15 ACID-BASE EQUILIBRIAgencheminkaist.pe.kr/Lecturenotes/CH101/Chap15_2021.pdf · 2021. 2. 15. ·...

General Chemistry II

1

ACID-BASE EQUILIBRIA15CHAPTER


15.1 Classifications of Acids and Bases

15.2 Properties of Acids and Bases in Aqueous

Solutions: The Brønsted-Lowry Scheme

15.3 Acid and Base Strength

15.4 Equilibria Involving Weak Acids and Bases

15.5 Buffer Solutions

15.6 Acid-Base Titration Curves

15.7 Polyprotic Acids

15.8 Organic Acids and Bases: Structure and Reactivity

15.9 Exact Treatment of Acid-Base Equilibria


2

Cyanidin is blue in the basic sap of the cornflower

and red in the acidic sap of the poppy.


3

Acid Base

Arrhenius [H3O+ ] > KW

1/2 [OH- ] > KW1/2

Brønsted-Lowry donates H+ accepts H+

Lewis accepts donates

lone-pair electrons lone-pair electrons

15.1 CLASSIFICATIONS OF ACIDS AND BASES


4

Arrhenius Acids and Bases

Acid: A substance that, when dissolved in water, increases the

concentration of hydronium ion (H3O+) above the value in

pure water.

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

Base: A substance that increases the concentration of hydroxide

ion (OH–).

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


5

Acid: A substance that can donate a proton

Base: A substance that can accept a proton

Brønsted-Lowry conjugated acid-base pairs:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

acid1 base2 base1 acid2

Brønsted-Lowry Acids and Bases


6


7

Lewis Acids and Bases

Acid: Any species that accepts lone-pair electrons

Base: Any species that donates lone-pair electrons

Competition between two bases for a proton by offering

electron pairs:

2 3

2 11 2

HF( ) H O ( )

acid

H O( ) F ( )

bas acid

e ba

se

aq a aql q

Reactions without proton transfers

~ Octet-deficient compound (BF3) ← strong Lewis acid


8


9

~ removing H2O from oxoacids or hydroxides

Acid anhydrides: Oxides of most of the nonmetals

N2O5(s) + H2O(l) HNO3(aq)

Base anhydrides: Oxides of Group I & II metals

CaO(s) + H2O(l) Ca(OH)2(s)

Amphoteric: Oxides of Group III & V metals

Al2O3(s) + 6 H3O+(aq) 2 Al3+(aq) + 9 H2O(l)

Al2O3(s) + 2 OH- (aq) + 3 H2O(l) 2 Al(OH)4- (aq)

Anhydrides of Acids and Bases


10


11

Fig. 15.2 Acidity and basicity of oxides of main group elements.


12

Autoionization of Water

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

acid1 base2 acid2 base1

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

14

w 3 [H O ][OH ] 1.0 10 K

[H3O+] = [OH-] = 1.0 x10–7 M

for pure water at 25°C

15.2 PROPERTIES OF ACIDS AND BASES IN

AQUEOUS SOLUTIONS: THE BRØ NSTED-

LOWRY SCHEME


13


14

Strong Acids and Bases

Strong Acids

~ ionizes fully in aqueous solution producing H3O+

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

base2 acid1 acid2 base1

Leveling Effect of water on HCl, HBr, HI, H2SO4, HNO3, HClO4

~ too strong to tell the difference in water


15

Strong Bases

~ ionizes fully in aqueous solution producing OH-,

amide ion (NH2-), hydride ion (H-), NaOH, ...

H2O(l) + NH2-(aq) OH-(aq) + NH3(aq)

acid1 base2 base1 acid2


16

The pH Function

10 3 pH log [H O ]

Fig. 15.4 pH’s of many everyday materials

aqueous solution at 25°C

pH + pOH = 14

pH < 7 acidic (can be negative)

pH = 7 neutral

pH > 7 basicFig. 15.3 A simple pH meter

with a digital readout.


17


18


19

Hydrolysis (ionization) of a weak acid

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

Acid ionization (hydrolysis) constant, Ka

The stronger an acid, the larger Ka (smaller pKa).

pKa of strong acids < 0

pKa of H3O+ = 0

pKa of weak acids > 0

pKa of H2O = 14

15.3 ACID AND BASE STRENGTH

𝐾𝑎 =𝐻3𝑂

+ 𝐴−

𝐻𝐴, 𝑝𝐾𝑎 = − log10 𝐾𝑎


20


21


22

Hydrolysis of a weak base

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

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

[HA] [OH ][H O ]

[H

[HA][OH ]=

[A ] [ O

[HA]

[H O ][A] ]A ]K

+

3

+

w

a3

b +

3

w

-

-

-

- -

K K

K

w a b w a b , p p p K K K K K K

Base hydrolysis constant, Kb

𝐾𝑏 =𝐻𝐴 𝑂𝐻−

𝐴−, 𝑝𝐾𝑏 = − log10 𝐾𝑏


23

Fig. 15.5 The relative strength

of some acids and their

conjugate bases.


24 Competition of two weak acids

~ Prediction of the direction of net hydrogen ion transfer

HF(aq) + H2O(l) H3O+(aq) + F-(aq), Ka = 6.6×10-4

HCN(aq) + H2O(l) H3O+(aq) + CN-(aq), Ka' = 6.17×10-10

HF a stronger acid than HCN

→ Equilibrium is strongly to the right.

HF(aq) + CN-(aq) HCN(aq) + F-(aq), K = Ka/Ka' =1.1×106

acid1 base2 acid2 base1

(strong) (strong) (weak) (weak)


25


26


27

Molecular Structure and Acid Strength

Fig. 15.6 (a) Basic compound, electropositive X, breaking X – O bond.

(b) Acidic compound, electronegative X, breaking O – H bond.

–X–O–H group ~ Oxoacid (electronegativity of X, pKa)

NaOH (0.93, basic)

HClO3 (3.16, –3) > HNO3 (3.04, –1.3) > HIO3 (2.66, 0.80)

H3PO4 (2.19, 2.12) > H3AsO4 (2.18, 2.30)

H2SO3 (2.58, 1.81) > H2CO3 (2.55, 6.37)


28

Fig. 15.7 Lewis diagram for H3PO3. (a) Wrong triprotic structure.

(b) Correct diprotic structure. Assignment of the formal charge to P

and the lone O. P – H bond is not breaking.


29

Indicators

Organic weak acid that has a different color from its

conjugate base

HIn(aq) + H2O(l) H3O+(aq) + In-(aq)

+[H O ][In ]

[HIn]3

a

K K

3

a

+[H O ][HIn]=

[In ]


30

Indicators

Organic weak acid that has a different color from its

conjugate base

HIn(aq) + H2O(l) H3O+(aq) + In-(aq)

+[H O ][In ]

[HIn]3

a

K K

3

a

+[H O ][HIn]=

[In ]

Range of color change: pH ~ pKa ± 1

Fig. 15.8 bromophenol red, thymolphthalein, phenolphthalein, bromocresol green


31

Fig. 15.9 Indicators change

their colors at very different

pH values.


32

Fig. 15.10 Natural indicator: Red cabbage extract in a natural pH indicator.

The color changes from red to violet to yellow as the solution becomes

less acidic.


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Weak Acids

HOAc H3O+ AcO–

--------------------------------------------------------------------------Initial 1.000 ~ 0 0Change –y + y + y

---------------- ------ -----Equilibrium 1.000 – y y y--------------------------------------------------------------------------

+[H O ][Ac ]1.76 10

[HAc] 1.000

253

a

Ky

y

→ y = 4.20×10–3

[H3O+] = y = 4.20×10–3 M → pH = 2.38

𝑦2

1000 − 𝑦≈

𝑦2

1000= 1.76 × 10−5

Fraction ionized = [Ac–] / [HAc]0 = y / 1.000 = 4.20×10–3 → 0.42%

EXAMPLE 15.6

15.4 EQUILIBRIA INVOLVING WEAK ACIDS

AND BASES

Calculate the pH and the fraction of HOAc ionized at equilibrium.

HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)

1 M


34

HOAc H3O+ AcO–


---------------- ------ -----Equilibrium 0.00100 – y y y--------------------------------------------------------------------------


35


36

Weak Bases

H2O(l) + NH3(aq) NH4+(aq) + OH–(aq)

--------------------------------------------------------------------------

Initial 0.0100 0 ~ 0

Change –y + y + y

---------------- ------ -----

Equilibrium 0.0100 – y y y

--------------------------------------------------------------------------

+[NH ][OH ]

1.8 10[NH ] 0.0100

254

b

3

Ky

y

y = 4.15 ×10–4 M = [OH–]

[H3O+] = Kw / [OH–] = 2.4 ×10–11 M → pH = 10.62

EXAMPLE 15.8 Calculate the pH of an aqueous solution of ammonia.0.01 M


37

Hydrolysis

AcO–(aq) + H2O(l) HOAc(aq) + OH–(aq) --------------------------------------------------------------------------

Initial 0.100 0 ~ 0

Change –y + y + y

---------------- ------ -----


--------------------------------------------------------------------------

EXAMPLE 15.9 Hydrolysis of NaOAc

NaOAc(s) + H2O(l) Na+(aq) + AcO–(aq)

0.1 M


38

Hydrolysis of a weak base

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

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

[HA] [OH ][H O ]

[H

[HA][OH ]=

[A ] [ O

[HA]

[H O ][A] ]A ]K

+

3

+

w

a3

b +

3

w

-

-

-

- -

K K

K

w a b w a b , p p p K K K K K K

Base hydrolysis constant, Kb

𝐾𝑏 =𝐻𝐴 𝑂𝐻−

𝐴−, 𝑝𝐾𝑏 = − log10 𝐾𝑏


39

Hydrolysis


Initial 0.100 0 ~ 0

Change –y + y + y

---------------- ------ -----


--------------------------------------------------------------------------

y = 7.5 × 10–6 M = [OH–]

[H3O+] = Kw / [OH–] = 1.3 × 10–9 M → pH = 8.89

[HAc][OH ]5.7 10

[Ac ] 0.100

2

w

a

10

b

KK

K

y

y



0.01 M


40

Buffer solution ~ maintains an approximately constant pH

Weak acid + Salt containing its conjugate base

(eg. HOAc/NaOAc)

15.5 BUFFER SOLUTIONS

- Controlling the solubility of ions

- Maintaining pH of biochemical and physiological

processes

blood pH 7.4 (7.0 – 7.8)


41

Calculations of Buffer Action

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)

---------------------------------------------------------------------------------------

Initial 1.00 ~ 0 0.500

Change –y + y + y

--------------- ------ -------------

Equilibrium 1.00 – y y 0.500 + y

---------------------------------------------------------------------------------------

EXAMPLE 15.10 Calculate the pH of a solution of HCOOH and NaHCOO.

Buffer solution of HCOOH (1.00 mol) / NaHCOO (0.500 mol) in 1L

[H O ][HCOO ]1.77 10

[HCOOH] 1.00

+43

a

0.500

Ky +y

y

1.77 101.00 1.00

40.500 0.500y +y y

y

y = [H3O+] = 3.54×10–4 M → pH = 3.45


42


43

Testing the buffer strength.


+ 0.10 mol of HCl

EXAMPLE 15.11

1. Before considering ionization of HCOOH….

HCl ionizes completely → reacts with HCOO– to give HCOOH

[HCOO–]0 = 0.500 – 0.10 = 0.40 M

[HCOOH]0 = 1.00 + 0.10 = 1.10 M

2. Now consider ionization of HCOOH


-----------------------------------------------------------------------------------------Initial 1.10 ~ 0 0.40Change –y + y + y

--------------- ------ -------------Equilibrium 1.10 – y y 0.40 + y-----------------------------------------------------------------------------------------


44

K+

43a

0.40[H O ][HCOO ]1.77 10

[HCOOH]

y +y

y1.10

40.40 0.401.77 10

y +y y

y1.10 1.10

y = [H3O+] = 4.9 ×10–4 M → pH = 3.31

Addition of 0.100 mol HCl to

Buffer solution of Ex. 15.10: pH = 3.45 → 3.31

Pure water: pH = 7 → 1


45


46

Designing Buffers

2 3HA( ) H O( ) H O ( ) A ( )aq l aq aq

3a

03

0

[H O ][A ] [H O ][A ]

[HA] [HA]K

a 100

0

[HA] pH p log

[A ]K

Determining pH of the buffer solution

1. Choose a weak acid whose pKa ≈ pH

2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0

Henderson-Hasselbalch Equation


47


48


49

a 100

0

[HA] pH p log

[A ]K


50

Capacity of the Buffer Solution

Fig. 15.12 pH change of buffer solutions as a strong base (NaOH) is added.

Red line: 100 mL of a buffer that is 0.1 M in both HAc and Ac–.

Blue line: 100 mL of a buffer that is 1.0 M in both HAc and Ac–.


51

Titration of a Strong Acid

with a Strong Base

Titration of 100.0 mL of

0.1000 M HCl with

0.1000 M NaOH at 25°C

H3O+(aq) + OH–(aq)

2 H2O(l)

15.6 ACID-BASE TITRATION CURVES


52

Region I: Before the equivalence point

The pH determined by the excess H3O+.

2. V = 30.00 mL NaOH added

n(OH–) = (0.1000 mol/L)(0.0300 L) = 3.000×10–3 mol

n(H3O+) = (1.000×10–2 – 3.000×10–3 ) mol = 7.00×10–3 mol

Volume increased: Vtot = 100.0 mL + 30.00 mL = 0.1300 L

[H3O+] = n(H3O

+) / Vtot = (7.00×10–3 mol) / (0.1300 L)

= 0.0538 M → pH = 1.27

1. V = 0 mL NaOH added

[H3O+] = 0.1000 M → pH = 1.00

n(H3O+) = (0.1000 mol/L)(0.1000 L) = 1.000×10–2 mol


53

Region II : At the equivalence point

The pH determined by the dissociation of water.

3. V = 100.0 mL NaOH added → pH = 7.00

Region III: Beyond the equivalence point

The pH determined by the excess OH–.

4. V = 100.05 mL NaOH added

n(OH–) = (0.1000 mol/L)(5×10–5 L) = 5×10–6 mol

Vtot = 0.1000 L (HCl) + 0.10005 L (NaOH)

= 0.20005 L

[OH–] = n(OH–) / Vtot = (5×10–5 mol) / (0.20005 L) = 2.5×10–5 M

→ [H3O+] = 4×10–10 M → pH = 9.4


54 Titration of a Weak Acid with a Strong Base

At the equivalence point,

c0V0 = ctVe (monoprotic acid)

c0 : concentration of weak acid

V0 : volume of acid originally present

ct : concentration of OH– in the base titrant

Ve : volume of the base at the equivalence point

Titration of 100.0 mL of 0.1000 M HOAc

with 0.1000 M NaOH at 25°C

H3O+(aq) + OH–(aq) 2 H2O(l)

Region I: Initial solution (Weak acid solution)

1. V = 0 mL NaOH added → pH of 0.100 M HOAc solution

[H3O+] = 1.32×10–3 → pH = 2.88


55

Weak Acids

HOAc H3O+ AcO–


---------------- ------ -----Equilibrium 1.000 – y y y--------------------------------------------------------------------------

+[H O ][Ac ]1.76 10

[HAc] 1.000

253

a

Ky

y

→ y = 4.20×10–3

[H3O+] = y = 4.20×10–3 M → pH = 2.38

𝑦2

1000 − 𝑦≈

𝑦2

1000= 1.76 × 10−5

Fraction ionized = [Ac–] / [HAc]0 = y / 1.000 = 4.20×10–3 → 0.42%

EXAMPLE 15.6

15.4 EQUILIBRIA INVOLVING WEAK ACIDS

AND BASES

Calculate the pH and the fraction of HOAc ionized at equilibrium.

HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)

1 M


56

HOAc H3O+ AcO–


---------------- ------ -----Equilibrium 0.00100 – y y y--------------------------------------------------------------------------


57 Titration of a Weak Acid with a Strong Base

At the equivalence point,

c0V0 = ctVe (monoprotic acid)

c0 : concentration of weak acid

V0 : volume of acid originally present

ct : concentration of OH– in the base titrant

Ve : volume of the base at the equivalence point

Titration of 100.0 mL of 0.1000 M HOAc

with 0.1000 M NaOH at 25°C

H3O+(aq) + OH–(aq) 2 H2O(l)

Region I: Initial solution (Weak acid solution)

1. V = 0 mL NaOH added → pH of 0.100 M HOAc solution

[H3O+] = 1.32×10–3 → pH = 2.88


58

Fig. 15.14 A titration curve for the titration of a weak acid by a strong base.

100. mL of 0.1000 M HOAc is titrated with 0.1000 M NaOH.


59

Originally,

n(HOAc) = (0.1000 mol/L)(0.1000 L) = 1.000×10–2 mol

n(AcO–) generated by adding OH– :

n(AcO–) = n(OH–) = (0.1000 mol/L)(0.03000 L) = 3.000×10–3 mol

Amount of HOAc unreacted :

n(HOAc) = 1.000×10–2 mol – 3.000×10–3 mol = 7.000×10–3 mol

Region II: Before the equivalence point (Buffer solution)

2. V = 30.00 mL NaOH added ( 0 < V < Ve )

HOAc(aq) + OH–(aq) AcO–(aq) + H2O(l)

K = 1/ Kb = Ka / Kw = 2×109 >> 1


60

Volume increased to 0.1300 L

Concentrations after adding 30.00 mL of NaOH:

[HOAc] = (7.000×10–3 mol) / (0.1300 L)

= 5.38×10–2 M

[AcO–] = (3.000×10–3 mol) / (0.1300 L)

= 2.31×10–2 M


61

Calculations of Buffer Action


---------------------------------------------------------------------------------------

Initial 1.00 ~ 0 0.500

Change –y + y + y

--------------- ------ -------------

Equilibrium 1.00 – y y 0.500 + y

---------------------------------------------------------------------------------------

EXAMPLE 15.10 Calculate the pH of a solution of HCOOH and NaHCOO.


[H O ][HCOO ]1.77 10

[HCOOH] 1.00

+43

a

0.500

Ky +y

y

1.77 101.00 1.00

40.500 0.500y +y y

y

y = [H3O+] = 3.54×10–4 M → pH = 3.45


62

Designing Buffers

2 3HA( ) H O( ) H O ( ) A ( )aq l aq aq

3a

03

0

[H O ][A ] [H O ][A ]

[HA] [HA]K

a 100

0

[HA] pH p log

[A ]K

Determining pH of the buffer solution

1. Choose a weak acid whose pKa ≈ pH

2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0

Henderson-Hasselbalch Equation


63

Volume increased to 0.1300 L

Concentrations after adding 30.00 mL of NaOH:

[HOAc] = (7.000×10–3 mol) / (0.1300 L)

= 5.38×10–2 M

[AcO–] = (3.000×10–3 mol) / (0.1300 L)

= 2.31×10–2 M

K

2

0a 10 10 2

0

[HAc] 5.38 10pH p log 4.75 log

[Ac ] 2.31 104.38

→ A buffer solution of

[HOAc]0 = 5.38×10–2 M and [NaOAc]0 = 2.31×10–2 M

At V = Ve/2, [HOAc]0 = [AcO–]0 → pH = pKa


64

Region III: At the equivalence point

(equivalent to Hydrolysis of salts)

AcO– + H2O HOAc + OH–

3. V = Ve pH = 8.73

Region IV: Beyond the equivalence point

The pH determined by the excess OH–.


65

Hydrolysis


Initial 0.100 0 ~ 0

Change –y + y + y

---------------- ------ -----


--------------------------------------------------------------------------

y = 7.5 × 10–6 M = [OH–]

[H3O+] = Kw / [OH–] = 1.3 × 10–9 M → pH = 8.89

[HAc][OH ]5.7 10

[Ac ] 0.100

2

w

a

10

b

KK

K

y

y



Now 0.05 M instead of 0.1 M due to dilution


66

Problem Sets

For Chapter 15,

10, 18, 24, 36, 50, 54, 94

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