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
1
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 amphiprotic substances: H2PO4 , HCO 3
Chapter 15 Acid-Base Chemistry
2
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.
HOAc, 1.8x10-5 > HCN, 4.9x10-10
See table of Kas on page 639
Chapter 15 Acid-Base Chemistry
4
Polyprotic Acids - more than one acidic proton.
H 2SO 4, H 3PO4,
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· Ka2 ·Ka3 = 1.7x10 –22
Strong Base
NaOH + H2 O
OH–(aq) + Na+(aq)
LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH)2
Weak Base
NH3 + H2O
OH– + NH4+
[NH4 + ][OH − ]
Kb =
= 1.8 × 10 −5
[ NH3 ]
Chapter 15 Acid-Base Chemistry
5
Ethylamine, N(C2H 5) 3, methylamine, N(CH3) 3, pyridine, C5H 5N, aniline,
C6H 5NH2.
See page 646 for some weak bases and their K b values.
Weak bases often have long formulas. Instead of writing out the formula
each time, substitute B for the formula.
BH + + OH -
B + H 2O
[ BH ][OH ]
=
+
Kb
−
[B ]
Percent Ionization
For a weak acid, HA
H + + A–
percent ionization =
ionized acid concentration at equilibrium
×100%
initial concentration of acid
An easy measure of the amount of ionized acid is the pH.
H+ ]
[
percent ionization =
×100%
[HA ]0
For a weak base,
OH − ]
[
percent ionization =
× 100%
[B]0
•
Increases with dilution (must be measured)
0.50 M HF
3.8% ionized
0.050 M HF
11.2% ionized
Chapter 15 Acid-Base Chemistry
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Acid Strength - extent of dissociation
Binary acids, 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
Right to left
HF » H2O > NH3 > CH4
HCl » H 2S > PH3 > SiH 4
Bottom to top
HI > HBr > HCl » HF
Oxoacids, 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
Chapter 15 Acid-Base Chemistry
7
Increasing Electronegativity of X increases acid strength
HClO4 > HBrO4 > HIO 4
(Cl more electroneg.)
HClO3 > HBrO3 > HIO 3
HClO2 > HBrO2 > HIO 2
Increasing Oxidation Number of non-metals, X
HNO2 <
HNO3
(N is
+5 in HNO3)
H 2SO 3 < H 2SO 4
HIO < HIO 2 < HIO 3 < HIO 4
Carboxylic Acids
H
H
O
C
C
O
H
H
H
Cl
O
C
C
O
H
H
Ka = 1.8x10–5
Ka = 1.4x10–3
Strength of Conjugate Acids and Conjugate Bases
The base strength of the conjugate base is inversely proportional to the
strength of the acid.
HCl
strong acid
Cl–
extremely weak base
HF
weak acid
F–
weak base
OH – strong base
HCl
H+ + Cl–
HF
H+ + F–
H2O
H+ + OH–
Chapter 15 Acid-Base Chemistry
conjugate base
strength
acid strength
H 2O extremely weak acid
8
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 2O
15.74
HBr
-9
HF
3.17
HI
-10
CH4
NH3
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
CH4 is an extremely weak acid, so its conjugate base is extremely strong.
Acid Strength:
HF > H2O > NH3 > CH4
Base Strength:
CH3– > NH2– > OH – > F–
Example:
CH3– + NH3 → CH4 + NH2–
Chapter 15 Acid-Base Chemistry
9
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
Now to salts of weak acids and weak bases
H + ][ A– ] [ HA][OH − ]
[
+
−
Kw = Ka Kb =
=
H
OH
[
]
[
]
−
[ HA ]
A
[ ]
Ka = Kw/Kb
Kb = Kw/Ka
Chapter 15 Acid-Base Chemistry
10
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
•
Metal ions which are small and highly charged, Al3+, Cr3+, Fe3+, Bi 3+,
and Be2+ react with water to produce acidic solutions.
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
11
Acidic Character and Ka Values of Common Cation in Water
Character
Examples
Ka
pKa
Conjugate acids of weak
Anilinium ion,
2.3 x10 -5
4.64
bases
C6H 5NH3+
5.6x10 -6
5.24
Ammonium ion, NH4+
5.6x10 -10
9.25
Methylammonium ion,
2.8x10 -11
10.56
Fe3+ as Fe(H2O) 63+
3.5x10 -3
2.46
Cr3+ as Cr(H 2O) 63+
1.3x10 -4
3.89
Al3+ as Al(H 2O) 63+
1.4x10 -5
4.85
Acidic
Pyridinium ion,
C5H 5NH+
CH3NH3+
Small highly charged
metal cations
Neutral
Group 1 and 2 cations;
Li+, Na+, K+, Mg 2+,
metal cations with
Ca2+, Ag+
charge +1
Basic
none
Chapter 15 Acid-Base Chemistry
12
(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
CN- + H 2O
NH3 + H 3O +
acidic
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
13
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+
+
X
Acid
H X
Base
Cl
Cl Al
Cl
Cl
+
Acid
electron pair acceptor
Fe2+ +
C
H
Donor atom
Al
H
H
Fe2+ C
O
Ag(NH3)2
H
H
N
Cl
Cl
Base
electron pair donor
Ag+ + 2 NH3
H3N
N
H
O
+
Ag+ NH3
Donor atom
Central metal ion
Chapter 15 Acid-Base Chemistry
14
Lewis Acids: Group 3A halides, Nonmetal oxides, CO2, SO 3, transition
metal ions
Lewis Bases : Ammonia and 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
Calculations
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?
Chapter 15 Acid-Base Chemistry
15
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?
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], this will
avoid using the quadratic equation.
[H3O+ ][A− ]
Ka =
[ HA]
x2
Ka =
[HA]
x = Ka [ HA]
Chapter 15 Acid-Base Chemistry
16
Example: What is the pH of a 0.100 M solution of hypobromous acid, Ka =
2.0 x 10 -9 ?
Percent Dissociation
%dissociation =
[ HA]dissociated
× 100
[ HA]undissociated
What is the percent dissociation in the example above?
Chapter 15 Acid-Base Chemistry
17
Determining acid concentration and K a 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
18
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
19
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 (Hydrolysis)
What is the pH of a 0.015 M solution of NaOCl? HOCl, Ka = 3.5x10 -8 .
Chapter 15 Acid-Base Chemistry
20