Acids and bases

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Acids and Bases

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Arrhenius Acid-Base Definition

An acid is a substance that increase H+ when dissolved in water.

A base is a substance that increase OH- ions when dissolved in water.

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

The ionization of acids actually produces H3O+ but H3O+ H2O + H+

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

Limitation: this theory cannot explain why NH3 is a base

18-4


Brønsted-Lowry Acid-Base Definition

An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.

An acid is a proton donor, any species which donates a H+.

A base is a proton acceptor, any species which accepts a H+.


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An acid and base that differ only in the presence or absence of a proton

The Conjugate Acid-Base Pairs

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Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions

Base Acid+Acid Base+

Conjugate Pair

Conjugate Pair

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

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

(a) H2PO4-(aq) + CO3

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

-(aq)

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

-(aq)

SOLUTION:

PLAN: Identify proton donors (acids) and proton acceptors (bases).

(a) H2PO4-(aq) + CO3

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

-(aq)

proton donor

proton acceptor

proton acceptor

proton donor

conjugate pair1conjugate pair2

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

-(aq)

conjugate pair2conjugate pair1

proton donor

proton acceptor

proton acceptor

proton donor

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A substance that can both accept and donate a proton– i.e. act as an acid and as a base

Ex. H2O

Amphoteric substances


Water acts as a base

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

Water acts as an acid

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Figure 18.9

Strengths of conjugate acid-

base pairs

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In every acid-base reaction, the position of equilibrium favors the weaker acid

Relative Strengths of Acids and Bases


Since H3O+ is weaker, the forward reaction is favored over the reverse reaction and the equilibrium lies to the right

Stronger acid Weaker acid

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SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base Reaction

PROBLEM: Predict the net direction of equilibrium

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

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

2-(aq) + NH4+(aq)

SOLUTION:

PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10 (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate.

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

2-(aq) + NH4+(aq)

stronger acid weaker acid

Net direction is to the right with Kc > 1.

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

stronger acidweaker acidNet direction is to the left with Kc < 1.

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Kc = [H3O+][OH-]

[H3O+][OH-]

The Ion-Product Constant for Water

Kw =

A change in [H3O+] causes an inverse change in [OH-].

= 1.0 x 10-14 at 250C

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

In an acidic solution, [H3O+] > [OH-]

In a basic solution, [H3O+] < [OH-]

In a neutral solution, [H3O+] = [OH-]

Autoionization of Water

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Figure 18.4The relationship between [H3O+] and [OH-] and the

relative acidity of solutions.

[H3O+] [OH-]Divide into Kw

ACIDIC SOLUTION

BASIC SOLUTION

[H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-]

NEUTRAL SOLUTION

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SAMPLE PROBLEM 18.2: Calculating [H3O+] and [OH-] in an Aqueous

Solution

PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?

SOLUTION:

PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].

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

[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =

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

3.3x10-11M

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Solve the following:

a. [ H+] = 1.4 x 10-6 [OH-] =

b. [ OH-] = 1.0 x 10-7 [H+] =

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-Measures the concentration of hydrogen ions-the power/potential of the hydrogen ion-a way of expressing H+ in a given solution

The pH Scale

pH = - log [H+]

The higher the H+ concentration, the lower the pH

pOH = - log [OH-]

The higher the OH- concentration, the lower the pOH

pOH + pH = 14

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Figure 18.5

The pH values of some familiar

aqueous solutions

pH = -log [H3O+]

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Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.

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SAMPLE PROBLEM 18.3: Calculating [H3O+], pH, [OH-], and pOH

PROBLEM: Given [H3O+] concentrations: 2.0M, 0.30M, and 0.0063M, Calculate pH, [OH-], and pOH of the three solutions at 250C.

SOLUTION:

PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH.

For [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30

[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48

For [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH

[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80

For [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH

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

STRONG ACIDS and BASES

Strong acids and bases dissociate completely into ions in water.

100% ionization

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Strong Acids Strong Bases

HCl LiOH

HBr NaOH

HI KOH

HNO3 Ca(OH)2

H2 SO4 Sr(OH)2

HClO4 Ba(OH)2

All other acids and bases not listed here are considered weak.

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For STRONG ACIDS,

H+ or OH- are equal to the original molarity of the solution

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

0.5 M 0.5 M0.5 M

Problem: Calculate the pH of 0.5 M HCl and 0.10 M NaOH solution:

pH = - log (0.5 M) = 0.30

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

0.10 M 0.10 M 0.10 M

pOH = - log (0.10 M) = 1.00pH = 14 – 1.00 = 13.00

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SAMPLE PROBLEM 18.5:

Calculate the pH of a solution prepared by dissolving 5.00 g of KOH in enough water to get 500.0 mL of solution

MW of KOH = 56.108 g/mol

Answer: 0.178 M of [OH-] pOH = 0.74902 pH = 13.25

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WEAK ACIDS and WEAK BASES

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

Weak acids and bases dissociate very slightly into ions in water.

- About 1% ionization, the equilibrium lies far to the left

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The Acid-Dissociation Constant, Ka

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

Kc = [H3O+][A-]

[HA]

Ka =[H3O+][A-]

[HA]

stronger acid higher [H3O+]

larger Ka

smaller Ka lower [H3O+]

weaker acid

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SAMPLE PROBLEM :

Write the equilibrium constant expression for the following weak acid:

a. NH4 +

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

Ka =[H3O+][NH3]

[NH4 + ]

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Table 18.3 The Relationship Between Ka and pKa

Acid Name (Formula) Ka at 250C pKa

Hydrogen sulfate ion (HSO4-) 1.02x10-2

Nitrous acid (HNO2)

Acetic acid (CH3COOH)

Hypobromous acid (HBrO)

Phenol (C6H5OH)

7.1x10-4

1.8x10-5

2.3x10-9

1.0x10-10

1.991

3.15

4.74

8.64

10.00

pKa = - log Ka

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The Base-Dissociation Constant, Kb

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

Kc = [OH-][HA]

[A-] stronger base higher [OH-]

larger Kb

Kb = [OH-][HA]

[A-]

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SAMPLE PROBLEM :

Write the equilibrium constant expression for the following weak base:

a. C5H5N

C5H5N (aq) + H2O(l) OH-(aq) + C5H5NH+

Kb =[C5H5NH+ ][OH-]

[C5H5N]

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The Relationship Between Ka and Kb

Ka x Kb = Kw

Example: Calculate Kb if Ka = 1.5 x 10-5

Kb =1.0 x 10-14

1.5 x 10-5

= 6.7 x 10-10

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Lewis Definition of Acids and Bases

F

B

F F

H

N

H H

+

F

B

F F

H

N

H H

acid base adduct

An acid is an electron-pair acceptor.

A base is an electron-pair donor.

M2+

H2O(l)

M(H2O)42+(aq)

adduct

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SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases

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

(a) H+ + OH- H2O

(b) Cl- + BCl3 BCl4-

(c) K+ + 6H2O K(H2O)6+

SOLUTION:

PLAN: Look for electron pair acceptors (acids) and donors (bases).

(a) H+ + OH- H2Oacceptor

donor

(b) Cl- + BCl3 BCl4-

donor

acceptor

(c) K+ + 6H2O K(H2O)6+

acceptor

donor

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Acids and bases

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