chapter 14 acids and bases ap*. ap learning objectives lo 2.1 students can predict properties of...

Chapter 14

Acids and Bases

AP*

AP Learning Objectives

LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of their properties based on particle views. (Sec 14.8)

LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure, interparticle forces, and solution equilibrium. (Sec 14.1-14.6)

LO 3.7 The student is able to identify compounds as Bronsted-Lowry acids, bases, and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification.

Two other learning objectives apply to multiple sections and serve as over-arching principles of acid-base equilibria. (Sec 14.1)

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. (Sec 14.1-14.12)


LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium. (Sec 14.4-14.7)

LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration. (Sec 14.2)

LO. 6.14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H+] = [OH-] as opposed to requiring pH = 7, including especially the applications to biological systems. (Sec 14.2)

LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution. (Sec 14.4-14.7)


LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations. (Sec 14.5-14.6)

Section 14.1The Nature of Acids and Bases

AP Learning Objectives, Margin Notes and References Learning Objectives LO 2.2 The student is able to explain the relative strengths of acids and bases based on molecular structure,

interparticle forces, and solution equilibrium. LO 3.7 The student is able to identify compounds as Bronsted-Lowry acids, bases, and/or conjugate acid-base pairs,

using proton-transfer reactions to justify the identification. Two other learning objectives apply to multiple sections and serve as over-arching principles of acid-base

equilibria. 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.

LO 6.11 The student can generate or use a particulate representation of an acid (strong or weak or polyprotic) and a strong base to explain the species that will have large versus small concentrations at equilibrium.


Arrhenius An acid is a substance that, when dissolved in

water, increases the concentration of hydrogen ions.

A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.

Brønsted–Lowry An acid is a proton donor. A base is a proton acceptor.


Brønsted–Lowry Acid and Base

A Brønsted–Lowry acid must have at least one removable (acidic) proton (H+) to donate.A Brønsted–Lowry base must have at least one nonbonding pair of electrons to accept a proton (H+).


Brønsted–Lowry Reaction

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What Is Different about Water?

Water can act as a Brønsted–Lowry base and accept a proton (H+) from an acid, as on the previous slide.

It can also donate a proton and act as an acid, as is seen below.

This makes water amphoteric.


Conjugate Acids and Bases

The term conjugate means “joined together as a pair.”

Reactions between acids and bases always yield their conjugate bases and acids.


Relative Strengths of Acids and Bases Acids above the

line with H2O as a base are strong acids; their conjugate bases do not act as acids in water.

Bases below the line with H2O as an acid are strong bases; their conjugate acids do not act as acids in water.

• The substances between the lines with H2O are conjugate acid–base pairs in water.

Section 14.2Acid Strength


interparticle forces, and solution equilibrium. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or



LO 6.12 The student can reason about the distinction between strong and weak acid solutions with similar values of pH, including the percent ionization of the acids, the concentrations needed to achieve the same pH, and the amount of base needed to reach the equivalence point in a titration.

LO. 6.14 The student can, based on the dependence of Kw on temperature, reason that neutrality requires [H+] = [OH-] as opposed to requiring pH = 7, including especially the applications to biological systems.


Acid and Base Strength In every acid–base reaction, equilibrium favors transfer of

the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.

HCl(aq) + H2O(l) → H3O+(aq) + Cl(aq)

H2O is a much stronger base than Cl, so the equilibrium lies far to the right (K >> 1).

CH3COOH(aq) + H2O(l) H⇌ 3O+(aq) + CH3COO–(aq)

Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1).


Strong acid: Ionization equilibrium lies far to the right. Yields a weak conjugate base.

Weak acid: Ionization equilibrium lies far to the left. Weaker the acid, stronger its conjugate base.



Various Ways to Describe Acid Strength



Autoionization of Water Water is amphoteric. In pure water, a few molecules act as bases and a

few act as acids. This is referred to as autoionization.


Ion Product Constant

The equilibrium expression for this process isKc = [H3O+][OH]

This special equilibrium constant is referred to as the ion product constant for water, Kw.

At 25 °C, Kw = 1.0 1014

No matter what the solution contains, the product of [H+] and [OH–] must always equal 1.0 × 10–14 at 25°C.


Aqueous Solutions Can Be Acidic, Basic, or Neutral

If a solution is neutral, [H+] = [OH–]. If a solution is acidic, [H+] > [OH–]. If a solution is basic, [H+] < [OH–].


Self-Ionization of Water





Acid Ionization Equilibrium





HA(aq) + H2O(l) H3O+(aq) + A-(aq)

acid base conjugate conjugate acid base

What is the equilibrium constant expression for an acid acting in water?


3H O A

= HA

K

CONCEPT CHECK!CONCEPT CHECK!


If the equilibrium lies to the right, the value for Ka is __________.

large (or >1)

If the equilibrium lies to the left, the value for Ka is ___________.

small (or <1)




HA(aq) + H2O(l) H3O+(aq) + A–(aq)

If water is a better base than A–, do products or reactants dominate at equilibrium?

Does this mean HA is a strong or weak acid?

Is the value for Ka greater or less than 1?




Acetic acid (HC2H3O2) and HCN are both weak acids. Acetic acid is a stronger acid than HCN.

Arrange these bases from weakest to strongest and explain your answer:

H2O Cl– CN– C2H3O2–




Discuss whether the value of K for the reaction: HCN(aq) + F–(aq) CN–(aq) + HF(aq)

is >1 <1 =1

(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)

Explain your answer.




Calculate the value for K for the reaction:

HCN(aq) + F–(aq) CN–(aq) + HF(aq)

(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)

K = 8.6 × 10–7



Section 14.3The pH Scale





pH is a method of reporting hydrogen ion concentration. pH = –log[H+] pH changes by 1 for every power of 10 change in [H+]. A compact way to represent solution acidity. pH decreases as [H+] increases. Significant figures:

The number of decimal places in the log is equal to the number of significant figures in the original number.



pH Range

pH = 7; neutral pH > 7; basic

Higher the pH, more basic. pH < 7; acidic

Lower the pH, more acidic.



The pH Scale and pH Values of Some Common Substances



Other “p” Scales

The “p” in pH tells us to take the –log of a quantity (in this case, hydrogen ions).

Some other “p” systems arepOH: –log[OH]pKw: –log Kw

pKa: –log Ka

pKb: –log Kb


Relating pH and pOH

Because[H3O+][OH] = Kw = 1.0 1014

we can take the –log of the equation

–log[H3O+] + –log[OH] = –log Kw = 14.00

which results inpH + pOH = pKw = 14.00


How Do We Measure pH? Indicators, including litmus paper, are used for less

accurate measurements; an indicator is one color in its acid form and another color in its basic form.

pH meters are used for accurate measurement of pH; electrodes indicate small changes in voltage to detect pH.


Calculate the pH for each of the following solutions.

a) 1.0 × 10–4 M H+

pH = 4.00b) 0.040 M OH–

pH = 12.60


EXERCISE!EXERCISE!


The pH of a solution is 5.85. What is the [H+] for this solution?

[H+] = 1.4 × 10–6 M


EXERCISE!EXERCISE!


pH and pOH

Recall:Kw = [H+][OH–]

–log Kw = –log[H+] – log[OH–]

pKw = pH + pOH

14.00 = pH + pOHCopyright © Cengage Learning. All rights reserved 40


Calculate the pOH for each of the following solutions.

a) 1.0 × 10–4 M H+

pOH = 10.00b) 0.040 M OH–

pOH = 1.40


EXERCISE!EXERCISE!


The pH of a solution is 5.85. What is the [OH–] for this solution?

[OH–] = 7.1 × 10–9 M


EXERCISE!EXERCISE!

Section 14.4Calculating the pH of Strong Acid Solutions





LO 6.15 The student can identify a given solution as containing a mixture of strong acids and/or bases and calculate or estimate the pH (and concentrations of all chemical species) in the resulting solution.


Thinking About Acid–Base Problems

What are the major species in solution? What is the dominant reaction that will take place?

Is it an equilibrium reaction or a reaction that will go essentially to completion?

React all major species until you are left with an equilibrium reaction.

Solve for the pH if needed.



Strong Acids You will recall that the seven strong acids

are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

These are, by definition, strong electrolytes and exist totally as ions in aqueous solution; e.g.,

HA + H2O → H3O+ + A–

So, for the monoprotic strong acids,[H3O+] = [acid]

(Unless the [acid] is too low, such that the [H3O+] from the acid is less than 1 x 10-7M.)


Consider an aqueous solution of 2.0 × 10–3 M HCl.

What are the major species in solution? H+, Cl–, H2O

What is the pH? pH = 2.70




Calculate the pH of a 1.5 × 10–11 M solution of HCl.

pH = 7.00




Calculate the pH of a 1.5 × 10–2 M solution of HNO3.




Let’s Think About It…

When HNO3 is added to water, a reaction takes place immediately:

HNO3 + H2O H3O+ + NO3–




Why is this reaction not likely?

NO3–(aq) + H2O(l) HNO3(aq) + OH–(aq)




What reaction controls the pH? H2O(l) + H2O(l) H3O+(aq) + OH–(aq) In aqueous solutions, this reaction is always taking place. But is water the major contributor of H+ (H3O+)?

pH = 1.82


Section 14.5Calculating the pH of Weak Acid Solutions






LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations.

Section 14.5Calculating the pH of Weak Acid Solutions Weak Acids

For a weak acid, the equation for its dissociation is HA(aq) + H2O(l) H⇌ 3O+(aq) + A–(aq) Since it is an equilibrium, there is an equilibrium constant

related to it, called the acid-dissociation constant, Ka

Ka = [H3O+][A–] / [HA]• The greater the value

of Ka, the stronger is the acid.


Solving Weak Acid Equilibrium Problems

1. List the major species in the solution.2. Choose the species that can produce H+, and write

balanced equations for the reactions producing H+.3. Using the values of the equilibrium constants for the

reactions you have written, decide which equilibrium will dominate in producing H+.

4. Write the equilibrium expression for the dominant equilibrium.




5. List the initial concentrations of the species participating in the dominant equilibrium.

6. Define the change needed to achieve equilibrium; that is, define x.

7. Write the equilibrium concentrations in terms of x.8. Substitute the equilibrium concentrations into the

equilibrium expression.




9. Solve for x the “easy” way, that is, by assuming that [HA]0 – x about equals [HA]0.

10.Use the 5% rule to verify whether the approximation is valid.

11.Calculate [H+] and pH.



Consider a 0.80 M aqueous solution of the weak acid HCN (Ka = 6.2 × 10–10).

What are the major species in solution?

HCN, H2O





Why aren’t H+ or CN– major species?



Consider This

HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)

Ka = 6.2 × 10-10

H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

Kw = 1.0 × 10-14

Which reaction controls the pH? Explain.



Calculate the pH of a 0.50 M aqueous solution of the weak acid HF.

(Ka = 7.2 × 10–4)


EXERCISE!EXERCISE!




HF, H2O

Why aren’t H+ and F– major species?




What are the possibilities for the dominant reaction?

HF(aq) + H2O(l) H3O+(aq) + F–(aq)

Ka=7.2 × 10-4

H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw=1.0 × 10-14

Which reaction controls the pH? Why? Copyright © Cengage Learning. All rights reserved 62


HF(aq) + H2O H3O+(aq) + F–(aq)

Initial 0.50 M ~ 0 ~ 0

Change –x +x +x

Equilibrium 0.50–x x x

Steps Toward Solving for pH

Ka = 7.2 × 10–4

pH = 1.72



Calculating Ka from the pH

The pH of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate Ka for formic acid at this temperature.

We know that [H3O+][HCOO–][HCOOH]

Ka =

To calculate Ka, we need the equilibrium concentrations of all three things.

We can find [H3O+], which is the same as [HCOO–], from the pH.

[H3O+] = [HCOO–] = 10–2.38 = 4.2 × 10–3

Section 14.5Calculating the pH of Weak Acid SolutionsCalculating Ka from pH

Now we can set up a table for equilibrium concentrations. We know initial HCOOH (0.10 M) and ion concentrations (0 M); we found equilibrium ion concentrations (4.2 × 10–3 M); so we calculate the change, then the equilibrium HCOOH concentration.

[HCOOH], M [H3O+], M [HCOO], M

Initially 0.10 0 0

Change 4.2 103 +4.2 103 +4.2 103

At equilibrium 0.10 4.2 103

= 0.0958 = 0.10

4.2 103 4.2 103


Calculating Ka from pH

[4.2 103][4.2 103][0.10]

Ka =

= 1.8 104

• This allows us to calculate Ka by putting in the equilibrium concentrations.



Calculating Percent Ionization

Percent ionization = 100 In this example,

[H3O+]eq = 4.2 103 M

[HCOOH]initial = 0.10 M

[H3O+]eq

[HA]initial

Percent ionization = 1004.2 103

0.10= 4.2%


Strong vs. Weak Acids—Another Comparison

Strong Acid: [H+]eq = [HA]init

Weak Acid: [H+]eq < [HA]init

This creates a difference in conductivity and in rates of chemical reactions.


Strong vs. Weak Acids For a given strong acid, the properties of the acid

solution are directly related to the acid concentration.

For a given weak acid, this is not the case. As the concentration of a weak acid is increased in solution, the percent dissociation decreases as the acid becomes more concentrated.


A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water.

Calculate the Ka value for formic acid.

Ka = 1.8 × 10–4


EXERCISE!EXERCISE!


Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide to help you solve this problem.

pH = 1.42


EXERCISE!EXERCISE!


The value of Ka for a 4.00 M formic acid solution should be:

higher than lower than the same as

the value of Ka of an 8.00 M formic acid solution.

Explain.


EXERCISE!EXERCISE!


The percent ionization of a 4.00 M formic acid solution should be:

higher than lower than the same as

the percent ionization of an 8.00 M formic acid solution.

Explain.




The pH of a 4.00 M formic acid solution should be:

higher than lower than the same as the pH of an 8.00 M formic acid solution.

Explain.




Calculate the percent ionization of a 4.00 M formic acid solution in water.

% Ionization = 0.67%


EXERCISE!EXERCISE!


Calculate the pH of a 4.00 M solution of formic acid.

pH = 1.57


EXERCISE!EXERCISE!

Section 14.6Bases






LO 6.16 The student can identify a given solution as being the solution of a monoprotic weak acid or base (including salts in which one ion is a weak acid or base), calculate the pH and concentration of all species in the solution, and/or infer the relative strengths of the weak acids or bases from given equilibrium concentrations.

Section 14.6Bases

Arrhenius: bases produce OH– ions. Brønsted–Lowry: bases are proton acceptors. In a basic solution at 25°C, pH > 7. pOH = –log[OH–] pH = 14.00 – pOH


Section 14.6Bases


Strong Bases

Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

Again, these substances dissociate completely in aqueous solution; e.g.,

MOH(aq) → M+(aq) + OH–(aq) orM(OH)2(aq) → M2+(aq) + 2 OH–(aq)

Section 14.6Bases

Calculate the pH of a 1.0 × 10–3 M solution of sodium hydroxide.

pH = 11.00



Section 14.6Bases

Calculate the pH of a 1.0 × 10–3 M solution of calcium hydroxide.

pH = 11.30



Section 14.6Bases

Weak Bases

Ammonia, NH3, is a weak base. Like weak acids, weak bases have an

equilibrium constant called the base dissociation constant.

Equilibrium calculations work the same as for acids, using the base dissociation constant instead.

Section 14.6Bases

Base Dissociation Constants

Section 14.6Bases

Example What is the pH of 0.15 M NH3?

1) NH3 + H2O NH⇌ 4+ + OH–

2) Kb = [NH4+][OH–] / [NH3] = 1.8 × 10–5

3) NH3 (M) NH4+ (M) OH– (M)

Initial Concentration (M)

0.15 0 0

Change in Concentration (M)

–x +x +x

Equilibrium Concentration (M)

0.15 – x x x

Section 14.6Bases

Example (completed)

4) 1.8 × 10 – 5 = x2 / (0.15 – x)If we assume that x << 0.15, 0.15 – x = 0.15.Then: 1.8 × 10–5 = x2 / 0.15and: x = 1.6 × 10–3

Note: x is the molarity of OH–, so –log(x) will be the pOH (pOH = 2.80) and [14.00 – pOH] is pH (pH = 11.20).

Section 14.6Bases

Types of Weak Bases Two main categories1) Neutral substances with an Atom that

has a nonbonding pair of electrons that can accept H+ (like ammonia and the amines)

2) Anions of weak acidsClO-(aq) +H2O (l) ↔ HClO(aq) + OH–(aq)

Section 14.6Bases

A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solutionhas a pH of 10.50. Using the equation below, calculate the number of moles of NaClO added to the water.

ClO-(aq) +H2O (l) ↔ HClO(aq) + OH–(aq)

Using pH to Determine the Concentration of a Salt

Section 14.6Bases

Relationship between Ka and Kb

For a conjugate acid–base pair, Ka and Kb are related in this way:

Ka × Kb = Kw

Therefore, if you know one of them, you can calculate the other.

Calculate Kb for the fluoride ion.

Section 14.8Acid-Base Properties of Salts

HC2H3O2 Ka = 1.8 × 10-5

HCN Ka = 6.2 × 10-10

Calculate the Kb values for: C2H3O2− and CN−

Kb (C2H3O2-) = 5.6 × 10-10

Kb (CN-) = 1.6 × 10-5

Copyright © Cengage Learning. All rights reserved

EXERCISE!EXERCISE!

Section 14.6Bases

Calculate the pH of a 2.0 M solution of ammonia (NH3).

(Ka = 5.6 × 10–10)

pH = 11.78



Section 14.7Polyprotic Acids

AP Learning Objectives, Margin Notes and References Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or





Acids that can furnish more than one proton. Always dissociates in a stepwise manner, one proton at a

time. The conjugate base of the first dissociation equilibrium

becomes the acid in the second step. For a typical weak polyprotic acid:

Ka1 > Ka2 > Ka3

If the factor in the Ka values for the first and second dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation.



Calculate the pH of a 1.00 M solution of H3PO4.

Ka1 = 7.5 × 10-3

Ka2 = 6.2 × 10-8

Ka3 = 4.8 × 10-13

pH = 1.08 (if use quadratic)


EXERCISE!EXERCISE!


Calculate the equilibrium concentration of PO43- in a

1.00 M solution of H3PO4.

Ka1 = 7.5 × 10-3

Ka2 = 6.2 × 10-8

Ka3 = 4.8 × 10-13

[PO43-] = 3.6 × 10-19 M




Be sure to read page 683-685 to learn about problems concerning sulfuric acid where the acid is strong in the first ionization step and weak in the second.


AP Learning Objectives, Margin Notes and References Learning Objectives LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of

their properties based on particle views. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or



Salts

Ionic compounds. When dissolved in water, break up into its ions (which

can behave as acids or bases).


The salt of a strong acid and a strong base gives a neutral solution. KCl, NaNO3


Acid–Base Properties of Salts Many ions react with water to create H+ or

OH–. The reaction with water is often called hydrolysis.

To determine whether a salt is an acid or a base, you need to look at the cation and anion separately.

The cation can be acidic or neutral. The anion can be acidic, basic, or neutral.


Anions Anions of strong acids are neutral. For

example, Cl– will not react with water, so OH– can’t be formed.

Anions of weak acids are conjugate bases, so they create OH– in water; e.g., C2H3O2

– + H2O HC⇌ 2H3O2 + OH–

Protonated anions from polyprotic acids (HSO3

-) can be acids or bases: If Ka > Kb, the anion will be acidic; if Kb > Ka, the anion will be basic.


Cations

Group I or Group II (Ca2+, Sr2+, or Ba2+) metal cations are neutral.

Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH4

+. Transition and post-transition metal cations are

acidic. Why? (There are no H atoms in these cations!)

Section 14.8Acid-Base Properties of SaltsHydrated Cations Transition and post-transition metals form

hydrated cations. The water attached to the metal is more acidic

than free water molecules, making the hydrated ions acidic.


Salt Solutions—Acidic, Basic, or Neutral?1) Group I/II metal cation with anion of a strong acid:

neutral (NaCl, RbClO4)

2) Group I/II metal cation with anion of a weak acid: basic (like the anion) (NaClO, BaSO3)

3) Transition/Post-transition metal cation or polyatomic cation with anion of a strong acid: acidic (like the cation) (NH4NO3, AlCl3, Fe(NO3) 3

4) Transition/Post-transition metal cation or polyatomic cation with anion of a weak acid: compare Ka and Kb; whichever is greater dictates what the salt is. (NH4ClO, Al(CH3COO) 3, CrF 3


Qualitative Prediction of pH of Salt Solutions (from Weak Parents)



Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

Solution

Analyze We are asked to predict whether a solution of Na2HPO4 is acidic or basic. This substance is an ionic compound composed of Na+ and HPO4

2− ions.

Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na+ is a cation of group 1A, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO4

2− ion. We need to consider that HPO42− can act as either an acid or a

base:

As acid HPO42–(aq) H+(aq) + PO4

3–(aq) [16.46]As base HPO4

2–(aq) + H2O H2PO4–(aq) + OH–(aq) [16.47]

Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic.

Predict whether the salt Na2HPO4 forms an acidic solution or a basic solution when dissolved in water.



Solve The value of Ka for Equation 16.46 is Ka3 for H3PO4: 4.2 × 10−13 (Table 16.3). For Equation 16.47, we must calculate Kb for the base HPO4

2−

from the value of Ka for its conjugate acid, H2PO4

−, and the relationship Ka × Kb = Kw (Equation 16.40). The relevant value of Ka for H2PO4

− is Ka2 for H3PO4: 6.2 × 10–8 (from Table 16.3). We therefore have

This Kb value is more than 105 times larger than Ka for HPO42−; thus, the reaction in Equation 16.47

predominates over that in Equation 16.46, and the solution is basic.


Arrange the following 1.0 M solutions from lowest to highest pH.

HBr NaOH NH4Cl

NaCN NH3 HCN

NaCl HF

Justify your answer.HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH




Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral:

(a) Ba(CH3COO)2, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3.


Consider a 0.30 M solution of NaF. The Ka for HF is 7.2 × 10-4.

What are the major species?

Na+, F-, H2O





Why isn’t NaF considered a major species? What are the possibilities for the dominant reactions?




The possibilities for the dominant reactions are:

1. F–(aq) + H2O(l) HF(aq) + OH–(aq)

2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

3. Na+(aq) + H2O(l) NaOH + H+(aq)

4. Na+(aq) + F–(aq) NaF




How do we decide which reaction controls the pH?

F–(aq) + H2O(l) HF(aq) + OH–(aq)


Determine the equilibrium constant for each reaction.



Calculate the pH of a 0.75 M aqueous solution of NaCN. Ka for HCN is 6.2 × 10–10.


EXERCISE!EXERCISE!




Na+, CN–, H2O

Why isn’t NaCN considered a major species?




What are all possibilities for the dominant reaction? The possibilities for the dominant reaction are:

1. CN–(aq) + H2O(l) HCN(aq) + OH–(aq)

2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)

3. Na+(aq) + H2O(l) NaOH + H+(aq)

4. Na+(aq) + CN–(aq) NaCN

Which of these reactions really occur?Copyright © Cengage Learning. All rights reserved 117



How do we decide which reaction controls the pH?

CN–(aq) + H2O(l) HCN(aq) + OH–(aq)




Steps Toward Solving for pH

Kb = 1.6 × 10–5

pH = 11.54

CN–(aq) + H2O HCN(aq) + OH–(aq)

Initial 0.75 M 0 ~ 0

Change –x +x +x

Equilibrium 0.75–x x x

Section 14.9The Effect of Structure on Acid-Base Properties

AP Learning Objectives, Margin Notes and References Learning Objectives LO 2.1 Students can predict properties of substances based on their chemical formulas, and provide explanations of

their properties based on particle views. LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical biological, or



Factors that Affect Acid Strength

1) H—A bond must be polarized with δ+ on the H atom and δ– on the A atom

2) Bond strength: Weaker bonds can be broken more easily, making the acid stronger.

3) Stability of A–: More stable anion means stronger acid.


Binary Acids Binary acids consist of

H and one other element. Within a group, H—A bond

strength is generally the most important factor.

Within a period, bond polarity is the most important factor to determine acid strength.


Oxyacids

Contains the group H–O–X. For a given series the acid strength increases with an

increase in the number of oxygen atoms attached to the central atom.

The greater the ability of X to draw electrons toward itself, the greater the acidity of the molecule.



Several Series of Oxyacids and Their Ka Values



Comparison of Electronegativity of X and Ka Value for hypohalous acids



Arrange the compounds in each series in order of increasing acid strength: (a) AsH3, HBr, KH, H2Se; (b) H2SO4, H2SeO3, H2SeO4.



Carboxylic Acids Carboxylic acids are organic acids containing

the —COOH group. Factors contributing to their acidic behavior:Other O attached to C draws electron density

from O—H bond, increasing polarity.Its conjugate base (carboxylate anion) has

resonance forms to stabilize the anion.

Section 14.10Acid-Base Properties of Oxides




Oxides Acidic Oxides (Acid Anhydrides):

O—X bond is strong and covalent.

SO2, NO2, CO2

When H—O—X grouping is dissolved in water, the O—X bond will remain intact. It will be the polar and relatively weak H—O bond that will tend to break, releasing a proton.



Oxides Basic Oxides (Basic Anhydrides):

O—X bond is ionic.K2O, CaO

If X has a very low electronegativity, the O—X bond will be ionic and subject to being broken in polar water, producing a basic solution.


Section 14.11The Lewis Acid-Base Model




Three Models for Acids and Bases


Section 14.12Strategy for Solving Acid-Base Problems: A Summary



Section 14.12Strategy for Solving Acid-Base Problems: A Summary

When analyzing an acid-base equilibrium problem:

Ask this question: What are the major species in the solution and what is their chemical behavior? What major species are present? Does a reaction occur that can be assumed to go to completion? What equilibrium dominates the solution? Let the problem guide you. Be patient.


chapter 14 acids and bases ap*. ap learning objectives lo 2.1 students can predict properties of...

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