Acids and Bases Chapter 15
Dec 29, 2015
Properties of Acids • Sour to the taste.• React with a variety of metals.• Turn blue litmus pink.• Denature proteins.• Cancel the effect of a base (neutralization).
Properties of Bases • Bitter to the taste.• Slippery to the touch.• Turn pink litmus blue.• Denature proteins.• Cancel the effect of an acid (neutralization).
Acid-Base Properties
3
• Binary acids have acid hydrogens attached to a nonmetal atom: HCl, HF
• Carboxylic acids have COOH group
Structure of Acids
• Oxy acids have acid hydrogens attached to an oxygen atom: H2SO4, HNO3
4
Common AcidsChemical Name Formula Uses Strength
Nitric Acid HNO3 explosive, fertilizer, dye, glue Strong
Sulfuric Acid H2SO4 explosive, fertilizer, dye, glue,
batteries Strong
Hydrochloric Acid HCl metal cleaning, food prep, ore
refining, stomach acid Strong
Phosphoric Acid H3PO4 fertilizer, plastics & rubber,
food preservation Moderate
Acetic Acid HC2H3O2 plastics & rubber, food preservation, Vinegar
Weak
Hydrofluoric Acid HF metal cleaning, glass etching Weak
Carbonic Acid H2CO3 soda water Weak
Boric Acid H3BO3 eye wash Weak
5
Structure of Bases
• Most ionic bases contain OH ionsNaOH, Ca(OH)2
• Some contain CO32- ions
CaCO3 , NaHCO3
• Molecular bases contain structures that react with H+
mostly amine groups
6
Common BasesChemical
Name Formula
Common Name
Uses Strength
sodium hydroxide
NaOH lye,
caustic soda soap, plastic,
petrol refining Strong
potassium hydroxide
KOH caustic potash
soap, cotton, electroplating
Strong
calcium hydroxide
Ca(OH)2 slaked lime cement Strong
sodium bicarbonate
NaHCO3 baking soda cooking, antacid Weak
magnesium hydroxide
Mg(OH)2 milk of
magnesia antacid Weak
ammonium hydroxide
NH4OH, {NH3(aq)}
ammonia water
detergent, fertilizer,
explosives, fibers Weak
Acid-Base Concepts
• Concepts of acid-base theory including:
The Arrhenius concept
The Bronsted Lowry concept
The Lewis concept
8
Arrhenius Theory
HCl ionizes in water,producing H+ and Cl– ions
NaOH dissociates in water,producing Na+ and OH– ions
Strong Acids and Bases
• In the Arrhenius concept, a strong acid is a substance that ionizes completely in aqueous solution to give H3O+(aq) and an anion.
)aq(ClO)aq(OH)l(OH)aq(HClO 4324
• In the Arrhenius concept, a strong base is a substance that ionizes completely in aqueous solution to give OH-(aq) and a cation.
)aq(OH)aq(Na)s(NaOH OH 2
The STRONG ACIDS (all dissociate completely in
water)• Hydrochloric acid: HCl Hydrobromic acid: HBr Hydroiodic acid: HI Sulfuric acid: H2SO4
Nitric acid: HNO3
Perchloric acid: HClO4
Some other acids that are sometimes considered strong are: chloric acid (HClO3), bromic acid (HBrO3), perbromic acid (HBrO4), iodic acid (HIO3), and per-iodic acid (HIO4).
The STRONG BASES (all dissociate completely in
water)• Lithium hydroxide: LiOH
Sodium hydroxide: NaOH Potassium hydroxide: KOH Rubidium hydroxide: RbOH Cesium hydroxide: CsOH Magnesium hydroxide: Mg(OH)2
Calcium hydroxide: Ca(OH)2
Strontium hydroxide: Sr(OH)2
Barium hydroxide: Ba(OH)2
• Note: Alkali and some alkaline earth metals
Weak Acids and Bases
• Most other acids and bases that you encounter are weak. They are not completely ionized and exist in reversible reaction with the corresponding ions.
Ammonia, NH3, is a weak base.
(aq)OHC(aq)OH 2323
An example is acetic acid, HC2H3O2.
)l(OH)aq(OHHC 2232
)aq(OH )aq(NH )l(OH )aq(NH 423
13
Arrhenius Acid-Base Reactions
• The H+ from the acid combines with the OH- from the base to make a molecule of H2Oit is often helpful to think of H2O as H-OH
• The cation from the base combines with the anion from the acid to make a salt
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
14
Brønsted-Lowry Concept of Acids and Bases
• A base is the species accepting the proton in a proton-transfer reaction.
o In any reversible acid-base reaction, both forward and reverse reactions involve proton transfer.
• According to the Brønsted-Lowry concept, an acid is the species donating the proton in a proton-transfer reaction.
15
Brønsted-Lowry Concept of Acids and Bases
• Consider the reaction of NH3 and H2O.
o In the forward reaction, NH3 accepts a proton from H2O. Thus, NH3 is a base and H2O is an acid.
)aq(OH )aq(NH )l(OH )aq(NH 423
H+
base acid
16
Amphoteric Substances
• Amphoteric substances can act as either an acid or a basehave both transferable H and atom with lone
pair
• Water acts as base, accepting H+ from HCl
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
• Water acts as acid, donating H+ to NH3
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
17
Brønsted-Lowry Concept of Acids and Bases
Brønsted-Lowry Concept of Acids and Bases
In the Brønsted-Lowry concept:
2. Acids and bases can be ions as well as molecular substances.
3. Acid-base reactions are not restricted to aqueous solution.
4. Some species can act as either acids or bases depending on what the other reactant is.
1. A base is a species that accepts protons; OH- is only one example of a base.
18 18
Conjugated Acid-Base PairsConjugated Acid-Base Pairs
• The Brønsted-Lowry concept introduced the idea of conjugate acid-base pairs and proton-transfer reactions.
o If an acid loses its H+, the resulting anion is now in a position to reaccept a proton, making it a Brønsted-Lowry base.
o It is logical to assume that if an acid is considered strong, its conjugate base (that is, its anion) would be weak, since it is unlikely to accept a hydrogen ion.
conjugate acid-base pairs
(aq)OHC(aq)OH 2323 )l(OH)aq(OHHC 2232
acid acidbase base
Copyright © Houghton Mifflin Company. All rights reserved.
Label each species as an acid or base. Identify the conjugate acid-base pairs.
a. HCO3-(aq) + HF(aq) H2CO3(aq) + F-(aq)
b. HCO3-(aq) + OH-(aq) CO3
2-(aq) + H2O(l)
Base Acid Conjugate base
Conjugate acid
Acid Base Conjugate acid
Conjugate base
25
Practice – Write the formula for the conjugate acid of the following
H2O H3O+
NH3 NH4+
CO32− HCO3
−
H2PO41− H3PO4
27
Practice – Write the formula for the conjugate base of the following
H2O HO−
NH3 NH2−
CO32− since CO3
2− does not have an H, it cannot be an acid
H2PO41− HPO4
2−
28
Polyprotic Acids
• Acid molecules have more than one ionizable H
1 H = monoprotic, 2 H = diprotic, 3 H = triprotic
HCl , H2SO4, H3PO4
• Polyprotic acids ionize in steps
each ionizable H removed sequentially
• Removing of the first H automatically makes removal of the second H harder H2SO4 is a stronger acid than HSO4
29
Strengths of Acids & Bases• Commonly, acid or base strength is measured by
determining the equilibrium constant of a substance’s reaction with water
HAcid + H2O Acid-1 + H3O+1
Base: + H2O HBase+1 + OH-1
• The farther the equilibrium position lies to the products, the stronger the acid or base
• The position of equilibrium depends on the strength of attraction between the base form and the H+
stronger attraction means stronger base or weaker acid
30
Acid Ionization Constant, Ka
• Acid strength measured by the size of the equilibrium constant when react with H2O
HAcid + H2O Acid-1 + H3O+1
• The equilibrium constant is called the acid ionization constant, Ka
larger Ka = stronger acid
[HAcid]
]O[H][Acid 13
1
a
K
32
Autoionization of WaterH2O H+ + OH–
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)• All aqueous solutions contain both H3O+ and OH–
the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 10-7M @ 25°C
• The ion product of water and has the symbol Kw
• [H3O+] x [OH–] = Kw = 1 x 10-14 @ 25°C
Self-ionization of WaterSelf-ionization of Water
Example– Calculate the [OH] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral
The units are correct. The fact that the [H3O+] < [OH] means the solution is basic
[H3O+] = 1.5 x 10-9 M
[OH]
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
[H3O+] [OH]
]-][OHOH[ 3wK
]OH[]-OH[
]-OH][OH[
3
w
3w
K
K
M 107.6105.1
100.1]-[OH 6
9
14
34
pH• pH = -log[H3O+], [H3O+] = 10-pH
exponent on 10 with a positive signpHwater = -log[10-7] = 7
• pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral
• Another way of expressing the acidity/basicity of a solution is pOH
• pOH = -log[OH], [OH] = 10-pOH
• pH + pOH = 14.00
Example– Calculate the pH at 25°C when the [OH] = 1.3 x 10-2 M, and determine if the solution is acidic,
basic, or neutral
pH is unitless. The fact that the pH > 7 means the solution is basic
[OH] = 1.3 x 10-2 M
pH
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
]-][OHOH[ 3wK
2
14
3
3w
103.1
100.1]OH[
]-OH][OH[
K M 107.7]O[H 133
[H3O+][OH] pH
]OH[log pH 3-
12.11 pH
107.7log- pH 13
36
pK
• A way of expressing the strength of an acid or base is pK
• pKa = -log(Ka), Ka = 10-pKa
• pKb = -log(Kb), Kb = 10-pKb
• The stronger the acid, the smaller the pKa
larger Ka = smaller pKa
37
The pH of a Strong Acid
• There are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water
• The contribution of the water to the total [H3O+] is negligibleshifts the Kw equilibrium to the left so far that
[H3O+]water is too small to be significant
• For a monoprotic strong acid [H3O+] = [HAcid]for polyprotic acids, the other ionizations can
generally be ignored
• 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
3838
• Calculate the concentration of H+ ion in 0.010 M NaOH.
Because you started with 0.010 M NaOH (a strong base) the reaction will produce 0.010 M OH-(aq).
– Substituting [OH-]=0.010 into the ion-product expression, we get:
)aq(OH)aq(Na)s(NaOH OH 2
MH 12-14-
101.0 010.0101.0
][
The pH of a Strong BaseThe pH of a Strong Base
39
Acid-Ionization Equilibria
• For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant).
)aq(A)aq(OH )l(OH)aq(HA 32
]OH][HA[]A][OH[
K2
3c
40
Acid-Ionization Equilibria• Since the concentration of water remains
relatively constant, we rearrange the equation to get:
]HA[]A][OH[
K]OH[K 3c2a
• Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.
]HA[]A][OH[
K 3a
41
[HNO2] [NO2-] [H3O+]
initial
change
equilibrium
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HNO2 + H2O NO2 + H3O+
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change
equilibrium
Ka for HNO2 = 4.6 x 10-4
42
[HNO2] [NO2-] [H3O+]
initial 0.200 0 0
change
equilibrium
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.200 x x x
x
xxK
12
3-2
a 1000.2HNO
]OH][[NO
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
43
x
xxK
12
3-2
a 1000.2HNO
OHNO
determine the value of Ka from Table 15.5
since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x
1
2
3-2
a 1000.2HNO
OHNO
xxK
1
24
1000.2106.4
x
3
14
106.9
1000.2106.4
x
x
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x0.200 x
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
44
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200 x x
check if the approximation is valid by seeing if x < 5% of [HNO2]init
%5%8.4%1001000.2
106.91
3
the approximation is valid
x = 9.6 x 10-3
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
45
Ka for HNO2 = 4.6 x 10-4
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.200-x x x
x = 9.6 x 10-3
substitute x into the equilibrium concentration definitions and solve
M 190.0106.9200.0200.0HNO 32 x
M 106.9OHNO 33
-2
x
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
46
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+] into the formula for pH and solve
[HNO2] [NO2-] [H3O+]
initial 0.200 0 ≈ 0
change -x +x +x
equilibrium 0.190 0.0096 0.0096
02.2106.9log
OH-logpH3
3
Example- Find the pH of 0.200 M HNO2(aq) solution @ 25°C
A solution has a hydroxide ion concentration of 0.001 M. What is the pOH of the solution?
A. 3
B. 7
C. 11
D. 13
47
The pH of a solution is 5.50. What is its hydronium concentration?
A. 5.50 M
B. 3.2 x 10-5 M
C. 3.2 x 10-6 M
D. 3.2 x 10-7 M
48
49
Base-Ionization Equilibria• Equilibria involving weak bases are treated
similarly to those for weak acids.
– In general, a weak base B with the base ionization
)aq(OH)aq(HB )l(OH)aq(B 2
has a base ionization constant equal to
]B[
]OH][HB[Kb
50
[NH3] [NH4+] [OH]
initial
change
equilibrium
Example - Find the pH of 0.100 M NH3(aq) solution
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH] from water is ≈ 0
NH3 + H2O NH4+ + OH
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
Kb for NH3 = 1.76 x 10-5
51
[NH3] [NH4+] [OH]
initial 0.100 0 0
change
equilibrium
Example - Find the pH of 0.100 M NH3(aq) solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx
0.100 x x x
x
xxK
13
4b 1000.1NH
]OH][[NH
Kb for NH3 = 1.76 x 10-5
52
x
xxK
13
4b 1000.1NH
OHNH
determine the value of Kb from Table 15.8
since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x
1
3
4b 1000.1NH
OHNH
xxK
1
25
1000.11076.1
x
3
15
1033.1
1000.11076.1
x
x
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x0.100 x
Example - Find the pH of 0.100 M NH3(aq) solution
53
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x
check if the approximation is valid by seeing if x < 5% of [NH3]init
%5%33.1%1001000.1
1033.11
3
the approximation is valid
x = 1.33 x 10-3
Example - Find the pH of 0.100 M NH3(aq) solution
54
substitute x into the equilibrium concentration definitions and solve
M 099.01033.1100.0100.0NH 33 x
M 1033.1][OH]NH[ 3-4
x
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x x
x = 1.33 x 10-3
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.099 1.33E-3 1.33E-3
Example - Find the pH of 0.100 M NH3(aq) solution
55
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
124.111052.7log
OH-logpH12
3
Kb for NH3 = 1.76 x 10-5
[NH3] [NH4+] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.099 1.33E-3 1.33E-3
12-3
3-
14-
3
-3w
107.52]O[H
101.33
101.00]O[H
]][OHO[H
K
Example - Find the pH of 0.100 M NH3(aq) solution
57
• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.1. A salt of a strong base and a strong acid.
The salt gives a neutral aqueous solution. NaCl.
2. A salt of a strong base and a weak acid. The salt gives a basic solution. NaCN
3. A salt of a weak base and a strong acid. The salt gives an acidic solution. NH4Cl
4. A salt of a weak base and a weak acid.Ka > Kb the solution is acidic. Kb > Ka the solution is basic.
Classifying Salt Solutions asAcidic, Basic, or Neutral
58
Example- Determine whether a solution of the following salts is acidic, basic, or neutral
a) SrCl2
Sr2+ is the counterion of a strong base, pH neutralCl− is the conjugate base of a strong acid, pH neutralsolution will be pH neutral
b) CH3NH3NO3
CH3NH3+ is the conjugate acid of a weak base, acidic
NO3− is the conjugate base of a strong acid, pH neutral
solution will be acidic
59
Example- Determine whether a solution of the following salts is acidic, basic, or neutral
c) NaCHO2
Na+ is the counterion of a strong base, pH neutral
CHO2− is the conjugate base of a weak acid, basic
solution will be basic
d) NH4F
NH4+ is the conjugate acid of a weak base, acidic
F− is the conjugate base of a weak acid, basic
Ka(NH4+) > Kb(F−); solution will be acidic
60
•If Ka > Kb, the solution is acidic. If Ka < Kb, the solution is basic. If Ka = Kb, the solution is neutral.
105
14
b
wa4 10 5.6
10 1.8
10 1.0 :NH For
K
KK
114
14
a
wb 10 1.5
10 6.8
10 1.0 :F For
K
KK
F for KNH for K b4a
acidic. is FNH4
An aqueous solution of __________ will produce a basic solution.
A. LiCl
B. NH4ClO4
C. CaSO4
D. KBr
E. Na2CO3
61
62
Ka and Kb Relationship
• Reference tables only provide Ka values because Kb values can be found from them
][A
]OHA][H[ )(OH )HA( )O(H )(A
[HA]
]O][HA[ )(OH )(A )O(H )HA(
3b2
3a32
Kaqaqlaq
Kaqaqlaqwhen you add equations, you multiply the K’s )(OH )(OH )O(H 2 32 aqaql
w3ba
3ba
]OH][OH[
]A[
]OH][HA[
]HA[
]OH][A[
KKK
KK
63
Example- Find the pH of 0.100 M NaCHO2(aq) solution
Na+ is the cation of a strong base – pH neutral. The CHO2
− is the anion of a weak acid – pH basic
Write the reaction for the anion with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH] from water is ≈ 0
CHO2− + H2O HCHO2 + OH
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
64
0.100 x
Example- Find the pH of 0.100 M NaCHO2(aq) solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5
substitute into the equilibrium constant expression
+x+xxx x
x
xxK
12
2b 1000.1CHO
]OH][[HCHO
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change
equilibrium
114
14
b
wba
106.5108.1
100.1
K
KKK
65
Example- Find the pH of 0.100 M NaCHO2(aq) solution
since Kb is very small, approximate the [CHO2
−]eq = [CHO2−]init
and solve for x
1
211
1000.1106.5
x
6
111
104.2
1000.1106.5
x
x
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 x x0.100 x
x
xxK
12
2b 1000.1CHO
]OH][[HCHO
12
2b 1000.1CHO
]OH][[HCHO
xxK
66
substitute x into the equilibrium concentration definitions and solve
M 100.0104.2100.0100.0CHO 62 x
M 104.2][OH]HCHO[ 6-2
x
x = 2.4 x 10-6
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 −x x x
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
Example- Find the pH of 0.100 M NaCHO2(aq) solution
67
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
38.8102.4log
OH-logpH9
3
9-3
6-
14-
3
-3w
102.4]O[H
104.2
101.00]O[H
]][OHO[H
K
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
Example- Find the pH of 0.100 M NaCHO2(aq) solution
69
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
11
26
2
2b
108.5100.0
104.2
CHO
OHHCHO
Kthough not exact, the answer is reasonably close
Kb for CHO2− = 5.6 x 10-11
[CHO2−] [HCHO2] [OH]
initial 0.100 0 ≈ 0
change -x +x +x
equilibrium 0.100 2.4E-6 2.4E-6
Example- Find the pH of 0.100 M NaCHO2(aq) solution
70
Polyprotic Acids
• Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids.
– The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4
-.
– Sulfuric acid, for example, can lose two protons in aqueous solution.
)aq(HSO)aq(OH)l(OH)aq(SOH 43242
)aq(SO)aq(OH )l(OH)aq(HSO 24324
71
Polyprotic Acids• For a weak diprotic acid like carbonic acid,
H2CO3, two simultaneous equilibria must be considered.
)aq(HCO)aq(OH )l(OH)aq(COH 33232
)aq(CO)aq(OH )l(OH)aq(HCO 23323
7
32
331a 103.4
]COH[]HCO][OH[
K
11
3
233
2a 108.4]HCO[
]CO][OH[K
73
Strengths of Binary Acids
• The more + H-X - polarized the bond, the more acidic the bond
• The stronger the H-X bond, the weaker the acid
• Binary acid strength increases to the right across a periodH-C < H-N < H-O < H-F
• Binary acid strength increases down the columnH-F < H-Cl < H-Br < H-I
XH + -
74
Strengths of Oxyacids, H-O-Y
• The more electronegative the Y atom, the stronger the acidhelps weakens the H-O bond
• The more oxygens attached to Y, the stronger the acidfurther weakens and polarizes the H-O bond
HOIHOBrHOCl
432 HClOHClOHClOHClO
77
Lewis Acid - Base Theory
• Electron donor = Lewis Base = nucleophilemust have a lone pair of electrons
• Electron acceptor = Lewis Acid = electrophileelectron deficient
• Covalent bond forms between the moleculesNucleophile: + Electrophile Nucleophile:Electrophile
• Product called an adduct• Other acid-base reactions also Lewis
78
Example - Complete the Following Lewis Acid-Base ReactionsLabel the Nucleophile and Electrophile
OH
H C H
+ OH-1
OH
H C H
OH
OH
H C H
+ OH-1
ElectrophileNucleophile
••
••••
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Practice - Complete the Following Lewis Acid-Base Reactions
Label the Nucleophile and Electrophile
• CaO + SO3
• KI + I2