Chapter 15
Acid-Base Chemistry
Arrhenius Acid-Base Theory
Acid - dissociates in water to produce H+ ions.
Base - dissociate in water to produce OH– ions.
Brønsted-Lowry Acid-Base Theory
+
Acid - proton (H ) donor
Cl–(aq) + H3O+ (aq)
HCl + H2 O
conjugate
base
conjugate
acid
H 3O +(aq) is called the hydronium ion.
Base - proton acceptor
OH–(aq) + NH4 +(aq)
NH3 + H 2O
conjugate
base
conjugate
acid
Conjugate Acid-Base Pairs
An acid and a base that differ only in the presence of a proton are called a
conjugate acid-base pair
HX and X
-
+
NH3 and NH4
X is the conjugate base of the acid HX
+
NH4 is the conjugate acid of the base NH 3
Chapter 15 Acid-Base Chemistry
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Amphiprotic substances
can behave as either acids or bases
H+ + H2O
acid
H 3 O+
base
OH– + NH4+
H2O + NH 3
acid
base
Autoionization of Water
2 H2 O
H
O
+
H
H
OH– + H3 O+
H
O
H
O
H
–
+
O
+
H
H
[ H + ][OH − ]
Kc =
[ H2 O]
Kc [ H2 O] = K w = [ H + ][OH − ] = 1.0 ×10 −14
at 25°C
In pure water [H+] = [OH - ] = 1.0X10 –7 M
[ OH − ] = [HKw+ ]
other examples of ampiprotic substances: H2PO4 , HCO 3
Chapter 15 Acid-Base Chemistry
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Measuring the Concentration of H
+
or OH
-
pH and pOH
p ≡ − log
[ ]
pH = − log H + = log
[
1
H+
]
pOH = − log OH − = log
[ ]
Acidic Solutions: [H+] > 1.0X10 -7 M, pH < 7.00
Basic Solutions: [H+] < 1.0X10 -7 M, pH > 7.00
Neutral Solutions: [H +] = 1.0X10 -7 M, pH = 7.00
+
[H ] = [OH ] in pure water
+
-14
[H ][OH ] = 1.0 X 10
x
2
= 1.0 X 10
x = 1.0 X 10
-7
-14
+
= [H ] = [OH ]
+
pH = -log[H ] =
-log 1.0 X 10
pOH = -log[OH ] =
-log 1.0 X 10
+
pKw = -log([H ][OH ]) =
-7
-7
-log 1.0 X 10
= 7.0
= 7.0
-14
=14.0
+
+
Kw = [H ][OH ] → -logKw =(-log[H ]) + (-log[OH ])
*
pKw = pH + pOH *
14.0 = pH + pOH
pH = 14.0 - pOH
pOH = 14.0 - pH
Chapter 15 Acid-Base Chemistry
3
1
OH −
[
]
Strong acids - completely ionize in water
HClO4, H 2SO 4, HNO 3, HI, HBr, HCl
HI + H2O
I– + H 3O+
• Water "levels" the acids to the level of H 3O +
(H 3O + is the strongest
acid that can exist in aqueous solition.)
• I- , Cl- , Br - , NO3- , HSO4- , ClO4- are weaker bases than H2O.
Weak acids - partially dissociate, exists in solution as acid molecules and
component ions.
CH3COOH
CH3 COO– + H+
[H + ][CH3COO − ]
Ka =
= 1.8× 10 −5
[CH3COOH]
Ka, equilibrium constant of the acid, subscript a indicates it is an acid (K b
for bases)
• The smaller the Ka, the weaker the acid.
10-5 > 10-10.
Polyprotic Acids - more than one acidic proton.
H 2SO 4, H 3PO4,
Chapter 15 Acid-Base Chemistry
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Polyprotic weak acids have a Ka for each proton
–
H3PO4
–
2–
H2PO4
HPO4
2–
3–
HPO4
K for:
+
H2PO4 + H
PO4
H 3PO4
Ka1 = 7.5 x 10
+
+H
Ka2 = 6.3 x 10
+
+H
Ka3 = 3.6 x 10
3 H + + PO43–
-3
-8
-13
K = Ka1 X Ka2 X Ka3
Strong Base
OH–(aq) + Na+(aq)
NaOH + H2 O
LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH)2
Weak Base
OH– + NH4+
NH3 + H2O
[NH4 + ][OH − ]
Kb =
= 1.8 × 10 −5
[ NH3 ]
Ethylamine, methylamine, pyridine, aniline
B + H 2O
BH + + OH -
Chapter 15 Acid-Base Chemistry
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Percent Ionization
percent ionization =
ionized acid concentration at equilibrium
×100%
initial concentration of acid
H+ ]
[
percent ionization =
×100%
[HA ]0
•
Increases with dilution
0.50 M HF
3.8% ionized
0.050 M HF
11.2% ionized
Acid Strength - extent of dissociation
H-X
If H and X form a strong bond, the weaker the acid
H ( 1s)
covalent bond strength
increases
B
C
N O
Al Si P S
F
2s 2p
Cl
3s
Br
4s
I
5s
Chapter 15 Acid-Base Chemistry
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H-O-X
The strength of the O-X bond influences the strength of the acid or
whether it has basic character.
X
O H
bond b
bond a
Increasing Electronegativity increases acid strength
Increasing Oxidation Number of non-metals
i)
H 2CO3 < HNO3
(N more electroneg.)
ii)
HNO2 <
(N is
HNO3
+5 in HNO3)
Acid Strength and pK a
pKa = -log Ka
The greater the pKa, the weaker the acid
Acid
pKa
40-50
Acid
HF
pKa
3.17
34
HCl
-7
H 20
15.74
HBr
-9
HF
3.17
HI
-10
CH4
NH3
Chapter 15 Acid-Base Chemistry
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Acid
H 2SO 3
pKa
1.9, 7.21
H 2SO 4
-3.0, 1.99
HNO2
3.29
HNO3
-1.3
H 2CO3
6.37, 10.3
B(OH) 3 9.23
H 2O
15.7
H 2S
7.00
Many values were taken from: http://daecr1.harvard.edu/pKa/pka.html
Weak Acids and Their Conjugate Bases
H + ( aq) + OAc − ( aq)
HOAc( aq)
H + ][OAc − ]
[
Ka =
[ HOAc ]
OAc − (aq ) + H2O(l )
OH − ( aq ) +
HOAc( aq)
OH − ][ HOAc]
[
Kb =
[OAc− ]
HOAc( aq)
H + ( aq) + OAc − ( aq)
OAc− (aq) + H 2O(l)
H 2O(l)
Ka
OH − ( aq) + HOAc( aq ) Kb
H + ( aq) + OH − ( aq )
Kw
Ka K b = Kw
Ka =
Kw
Kb
Kb =
Kw
Ka
Chapter 15 Acid-Base Chemistry
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Now to salts of weak acids and weak bases
H + ][ A – ] [ HA][OH − ]
[
+
−
Kw = KaKb =
=
H
OH
[
]
[
]
−
[ HA]
A
[ ]
Ka = Kw/Kb
Kb = Kw/Ka
Hydrolysis - reaction of salts of weak acids and bases with water to
produce acidic or basic solutions.
Hydrolysis reaction:
X– + H 2O → HX + OH–
[H + ][ X − ]
Ka =
[ HX]
[OH − ][HX ] [OH − ][HX ]
Kh =
=
= Kb
[X − ][H2O]
[X − ]
A salt resulting from a strong acid and a strong base is a neutral salt. Ex:
NaCl, KBr.
A salt resulting from a weak acid and strong base is a basic salt. Ex:
NaCH3CO2, KCN
A salt resulting from a strong acid and a weak base is an acidic salt. Ex:
NH4Cl
Chapter 15 Acid-Base Chemistry
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(OAc– = acetate ion, CH 3CO2–)
NaOAc
–
OAc + H 2O = HOAc + OH
–
Kh
OH ][ HAOc ]
[
=
[OAc ]
Kh
[OH ][ HAOc ] = K
=
K
[OAc ]
−
−
−
w
−
a
1.0 ×10 −14
−10
=
=
5.6
×
10
1.8 × 10 −5
NH4CN Acidic or Basic?
NH4+ + H 2O = NH3 + H 3O +
acidic
CN- + H 2O = HCN + OH -
basic
Ka = Kw/Kb = (1.0x10
-14
)/(1.8x10 ) = 5.6x10
-5
Kb = Kw/Ka = (1.0x10
-14
)/(4x10
-10
) = 2.5x10
-10
-5
Kb > Ka , so solution is basic!
Chapter 15 Acid-Base Chemistry
10
Preparation of Acids and Bases
acids from oxides of non-metals:
SO 3 + H 2O → H 2SO 4
bases: from oxides of active metals:
CaO + H2O → Ca(OH)2
Lewis Acids and Bases
Lewis Acid - electron pair acceptor
Lewis Base - electron pair donor
H+
+
Acid
Cl
Cl Al
X
H X
Base
Cl
Cl
+
Acid
electron pair acceptor
Fe2+ +
C
O
H
N
Al
H
H
N
Cl
Cl
Base
electron pair donor
Fe2+ C
H
H
H
O
Chapter 15 Acid-Base Chemistry
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Ag+ + 2 NH3
H3N
Ag(NH3)2
+
Ag+ NH3
Donor atom
Donor atom
Central metal ion
Coordination sphere: [Ag(NH3) 2]
Coordination number = number of donor atoms
Lewis Acids: Group 3A halides, Nonmetal oxides, transition metal ions
Lewis Bases: Amines, water, halide ions, CO, CN- , OH–.
Amphoterism
metal oxides and metal hydroxides that dissolve in strongly acidic and
strongly basic solutions because they behave as either acids or bases.
They are amphoteric.
Al(OH)3, Sn(OH)2, Zn(OH)2
H
Al
+3
O
H
Al(H 2O) 3+
6 ( aq) + H2 O(l )
+
Al (OH)( H2 O)2+
5 ( aq) + H3O (aq)
Chapter 15 Acid-Base Chemistry
12
Overview
A.
Calculating pH of systems (solutions)
I.
Strong acids-bases
II.
Weak acids-bases
III.
Salt of a weak acid or base (Hydrolysis)
IV.
Buffers: combinations of a weak acid or base with it salt
B.
Titrations
Combination of strong (or weak) acid with strong (or weak) base (4
combinations to consider)
Chapter 15 Acid-Base Chemistry
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The pH of Strong Acids and Strong Bases
Assume 100% dissociation.
Strong Acids : HCl, HBr, HI, HNO3, H 2SO 4, HClO 4
[H 3O +] = [A–] = initial concentration of the acid.
pH = –log[H 3O +]
Example: What is the pH of at 0.650 M solution of nitric acid?
Strong Bases - alkali metal hydroxides
[OH –] = initial concentration of base
pOH = –log[OH–]
pH = 14 – pOH
Example: What is the pH of a 0.0250 M solution of KOH?
Chapter 15 Acid-Base Chemistry
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Equilibrium in Solutions of Weak Acids
HA(aq) + H2O(l) → H 3O +(aq) + A–(aq)
[H3O+ ][A− ]
Ka =
[ HA]
[H 30+] = [A–] = x
If [HA] > 100•Ka, then use the approximation, [HA – x] ≈ [HA]
[H3O+ ][A− ]
Ka =
[ HA]
x2
Ka =
[HA]
x = Ka [ HA]
Example: What is the pH of a 0.100 M solution of hypobromous acid, Ka =
2.0 x 10 -9 ?
Chapter 15 Acid-Base Chemistry
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Percent Dissociation
%dissociation =
[ HA]dissociated
× 100
[ HA]undissociated
Determining acid concentration and Ka from pH.
Since pH = -log[H+], then [H +] = antilog(pH)
[ HA] =
[ ]
H+
2
Ka
Example: A 0.10 M aqueous solution of lactic acid has a pH of 2.43. What
is the value of Ka for lactic acid?
Chapter 15 Acid-Base Chemistry
16
Polyprotic Acids
For diprotic acids, H 2A → 2 H + + A2–
Calculate the first ionization as usual. Use this for the pH of the solution.
The second ionization is much smaller. It has little effect on pH.
The concentration of the dianion, A2- , is approximately equal to the
second dissociation constant, Ka2.
H + ][ HA − ]
[
Ka1 =
[ H 2 A]
[ H + ] ≈ [ HA− ]
H + ][ A 2− ]
[
2−
Ka2 =
≈
A
[
]
−
HA
[ ]
What is the pH of 0.10 M solution of carbonic acid? What is the
concentration of carbonate ion, CO32- at equilibrium?
Chapter 15 Acid-Base Chemistry
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Weak Bases
B(aq) + H2O(l)
BH +(aq) + OH–(aq)
Weak bases are commonly amines.
Amines are derivatives of ammonia, NH3, where one or more hydrogen is
replaced by another group.
Ex: Determine the pH of a 0.075 M trimethylamine, (CH 3) 3N, solution. Kb
= 6.5 x 10 –5.
Salt of a weak acid
What is the pH of a 0.015 M solution of NaOCl? HOCl, Ka = 3.5x10 -8 .
Chapter 15 Acid-Base Chemistry
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