18-1 CHEM 1B General Chemistry Ch. 18 Acid-Base Equilibria Instructor: Dr. Orlando E. Raola Santa...

18-1

CHEM 1BGeneral Chemistry

Ch. 18 Acid-Base Equilibria

Instructor: Dr. Orlando E. RaolaSanta Rosa Junior College

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Chapter 18

Acid-Base Equilibria

18-3

Acid-Base Equilibria

18.1 Acids and Bases in Water

18.2 Autoionization of Water and the pH Scale

18.3 Proton Transfer and the Brønsted-Lowry Acid-Base Definition

18.4 Solving Problems Involving Weak-Acid Equilibria

18.5 Weak Bases and Their Relation to Weak Acids

18.6 Molecular Properties and Acid Strength

18.7 Acid-Base Properties of Salt Solutions

18.8 Generalizing the Brønsted-Lowry Concept: The Leveling Effect

18.9 Electron-Pair Donation and the Lewis Acid-Base Definition

18-4

Table 18.1 Some Common Acids and Bases and their Household Uses.

18-5

Arrhenius Acid-Base Definition

An acid is a substance with H in its formula that dissociates to yield H3O+.

This is the earliest acid-base definition, which classifies these substances in terms of their behavior in water.

A base is a substance with OH in its formula that dissociates to yield OH-.

When an acid reacts with a base, they undergo neutralization:

H+(aq) + OH-(aq) → H2O(l) H°rxn = -55.9 kJ

18-6

Strong and Weak Acids

A strong acid dissociates completely into ions in water:HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)

A dilute solution of a strong acid contains no HA molecules.

A weak acid dissociates slightly to form ions in water:

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

In a dilute solution of a weak acid, most HA molecules are undissociated.

[H3O+][A-]

[HA][H2O]Kc = has a very small value.

18-7

Strong acid: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)

Figure 18.1A The extent of dissociation for strong acids.

There are no HA molecules in solution.

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Figure 18.1B The extent of dissociation for weak acids.

Weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Most HA molecules are undissociated.

18-9

Figure 18.2 Reaction of zinc with a strong acid (left) and a weak acid (right).

1 mol/L HCl(aq)

1 mol/L CH3COOH(aq)

Zinc reacts rapidly with the strong acid, since [H3O+] is much higher.

18-10

The Acid Dissociation Constant, Ka

[H3O+][A-]

[HA][H2O]Kc = Kc[H2O] = Ka =

[H3O+][A-]

[HA]

The value of Ka is an indication of acid strength.

Stronger acid larger Kahigher [H3O+]

Weaker acid smaller Kalower % dissociation of HA


18-11

Table 18.2 Ka Values for some Monoprotic Acids at 25°C

18-12

Classifying the Relative Strengths of Acids

• Strong acids include

– the hydrohalic acids (HCl, HBr, and HI) and

– oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more (eg., HNO3, H2SO4, HClO4.)

• Weak acids include

– the hydrohalic acid HF,

– acids in which H is not bonded to O or to a halogen (eg., HCN),

– oxoacids in which the number of O atoms equals or exceeds the number of ionizable protons by one (eg., HClO, HNO2), and

– carboxylic acids, which have the general formula RCOOH (eg., CH3COOH and C6H5COOH.)

18-13

Classifying the Relative Strengths of Bases

• Strong bases include

– water-soluble compounds containing O2- or OH- ions.

– The cations are usually those of the most active metals:

• M2O or MOH, where mol/L = Group 1A(1) metal (Li, Na, K, Rb, Cs)

• MO or M(OH)2 where mol/L = group 2A(2) metal (Ca, Sr, Ba).

• Weak bases include

– ammonia (NH3),

– amines, which have the general formula

– The common structural feature is an N atom with a lone electron pair.

RNH2, R2NH, or R3N

18-14

Sample Problem 18.1 Classifying Acid and Base Strength from the Chemical Formula

PROBLEM: Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base.

(a) KOH (b) (CH3)2CHCOOH

(c) H2SeO4 (d) (CH3)2CHNH2

PLAN: We examine the formula and classify each acid or base, using the text descriptions. Particular points to note for acids are the numbers of O atoms relative to ionizable H atoms and the presence of the –COOH group. For bases, note the nature of the cation or the presence of an N atom that has a lone pair.

SOLUTION:

(a) Strong base: KOH is one of the group 1A(1) hydroxides.

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Sample Problem 18.1

(b) Weak acid: (CH3)2CHCOOH is a carboxylic acid, as indicated by the –COOH group. The –COOH proton is the only ionizable proton in this compound.

(c) Strong acid: He2SO4 is an oxoacid in which the number of atoms exceeds the number of ionizable protons by two.

(d) Weak base: (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.

18-16

Autoionization of Water

Water dissociates very slightly into ions in an equilibrium process known as autoionization or self-ionization.

2H2O (l) H3O+ (aq) + OH- (aq)

18-17

The Ion-Product Constant for Water (Kw)

2H2O (l) H3O+ (aq) + OH- (aq)

[H3O+][A-]

[H2O]2Kc =

Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0x10-14 (at 25°C)

= 1.0x10-7 (at 25°C)In pure water,[H3O+] = [OH-] = 1.0x10-14

Both ions are present in all aqueous systems.

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Higher [H3O+] lower [OH-]

Higher [OH-] lower [H3O+]

A change in [H3O+] causes an inverse change in [OH-], and vice versa.

We can define the terms “acidic” and “basic” in terms of the relative concentrations of H3O+ and OH- ions:

In an acidic solution, [H3O+] > [OH-]In a neutral solution, [H3O+] = [OH-]In basic solution, [H3O+] < [OH-]

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Figure 18.3 The relationship between [H3O+] and [OH-] and the relative acidity of solutions.

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Sample Problem 18.2 Calculating [H3O+] or [OH-] in an Aqueous Solution

PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 25°C and obtains a solution with [H3O+] = 3.0x10-4 mol/L. Calculate [OH-]. Is the solution neutral, acidic, or basic?

SOLUTION:

PLAN: We use the known value of Kw at 25°C (1.0x10-14) and the given [H3O+] to solve for [OH-]. We can then compare [H3O+] with [OH-] to determine whether the solution is acidic, basic, or neutral.

Kw = 1.0x10-14 = [H3O+] [OH-] so

[H3O+] is > [OH-] and the solution is acidic.

= 3.3x10-11 mol/L[OH-] =Kw

[H3O+]

1.0x10-14

3.0x10-4=

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The pH Scale

pH = -log[H3O+]

The pH of a solution indicates its relative acidity:

In an acidic solution, pH < 7.00In a neutral solution, pH = 7.00In basic solution, pH > 7.00

The higher the pH, the lower the [H3O+] and the less acidic the solution.

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Figure 18.4

The pH values of some familiar aqueous solutions.

pH = -log [H3O+]

18-23

Table 18.3 The Relationship between Ka and pKa

Acid Name (Formula) Ka at 25°C pKa

Hydrogen sulfate ion (HSO4-) 1.0x10-2 1.99

Nitrous acid (HNO2) 7.1x10-4 3.15

Acetic acid (CH3COOH) 1.8x10-5 4.75

Hypobromous acid (HBrO) 2.3x10-9 8.64

Phenol (C6H5OH) 1.0x10-10 10.00

pKa = -logKa

A low pKa corresponds to a high Ka.

18-24

pH, pOH, and pKw

pH = -log[H3O+]

pOH = -log[OH-]

pH + pOH = pKw for any aqueous solution at any temperature.

pKw = pH + pOH = 14.00 at 25°C

Kw = [H3O+][OH-] = 1.0x10-14 at 25°C

Since Kw is a constant, the values of pH, pOH, [H3O+], and [OH-] are interrelated:• If [H3O+] increases, [OH-] decreases (and vice versa).• If pH increases, pOH decreases (and vice versa).

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Figure 18.5 The relations among [H3O+], pH, [OH-], and pOH.

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Sample Problem 18.3 Calculating [H3O+], pH, [OH-], and pOH

PROBLEM: In an art restoration project, a conservator prepares copper-plate etching solutions by diluting concentrated HNO3 to 2.0 mol/L, 0.30 mol/L, and 0.0063 mol/L HNO3. Calculate [H3O+], pH, [OH-], and pOH of the three solutions at 25°C.

SOLUTION:

PLAN: HNO3 is a strong acid so it dissociates completely, and [H3O+] = [HNO3]init. We use the given concentrations and the value of Kw at 25°C to find [H3O+] and [OH-]. We can then calculate pH and pOH.

[H3O+] = 2.0 mol/L

Calculating the values for 2.0 mol/L HNO3:

pH = -log[H3O+] = -log(2.0) = -0.30

Kw

[H3O+][OH-] = =

1.0x10-14

2.0= 5.0x10-15 mol/L

pOH = -log[OH-] = -log(5.0x10-15) = 14.30

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Sample Problem 18.3

[H3O+] = 0.30 mol/L


pH = -log[H3O+] = -log(0.30) = 0.52

Kw

[H3O+][OH-] = =

1.0x10-14

0.30= 3.3x10-14 mol/L

pOH = -log[OH-] = -log(3.3x10-14) = 13.48

[H3O+] = 0.0063 mol/L


pH = -log[H3O+] = -log(0.30) = 2.20

Kw

[H3O+][OH-] = =

1.0x10-14

0.0063= 1.6x10-12 mol/L

pOH = -log[OH-] = -log(1.6x10-12) = 11.80

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Figure 18.6 Methods for measuring the pH of an aqueous solution.

pH paper

pH meter

18-29

Brønsted-Lowry Acid-Base Definition

An acid is a proton donor, any species that donates an H+ ion.• An acid must contain H in its formula.

A base is a proton acceptor, any species that accepts an H+ ion.• A base must contain a lone pair of electrons to bond to H+.

An acid-base reaction is a proton-transfer process.

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Figure 18.7 Dissolving of an acid or base in water as a Brønsted-Lowry acid-base reaction.

(acid, H+ donor) (base, H+ acceptor)

Lone pair binds H+

(base, H+ acceptor) (acid, H+ donor)

Lone pair binds H+

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Conjugate Acid-Base Pairs

H2S + NH3 HS- + NH4+

NH3 accepts a H+ to form NH4+.

In the forward reaction:

In the reverse reaction:


HS- accepts a H+ to form H2S.

H2S donates a H+ to form HS-.

NH4+ donates a H+ to form NH3.

18-32

Conjugate Acid-Base Pairs


H2S and HS- are a conjugate acid-base pair:HS- is the conjugate base of the acid H2S.

A Brønsted-Lowry acid-base reaction occurs when an acid and a base react to form their conjugate base and conjugate acid, respectively.

NH3 and NH4+ are a conjugate acid-base pair:

NH4+ is the conjugate acid of the base NH3.

acid1 + base2 base1 + acid2

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Base Acid+Acid Base+

Conjugate Pair

Conjugate Pair

Table 18.4 The Conjugate Pairs in some Acid-Base Reactions

Reaction 4 H2PO4- OH-+

Reaction 5 H2SO4 N2H5++

Reaction 6 HPO42- SO3

2-+

Reaction 1 HF H2O+ F- H3O++

Reaction 3 NH4+ CO3

2-+

Reaction 2 HCOOH CN-+ HCOO- HCN+

NH3 HCO3-+

HPO42- H2O+

HSO4- N2H6

2++

PO43- HSO3

-+

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Sample Problem 18.4 Identifying Conjugate Acid-Base Pairs

PLAN: To find the conjugate pairs, we find the species that donated an H+ (acid) and the species that accepted it (base). The acid donates an H+ to becomes its conjugate base, and the base accepts an H+ to becomes it conjugate acid.

PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs.

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)

SOLUTION:

(a) H2PO4-(aq) + CO3

2-(aq) HPO42-(aq) + HCO3

-(aq)acid1 base1 base2 acid2

The conjugate acid-base pairs are H2PO4-/HPO4

2- and CO32-/HCO3

-.

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Sample Problem 18.4

(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3

-(aq)

acid1 base1 acid2 base2

The conjugate acid-base pairs are H2O/OH- and SO32-/HSO3

-.

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Net Direction of Reaction

The net direction of an acid-base reaction depends on the relative strength of the acids and bases involved.

A reaction will favor the formation of the weaker acid and base.


stronger baseweaker basestronger acid

weaker acid

This reaction favors the formation of the products.

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Figure 18.8 Strengths of conjugate acid-base pairs.

The stronger the acid is, the weaker its conjugate base. When an acid reacts with a base that is farther down the list, the reaction proceeds to the right (Kc > 1).

18-38

Sample Problem 18.5 Predicting the Net Direction of an Acid-Base Reaction

SOLUTION:

PLAN: We identify the conjugate acid-base pairs and consult figure 18.8 to see which acid and base are stronger. The reaction favors the formation of the weaker acid and base.

stronger acid weaker acidstronger base weaker base

PROBLEM: Predict the net direction and whether Kc is greater or less than 1 for each of the following reactions (assume equal initial concentrations of all species):

(a) H2PO4-(aq) + NH3(aq) HPO4

2-(aq) + NH4+(aq)

(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)

(a) H2PO4-(aq) + NH3(aq) HPO4

2-(aq) + NH4+(aq)

The net direction for this reaction is to the right, so Kc > 1.

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Sample Problem 18.5

(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)

stronger acidweaker acid stronger baseweaker base

The net direction for this reaction is to the left, so Kc < 1.

18-40

Sample Problem 18.6 Using Molecular Scenes to Predict the Net Direction of an Acid-Base Reaction

PROBLEM: Given that 0.10 mol/L of HX (blue and green) has a pH of 2.88, and 0.10 mol/L HY (blue and orange) has a pH 3.52, which scene best represents the final mixture after equimolar solutions of HX and Y- are mixed?

PLAN: A stronger acid and base yield a weaker acid and base, so we have to determine the relative acid strengths of HX and HY to choose the correct molecular scene. The concentrations of the acid solutions are equal, so we can recognize the stronger acid by comparing the pH values of the two solutions.

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Sample Problem 18.6

SOLUTION:

The HX solution has a lower pH than the HY solution, so HX is the stronger acid and Y- is the stronger base. The reaction of HX and Y- has a Kc > 1, which means the equilibrium mixture will contain more HY than HX. Scene 1 has equal numbers of HX and HY, which could occur if the acids were of equal strength. Scene 2 shows fewer HY than HX, which would occur if HY were the stronger acid.

Scene 3 is consistent with the relative acid strengths, because it contains more HY than HX.

18-42

Solving Problems Involving Weak-Acid Equilibria

Problem-solving approach

1.Write a balanced equation.

2.Write an expression for Ka.

3.Define x as the change in concentration that occurs during the reaction.

4.Construct a reaction table in terms of x.

5.Make assumptions that simplify the calculation.

6.Substitute values into the Ka expression and solve for x.

7.Check that the assumptions are justified.

18-43

Solving Problems Involving Weak-Acid Equilibria

The notation system

• Molar concentrations are indicated by [ ].

• A bracketed formula with no subscript indicates an equilibrium concentration.

The assumptions

• [H3O+] from the autoionization of H2O is negligible.

• A weak acid has a small Ka and its dissociation is negligible. [HA] ≈ [HA]init.

18-44

Sample Problem 18.7 Finding Ka of a Weak Acid from the Solution pH

PROBLEM: Phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) builds up in the blood of persons with phenylketonuria, an inherited disorder that, if untreated, causes mental retardation and death. A study of the acid shows that the pH of 0.12 mol/L HPAc is 2.62. What is the Ka of phenylacetic acid?

PLAN: We start with the balanced dissociation equation and write the expression for Ka. We assume that [H3O+] from H2 is negligible and use the given pH to find [H3O+], which equals [PAc-] and [HPAc]dissoc. We assume that [HPAc] ≈ [HPAc]init because HPAc is a weak acid.

SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)

Ka = [H3O+][PAc-]

[HPAc]

18-45

Sample Problem 18.7

Initial 0.12 - 0 0

Change -- x + x +x

Concentration (mol/L)

HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)

[H3O+] = 10-pH = 2.4x10-3 mol/L which is >> 10-7 (the [H3O+] from water)x ≈ 2.4x10-3 mol/L ≈ [H3O+] ≈ [PAc-]

[HPAc] = 0.12 - x ≈ 0.12 mol/L

So Ka =(2.4x10-3) (2.4x10-3)

0.12= 4.8x10-5

Checking the assumptions by finding the percent error in concentration:

x 100[HPAc]dissoc =

2.4x10-3

mol/L0.12 mol/L

= 2.0 % (<5%; assumption is justified).

Equilibrium -0.12 - x xx

= 4x10-3 % (<5%; assumption is justified).[H3O+] =1x10-7

mol/L2.4x10-3

mol/L

x 100from H2O

18-46

Sample Problem 18.8 Determining Concentration from Ka and Initial [HA]

PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify as HPr) is a carboxylic acid whose salts are used to retard mold growth in foods. What is the [H3O+] of 0.10 mol/L HPr (Ka = 1.3x10−5)?

PLAN: We write a balanced equation and the expression for Ka. We know [HPr]init but not [HPr] (i.e., the concentration at equilibrium). We define x as [HPr]dissoc and set up a reaction table. We assume that, since HPr has a small Ka value, it dissociates very little and therefore [HPr] ≈ [HPr]init.

Ka = [H3O+][Pr−]

[HPr]

SOLUTION: HPr(aq) + H2O(l) H3O+(aq) + Pr−(aq)

18-47

Sample Problem 18.8

Initial 0.10 -

0 0

Change -−x +x +x

HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)Concentration (mol/L)

Since Ka is small, we will assume that x << 0.10 and [HPr] ≈ 0.10 mol/L.

Equilibrium - 0.10 - x xx

x2

0.10Ka = 1.3x10-5 =

[H3O+][Pr-]

[HPr]=

= 1.1x10-3 mol/L = [H3O+]

x = (0.10)(1.3x10-5)

Check: [HPr]diss =1.1x10-3 mol/L

0.10 mol/Lx 100 = 1.1% (< 5%; assumption is justified.)

18-48

Concentration and Extent of Dissociation

Percent HA dissociated = [HA]dissoc

[HA]init

x 100

As the initial acid concentration decreases, the percent dissociation of the acid increases.


A decrease in [HA]init means a decrease in [HA]dissoc = [H3O+] = [A-], causing a shift towards the products.

The fraction of ions present increases, even though the actual [HA]dissoc decreases.

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Sample Problem 18.9 Using Molecular Scenes to Determine the Extent of HA Dissociation

PROBLEM: A 0.15 mol/L solution of acid HA (blue and green) is 33% dissociated. Which scene best represents a sample of the solution after it is diluted with water?

PLAN: We are given the percent dissociation of the original HA solution (33%), and we know that the percent dissociation increases as the acid is diluted. Thus, we calculate the percent dissociation of each diluted sample and see which is greater than 33%.

18-50

Sample Problem 18.9

SOLUTION:

Solution 1. % dissociated = 4/(5 + 4) x 100 = 44%Solution 2. % dissociated = 2/(7 + 2) x 100 = 22%Solution 3. % dissociated = 3/(6 + 3) x 100 = 33%

% dissociation =[H3O+]

[HA] + [H3O+]x 100

In each case, remember that each unit of H3O+ is produced from one unit of HA, so [HA]init = [HA] + [H3O+].

Scene 1 represents the diluted solution.

18-51

A polyprotic acid is an acid with more than one ionizable proton. In solution, each dissociation step has a different value for Ka:

Ka1 > Ka2 > Ka3

Polyprotic Acids

Ka1 =[H3O+][H2PO4

-]

[H3PO4]= 7.2x10-3H3PO4(aq) + H2O(l) H2PO4

-(aq) + H3O+(aq)

Ka2 =[H3O+][HPO4

2-]

[H2PO4-]

= 6.3x10-8H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+(aq)

Ka3 =[H3O+][PO4

3-]

[HPO42-]

= 4.2x10-13HPO42-(aq) + H2O(l) PO4

3-(aq) + H3O+(aq)

We usually neglect [H3O+] produced after the first dissociation.

18-52

Table 18.5 Successive Ka values for Some Polyprotic Acids at 25°C

18-53

Sample Problem 18.10 Calculating Equilibrium Concentrations for a Polyprotic Acid

PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1 = 1.0x10-5 and Ka2 = 5x10-12) found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the pH of 0.050 mol/L H2Asc.

PLAN: We first write the dissociation equations and the associated Ka expressions. Since Ka1 >> Ka2, we can assume that the first dissociation produces almost all the H3O+. Also, since Ka1 is small, the amount of H2ASc that dissociates can be neglected. We set up a reaction table for the first dissociation, with x equal to [H2Asc]dissoc, and solve for [H3O+] and [HAsc-].

SOLUTION:

H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq) Ka1 = [HAsc-][H3O+]

[H2Asc]= 1.0x10-5

Ka2 = [Asc2-][H3O+]

[HAsc-]= 5x10-12HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)

18-54

Sample Problem 18.10

Initial 0.050 - 0 0

pH = -log[H3O+] = -log(7.1x10-4) = 3.15

Change -x - +x +x

Concentration (mol/L)

H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)

Equilibrium 0.050 - x - x x

Ka1 = [HAsc-][H3O+]

[H2Asc]= 1.0x10-5 =

x2

0.050 - x≈

x2

0.050

x = [H3O+] = [Asc-] = (0.050)(1.0x10-5) = 7.1x10-4 mol/L

18-55


Checking assumptions:

= (7.1x10-4)(5x10-12) = 6x10-8 mol/L[H3O+]from HAsc-

≈ [HAsc-](Ka2)

1. [H3O+]from HAsc-

<< [H3O+]from H2Asc : For any second dissocation,

This is even less than [H3O+]from H2O

, so the assumption is justifed.

2. [H2Asc]dissoc << [H2Asc]init:

7.1x10-4 mol/L

0.050 mol/Lx 100 = 1.4% (< 5%; assumption is justified).

18-56

Weak Bases

A Brønsted-Lowry base is a species that accepts an H+. For a weak base that dissolves in water:

B(aq) + H2O(l) BH+(aq) + OH-(aq)

The base-dissociation or base-ionization constant is given by:

[BH+][OH-]

[B]Kb =

Note that no base actually dissociates in solution, but ions are produced when the base reacts with H2O.

18-57

Figure 18.9 Abstraction of a proton from water by the base methylamine.

Lone pair of N pair binds H+

18-58

Table 18.6 Kb Values for Some Molecular (Amine) Bases at 25°C

18-59

Sample Problem 18.11 Determining pH from Kb and Initial [B]

PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent manufacture, has a Kb of 5.9x10-4. What is the pH of 1.5 mol/L (CH3)2NH?

SOLUTION:

PLAN: We start with the balanced equation for the reaction of the amine with H2O, remembering that it is a weak base. We then write the expression for Kb, set up a reaction table and solve for [OH-]. From [OH-] we can calculate [H3O+] and pH.We make similar assumptions to those made for weak acids. Since Kb >> Kw, the [OH-] from H2O is neglible. Since Kb is small, we can assume that the amount of amine reacting is also small, so [(CH3)2NH] ≈ [(CH3)2NH]init.

(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)

Kb = [(CH3)2NH2

+][OH-]

[(CH3)2NH]

18-60


Initial 1.50 0 0-

Change -x - +x +x

(CH3)2NH(aq) + H2O(l) (CH3)2NH2+(aq) + OH-(aq)Concentration

(mol/L)

Equilibrium 1.50 - x - x x

Kb = [(CH3)2NH2

+][OH-]

[(CH3)2NH]= 5.9x10-4 ≈

x2

1.5

Since Kb is small, x << 1.50 and 1.50 – x ≈ 1.50

x = [OH-] = 3.0x10-2

mol/LCheck assumption:

3.0x10-2 mol/L

1.5 mol/Lx 100 = 2.0% (< 5%; assumption is justified).

18-61

pH = -log (3.3x10-13) = 12.48


[H3O+] =Kw

[OH-]=

1.0x10-14

3.0x10-2= 3.3x10-13 mol/L

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Anions of Weak Acids as Weak Bases

The anions of weak acids often function as weak bases.

A-(aq) + H2O(l) HA(aq) + OH-(aq) Kb = [HA][OH-]

[A-]

A solution of HA is acidic, while a solution of A- is basic.

HF(aq) + H2O(l) H3O+(aq) + F-(aq)

HF is a weak acid, so this equilibrium lies to the left.[HF] >> [F-], and [H3O+]from HF >> [OH-] ;the solution is therefore acidic.

from H2O

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F-(aq) + H2O(l) HF(aq) + OH-(aq)

If NaF is dissolved in H2O, it dissolves completely, and F- can act as a weak base:

HF is a weak acid, so this equilibrium also lies to the left.[F- ] >> [HF], and [OH- ] >> [H3O+ ] ;the solution is therefore basic.

from H2Ofrom F-

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Ka and Kb for a Conjugate Acid-Base Pair

HA + H2O H3O+ + A-

A- + H2O HA + OH-

2H2O H3O+ + OH-

= [H3O+][OH-]

Kc for the overall equation = K1 x K2, so

[H3O+][A-]

[HA]

[HA][OH-]

[A-]x

Ka x Kb = Kw

This relationship is true for any conjugate acid-base pair.

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Sample Problem 18.12 Determining the pH of a Solution of A-

PROBLEM: Sodium acetate (CH3COONa, or NaAc for this problem) has applications in photographic development and textile dyeing. What is the pH of 0.25 mol/L NaAc? Ka of acetic acid (HAc) is 1.8x10-5.

SOLUTION:

PLAN: Sodium salts are soluble in water and acetate is the anion of HAc so it acts as a weak base. We write the base dissociation equation and the expression for Kb, and solve for [OH-]. We recall that any soluble ionic salt dissociates completely in solution, so [Ac-]init = 0.25 mol/L.

Kb = [HAc][OH-]

[Ac-]

Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)

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Initial 0.25 - 0 0

Change -x +x +x-

Kb =Kw

Ka

= 5.6x10-10

mol/L=

1.0x10-14

1.8x10-5

Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)Concentration (mol/L)

Equilibrium -0.25 - x x x

Kb = 5.6x10-10 = [HAc][OH-]

[Ac-]≈

x2

0.25 so x = [OH-] = 1.2x10-5 mol/L

Checking the assumption:

1.0x10-14

1.2x10-5

x 100 = 4.8x10-3% (<5%; assumption is justified)

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pH = -log (8.3x10-10) = 9.08

[H3O+] =Kw

[OH-]=

1.0x10-14

1.2x10-5= 8.3x10-10 mol/L

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Acid Strength of Nonmetal Hydrides

For nonmetal hydrides (E-H), acid strength depends on:

• the electronegativity of the central nonmetal (E), and

• the strength of the E-H bond.

Across a period, acid strength increases.

Electronegativity increases across a period, so the acidity of E-H increases.

Down a group, acid strength increases.

The length of the E-H bond increases down a group and its bond strength therefore decreases.

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6A(16)

H2O

H2S

H2Se

H2Te

7A(17)

HF

HCl

HBr

HI

Electronegativity increases, so acidity increases

Bo

nd

str

en

gth

d

ecr

ea

ses

, so

ac

idit

y

incr

ea

ses

Figure 18.10 The effect of atomic and molecular properties on nonmetal hydride acidity.

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Acid Strength of Oxoacids

All oxoacids have the acidic H bonded to an O atom.

Acid strength of oxoacids depends on:

• the electronegativity of the central nonmetal (E), and

• the number of O atoms around E.

For oxoacids with the same number of O atoms, acid strength increases as the electronegativity of E increases.

For oxoacids with different numbers of O atoms, acid strength increases with the number of O atoms.

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Figure 18.11 The relative strengths of oxoacids.

H O I H O Br< < H O Cl+ - + - + -

A

Electronegativity increases, so acidity increases.

H O Cl <+ - + -

H O Cl

O

O

O

B

Number of O atoms increases, so acidity increases.

18-72

Hydrated Metal Ions

Some hydrated metal ions are able to transfer an H+ to H2O. These metal ions will form acidic solutions.

M(H2O)xn+(aq) + H2O(l) M(H2O)x-1OH(n-1)

(aq) + H3O+(aq)

Mn+(aq) + H2O(l) → M(H2O)xn+(aq)

Consider a metal ion in solution, Mn+:

If Mn+ is small and highly charged, it will withdraw enough e- density from the O-H bonds of the bound H2O molecules to release H+:

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Figure 18.12 The acidic behavior of the hydrated Al3+ ion.

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Salts that Yield Neutral Solutions

A salt that consists of the anion of a strong acid and the cation of a strong base yields a neutral solution.

NaNO3

Na+ is the cation of NaOH, a strong base.

NO3- is the anion of

HNO3, a strong acid.

This solution will be neutral, because neither Na+ nor NO3

- will react with H2O to any great extent.

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A salt that consists of the anion of a strong acid and the cation of a weak base yields an acidic solution.

NH4Cl

NH4 + is the cation of NH3, a weak base.

Cl- is the anion of HCl, a strong acid.

This solution will be acidic, because NH4+ will react with

H2O to produce H3O+:

NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

Salts that Yield Acidic Solutions

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A salt that consists of the anion of a weak acid and the cation of a strong base yields a basic solution.

CH3COONa

CH3COO- is the anion of CH3COOH, a weak acid.

Na+ is the cation of NaOH, a strong base.

This solution will be basic, because CH3COO- will react with H2O to produce OH-:

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

Salts that Yield Basic Solutions

18-77

Sample Problem 18.13 Predicting Relative Acidity of Salt Solutions from Reactions of the Ions with Water

PROBLEM: Predict whether aqueous solutions of the following are acidic, basic, or neutral, and write an equation for the reaction of any ion with water:

PLAN: We identify the cation and anion from the formula for each salt. Depending on an ion’s ability to react with water, the solution will be neutral (strong-acid anion with strong-base cation), acidic (weak-base cation with strong-acid anion), or basic (weak-acid anion and strong-base cation).

(a) Potassium perchlorate, KClO4 (b) Sodium benzoate, C6H5COONa

(c) Chromium(III) nitrate, Cr(NO3)3

SOLUTION:

(a) K+ is the cation of a strong base (KOH) while ClO4- is the anion of

a strong acid (HClO4). This solution will be neutral.

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(b) Na+ is the cation of a strong base (NaOH) while the benzoate anion (C6H5COO-) is the anion of a weak acid (benzoic acid). The benzoate ion will react with H2O to produce OH- ions:

C6H5COO-(aq) + H2O(l) C6H5COOH(aq) + OH-(aq)

This solution will be basic.

(c) NO3- is the anion of a strong acid (HNO3) and will not react with

H2O to any great extent. Cr3+ is a small metal cation with a fairly high charge density. It will become hydrated and the hydrated ion will react with H2O to form H3O+ ions:

Cr(H2O)63+(aq) + H2O(l) Cr(H2O)5OH2+(aq) + H3O+(aq)

This solution will be acidic.

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If a salt that consists of the anion of a weak acid and the cation of a weak base, the pH of the solution will depend on the relative acid strength or base strength of the ions.

NH4CN

NH4+ is the cation of a weak base, NH3.

CN- is the anion of a weak acid, HCN.

Salts of Weakly Acidic Cations and Weakly Basic Anions

CN-(aq) + H2O(l) HCN(aq) + OH-(aq)

NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)

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Ka of NH4+ = Kw

Kb of NH3

1.0x10-14

1.76x10-5 = = 5.7x10-10

Kb of CN- = Kw

Ka of HCN

1.0x10-14

6.2x10-10 = = 1.6x10-5

The reaction that proceeds farther to the right determines the pH of the solution, so we need to compare the Ka of NH4

+ with the Kb of CN-.

Since Kb of CN- > Ka of NH4+, CN- is a stronger base than

NH4+ is an acid. A solution of NH4CN will be basic.

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Table 18.7 The Acid-Base Behavior of Salts in Water

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Sample Problem 18.14 Predicting the Relative Acidity of Salt Solutions from Ka and Kb of the Ions

PROBLEM: Determine whether an aqueous solution of zinc formate, Zn(HCOO)2, at 25°C is acidic, basic, or neutral.

SOLUTION:

PLAN: Zn2+ is a small, highly charged metal cation, while HCOO- is the anion of a weak acid. Both will react with H2O, so to determine the acidity of the solution we must compare the Ka of the hydrated Zn2+ ion with the Kb of the HCOO- ion.

Ka of Zn(H2O)62+ = 1x10-9 (from Appendix C.)

Ka for Zn(H2O)62+ >> Kb HCOO-, therefore the solution is acidic.

Zn(H2O)62+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq)

HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)

Kb of HCOO- =Kw

Ka of HCOOH

1.0x10-14

1.8x10-4 = = 5.6x10-11

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The Leveling Effect

H2O exerts a leveling effect on any strong acid or base.

All strong acids and bases are equally strong in water.

All strong acids dissociate completely to form H3O+, while all strong bases dissociate completely to form OH-.

In water, the strongest acid possible is H3O+ and the strongest base possible is OH-.

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The Lewis Acid-Base Definition

A Lewis base is any species that donates an electron pair to form a bond.

A Lewis acid is any species that accepts an electron pair to form a bond.

The Lewis definition views an acid-base reaction as the donation and acceptance of an electron pair to form a covalent bond.

B + H+ B H+

base acid adduct

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Lewis Acids and Bases

The Lewis definition expands the classes of acids.

A Lewis base must have a lone pair of electrons to donate.Any substance that is a Brønsted-Lowry base is also a Lewis base.

A Lewis acid must have a vacant orbital (or be able to rearrange its bonds to form one) to accept a lone pair and form a new bond.Many substances that are not Brønsted-Lowry acids are Lewis acids.

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Electron-Deficient Molecules as Lewis Acids

B and Al often form electron-deficient molecules, and these atoms have an unoccupied p orbital that can accept a pair of electrons:

BF3 accepts an electron pair from ammonia to form a covalent bond.

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Lewis Acids with Polar Multiple Bonds

Molecules that contain a polar multiple bond often function as Lewis acids:

The O atom of an H2O molecule donates a lone pair to the S of SO2, forming a new S‒O σ bond and breaking one of the S‒O bonds.

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Metal Cations as Lewis Acids

A metal cation acts as a Lewis acid when it dissolves in water to form a hydrated ion:

The O atom of an H2O molecule donates a lone pair to an available orbital on the metal cation.

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Figure 18.13 The Mg2+ ion as a Lewis acid in chlorophyll.

18-90

Sample Problem 18.15 Identifying Lewis Acids and Bases

SOLUTION:

PLAN: We examine the formulas to see which species accepts the electron pair (Lewis acid) and which donates it (Lewis base) in forming the adduct.

PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:

(a) H+ + OH- H2O (b) Cl- + BCl3 BCl4- (c) K+ + 6H2O K(H2O6)+

(a) The H+ ion accepts the electron pair from OH-. H+ is the Lewis acid and OH- is the Lewis base.

(b) BCl3 accepts an electron pair from Cl-. Cl- is the Lewis base and BCl3 is the Lewis acid.

(c) An O atom from each H2O molecule donates an electron pair to K+. H2O is therefore the Lewis base, and K+ is the Lewis acid.

18-1 CHEM 1B General Chemistry Ch. 18 Acid-Base Equilibria Instructor: Dr. Orlando E. Raola Santa...

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Transcript of 18-1 CHEM 1B General Chemistry Ch. 18 Acid-Base Equilibria Instructor: Dr. Orlando E. Raola Santa...