Advanced Placement Chemistry
Chapters 14 – 16 Syllabus
As you work through the chapter, you should be able to:
Chapter 14 – Acids and Bases
1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and Lewis models for acids and bases.
2. Understand the difference between strong and weak acids or bases in terms of percent dissociation.
3. Determine conjugate acid-base pairs for an acid base reaction.
4. Write an equation for an acid or base dissociation constant.
5. Explain how polyprotic acids dissociate.
6. Explain what the term amphoteric means.
7. Know the relationship between Ka, Kb, and Kw as defined by the autoionization of water.
8. Describe the relationship between [OH-] and [H+] in an acidic, basic, and neutral solution.
9. Calculate pH and pOH for strong and weak acids and bases.
10. Calculate Ka and Kb values.
11. Calculate percent dissociation of a species in an acidic or basic solution.
12. Explain why different salts will produce acidic, basic, or neutral solutions
13. Explain the effect of structure on acid-base properties.
Chapter 15 – Acid-Base Equilibria
1. Determine the equilibrium position of a solution involving a common ion.
2. Describe the components of a buffered solution.
3. Explain how a buffered solution works and how to prepare a buffered solution.
4. Determine the buffer capacity of a buffered solution.
5. Use the Henderson-Hasselbalch to determine the pH of a buffered solution.
6. Determine the equilibrium position of a buffered solution.
7. Monitor the progress of an acid-base titration using a titration curve.
8. Explain what occurs at the stoichiometric point or equivalence point.
9. Determine the equilibrium position at various points during an acid base titration.
10. Choose an appropriate indicator for a particular acid base titration.
Chapter 16 – Solubility Equilibria
1. Use the solubility product to determine the equilibrium position of an aqueous solution.
2. Use the relationship between the ion product and solubility product to determine whether or not a precipitate
will form.
3. Use the relationship between the ion product and solubility product to develop a scheme to qualitatively
analyze a solution.
Lab Experience:
1. Determination of the Ka for a weak acid
2. Preparation and properties of buffered solutions
Assignments:
HW1: Ch 14 Handout
HW2: Ch 14 Text Q’s
HW3: Ch 14 Quiz w/ notes check
HW4: Ch 15 Handout
HW5: Ch 15 Text Q’s
HW6: Ch 15 Quiz w/ notes check
HW7: Ch 16 Handout
HW8: Ch 16 Text Q’s
HW9: Ch 16 Quiz w/ notes check
Week of February 23th
Day
Concepts
M
1-6
1st
CH 14 Introduction
Block
Goals 1 – 8
2nd
Goals 1-13
Block
F
Goal 9 - 13
1–6
Week of March 2nd
Day
Concepts
M
Goals 1 – 6
1–6
1st
Goals 1 – 13
Block
R
Goals 7 – 9
1–6
F
Goal 10
1–6
Week of March 9th
Day
Concepts
M
Lab: Preparation and
1 – 6 Properties of Buffered
Solutions
1st
Block
2nd
Block
F
1–6
-Bring Text Books
-Notes: CH 14.8 – 14.12
Class Activities
-Notes CH 15.1 – 15.3
-Common Ion Effect Activity
-Notes: CH 15.4
-Notes: CH 15.5
Class Activities
-Preparation and Properties of Buffered
Solutions:
• Pre-lab
-Preparation and Properties of Buffer
Solutions:
• Collect Data
Ch 15 Quiz and Note Check
CH 16 Introduction
Goals 1 – 3
-Notes: CH 16.1 – 16.2
CH 17 Introduction
Homework
-Notes: CH 14.1 – 14.7
Lab: Preparation and
Properties of Buffered
Solutions
Goals 1 – 10
Week of March 16th
Day
Concepts
M
Goals 1 – 3
1–6
1st
CH 14 – 15 Review
Block
2nd
CH 14 – 16 Summative
Block Assessment
F
1–6
Class Activities
-Unit 5 Test Day 2
Class Activities
- Review time and Activity
- Ch 16 Quiz and Note Check
-CH 14 – 16 Examination
-Pass out Unit 7 Materials
-Ch 14 – 16 Examination Day 2
Homework
Due:
HW1
Due:
HW2
Due:
HW3
Due:
Common Ion Activity
Homework
Due:
HW4
Due:
HW5
Due:
HW6
Due:
HW7
Homework
Due:
HW8
Due:
HW9
Due:
Preparation and Properties of
Buffer Solutions
Chapter 14 – Acids and Bases
The Nature of Acids and Bases
• Acids – sour, burn
• Bases – Bitter, slippery
1. Arrhenius Concepts:
– Acid – produces H+ cation in aqueous solution
– Base – produces OH- anion in aqueous solution
2. Brønsted-Lowry Model:
– Acid – proton donor
– Base – proton acceptor
Brønsted-Lowry Example
• HCl + H2O H3O+ + Cl-, where H3O+ is called the hydronium ion
• The general reaction that occurs when an acid dissolves in water:
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
acid
base
conjugate
conjugate
acid
base
• A conjugate acid is formed when the proton is transferred to the base.
• A conjugate base is everything that remains of an acid molecule once the proton is lost
• A conjugate acid-base pair consists of two substances related to each other by the donating and accepting of
protons.
Acid-Base Equilibrium
• For the reaction:
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
Ka = ([H3O+][A-])/[HA] = ([H+][A-])/[HA]
where Ka is called the acid dissociation constant. H+ and H3O+ both represent the hydronium ion. Water is left
out of the expression because its concentration remains relatively constant.
Acid Strength
• Strong acid:
– Equilibrium lies far to the right
– Almost all of the original HA has dissociated
– Strong acids yield weak conjugate bases that have a low affinity for protons (conjugate base is a much
weaker base than water)
• Weak acid:
– Equilibrium lies far to the left
– Most of the original acid still exists as HA at equilibrium, meaning it dissociates to a very small extent
– Weak acids yield conjugate bases that are much stronger than water
Various Ways to Describe an Acid
Property
Strong Acid
Weak Acid
Ka value
Ka is large
Ka is small
Position of the dissociation at
equilibrium
Far to the right
Far to the left
Equilibrium concentration of H+
compared with the original
concentration of HA
[H+] ≈ [HA]0
[H]+ <<< [HA]0
Strength of conjugate base
compared with that of water
A- much weaker than H2O
A- much stronger than H2O
Common Strong Acids
• Sulfuric (H2SO4), hydrochloric (HCl), nitric (HNO3), perchloric (HClO4)
• Can’t calculate Ka values for strong acids because [HA] is so small.
Diprotic Acids
• An acid having two acidic protons.
• H2SO4:
H2SO4(aq) H+(aq) + HSO4-(aq) (strong)
HSO4-(aq) ↔ H+(aq) + SO42-(aq) (weak)
Oxyacids
• Acidic proton is attached to an oxygen atom.
• Organic acids – carbon backbone and contain a carboxyl group (COOH)
Water as an Acid and a Base
• Amphoteric – behaves as an acid or a base
• Autoionization of water:
2H2O(l) ↔ H3O+(aq) + OH-(aq)
Kw = [H3O+][OH-], where Kw is the ion-product constant.
• At 25oC [H+] = [OH-] = 1.0x10-7 M
• Kw = [H3O+][OH-] = (1.0x10-7 M)(1.0x10-7 M)
Kw = 1.0x10-14 (increases slightly with temperature)
Significance of Kw
• In any aqueous solution at 25oC, no matter what it contains, the product [H+] and [OH-] must equal 1.0x10-14.
• Three results follow:
1. A neutral solution, where [H+] = [OH-]
2. An acidic solution, where [H+] > [OH-]
3. A basic solution, where [H+] < [OH-]
The pH Scale
• Aqueous solutions can have very small H+ concentrations.
• pH is measured using a logarithmic scale to conveniently represent solution acidity.
• pH = -log[H+]
Significant Figures for Logarithms:
• The number of decimal places in the log is equal to the number of significant figures in the original number
pH continued
• Since pH is a log scale, the pH changes by one for every power of ten change in [H+].
• The pH decreases as [H+] increases.
• pH can be measured using a probe that can measure potential differences between a test solution and the
solution in the probe.
Relationship Between pH and pOH
• Kw = [H+][OH-]
log Kw = log[H+] + log[OH-]
-log Kw = -log[H+] - log[OH-]
pKw = pH + pOH
Since pKw = -log(1.0x10-14) = 14.00
• pH + pOH = 14.00
pH Example
• The pH of a sample was measured to be 8.23 at 25oC. Calculate the pOH, [H+], and [OH-] for the sample.
Calculating the pH of Strong Acid Solutions
• Determine what is the major species providing H+ in solution.
• If 1.0 M HCl is allowed to dissociate in water,
HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)
• Two species provide H+ in this example HCl and H2O through autoionization. However, HCl will dissociate into
~1.0 M H+ and H2O will only make ~1.0x10-7 M H+. This means the contribution from H2O can be neglected.
• pH = -log[H+] = -log(1.0) = 0
Calculating the pH of Weak Acid Solutions
• List the major species in the solution.
• Choose the species that can produce H+, and write balanced equations for the reactions producing H+.
• Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will
dominate in producing H+.
• Write the equilibrium expression for the dominant equilibrium.
• List the initial concentrations of the species participating in the dominant equilibrium.
• Define the change needed to achieve equilibrium; that is, define x.
• Write the equilibrium concentrations in terms of x.
• Substitute the equilibrium concentrations into the equilibrium expression.
• Solve for x the “easy” way; that is, by assuming [HA]0 – x ≈ [HA]0.
• Use the 5% rule to verify whether the approximation is valid.
• Calculate [H+] and pH.
pH of a Weak Acid Example
• What is the pH of a 1.00 M solution of HNO2 at equilibrium (Ka = 4.0x10-4)?
Percent Dissociation
• Percent Dissociation:
[(amount dissociated (M))/(initial concentration (M))]x100%
• For solutions of any weak acid HA, [H+] decreases as [HA]0 decreases, but the percent dissociation increases as
[HA]0 decreases.
Bases
• Strong Bases:
– Complete dissociation
– All group 1A, Ca, Ba, and Sr hydroxides are strong bases
– Low solubility of OH- useful in products like antacids
•
Weak Bases:
– Incomplete dissociation
– Many electron-pair donors are weak bases
Base Equilibrium
• For the reaction of a base with water:
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq)
base acid
conjugate conjugate
acid
base
Kb = ([BH+][OH-])/[B]
pH of a Strong Base
• Calculate the pH of a 2.7x10-3 M NaOH solution.
pH of a Weak Base
• Calculate the pH for a 10.0 M solution of NH3 (Kb = 1.8x10-5)
Polyprotic Acids
• Acids that can provide more than one proton.
• Examples:
– H2CO3
– H2SO4
– H3PO4
• Each proton dissociation has a corresponding Ka, where Ka1>Ka2>Ka3.
• Therefore equilibrium concentrations and pH can be determined by examining the first proton dissociation.
Acid-Base Properties of Salts
• Salts are ionic compounds that will dissolve in water.
• Under certain conditions these solutions can behave as acids or bases.
Salts That Produce Neutral Solutions
• Salts that consist of cations of strong bases and anions of strong acids have no effect on pH.
• Examples:
– KCl
– NaCl
– NaNO3
Salts That Produce Basic Solutions
• Salts whose cation has neutral properties (such as Na+ and K+) and whose anion is the conjugate base of a weak
acid, the aqueous solution will be basic.
• KaKb=Kw
• Example:
When NaC2H3O2 is dissolved in water
C2H3O2-(aq) + H2O(l) ↔ HC2H3O2(aq) + OH-(aq)
Salts That Produce Acidic Solutions
• Salts in which the anion is not a base and the cation is the conjugate acid of a weak base produce acidic
solutions.
• Salts containing highly charged metal ions will also produce acidic solutions.
• Example:
When solid NH4Cl is dissolved in water
NH4+(aq) ↔ NH3(aq) + H+(aq)
Example
• Calculate the pH of a 0.20 M NH4Cl solution. The Kb value for NH3 is 1.8x10-5.
Effects of Structure on Acid-Base Properties
• Electronegativity influences the strength of an acid.
• Acidity generally increases as the electronegativity of the atoms on the molecule increase.
• Acidity generally increases with the number of oxygen atoms attached to the central atom.
Acid-Base Properties of Oxides
• When Covalent oxides dissolve in water, an acidic solution forms.
• Example:
SO3(g) + H2O(l) H2SO4(aq)
• When ionic oxides are dissolved in water, a basic solution forms.
• Example:
CaO(s) + H2O(l) Ca(OH)2(aq)
Lewis Acid-Base Model
• Lewis Acid – electron-pair acceptor
– H+, BF3, metal cations
• Lewis Base – electron-pair donor
– OH-, H2O, NH3
Chapter 15 – Acid-Base Equilibria
Common Ion Effect
• The shift in an equilibrium caused by the addition or presence of an ion involved in the equilibrium reaction.
Solutions of Acids Containing a Common Ion
• Example:
Add NaF to a solution of HF
NaF Na+ + FHF ↔ H+ + FAccording to Le Châtelier’s principle, the additional F- in solution will drive the equilibrium position towards the
reactants. The solution will become less acidic.
Solutions of Bases Containing a Common Ion
• Example:
Add NH4Cl to a solution of NH3
NH4Cl NH4+ + ClNH3 + H2O ↔ NH4+ + OHAccording to Le Châtelier’s principle, the additional NH4+ in solution will drive the equilibrium position towards
the reactants. The solution will become less basic.
Equilibrium Calculations
• Use the same basic problem solving strategy as all equilibrium calculations. The only difference is that the salt
(anion or cation) is adding to the initial concentration of one of the products.
Common Ion Example
• The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2 M, and the percent dissociation of HF is
2.7%. Calculate the [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4) and
1.0 M NaF.
Buffered Solutions
• Resists changes in pH upon addition of an acid or a base.
• May contain:
– Weak acid and its conjugate base salt (HF and NaF)
– Weak base and its conjugate acid salt (NH3 and NH4Cl)
Solving Buffered Solution Problems
• Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on
buffered solutions require exactly the same procedures as introduced previously.
• When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the
resulting reaction first (assuming reaction with H+/OH- goes to completion). After the stoichiometric calculations
are completed, then consider the equilibrium calculations.
The pH of a Buffered Solution
• A buffered solution contains 0.50 M acetic acid (CH3COOH, Ka = 1.8x10-5) and 0.50 M sodium acetate
(NaCH3COO). Calculate the pH.
Change in pH of a Buffered Solution
• Calculate the change in pH that occurs when 0.010 mol of solid NaOH is added to 1.0 L of the buffered solution
from the previous problem. Compare this change with that which occurs when 0.010 mol of solid NaOH is
added to 1.0 L of H2O. (Hint: stoichiometry first then equilibrium)
How Buffering Works
• Buffering works by having a large excess of a weak acid and its conjugate base to minimize the effect of adding
H+ or OH- ions.
• For the reaction HA ↔ H+ + A-, the acidity of the solution, [H+] = Ka([HA]/[A-]) is dependent upon the ratio of
[HA]/[A-]. If this ratio is made so it resists change (large excess) a buffered solution can be made.
• This also applies to solutions of a weak base and a conjugate acid made to be in large excess compared to any
added H+/OH- ions.
Henderson-Hasselbalch Equation
• For a particular buffered system (acid-conjugate base pair), all solutions that have the same ratio [A-]/[HA] will
have the same pH.
• pH = pKa + log([A-]/[HA]), when using this equation one assumes that [A-] and [HA] are ≈ [A-]0 and [HA]0 .
Example
• Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4x10-4) and 0.25 M lactate.
Buffering Capacity
• Represents the amount of proton or hydroxide ions a buffered solution can absorb without a significant change
in pH.
• A solution with a large buffer capacity contains a large concentration of buffering components.
• The pH of a buffered solution is determined by the ratio
[A-]/[HA].
• The capacity of a buffered solution is determined by the magnitude of [HA] and [A-].
Optimal Buffering
• Large changes in the ratio [A-]/[HA] will produce large changes in pH.
• Optimal buffering occurs when [HA] is equal to [A-].
•
•
When this ratio is achieved:
pH = pKa + log([A-]/[HA])
pH = pKa + log(1)
pH = pKa
The pKa of a weak acid used in a buffer should be as close as possible to the desired pH.
Choosing a Buffer
• Suppose a buffered solution with a pH of 4.00 is needed.
• From the Henderson-Hasselbalch Equation:
pH = pKa + log([A-]/[HA])
4.00 = pKa + log(1)
pKa = 4.00
Ka = antilog(-4.00) = 1.0x10-4
Example
• A chemist needs a solution at a pH of 4.30 and can choose from the following acids.
– Chloroacetic acid (Ka = 1.35x10-3)
– Propanoic acid (Ka = 1.3x10-5)
– Benzoic acid (Ka = 6.4x10-5)
– Hypochlorous acid (Ka = 3.5x10-8)
•
Solution
Acid
Chloroacetic
Propanoic
Benzoic
Hypochlorous
•
•
•
•
[H+] = Ka([HA]/[A-])
5.0x10-5 = 1.35x10-3([HA]/[A-])
5.0x10-5 = 1.3x10-5 ([HA]/[A-])
5.0x10-5 = 6.4x10-5 ([HA]/[A-])
5.0x10-5 = 3.5x10-8 ([HA]/[A-])
[HA]/[A-]
3.7x10-2
3.8
0.78
1.4x103
Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30.
[H+] = 10-4.30 = antilog(-4.30) = 5.0x10-5 M
[H+] = Ka([HA]/[A-])
Benzoic acid is the best choice since the ratio of [HA]/[A] is closest to 1. A solution with a ~1:1 ratio of benzoic
acid and sodium benzoate would make an optimal buffered solution of 4.30.
Titrations and pH Curves
• This process involves a solution of known concentration (the titrant) delivered from a buret to a solution of
unknown concentration until the substance being analyzed is just consumed (moles of titrant are equal to moles
of unknown).
• The stoichiometric equivalence point is often signaled by a color change of an indicator.
• The progress of an acid-base titration can be monitored by analyzing a pH curve.
• pH curves are a plot of solution pH as a function of titrant volume.
Strong Acid-Strong Base Titrations
• Net ionic equation:
H+(aq) + OH-(aq) H2O(l)
• Often useful to use mmol (10-3 mol) instead of moles. (mmol = mLxM)
• Before the equivalence point, [H+] can be calculated by dividing the number of millimoles of H+ remaining by the
total volume of the solution in milliliters.
• A pH of 7.00 at the equivalence point is characteristic of a strong acid-strong base titration.
• After the equivalence point, [OH-] can be calculated by dividing the number of millimoles of excess OH- by the
total volume of the solution in milliliters. Then use Kw to find [H+].
• Plot pH versus titrant volume to construct a pH curve (titration curve).
Example I: Strong Acid Titrated with a Strong Base
• 50.0 mL of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate pH after the following volumes of NaOH have
been added: 0.0, 10.0, 20.0, 50.0, 100.0, 150.0 mL
Titrations of Weak Acids with Strong Bases
• Treat as a series of equilibrium problems.
• Weak acids will react with a strong base completely.
1. Determine moles of H+ from stoichiometry
2. Find equilibrium [H+] and pH
• The pH at the equivalence point is always greater than 7 due to the presence of the weak acid’s conjugate base.
• Example II: Weak Acid Titrated with a Strong Base
• 50.0 mL of 0.10 M HC2H3O2 (Ka=1.8x10-5) is titrated with 0.10 M NaOH. Calculate pH after the following volumes
of NaOH have been added:
0.0, 10.0, 25.0, 50.0, 75.0 mL(Hint: may use Henderson-Hasselbalch equation to determine pH)
Conclusions
• Amount of acid determines equivalence point, not the acid strength.
• pH value of the equivalence point is affected by acid strength.
Titrations of Weak Bases with Strong Acids
• Treat the same way as weak acid strong base.
• Before the equivalence point:
1. List major species before reaction.
2. Base will react to completion with H+.
3. Stoichiometry first.
4. Weak base equilibrium second to determine pH.
• At the equivalence point:
1. Reaction won’t go to completion.
2. Equilibrium is dominated by the weak base’s conjugate acid.
3. Solution will be acidic.
• Beyond equivalence point:
1. Excess H+ dominates equilibrium.
2. pH is determined by excess H+.
3. Acid-Base Indicators
• Two methods for determining the equivalence point of an acid-base titration:
1. Use a pH probe to monitor the change in pH as titrant is added. The center of the vertical region of the
pH curve indicates the equivalence point.
2. Use an acid-base indicator, which marks the end point of a titration by changing color. Although the
equivalence point of a titration is not necessarily the same as the end point.
Indicators
• Usually a weak organic acid (HIn)
• Example:
1. Phenolpthalein
2. Colorless in HIn form
3. Pink in In- form
Ka ≈ 1.0x10-8
HIn ↔ H+ + In• Color change occurs when ([In ]/[HIn]) ≈ (1/10)
Example: Indicators
• Bromothymol blue, an indicator with a Ka value of 1.0x10-7, is yellow in its HIn form and blue in its In- form.
Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with
NaOH, at about what pH will the indicator color change first be visible?
Chapter 16 – Solubility Equilibria
Solubility Equilibria and the Solubility Product
• When solids dissolve in water two processes occur:
• Dissolution of the solid
• Reformation of the solid
Example:
CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
• Eventually a dynamic equilibrium is reached, and the solution is saturated.
• Ksp = [Ca2+][F-]2
• Ksp is called the solubility product constant.
Solubility Examples
• Calculate the Ksp for bismuth sulfide (Bi2S3), which has a solubility of 1.0x10-15 M at 25oC.
• The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4x10-7 at 25oC. Calculate its solubility at 25oC.
Relative Solubilities
• A salt’s Ksp value provides information about its solubility.
• Relative solubilities of salts that produce the same number of ions in solution can be compared.
Salt
Ksp
AgI
1.5x10-16
CuI
5.0x10-12
CaSO4
6.1x10-5
In all cases a Salt ↔ Cation + Anion
CaSO4 > CuI > AgI
• Ksp values for salts that produce different numbers of ions (NaCl and MgCl2) cannot be directly compared.
Common Ion Effect
• The solubility of a solid is lowered if the solution already contains ions common to the solid.
pH and Solubility
• pH can greatly affect a salt’s solubility.
• If an anion, X-, is an effective base (meaning HX is a weak acid) the salt MX will show increased solubility in an
acidic solution.
Precipitation and the Ion Product
• When solutions are mixed a solid precipitate may form.
• Ion Product (Q) – defined like Ksp only using initial concentrations
• Example:
CaF2(s) ↔ Ca2+(aq) + 2F-(aq)
Q = [Ca2+]0[F-]02
• Compare Q and Ksp to predict whether or not a precipitate will form:
– If Q > Ksp, precipitation will occur.
– If Q < Ksp, no precipitation occurs.
Determining Precipitation Conditions
• A solution is prepared by adding 750.0 mL of 4.00x10-3 M Ce(NO3)3 to 300.0 mL of 2.00x10-2 M KIO3.
Will Ce(IO3)3 (Ksp = 1.9x10-10) precipitate from this solution?
Precipitation Reactions
• Determine the concentrations of the species in solution.
• Find Q to see if a precipitate will form.
• Perform stoichiometry on the initial amounts of reactants in solution (run the precipitation reaction to
completion).
• Perform an equilibrium calculation to determine the reactant concentrations at equilibrium using ICE and the Ksp
value for the particular solid.
Precipitation Reaction Example
• A solution is prepared by mixing 150.0 mL of 1.00x10-2 M Mg(NO3)2 and 250.0 mL of 1.00x10-1 M NaF. Calculate
the concentration of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4x10-9).