Acids and Bases Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for...

Acids and Bases

Chapter 14

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon dioxide gas

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Bases

4.3

Arrhenius acid is a substance that produces H+ (H3O+) in water

A Brønsted acid is a proton donor

A Lewis acid is a substance that can accept a pair of electrons

A Lewis base is a substance that can donate a pair of electrons

Definition of An Acid

H+ H O H••••

+ OH-••••••

acid base

N H••

H

H

H+ +

acid base15.12

N H

H

H

H+

Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

4.3

A Brønsted acid is a proton donorA Brønsted base is a proton acceptor

acidbase acid base

15.1

acidconjugate

basebase conjugate

acid

Lewis Acids and Bases

N H••

H

H

acid base

F B

F

F

+ F B

F

F

N H

H

H

No protons donated or accepted!

15.12

Water as an Acid and a Base

Water is amphoteric (it can behave either as an acid or a base).

H2O + H2O H3O+ + OH

conj conj acid 1 base 2 acid 2 base 1

Kw = 1 × 1014 at 25°C

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

7

O

H

H + O

H

H O

H

H H OH-+[ ] +

Acid-Base Properties of Water

H2O (l) H+ (aq) + OH- (aq)

H2O + H2O H3O+ + OH-

acid conjugate

base

base conjugate

acid

15.2

autoionization of water

H2O (l) H+ (aq) + OH- (aq)

The Ion Product of Water

Kc =[H+][OH-]

[H2O][H2O] = constant

Kc[H2O] = Kw = [H+][OH-]

The ion-product constant (Kw) is the product of the molar concentrations of H+ and OH- ions at a particular temperature.

At 250CKw = [H+][OH-] = 1.0 x 10-14

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

15.2

What is the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3 M?

Kw = [H+][OH-] = 1.0 x 10-14

[H+] = 1.3 M

[OH-] =Kw

[H+]

1 x 10-14

1.3= = 7.7 x 10-15 M

15.2

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-]

[H+] > [OH-]

[H+] < [OH-]

Solution Is

neutral

acidic

basic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7

pH < 7

pH > 7

At 250C

pH [H+]

15.3

15.3

pOH = -log [OH-]

[H+][OH-] = Kw = 1.0 x 10-14

-log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

15.3

pH calculations – Solving for H+pH calculations – Solving for H+pH calculations – Solving for H+pH calculations – Solving for H+

If the pH of Coke is 3.12, [HIf the pH of Coke is 3.12, [H++] = ???] = ???

Because pH = - log [HBecause pH = - log [H++] then] then

- pH = log [H- pH = log [H++]]

Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get

1010-pH -pH == [H[H++]]

[H[H++] = 10] = 10-3.12-3.12 = 7.6 x 10 = 7.6 x 10-4-4 M M

*** to find antilog on your calculator, look for “Shift” *** to find antilog on your calculator, look for “Shift” or “2or “2nd nd function” and then the log buttonfunction” and then the log button

Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

15.4

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

Weak Acids are weak electrolytes


HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

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

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

15.4

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

Weak Bases are weak electrolytes

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Conjugate acid-base pairs:

• The conjugate base of a strong acid has no measurable strength.

• H3O+ is the strongest acid that can exist in aqueous solution.

• The OH- ion is the strongest base that can exist in aqeous solution.

15.4

Strong Acid Weak Acid

15.4

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

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-][HA]

Ka is the acid ionization constant

Ka

weak acidstrength

15.5

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka ≈x2

0.50= 7.1 x 10-4

0.50 – x ≈ 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M15.5

When can I use the approximation?

0.50 – x ≈ 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0190.019 M0.50 M

x 100% = 3.8%Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka ≈x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or method of successive approximation. 15.5

Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization of water.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

15.5

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka ≈x2

0.122= 5.7 x 10-4

0.122 – x ≈ 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 M

x 100% = 6.8%More than 5%

Approximation not ok.

15.5

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac √

2ax =

x = 0.0081 x = - 0.0081


Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

15.5

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

15.5

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.


pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.615.4

Weak BasesWeak Bases

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Kb

weak basestrength

15.6

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Equilibrium Equilibrium Constants Constants

for Weak Basesfor Weak Bases

Equilibrium Equilibrium Constants Constants

for Weak Basesfor Weak Bases

Weak base has KWeak base has Kbb < 1 < 1

Leads to small [OHLeads to small [OH--] and a pH of 12 - 7] and a pH of 12 - 7

Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O ↔ ↔ NHNH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx

Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44

++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M

The approximation is validThe approximation is valid !!

Equilibria Involving A Equilibria Involving A Weak BaseWeak Base

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 3.Step 3. Calculate pHCalculate pH

[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M

so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37

Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

15.7

Ionization Constants of Conjugate Acid-Base Pairs


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

Ka

Kb

H2O (l) H+ (aq) + OH- (aq) Kw

KaKb = Kw

Weak Acid and Its Conjugate Base

Ka = Kw

Kb

Kb = Kw

Ka

Molecular Structure and Acid Strength

H X H+ + X-

The stronger the bond

The weaker the acid

HF << HCl < HBr < HI

15.9


Z O H Z O- + H+- +

The O-H bond will be more polar and easier to break if:

• Z is very electronegative or

• Z is in a high oxidation state

15.9


1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.

Acid strength increases with increasing electronegativity of Z

H O Cl O

O••

••••••

••

••••

••••

H O Br O

O••

••••••

••

••••

••••Cl is more electronegative than Br

HClO3 > HBrO3

15.9


2. Oxoacids having the same central atom (Z) but different numbers of attached groups.

Acid strength increases as the oxidation number of Z increases.

HClO4 > HClO3 > HClO2 > HClO

15.9

Acid-Base Properties of SaltsEXAMINE CATION OF SALT:If CATION came from a strong BASE it would make a neutral soln

weak BASE it would make an Acidic soln

EXAMINE ANION OF SALT:If ANION came from a strong ACID it would make a neutral soln

weak ACID it would make an Basic soln

Acid-Base Properties of Salts


43

Salt comes from SA & SB makes Neutral Salt

Salt comes from SA & WB makes Acidic Salt

Salt comes from WA & SB makes Basic Salt

Salt comes from WA & WB makes ? Saltneed to check

Ka vs Kb

CATION ANION ACIDIC OR BASIC

EXAMPLE

NEUTRAL NEUTRAL NEUTRAL NaCl

NEUTRAL CONJUGATE BASE OF A WEAK ACID

BASIC NaF

CONJUGATE ACID OF A WEAK BASE

NEUTRAL ACIDIC NH4Cl

CONJUGATE ACID OF A WEAK BASE

CONJUGATE BASE OF A WEAK ACID

DEPENDS ON Ka & Kb VALUES

Al2(SO4)3

Acid-Base Properties of SaltsNeutral Solutions:

Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3

-).

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Basic Solutions:

Salts derived from a strong base and a weak acid.

NaCH3COO (s) Na+ (aq) + CH3COO- (aq)H2O

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

15.10


Acid Solutions:

Salts derived from a strong acid and a weak base.

NH4Cl (s) NH4+ (aq) + Cl- (aq)

H2O

NH4+ (aq) NH3 (aq) + H+ (aq)

Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.

Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+

15.10


Solutions in which both the cation and the anion hydrolyze:

• Kb for the anion > Ka for the cation, solution will be basic

• Kb for the anion < Ka for the cation, solution will be acidic

• Kb for the anion ≈ Ka for the cation, solution will be neutral

15.10

Polyprotic Acids

. . . can furnish more than one proton (H+) to the solution.


48

H CO H HCO ( )

HCO H CO ( )

2 3 3 a

3 32

a

1

2

K

K

Calculating Concentrations of Species in a Solution of a Diprotic Acid

• Ascorbic acid (vitamin C) is a diprotic acid, H2C6H6O6. What is the pH of a 0.10 M solution? What is the concentration of ascorbate ion, C6H6O6

2-?

• The acid ionization constants are Ka1 = 7.9 x 10-5 and Ka2 = 1.6 x 10-12.

Consider first ionization• H2Asc(aq) <==> H+(aq) + Hasc-(aq)

• I 0.10 0 0• C -x +x +x • E 0.10 - x x x

[H3O+] [HAsc-] = Ka1 [H2Asc] x2 = 7.9 x 10-5 0.10 - x

x = 2.8 x 10-3 = 0.0028 = [H+] = [HAsc-] pH = -log [H+] = -log(0.0028) = 2.55

• Hasc- (aq) <==> H+ (aq) + Asc- (aq)

• I 0.0028 0.0028 0• C -x +x +x • E 0.0028- x 0.0028+ x x

[H3O+][Asc2-] = Ka2

[HAsc-](0.0028) x = 1.6 x 10-12

0.0028

x = 1.6 x 10-12 = [H+]

The 2nd [H+] always equals Ka2

• From 1st Ionization we get [H+] = 2.8 x 10-3 M• From 2nd Ionization we get [H+] = 1.6 x 10-12 M

• If you add together: • [H+] = 2.8000000016 x 10-3 M• Same pH, 1st Ionization always contributes

most [H+]

OXIDES

• Acidic Oxides (Acid Anhydrides): OX bond is strong and covalent.

• SO2, NO2, CrO3

• Basic Oxides (Basic Anhydrides): OX bond is ionic.

• K2O, CaO

Oxides of the Representative ElementsIn Their Highest Oxidation States

15.11

CO2 (g) + H2O (l) H2CO3 (aq)

N2O5 (g) + H2O (l) 2HNO3 (aq)

Chemistry In Action: Antacids and the Stomach pH Balance

NaHCO3 (aq) + HCl (aq)

NaCl (aq) + H2O (l) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq)

MgCl2 (aq) + 2H2O (l)

Acids and Bases Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for...

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Transcript of Acids and Bases Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for...