15 acid-base equilibria · 2020-06-04 · acid-base titration curves. 15.7. polyprotic acids. 15.8....

General Chemistry II

ACID-BASE EQUILIBRIA15CHAPTER


15.1 Classifications of Acids and Bases15.2 Properties of Acids and Bases in Aqueous

Solutions: The Brønsted-Lowry Scheme15.3 Acid and Base Strength15.4 Equilibria Involving Weak Acids and Bases15.5 Buffer Solutions15.6 Acid-Base Titration Curves15.7 Polyprotic Acids15.8 Organic Acids and Bases: Structure and Reactivity15.9 Exact Treatment of Acid-Base Equilibria


Cyanidin is blue in the basic sap of the cornflower and red in the acidic sap of the poppy.

669


Acid Base

Arrhenius [H3O+ ] > KW1/2 [OH- ] > KW

1/2

Brønsted-Lowry donates H+ accepts H+

Lewis accepts donates

lone-pair electrons lone-pair electrons

15.1 CLASSIFICATIONS OF ACIDS AND BASES

670


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)

670


Acid: A substance that can donate a protonBase: 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 671


Lewis Acids and Bases Acid: Any species that accepts lone-pair electronsBase: 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

674


~ removing H2O from oxoacids or hydroxides

Acid anhydrides: Oxides of most of the nonmetalsN2O5(s) + H2O(l) → HNO3(aq)

Base anhydrides: Oxides of Group I & II metalsCaO(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

675


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

676


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 BRØNSTED-

LOWRY SCHEME

677


677


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

678


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

679


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 = 14pH < 7 acidic (can be negative)pH = 7 neutralpH > 7 basic

679

Fig. 15.3 A simple pH meterwith a digital readout.


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 < 0pKa of H3O+ = 0pKa of weak acids > 0pKa of H2O = 14

15.3 ACID AND BASE STRENGTH681

𝐾𝐾𝑎𝑎 =𝐻𝐻3𝑂𝑂+ 𝐴𝐴−

𝐻𝐻𝐴𝐴, 𝑝𝑝𝐾𝐾𝑎𝑎 = − log10 𝐾𝐾𝑎𝑎


682


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

a3b +

3

w-

-

-

- -K K

K

w a b w a b , p p p K K K K K K= = +

Base hydrolysis constant, Kb

682

𝐾𝐾𝑏𝑏 =𝐻𝐻𝐴𝐴 𝑂𝑂𝐻𝐻−

𝐴𝐴−, 𝑝𝑝𝐾𝐾𝑏𝑏 = − log10 𝐾𝐾𝑏𝑏


Fig. 15.5 The relative strengthof some acids and theirconjugate bases.

683


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.

684

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

acid1 base2 acid2 base1(strong) (strong) (weak) (weak)


682


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)

685


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.

686


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 ]

→

687


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

687


Fig. 15.9 Indicators changetheir colors at very differentpH values.

688


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 becomesless acidic.

688


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

−

K yy

→ y = 4.20×10–3

[H3O+] = y = 4.20×10–3 M → pH = 2.38

𝑦𝑦2

1000 − 𝑦𝑦 ≈𝑦𝑦2

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 ACIDSAND BASES

Calculate the pH and the fraction of HOAc ionized at equilibrium.HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)

689

1 M


HOAc H3O+ AcO–


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


Weak Bases

H2O(l) + NH3(aq) NH4+(aq) + OH–(aq)

--------------------------------------------------------------------------Initial 0.0100 0 ~ 0Change –y + y + y

---------------- ------ -----Equilibrium 0.0100 – y y y--------------------------------------------------------------------------

−= = = ×−

+[NH ][OH ] 1.8 10[NH ] 0.0100

254

b3

−

K yy

y = 4.15 ×10–4 M = [OH–]

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

692

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


Hydrolysis

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

Initial 0.100 0 ~ 0Change –y + y + y

---------------- ------ -----Equilibrium 0.100 – y y y

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

693

EXAMPLE 15.9 Hydrolysis of NaOAc

NaOAc(s) + H2O(l) → Na+(aq) + AcO–(aq)

0.1 M


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

a3b +

3

w-

-

-

- -K K

K

w a b w a b , p p p K K K K K K= = +

Base hydrolysis constant, Kb

682

𝐾𝐾𝑏𝑏 =𝐻𝐻𝐴𝐴 𝑂𝑂𝐻𝐻−

𝐴𝐴−, 𝑝𝑝𝐾𝐾𝑏𝑏 = − log10 𝐾𝐾𝑏𝑏


Hydrolysis



---------------- ------ -----Equilibrium 0.100 – 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

2w

a

10b

−

K KK

yy

693



0.01 M


Buffer solution ~ maintains an approximately constant pHWeak acid + Salt containing its conjugate base(eg. HOAc/NaOAc)

15.5 BUFFER SOLUTIONS

694

- Controlling the solubility of ions- Maintaining pH of biochemical and physiological

processesblood pH 7.4 (7.0 – 7.8)


Calculations of Buffer Action

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq) ---------------------------------------------------------------------------------------Initial 1.00 ~ 0 0.500Change –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

a0.500−

Ky + y

y

( ) ( ) −≈ ≈ ×−

1.77 101.00 1.00

40.500 0.500y + y yy

y = [H3O+] = 3.54×10–4 M → pH = 3.45

695


Testing the buffer strength.

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

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

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

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

696


( ) −= = = ×−

K+

43a

0.40[H O ][HCOO ] 1.77 10[HCOOH]

− y + yy1.10

( ) ( ) −≈ ≈ ×−

40.40 0.401.77 10

y + y yy1.10 1.10

y = [H3O+] = 4.9 ×10–4 M → pH = 3.31

Addition of 0.100 mol HCl toBuffer solution of Ex. 15.10: pH = 3.45 → 3.31Pure water: pH = 7 → 1

696


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 solution1. Choose a weak acid whose pKa ≈ pH2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0

Henderson-Hasselbalch Equation

697


682


a 100

0

[HA] pH p log [A ]

K −≈ −


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–.

698


Titration of a Strong Acidwith 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 CURVES699


Region I: Before the equivalence point

The pH determined by the excess H3O+.

2. V = 30.00 mL NaOH addedn(OH–) = (0.1000 mol/L)(0.0300 L) = 3.000×10–3 moln(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 = 1.00n(H3O+) = (0.1000 mol/L)(0.1000 L) = 1.000×10–2 mol

699


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

700


Titration of a Weak Acid with a Strong Base

At the equivalence point,

c0V0 = ctVe (monoprotic acid)c0 : concentration of weak acidV0 : volume of acid originally presentct : concentration of OH– in the base titrantVe : volume of the base at the equivalence point

Titration of 100.0 mL of 0.1000 M HOAcwith 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

701


Weak Acids

HOAc H3O+ AcO–


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

→←

−= = = ×−

+[H O ][Ac ] 1.76 10[HAc] 1.000

253

a

−

K yy

→ y = 4.20×10–3

[H3O+] = y = 4.20×10–3 M → pH = 2.38

𝑦𝑦2

1000 − 𝑦𝑦 ≈𝑦𝑦2

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 ACIDSAND BASES

Calculate the pH and the fraction of HOAc ionized at equilibrium.HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)

689

1 M


HOAc H3O+ AcO–


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


Titration of a Weak Acid with a Strong Base

At the equivalence point,

c0V0 = ctVe (monoprotic acid)c0 : concentration of weak acidV0 : volume of acid originally presentct : concentration of OH– in the base titrantVe : volume of the base at the equivalence point

Titration of 100.0 mL of 0.1000 M HOAcwith 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

701


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.

702


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

701


Volume increased to 0.1300 LConcentrations 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

702


Calculations of Buffer Action

HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq) ---------------------------------------------------------------------------------------Initial 1.00 ~ 0 0.500Change –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

a0.500−

Ky + y

y

( ) ( ) −≈ ≈ ×−

1.77 101.00 1.00

40.500 0.500y + y yy

y = [H3O+] = 3.54×10–4 M → pH = 3.45

695


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 solution1. Choose a weak acid whose pKa ≈ pH2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0

Henderson-Hasselbalch Equation

697


Volume increased to 0.1300 LConcentrations 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

20

a 10 10 20

[HAc] 5.38 10pH p log 4.75 log[Ac ] 2.31 10

4.38−

702

→ 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


702

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 pointThe pH determined by the excess OH–.


Hydrolysis



---------------- ------ -----Equilibrium 0.100 – 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

2w

a

10b

−

K KK

yy

693



Now 0.05 M instead of 0.1 M due to dilution


Problem Sets

For Chapter 15,

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

15 acid-base equilibria · 2020-06-04 · acid-base titration curves. 15.7. polyprotic acids. 15.8....

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