15 acid-base equilibria · general chemistry ii 15. acid-base equilibria. chapter. 15.1....
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General Chemistry II
ACID-BASE EQUILIBRIA15CHAPTER
General Chemistry II
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
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General Chemistry II
Cyanidin is blue in the basic sap of the cornflower and red in the acidic sap of the poppy.
669
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
~ 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
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General Chemistry II
Fig. 15.2 Acidity and basicity of oxides of main group elements.
676
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General Chemistry II
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
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General Chemistry II
677
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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.
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General Chemistry II
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
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General Chemistry II
682
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General Chemistry II
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
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General Chemistry II
Fig. 15.5 The relative strengthof some acids and theirconjugate bases.
683
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General Chemistry II
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)
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
Fig. 15.9 Indicators changetheir colors at very differentpH values.
688
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
Hydrolysis
AcO–(aq) + H2O(l) HOAc(aq) + OH–(aq) --------------------------------------------------------------------------
Initial 0.100 0 ~ 0Change –y + y + y
---------------- ------ -----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
EXAMPLE 15.9 Hydrolysis of NaOAc
NaOAc(s) + H2O(l) → Na+(aq) + AcO–(aq)
0.01 M
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General Chemistry II
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)
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
( ) −= = = ×−
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
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
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General Chemistry II
702
Region III: At the equivalence point(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–.
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General Chemistry II
Weak Polyprotic Acids H2CO3(aq) + H2O(l) H3O+(aq) + HCO3
– (aq), Ka1 = 4.3×10–7
HCO3– (aq) + H2O(l) H3O+(aq) + CO3
2– (aq), Ka2 = 4.8×10–11
1. [H3O+] in the two equilibria are one and the same.2. Ka1 >> Ka2
15.7 POLYPROTIC ACIDS704
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General Chemistry II
704
H2CO3(aq) + H2O(l) H3O+(aq) + HCO3– (aq)
---------------------------------------------------------------------------------------Initial 0.034 ~ 0 0Change –y + y + y
--------------- ------ -------------Equilibrium 0.034 – y y y---------------------------------------------------------------------------------------
EXAMPLE 15.15 Saturated aqueous solution of CO2 → [H2CO3] = 0.034 MCalculate equilibrium concentrations of H2CO3, HCO3
–, CO32–, H3O+.
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General Chemistry II
705
Assume that the effect of the second ionization on [HCO3–] and
[H3O+] is negligible.
−= = = ×−
K+ 2
73 3a1
2 3
[H O ][HCO ] 4.3 10[H CO ] 0.034
− yy
y = [H3O+] = [HCO3–] = 1.2×10–4 M
( )−−
−
×= = = ×
×K
4 2+ 23 113 3
a2 43
1.2 10 [CO ][H O ][CO ] 4.8 10[HCO ] 1.2 10
−−
−
[CO32–] = Ka2 = 4.8×10–11 M << [HCO3
–]
HCO3– (aq) + H2O(l) H3O+(aq) + CO3
2– (aq), Ka2 = 4.8×10–11
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General Chemistry II
Effect of pH on Solution Composition
Change in pH shifts the position of all acid-base equilibria
=K3 a1
+2 3 3
[HCO ][H CO ] [H O ]
− −
− =K2
3 a2+
3 3
[CO ][HCO ] [H O ]
706
Fractions of H2CO3, HCO3–, CO3
2– at pH 10.00 at 25°C
−
−
×= = ×
×K 7
33 a1+ 10
2 3 3
[HCO ] 4.3 10 = 4.3 10[H CO ] [H O ] 1.0 10
−
−
×→ = ×
[H CO ] 1 = 2.3 104.3 10[HCO ]
42 33
3−
− −
− −
×= =
×K2 11
3 a2+ 10
3 3
[CO ] 4.8 10 = 0.48[HCO ] [H O ] 1.0 10
EXAMPLE 15.16
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General Chemistry II
Fractions of H2CO3, HCO3–, CO3
2–
]− −=2 3
2 3H CO 2
2 3 3 3
[H CO ][H CO ]+[HCO ]+[CO
α1 ] /
−
− − −= 2 3 32
2 3 3 3 3
[H CO ]/[HCO ][H CO ]/[HCO ]+ +[CO [HCO ]
1
−−
−= =4
44
2.3 10 1.6 10(2.3 10 )+ +0.48
××
×
Similarly, and − =3HCO 0.68α − =2
3CO 0.32α
706
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General Chemistry II
Electronegativity
Relative acid strength of hydrocarbons, amines, and alcoholsAcid constant: pKa (C2H6) = 50, pKa (C2H5NH2) = 35, pKa (C2H5OH) = 16
15.8 ORGANIC ACIDS AND BASES: STRUCTURE AND REACTIVITY
710
“Inductive effect” involving bond dipoles:Acid constant: pKa (HCH2COOH) = 4.8, pKa (ClCH2COOH) = 2.9Electronegativity: 2.20 3.16
Conjugate base: CH3-CH2- C2H5NH- C2H5O–
Electronegativity: 2.55 3.04 3.44
→ The higher the electronegativity of X in the conjugate base–X–, the stronger the acid strength.
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General Chemistry II
711
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General Chemistry II
Effect of the position of electronegative substituentpKa (ClCH2CH2CH2COOH) = 4.5 ← 4-chlorobutanoic acidpKa (CH3CH2CHClCOOH) = 2.9 ← 2-chlorobutanoic acid
Steric Hindrance Acid constant: pKa (CH3OH) = 15, pKa (CH3)3COH = 18 ← tert-butanol
Conjugate base: CH3O– (CH3)3CO–
CH3O– is more stable in solution due to easier access of solvent
molecules to the negatively charged site.
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General Chemistry II
Resonance - Acid constant: pKa (CH3COOH) = 4.8, pKa (CH3CH2OH) = 16
pKa difference is too big for the inductive effect by O w.r.t. H2
Conjugate base
- Acid constant: pKa (C6H5OH) = 10, pKa (CH3CH2OH) = 16
Conjugate base: C6H5O– C2H5O–
Resonance structure of phenoxide:
C6H5OH + NaOH C6H5O–Na+ + H2O
C2H5OH + Na C2H5O–Na+ + ½ H2
→
→
713
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General Chemistry II
Problem Sets
For Chapter 15,
10, 18, 24, 36, 50, 54, 68, 76, 94, 106