half the equivalence point of a weak acid. Perform at least two labs

involving titration so that students understand the physical manipulation that

accompanies this method of analysis since this is often part of the laboratory question

on the AP exam. Discuss the weak acid nature of most indicators and use this to

review equilibrium as well as the role of electron configuration in producing color.

4. When introducing solubility equilibrium, be certain to emphasize that this is still

equilibrium and all the concepts that have been presented so far will be used here as


well. These equilibria calculations are simpler than most, since the use of pure water

and a solid as reactants leaves only the aqueous ions present in the equilibrium

expression. The only variation in the equilibrium expression for these types of

problems is associated with the stoichiometry of the balanced equation. Make certain

that students know names, formulas, and charges on ions, especially polyatomic ions.

Give them problems involving the question of whether or not precipitation occurs,

since this is a common question in industry. Demonstrate the usefulness of separating

ions based on solubility equilibria and use this concept to review the reaction quotient

(Q) and how it relates to equilibrium and precipitation. Because equilibrium is always

the first question on the AP exam, it is common to see equilibrium involving slightly

soluble salts.

5. Students need to be familiar with the use of molar solubility and solubility using g/L

or g/100 g H2O. They should be able to calculate these values as well as explain the

meaning of their numeric value. Review the rules for solubility of ions stressing the

relationship between charge and solubility. Stress how the Ksp value can be used to

determine solubility and vice-versa. Emphasize that solubilities are given for

saturated solutions and demonstrate how a saturated solution is produced. These

concepts are found in the free response as well as the multiple-choice section of the

AP exam.

6. The effect of common ions and pH are changes in equilibrium using Le

Châtelier‘s principle in much the same way as in the previous chapters. Both these

factors normally involve adding or removing a product ion so that the equilibrium

shifts to the reactant (left) or product (right) side of the equation. Complex ions are a

little different in that they tend to ―remove‖ ions or molecules by making complexes

causing a shift in equilibrium to the right and ‗dissolving‖ the precipitate. Note that

these reactions demonstrate another specific case of equilibrium as evidenced by the

use of a new constant, the formation constant, (Kf). In cases involving the formation

of complex ions make sure that the students realize that complex ions (1) have ions or

molecules bound to a central metallic ion, (2) these species are no longer considered

to be ―in solution‖ and thus are effectively removed, and (3) that complex ions are

generally soluble. The disappearance of a precipitate through the formation of a

complex ion is an excellent way to review and emphasize the dynamic nature of

chemical equilibrium. Activity 2 addresses this concept.

Complex ions are one of the most illuminating examples of Lewis acid-base theory.

Emphasize this while reviewing Lewis structures so that the donation and acceptance

of electron pairs can be readily observed. Common ion and pH problems are more

often found on the free-response section of the AP exam. However, the formation of

complex ions is sometimes found on question four (the reaction question) and in the

multiple choice section of the exam.


Activities

Activity 1:

This activity explores the concept of buffer capacity. Additionally it reviews the

Henderson-Hasselbalch equation, titrations, pH, and acid-base indicators.

Procedure:

1. Prepare a buffer solution that contains 2 mL each of 1M CH3COOH and 1M

NaC2H3O2.

2. Prepare another buffer solution that contains 2 mL each of 0.1M CH3COOH and

0.1M NaC2H3O2.

3. Use a reference table to find the Ka value for CH3COOH. Record the value. 4. Use the Henderson-Hasselbalch equation to calculate the pH for each of your buffers.

Show your work.

5. Place 2 mL of each solution into two small test tubes along with 2 drops of universal

indicator. Record the initial color and the pH indicated by the. Also record this as the

initial pH in Table 1.

6. To each test tube, add 6 M HCl drop by drop. Count the number of drops needed to

cause a color change in each buffer solution that indicates a change of one pH unit

(e.g. a change from pH 5 to pH 4). Record the number of drops and the resulting pH

indicated by the color in Table 1. Stop adding drops when each test tube has reached

a pH of 1.

Activity 2:

This activity illustrates the formation of complex ions on the solubility of precipitates.

Additionally it reviews Le Châtelier‘s principle, differences between d and s sublevels,

and net ionic equations

Introduction:

The Ksp value for zinc hydroxide {Zn(OH)2} is 5 x 10

-17. This small number indicates a

very low solubility and we would correctly assume that zinc hydroxide would precipitate

out of solution at very low concentrations. The equation showing this reaction can be

expressed as follows:

Zn(OH)2(s) ↔ Zn

2+(aq) + 2 OH

-(aq) Ksp = 5 x 10

-17

However, Le Châtelier‘s principle indicates that if we could remove either the Zn2+

or

OH- (or both) from solution we could get more of the Zn(OH)2(s) to dissolve.

In this experiment, we will form precipitates and then determine if we can shift the

equilibrium to the right in order to get the precipitate to dissolve.


Procedure:

Note: Some experiments may need 20 or more drops to see a change. If you don’t

notice a change after 30 drops have been added, you may stop the experiment.

Step 1 Record your observations in Table 1.

Step 2 To each of three small test tubes add about 2 mL of 0.1 M Zn(NO3)2. To each test

tube add one drop of 6 M NaOH and stir [CAUTION: 6 M NaOH is dangerous to

human tissue]. Record your observations in Table 1.

Step 3 To test tube 1 add 6 M HCl drop by drop with stirring until you notice a definite

change in appearance [CAUTION: 6M HCl is dangerous to human tissue!]. To

test tube 2 add 6M NaOH, again drop by drop with stirring. To test tube 3, add 6M

NH3 in the same manner as before. Note what happens in each case and record

your observations in Table 1.

Step 4 Repeat steps 2 and 3 for test tubes 4-6 using a solution of 0.1 M Mg(NO3)2. Record your observations in Table 1.

Waste: The products of these reactions may be poured down the sink.

Have students record observations in Table 1 and complete the questions on the

worksheet found at the end of this chapter of the teacher‘s manual.

Common Mistakes and Misconceptions • Students often interpret the term ―resists changes in pH‖ to mean that buffered

solutions do not change pH when an acid or base is added to them. Do several

problems that illustrate that pH change does occur, but only very little until the buffer

is exhausted. Use weak acid–strong base or strong acid–weak base titration curves to

illustrate the slow initial change due to the shift in equilibrium. Relate this to buffer

activity.

• It is common for students to use the terms end point and equivalence point

incorrectly, often substituting one for the other. Avoid using end point as much as

possible and stress the idea behind the equivalence point.

• Many students see the list of special equilibrium cases as separate topics and fail to

realize that the basic tenets of chemical equilibrium initially discussed apply to all these cases. Emphasize that ―equilibrium is equilibrium‖ and that Ka, Kb, Kw, Kf, and

Ksp are simply equilibrium constants with letters to indicate when they are to be used.


Math Skills

• Computations involving the Henderson-Hasselbalch equation

• Determining pH and pKa values from concentrations of reactants and products

• Determining Ksp value and using them to determine concentrations of ions, solubility,

and whether precipitation will occur


Practice Questions Multiple-Choice:

1. Several drops of concentrated HCl (12M) are added to a test tube containing an

aqueous saturated solution of NaCl. Which of the following best describes what you

would observe and the reasoning behind the observation?

(A) A white precipitate is formed because sodium hydride is insoluble

(B) A white precipitate is formed because a common ion shifted the equilibrium to

form solid sodium chloride

(C) Bubbles are formed due to the production of hydrogen gas

(D) No change is observed because sodium chloride is a neutral salt

2. The ionization constant for phenol is 1.3 x 10

-10. The pKa and the pH will both equal

10 when

(A) The phenol is dissolved in a basic solution

(B) The phenol is dissolved in an acidic solution

(C) The concentration of conjugate base equals the concentration of acid [conjugate

base] = [acid]

(D) The pKa will never equal the pH

3. Which of the following combinations constitutes a buffer?

I. A weak acid and a weak base

II. A weak acid and its salt

III. A weak base and its salt

(A) I only

(B) II only

(C) I and II

(D) II and III

4. An amount of acid is added to a buffer solution. Which of the following represents

the most likely observation?

(A) The pH doesn‘t change

(B) The pH is slightly lowered

(C) The pH is slightly raised

(D) Bubbles form


5. A titration is performed by adding a strong base to neutralize an unknown weak acid.

Which of the following statements is most likely true regarding this titration?

(A) The color of the solution will change from clear to pink at the end point

(B) The color of the solution will change from pink to clear at the end point

(C) The pH of the resulting solution will be 7.0 at its equivalence point

(D) The pH of the resulting solution will be 9.1 at its equivalence point

6. A titration is performed involving the addition of a strong acid to an unknown strong

base. Which of the following statements regarding the change in pH is true?

(A) The pH changes rapidly from pH 1 to pH 2

(B) The pH changes rapidly from pH 3 to pH 7

(C) The pH changes rapidly from pH 7 to pH 11

(D) The pH changes rapidly from pH 11 to pH 3

7. A titration is performed involving the addition of a certain volume of a strong acid

that has a concentration of 1.0 M to 25.0 mL of a weak base that has a concentration

of 2.0 M. Which of the following statements regarding the volume of acid needed to

reach the equivalence point is most likely true?

(A) 12.5 mL of acid will be needed because the acid is a strong acid

(B) 12.5 mL of acid will be needed because the pH at the equivalence point of a

strong acid and weak base is greater than 7.0.

(C) 25.0 mL of acid will be needed because the equivalence point is where the two

volumes are equal

(D) 50.0 mL of acid will be needed because the concentration of acid is half the

concentration of the base

8. Which of the following best describes the end point in a titration?

(A) The end point is the point where the indicator changes color (B) The end point is the point where the pH = 7.0

(C) The end point is the point where all the H3O+

has reacted with the OH- to form

water

(D) The end point is the point where the volume of acid equals the volume of base

9. Lead(II) fluoride is a slightly soluble salt with a Ksp = 4.1 x 10

-8. Which of the

following expressions represents the concentration of the aqueous fluoride ion in

equilibrium with solid lead(II) fluoride?

(A) (4.1 x 10-8

)1/2

(B) (4.1 x 10-8

)1/2

4

(C) (4.1 x 10-8

)1/3

2

(D) (4.1 x 10-8

)1/3

4

10. Which of the following solutions will form a precipitate?


I. Q < Ksp

II. Q = Ksp

III. Q > Ksp

(A) I only

(B) II only

(C) III only

(D) II and III

Free Response:

1. Calcium hydroxide has many practical uses as indicated by its many different names

such as slaked lime, hydrated lime, builders lime, and pickling lime to name a few. It

is a slightly soluble material with a Ksp = 5.5 x 10-5

.

(a) Give a balanced chemical equation that shows an aqueous solution of calcium

hydroxide in equilibrium with its solid form

(b) Give Ksp expression for an aqueous solution of calcium hydroxide in equilibrium

with its solid form

(c) 50.0 mL of a 0.0050 M solution of calcium nitrate is mixed with 50.0 mL of a

0.50M solution of sodium hydroxide.

(i) Does precipitation of calcium hydroxide occur in the resulting solution?

Defend your answer with a calculation and reasoning.

(ii) What is the pH of the resulting solution?

(d) A different solution contains a saturated solution of calcium hydroxide in

equilibrium with its solid form. Describe one method that you could use that

would get the sodium hydroxide to dissolve. Explain your reasoning.


Answers to the Practice Questions Multiple-Choice:

1. (B)

2. (C)

3. (D)

4. (B)

5. (D)

6. (D)

7. (D)

8. (A)

9. (C)

10. (C)

Free Response:

1. (a) Ca(OH)2 ↔ Ca2+

+ 2 OH-

[One point is given for a correct balanced equation. NOTE: the double arrow must

be indicated but states of matter do NOT.]

(b) Ksp = [Ca2+

] [OH-]2

[One point given for a completely correct expression]

(c)

(i) Since calcium nitrate is soluble 50.0 mL will contain 2.5 x 10-4

mol Ca2+

Since sodium nitrate is soluble 50.0 mL will contain 0.025 mol OH

-

[Ca2+] = 2.5 x 10-4

mol / .100 L = 2.5 x 10-3

M

[OH-] = .025 mol / .100 L = .25M

Ksp = [Ca2+

][OH-] = [.0025][.25]

2 = 1.56 x 10

-4

Yes, the product of the ion concentrations is greater than the Ksp value, so

precipitation will occur.

(ii) Since the Ca

2+ is the limiting reactant, all of it will be precipitated and will

require 2 (2.5 x 10-4

) mol of OH- = 5.0 x 10

-4 mol of OH

- used.

The number of moles of OH- left will be .025 mol – 5 x 10-4

mol = .0245 mol

[OH-] = .0245 mol / .100 L = .245 M


pOH = - log[OH-] = - log [.245] = 0.61

pH = 14- pOH = 14 – 0.61 = 13.39

[Two points for correctly calculating the pH from any answer given in ‗i‘. One

point is given if the process is essentially correct, but a math error is made.

One point is given for the correct calculation of pH from an incorrect [OH-].]

(d) You can get the precipitate to dissolve by effectively reducing the concentration

of Ca2+

and/or OH- to the point where [Ca

2+][OH

-] < 5.5 x 10

-5. This can be

accomplished in a number of ways such as

• Diluting the solution with water so that all concentrations decrease

• Adding H+

ions to react with OH- to form water. This removes the OH

-

from solution

• Forming a soluble complex such as Al(OH)4-, thus ―removing‖ the ions

from solution

[One point is given for correctly describing any method that lowers the product of

concentrations below the Ksp value.]


Worksheet 16.1

AP Chemistry

Activity 1

Procedure Step 3: Ka value for CH3COOH:

Ka =

Procedure Step 4: pH calculation of buffers using Henderson-Hasselbalch equation:

Calculated pH for buffer 1: Calculated pH for buffer 2:

Procedure Step 5: Record initial color and pH indicated by color

Indicator pH for buffer 1: Indicator pH for buffer 2:

Table 1

First Color Change

(# drops- pH)

Second Color Change

(# drops- pH)

Third Color Change

(# drops- pH)

Fourth Color Change

(# drops- pH)

Fifth Color Change

(# drops- pH)

Buffer 1 Initial pH=

Buffer 2 Initial pH=

Questions:

1. Was there a difference between the calculated pH and the pH specified by the

indicator color? Explain any difference.


2. What is the relationship between the pH and the pKa for the initial buffer solutions? Is

it the same for both buffer solutions? Explain this relationship.

3. Was there a linear relationship between the color change and the number of drops

needed to cause the change throughout the entire experiment? In other words, was the

same number of drops always required to change the color? Explain your

observations in light of what you know about buffers.

4. Was the same number of drops needed to cause a change of one pH unit in both

buffers? Explain your observations in terms of what you know about buffer capacity.

5. The buffer capacity of the oceans has been reduced by the addition of enormous

amounts of CO2 due to the burning of fossil fuels. CO2 combines with water to form

carbonic acid. Use the results of this experiment to discuss how fast the pH of the

oceans will change in the future compared to how fast they have been changing in the

past.


Worksheet 16.2

AP Chemistry

Activity 2

Table 1

Test Tube Initial Solution

+ 6M NaOH + 6M HCl + 6M NaOH

(2nd

addition) + 6M NH3

1 Zn(NO3)2

2 Zn(NO3)2

3 Zn(NO3)2

4 Mg(NO3)2

5 Mg(NO3)2

6 Mg(NO3)2

Questions:

1. Give net ionic equations for any reactions that you observed.

2. Use Le Châtelier‘s principle and your equations from question 1 to explain your

observations when HCl is added to the test tube containing Zn(OH)2.

3. What is a complex ion?


4. Use Le Châtalier‘s principle and the formation of complex ions to explain your

observations when excess NaOH and excess NH3 are added to the test tubes with

Zn(OH)2.

5. In step 4, you probably found that the reactions involving Mg(OH)2 were similar in

some ways to the behavior of Zn(OH)2, but different in others.

a. How was it similar? Explain that similarity.

b. How was it different? Explain the differences.

6. Give possible reasons for your observations between Zn(OH)2 and Mg(OH)2 in terms of the

electron structure of the Zn2+

and Mg2+

cations.

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