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Acid-Base Concepts: The Brønsted-Lowry Theory
Arrhenius Acid: A substance that dissociates in water to
produce hydrogen ions, H+
HA(aq)
H+(aq) + A–(aq)
Arrhenius Base: A substance that dissociates in water to
produce hydroxide ions, OH–
MOH(aq)
1
M+(aq) + OH–(aq)
‫חומצות ובסיסים‬-12
Acid-Base Concepts: The Brønsted-Lowry Theory
Brønsted-Lowry Acid: A substance that can transfer
hydrogen ions, H+. In other words, a proton donor
Brønsted-Lowry Base: A substance that can accept
hydrogen ions, H+. In other words, a proton acceptor
Conjugate Acid-Base Pairs: Chemical species whose
formulas differ only by one hydrogen ion, H+
2
‫חומצות ובסיסים‬-12
Acid-Base Concepts: The Brønsted-Lowry Theory
Acid-Dissociation Equilibrium
3
‫חומצות ובסיסים‬-12
Acid-Base Concepts: The Brønsted-Lowry Theory
Base-Dissociation Equilibrium
4
‫חומצות ובסיסים‬-12
Brønsted-Lowry Concept
Brønsted-Lowry Acid = Proton donor
Brønsted-Lowry Base = Proton acceptor
Works for non-aqueous solutions and explains why NH3
is basic:
NH3(g) + H2O(ℓ)
Base:
H+ acceptor
5
NH4+(aq) + OH-(aq)
Acid:
H+ donor
‫חומצות ובסיסים‬-12
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are molecules or ions
related by the loss/gain of one H+:
Conjugate Acid
H3O+
Conjugate Base
H2O
CH3COO-
CH3COOH
NH4
+
H2SO4
donate H+
accept H+
NH3
HSO4-
HSO4-
SO42-
HCl
Cl-
NH4+ and NH2- are not conjugate, conversion requires
2 H+.
6
‫חומצות ובסיסים‬-12
Conjugate Acid-Base Pairs
Identify the base conjugate to HF(aq) and the acid
conjugate to HCO3-(aq).
H+ + F-
HF
Conjugate
Base
Acid:
H+ donor
HCO3Base:
H+ acceptor
+ H+
H2CO3
Conjugate
Acid
( CO32- is the base conjugate to HCO3- )
7
‫חומצות ובסיסים‬-12
Water’s Role as Acid or Base
Water acts as a base when an acid dissolves in water:
HBr(aq) + H2O(ℓ)
acid
base
H3O+(aq) + Br-(aq)
acid
base
But water acts as an acid for some bases:
H2O(ℓ) + NH3(aq)
acid
base
NH4+(aq) + OH-(aq)
acid
base
Water is amphiprotic - it can donate or accept a proton
(act as acid or base).
8
‫חומצות ובסיסים‬-12
Lewis Acids & Bases
The most general acid-base definition:
A Lewis acid accepts a pair of e- to form a bond.
A Lewis base donates a pair of e- to form a bond.
Acid
e- pair acceptor
+
..
A
B
Base
e- pair donor
A
B
Coordinate
covalent bond
Coordinate covalent bond: a shared e- pair, with both edonated by the same atom.
9
‫חומצות ובסיסים‬-12
Lewis Acids & Bases
Examples:
H
+
Lewis acid
H+
+
B
F
10
Lewis acid
H
O H
O H
Lewis base
H
H2O
H3O+
H
F
F
H
+
N
H
H
Lewis base
F
F
H
B
N
F
H
H
‫חומצות ובסיסים‬-12
Positive Metal Ions as Lewis Acids
Metal ions can act as Lewis acids. They have:
• Missing e-.
• Empty valence orbitals.
• Metal ion + Lewis base → complex ion
Ag+ (aq) + 2 :NH3 (aq)
[H3N:Ag:NH3]+ (aq)
11
δ- δ+
:O–H -
: :
Hydroxide ions are Lewis bases.
• O is e- rich (has large electronegativity):
• Can easily donate an e- pair to a bond.
‫חומצות ובסיסים‬-12
Positive Metal Ions as Lewis Acids
Many metal hydroxides are amphoteric – they react
with acids and bases.
Aluminum hydroxide can act as a Lewis acid:
Al(OH)3(s) + OH-(aq)
[Al(OH)4]-(aq)
or as a Brønsted-Lowry base:
Al(OH)3(s) + 3 H3O+(aq)
12
Al3+(aq) + 6 H2O(ℓ)
‫חומצות ובסיסים‬-12
Neutral Molecules as Lewis Acids
Non-metal oxides act as Lewis acids.
• O attracts e- from any multiple bond.
• Leaves the other non-metal e- deficient.
δ+
δ+
δ- O
δ-
S
S
O δ-
δ- O
δ+
δ-
O=C=O
O δ-
• Susceptible to attack by Lewis base:
-
H
O
H
O
C=O
O=C=O
13
O
‫חומצות ובסיסים‬-12
Acid Strength and Base Strength
Weak Acid: An acid that is only partially dissociated in
water and is thus a weak electrolyte
14
‫חומצות ובסיסים‬-12
Acid Strength and Base Strength
15
‫חומצות ובסיסים‬-12
Hydrated Protons and Hydronium Ions
HA(aq)
H+(aq) + A(aq)
Due to high reactivity of the hydrogen ion, it is actually
hydrated by one or more water molecules.
[H(H2O)n]+
n=1
H3O+
n=2
H5O2+
n=3
H7O3+
n=4
H9O4+
For our purposes, H+ is equivalent to H3O+.
16
‫חומצות ובסיסים‬-12
Acid Strength and Base Strength
H3O+(aq) + A(aq)
HA(aq) + H2O(l)
Acid
Base
Acid
Base
With equal concentrations of reactants and
products, what will be the direction of reaction?
Stronger acid + Stronger base
17
Weaker acid + Weaker base
‫חומצות ובסיסים‬-12
Relative Strength of Acids & Bases
Strong acids are better H+ donors than weak acids
Strong bases are better H+ acceptors than weak bases
• Stronger acids have weaker conjugate bases.
• Weaker acids have stronger conjugate bases.
Strong acid + H2O
H3O+ + conjugate base
Fully ionized, reverse reaction essentially does not occur.
The conjugate base is extremely weak.
Weak acid + H2O
H3O+ + conjugate base
Weakly ionized, reverse reaction readily occurs.
The conjugate base is relatively strong.
18
‫חומצות ובסיסים‬-12
Relative Strength of Acids & Bases
NH4+ + F-
NH3 + HF
HF has greater tendency to ionize
than NH4+.
Reactant Favored
19
Acid strength increasing
Acid strength:
HF > NH4+.
(Base strength: NH3 > F-)
Conj acid. Conj. base
HSO4- H2SO4
BrHBr
ClHCl
F-
HF
NH3
NH4+
O2HCH3-
OHH2
CH4
Base strength increasing
Problem Is the following aqueous reaction product or reactant
favored?
‫חומצות ובסיסים‬-12
Factors That Affect Acid Strength
Bond Polarity
20
‫חומצות ובסיסים‬-12
Factors That Affect Acid Strength
Bond Strength
21
‫חומצות ובסיסים‬-12
Factors That Affect Acid Strength
Oxoacids
22
‫חומצות ובסיסים‬-12
Factors That Affect Acid Strength
Oxoacids
23
‫חומצות ובסיסים‬-12
Carboxylic Acids & Amines
Organic acids
Contain the carboxylic acid functional group:
24
‫חומצות ובסיסים‬-12
Carboxylic Acids & Amines
The carboxylic acid H is acidic (will ionize). All other
protons are non acidic:
O
O
+ H2O
CH3 – C
+ H3O+
CH3 – C
O-
OH
• O is very electronegative.
• O withdraws e- from the O–H bond (weakens O-H).
• Makes it easier for H+ to leave.
• Once formed, the anion is stabilized by resonance:
O-
O
CH3 – C
25
CH3 – C
O-
O
‫חומצות ובסיסים‬-12
Strengths of Carboxylic Acids
R-COOH
Acetic
Hexanoic
Chloroacetic
Trichloroacetic
Trifluoroacetic
Acid
CH3COOH
CH3CH2CH2CH2CH2COOH
CH2ClCOOH
CCl3COOH
CF3COOH
Ka
1.8 x 10-5
1.4 x 10-5
0.0013
0.3
0.6
All hydrocarbon R’s “pull” equally… ~equal strength.
Add a halogen to R = more e- withdrawing
= stronger acid
26
‫חומצות ובסיסים‬-12
Carboxylic Acids & Amines
..
..
..
Amine groups are basic: R–NH2, R–NHR or R–NR2
• R = any hydrocarbon.
• Lone-pair on N accepts a proton (like NH3).
27
‫חומצות ובסיסים‬-12
Amino Acids & Zwitterions
Amino acids contain acid and amine functional groups.
Alpha
carbon
Name
R
glycine
H
alanine
CH3
valine
(CH3)2CH
serine
HOCH2
… and others
=
−
−
H O
R–C–C–O–H
NH2
They are crystalline solids (m.p. > 200°C). Why?
• The solid is a zwitterion (has multiple charges)
−
28
=
−
H O
R–C–C–ONH3+
The acid H+ has
transferred to the
amine (base)
‫חומצות ובסיסים‬-12
Amino Acids & Zwitterions
Amino acid in solution have pH dependent structures.
Alanine:
Positively
charged
29
base
added
=
−
=
Zwitterion
(no net charge)
OH-
H O
CH3– C– C–ONH2
−
acid
added
−
H+
H O
CH3– C– C–ONH3+
−
=
−
−
H O
CH3– C– C–O–H
NH3+
Negatively
charged
‫חומצות ובסיסים‬-12
Dissociation of Water
Dissociation of Water: 2 H2O(l)
H3O+(aq) + OH(aq)
Ion-Product Constant for Water: Kw = [H3O+][OH]
at 25 oC: [H3O+] = [OH] = 1.0 × 107 M
Kw = (1.0 × 107)(1.0 × 107) = 1.0 × 1014
30
‫חומצות ובסיסים‬-12
Dissociation of Water
Kw = [H3O+][OH] = 1.0 × 1014
14
1.0
×
10
[H3O+] =
[OH]
31
14
1.0
×
10
[OH] =
[H3O+]
‫חומצות ובסיסים‬-12
Dissociation of Water
32
‫חומצות ובסיסים‬-12
Dissociation of Water
33
Acidic:
[H3O+] > [OH]
Neutral:
[H3O+] = [OH]
Basic:
[H3O+] < [OH]
‫חומצות ובסיסים‬-12
Autoionization of Water
Example
Calculate the hydronium and hydroxide ion
concentrations at 25 °C in a 6.0 M aqueous sodium
hydroxide solution.
Kw = [H3O+][OH-] = 1.0 x 10-14
NaOH(aq) is strong (100% ionized) so [OH-] = 6.0 M.
[H3O+](6.0) = 1.0 x 10-14
[H3O+] = 1.7 x 10-15 M
[OH-] = 6.0 M
34
‫חומצות ובסיסים‬-12
The pH Scale
pH =
35
log[H3O+]
Acidic:
pH < 7
Neutral:
pH = 7
Basic:
pH > 7
‫חומצות ובסיסים‬-12
[H3O+]
pH
= 10
The pH Scale
The hydronium ion concentration for lemon juice is
approximately 0.0025 M. What is the pH when [H3O+] =
0.0025 M?
2 significant figures
pH = log(0.0025) = 2.60
2 decimal places
36
‫חומצות ובסיסים‬-12
The pOH scale
Base concentrations:
pOH = −log10[OH-]
A neutral solution (25°C) has:
pOH = −log10[1.0 x 10-7] = −(−7.00) = 7.00
Since
Kw = [ H3O+ ][ OH- ] = 1.0 x 10-14
−log(KW)= −log[H3O+] + (−log[OH-]) = −log(1.0 x 10-14)
pKw = pH + pOH = 14.00
(Valid in all aq. solns. at 25°C: acidic, neutral or basic)
37
‫חומצות ובסיסים‬-12
pH Calculations
Given two aqueous solutions (25°C).
Solution A: [OH-] = 4.3 x 10-4 M,
Solution B: [H3O+] = 7.5 x 10-9 M.
Which has the higher pH? Which is more acidic?
Solution A:
pOH = −log[ OH-] = 3.37
pH + pOH = pKw = 14.00
pH = 10.63
Solution B:
pH = −log[ H3O+] = 8.12
A has higher pH, B is more acidic
38
‫חומצות ובסיסים‬-12
Measuring pH
H3O+ concentrations can be measured with an:
Electronic pH meter:
• fast and accurate.
• preferred method.
39
Acid-base indicator:
• substance changes color over a small pH range.
• may have multiple colors (e.g. bromthymol blue).
• one “color” may be colorless (e.g. phenolphthalein).
• cheap and convenient.
‫חומצות ובסיסים‬-12
Measuring pH
Acid–Base Indicator: A substance that changes color in a specific pH range.
Indicators exhibit pH-dependent color changes because they are weak acids
and have different colors in their acid (HIn) and conjugate base (In) forms.
HIn(aq) + H2O(l)
Color A
40
H3O+(aq) + In(aq)
Color B
‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Acids
HA(aq) + H2O(l)
H3O+(aq) + A(aq)
[H3O+][A]
Acid-Dissociation Constant: Ka =
[HA]
41
‫חומצות ובסיסים‬-12
‫‪-12‬חומצות ובסיסים‬
‫‪42‬‬
Calculation of [H+] in a Water Solution of a Weak Acid
x
 0.05, then a  x  a
• If
a
• Note x = [H+]eq and a = [HB]o
+
[H
]eq
x
=
a [HB]o
If the % ionization is less than 5%, your
assumption is valid. If not, solve for x
using the quadratic equation
• % ionization can be obtained by multiplying 100%
[H + ]eq
x
%=
× 100%
a
[HB] o
43
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‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Acids
The pH of 0.250 M HF is 2.036. What are the values of Ka and
pKa for hydrofluoric acid?
HF(aq) + H2O(l)
H3O+(aq) + F(aq)
0.250
≈0
0
x
+x
+x
0.250  x
x
x
x = [H3O+] = 102.036 = 0.00920 M
44
‫חומצות ובסיסים‬-12
pKa =  log(Ka)
Equilibria in Solutions of Weak Acids
Ka =
[H3O+][F]
[HF]
[F] = [H3O+] = 0.00920 M
[HF] = 0.250  x = 0.250  0.00920 = 0.241 M
Ka =
[H3O+][F]
[HF]
(0.00920)(0.00920)
=
0.241
= 3.51 × 104
pKa =  log(Ka) = log(3.51 × 104) = 3.455
45
‫חומצות ובסיסים‬-12
Calculating Equilibrium Concentrations of Weak Acids
Calculate the pH of a 0.10 M HCN solution. At 25 oC, Ka = 4.9 × 1010.
HCN(aq) + H2O(l)
0.10
≈0
0
x
+x
+x
0.10  x
x
x
Ka =
46
H3O+(aq) + CN(aq)
[H3O+][CN]
[HCN]
‫חומצות ובסיסים‬-12
Calculating Equilibrium Concentrations of Weak Acids
(x)(x)
x2
≈
4.9 × 1010 =
(0.10  x) 0.10
x = [H3O+] = 7.0 × 106 M
pH =log([H3O+]) = log(7.0 × 106) = 5.15
47
‫חומצות ובסיסים‬-12
Percent Dissociation in Solutions of Weak Acids
Percent dissociation =
48
[HA]dissociated
[HA]initial
‫חומצות ובסיסים‬-12
× 100%
Polyprotic Acids
Some acids “release” multiple H+
Acid(aq)
H2S
Name
Hydrosulfuric Acid
H3PO4
H2CO3
HOOC-COOH
C3H5(COOH)3
Phosphoric Acid
Carbonic Acid
Oxalic Acid
Citric Acid
Acidic H
2
3
2
2
3
Each H+ ionization has a different Ka.
• 1st proton is easiest to remove
• 2nd is harder, etc.
49
‫חומצות ובסיסים‬-12
Polyprotic Acids
50
‫חומצות ובסיסים‬-12
Ka Values for Polyprotic Acids
Phosphoric acid (H3PO4) has three acidic protons:
H3O+(aq) + H2PO4- (aq)
Ka = 7.2 x 10-3
H2PO4-(aq) + H2O(ℓ)
H3O+(aq) + HPO42- (aq)
Ka = 6.3 x 10-8
HPO42-(aq) + H2O(ℓ)
H+ harder to remove
H3PO4(aq) + H2O(ℓ)
H3O+(aq) + PO43- (aq)
Ka = 4.6 x 10-13
51
‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Bases
B(aq) + H2O(l)
Base
Acid
BH+(aq) + OH(aq)
Acid
Base
[BH+][OH]
Base-Dissociation Constant: Kb =
[B]
NH3(aq) + H2O(l)
NH4+(aq) + OH(aq)
[NH4+][OH]
Kb =
[NH3]
52
‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Bases
53
‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Bases
Calculate the pH of a 0.40 M NH3 solution. At 25 oC, Kb = 1.8 × 105.
NH3(aq) + H2O(l)
NH4+(aq) + OH(aq)
0.40
0
≈0
x
+x
+x
0.40  x
x
x
[NH4+][OH]
Kb =
[NH3]
54
‫חומצות ובסיסים‬-12
Equilibria in Solutions of Weak Bases
(x)(x)
x2
≈
1.8 × 105 =
(0.40  x)
0.40
x = [OH] = 0.0027 M
14
1.0
×
10
[H3O+] =
= 3.7 × 1012 M
0.0027
pH =log([H3O+]) = log(3.7 × 1012) = 11.43
55
‫חומצות ובסיסים‬-12
Relation Between Ka and Kb
NH4+(aq) + H2O(l)
H3O+(aq) + NH3(aq)
Ka
NH3(aq) + H2O(l)
NH4+(aq) + OH(aq)
Kb
H3O+(aq) + OH(aq)
Kw
2 H2O(l)
Ka × Kb =
[H3O+][NH3]
[NH4+]
×
[NH4+][OH]
[NH3]
= [H3O+][OH] = Kw
= (5.6 × 1010)(1.8 × 105) = 1.0 × 1014
56
‫חומצות ובסיסים‬-12
Relation Between Ka and Kb
Ka × Kb = Kw
conjugate acid-base pair
Ka =
Kw
Kb
Kb =
Kw
Ka
pKa + pKb = pKw = 14.00
57
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Yield Neutral Solutions
The following ions do not react appreciably with water to
produce either H3O+ or OH ions:
58
•
Cations from strong bases:
• Alkali metal cations of group 1a (Li+, Na+, K+)
• Alkaline earth metal cations of group 2a
(Mg2+, Ca2+, Sr2+, Ba2+), except for Be2+
•
Anions from strong monoprotic acids:
• Cl, Br, I, NO3, and ClO4
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Yield Acidic Solutions
Salts such as NH4Cl that are derived from a weak base
(NH3) and a strong acid (HCl) yield acidic solutions.
NH4+(aq) + H2O(l)
H3O+(aq) + NH3(aq)
Ammonium ion (NH4+) is the conjugate acid of the weak
base ammonia (NH3) while chloride ion (Cl) is neither
acidic nor basic.
59
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Yield Acidic Solutions
Hydrated cations of small, highly charged metal
ions, such as Al3+.
60
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Yield Acidic Solutions
61
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Yield Basic Solutions
Salts such as NaCN that are derived from a strong base
(NaOH) and a weak acid (HCN) yield basic solutions.
CN(aq) + H2O(l)
HCN(aq) + OH(aq)
Cyanide ion (CN) is the conjugate base of the weak acid
hydrocyanic acid (HCN) while sodium ion (Na+) is neither
acidic nor basic.
62
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Contain Acidic Cations and Basic Anions
The pH of an ammonium carbonate solution, (NH4)2CO3,
depends on the relative acid strength of the cation and
the relative base strength of the anion.
Is it acidic or basic?
63
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Contain Acidic Cations and Basic Anions
(NH4)2CO3:
NH4+(aq) + H2O(l)
CO32(aq) + H2O(l)
64
H3O+(aq) + NH3(aq)
Ka
HCO3(aq) + OH(aq)
Kb
Three possibilities:
• Ka > Kb: The solution will contain an excess of
H3O+ ions (pH < 7).
• Ka < Kb: The solution will contain an excess of
OH ions (pH > 7).
• Ka ≈ Kb: The solution will contain approximately
equal concentrations of H3O+ and OH ions
(pH ≈ 7).
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
Salts That Contain Acidic Cations and Basic Anions
(NH4)2CO3:
NH4+(aq) + H2O(l)
H3O+(aq) + NH3(aq)
Ka
CO32(aq) + H2O(l)
HCO3(aq) + OH(aq)
Kb
1.0 × 1014
10
Ka for NH4+ =
=
=
5.6
×
10
Kb for NH3 1.8 × 105
Kw
Kb for CO32 =
65
Kw
Ka for HCO3
1.0 × 1014
4
=
=
1.8
×
10
5.6 × 1011
Basic, Ka < Kb
‫חומצות ובסיסים‬-12
Acid–Base Properties of Salts
66
‫חומצות ובסיסים‬-12
pH of a Salt Solution
What is the pH of a 1.50 M Na2CO3(aq)? Kb = 2.1 x 10-4
CO32-(aq) + H2O(ℓ)
[ ]initial
[ ]change
[ ]equil
1.50
-x
1.50 - x
HCO3-(aq) + OH-(aq)
0
+x
x
0
+x
x
2
2
-][OH-]
x
x
[HCO
3
Kb = 2.1 x 10-4 =
=
≈
2(1.50 – x)
1.50
[CO3 ]
x = 1.77 x 10-2
pOH = −log(1.77 x 10-2) = 1.75
pH = 14.00 - 1.75 = 12.25
67
‫חומצות ובסיסים‬-12
pH of a Salt Solution
0.132 M NH4Br? Acidic, basic or neutral? Ka(NH4+) = 5.6 x 10-10.
NH4+(aq) + H2O(ℓ)
[ ]initial
[ ]change
[ ]equil
0.132
-x
0.132 - x
0
+x
x
O+]
Ka = 5.6 x 10-10 = [NH3][H+3
[NH4 ]
x = 8.57 x 10-6
68
NH3(aq) + H3O+(aq)
0
+x
x
x2
x2
=
≈
(0.132 – x)
0.132
pH = −log(8.57 x 10-6) = 5.07
Acidic!
‫חומצות ובסיסים‬-12
The Common-Ion Effect
Common-Ion Effect: The shift in the position of an
equilibrium on addition of a substance that provides an
ion in common with one of the ions already involved in
the equilibrium
CH3CO2H(aq) + H2O(l)
69
H3O+(aq) + CH3CO2–(aq)
‫חומצות ובסיסים‬-12
The Common-Ion Effect
The pH of 0.10 M acetic acid is 2.89. Calculate the pH
of a solution that is prepared by dissolved 0.10 mol of
acetic acid and 0.10 mol sodium acetate in enough
water to make 1.00 L of solution.
CH3CO2H(aq) + H2O(l)
0.10
≈0
0.10
–x
+x
+x
0.10 – x
x
0.10 + x
Ka =
70
H3O+(aq) + CH3CO2–(aq)
[H3O+][CH3CO2–]
[CH3CO2H]
‫חומצות ובסיסים‬-12
The Common-Ion Effect
(x)(0.10 + x) x(0.10)
≈
1.8 × 10–5 =
0.10
(0.10 – x)
x = [H3O+] = 1.8 × 10–5 M
pH = –log([H3O+]) = –log(1.8 × 10–5) = 4.74
71
‫חומצות ובסיסים‬-12
The Common-Ion Effect
72
‫חומצות ובסיסים‬-12
The Common-Ion Effect
Le Châtelier’s Principle
CH3CO2H(aq) + H2O(l)
H3O+(aq) + CH3CO2–(aq)
The addition of acetate ion to a solution of acetic acid
suppresses the dissociation of the acid. The equilibrium
shifts to the left.
73
‫חומצות ובסיסים‬-12
The Common-Ion Effect
74
‫חומצות ובסיסים‬-12
Buffer Solutions
Buffer Solution: A solution that contains a weak acid and
its conjugate base and resists drastic changes in pH
Weak acid
+
Conjugate base
75
For
example:
CH3CO2H + CH3CO2–
HF + F–
NH4+ + NH3
H2PO4– + HPO42–
‫חומצות ובסיסים‬-12
Buffer Solutions
Buffer = chemical system that resists changes in pH.
Example
Add 0.010 mol of HCl or NaOH to:
1 L of solution
Pure water
Initial
7.00
[CH3COOH] = 0.5 M +
[CH3COONa] = 0.5 M
4.74
pH
after HCl
2.00
4.72
after NaOH
12.00
4.76
A buffered solution
76
‫חומצות ובסיסים‬-12
Buffer Action
Buffers must contain:
• A weak acid to react with any added base.
• A weak base to react with any added acid.
• These components must not react with each other.
Buffers are made with ~equal quantities of a
conjugate acid-base pair.
(e.g. CH3COOH + CH3COONa)
77
‫חומצות ובסיסים‬-12
Buffer Action
Added OH- removed by the acid:
CH3COOH + OHCH3COO- + H2O
Added H3O+ removed by the conjugate base:
CH3COO- + H3O+
CH3COOH + H2O
78
‫חומצות ובסיסים‬-12
Buffer Action
The equilibrium maintains the acid/base ratio.
CH3COOH + H2O
CH3COO- + H3O+
pH remains stable.
79
‫חומצות ובסיסים‬-12
Buffer Action
Blood pH
• Blood is buffered at pH = 7.40 ± 0.05.
 pH too low: acidosis.
 pH too high: alkalosis.
• CO2 generates the most important blood buffer.
CO2(aq) + H2O(ℓ)
H2CO3(aq) + H2O(ℓ)
80
H2CO3(aq)
H3O+(aq) + HCO3-(aq)
‫חומצות ובסיסים‬-12
The pH of Buffer Solutions
Depends on the [acid]/[base] – not absolute amounts.
Henderson-Hasselbalch equation
[H3O+] = Ka
log [H3
O+]
[HA]
= log Ka + log [A-]
-]
[A
pH = pKa + log
[HA]
Note: pH = pKa
81
[HA]
[A-]
With pKa = -log Ka
when [HA] = [A-]
+]
[BH
pOH = pKb + log
[B]
B: weak base
‫חומצות ובסיסים‬-12
The pH of Buffer Solutions
What is the pH of a 0.050 M (monoprotic) pyruvic acid +
0.060 M sodium pyruvate buffer? Ka = 3.2 x 10-3.
pH = – log(3.2 x 10-3) + log(0.060/0.050)
= 2.49 + 0.08
= 2.57
What is the HPO42-/H2PO4- ratio in blood at pH =7.40.
Ka(H2PO4- ) = Ka,2(H3PO4) = 6.2 x 10-8.
7.40 = −log(6.2 x 10-8) + log([HPO42-]/[H2PO4-])
log([HPO42-]/[H2PO4-]) = 0.192
[HPO42-]/[H2PO4-] = 1.5
82
‫חומצות ובסיסים‬-12
Common Buffers
pH
weak acid
weak base
Ka(weak acid) pKa
4 lactic acid
lactate ion
1.4 x 10-4
3.85
5 acetic acid
acetate ion
1.8 x 10-5
4.74
6 carbonic acid
hydrogen carbonate ion
4.3 x 10-7
6.37
7 dihydrogen phosphate hydrogen phosphate ion
6.3 x 10-8
7.20
8 hypochlorous acid
hypochlorite ion
6.8 x 10-8
7.17
9 ammonium ion
ammonia
5.6 x 10-10
9.25
carbonate ion
4.7 x 10-11 10.33
10 hydrogen carbonate
Useful buffer range: pH = pKa ±1
(10:1 or 1:10 ratio of [A-]/[HA]).
83
Buffer Capacity: The amount of
acid (or base) that can be added
without large pH changes.
‫חומצות ובסיסים‬-12
Addition of Acid or Base to a Buffer
1.0 L of buffer is prepared with [NaH2PO4] = 0.40 M and
[Na2HPO4] = 0.25 M. Calculate the pH of: (a) the buffer
(b) after 0.10 mol of NaOH is added.
Ka (H2PO4-) = 6.3 x 10-8
pKa = -log(6.3 x 10-8) = 7.20
(a) No base added:
[A-]
pH = pKa + log
[HA]
84
pH = 7.20 + log 0.25 = 7.00
0.40
‫חומצות ובסיסים‬-12
Addition of Acid or Base to a Buffer
1.00 L Buffer: [NaH2PO4]= 0.40 M ; [Na2HPO4] = 0.25 M. (b) calculate pH after 0.10 mol
of NaOH is added. Ka (H2PO4-) = 6.3 x 10-8
(b) 0.10 mol of NaOH, converts conj. acid to base:
ninitial
nadded
nafter
H2PO4- + OH0.40
0.10
(0.40 – 0.10) 0
[A-]
pH = pKa + log
[HA]
→ HPO42- + H2O
0.25
(0.25 + 0.10)
Use n directly. [ ] = n/ V and V is
the same for both (cancels)
pH = 7.20 + log 0.35 = 7.27
0.30
85
‫חומצות ובסיסים‬-12
Acid-Base Titrations
• Standard solution (titrant) is added from a buret.
• Equivalence occurs when a stoichiometric amount of
titrant has been added.
86
‫חומצות ובסיסים‬-12
Acid-Base Titrations
An indicator is used to find the end point.
• Color change observed.
• End point ≠ equivalence point (should be close...).
At the equivalence point:
ntitrant = nanalyte
ntitrant = Vtitrant [titrant]
nanalyte= Vanalyte[analyte]
87
‫חומצות ובסיסים‬-12
Detection of the Equivalence Point
The acid-Base indicator is a weak acid that changes
color with changes in pH.
HIn(aq) + H2O(ℓ)
H3O+(aq) + In-(aq)
color in acid
color in base
Observed color will vary (depends on the relative
amounts of HIn and In- in solution).
Ka =
88
[H3O+][In-]
[HIn]
‫חומצות ובסיסים‬-12
Detection of the Equivalence Point
The acidic form dominates when the [HIn] >> [In-]
If
[HIn] = 10
[In-]
Ka =
-pKa = -pH - 1
[H3O+][In-]
[H3O+]
=
[HIn]
10
pH = pKa -1
The basic color dominates when [In-] >> [HIn]
If
89
[In-] = 10
[HIn]
-pKa = 1 - pH
Ka =
[H3O+][In-]
= 10 [H3O+]
[HIn]
pH = pKa +1
‫חומצות ובסיסים‬-12
Detection of the Equivalence Point
90
‫חומצות ובסיסים‬-12
Titration of Strong Acid with Strong Base
Acid-base titration curve = plot of pH vs Vtitrant added.
Titrate 50.0 mL of 0.100 M HCl with 0.100 M NaOH
Initial pH = -log(0.100) = 1.000 (fully ionized acid)
0.100 mol
nacid = 0.0500 L
L
= 5.00 x 10-3 mol
91
‫חומצות ובסיסים‬-12
Titration of Strong Acid with Strong Base
After 40.0 mL of base added (before equivalence)
Base removes H3O+ :
H O+ + OH-  2 H O
3
2
Acid concentration becomes:
[H3
O+]
nacid – nbase added
=
Vacid(ℓ) + Vbase(ℓ) added
-3 – 4.00 x 10-3) mol
(5.00
x
10
= 0.0111 M
[H3O+] =
(0.0500 + 0.0400) L
pH = -log(0.0111) = 1.95
At Equivalence (50.0 mL added)
All acid and base react. Neutral salt. pH = 7.00.
92
‫חומצות ובסיסים‬-12
Titration of Strong Acid with Strong Base
After Equivalence (50.2 mL added)
All acid consumed.
nOH- added = 0.0502 L
0.100 mol
= 5.02 x 10-3 mol
L
original nacid = 0.0500 L 0.100 mol = 5.00 x 10-3 mol
L
nOH- remaining = 0.02 x 10-3 mol
Vtotal = (0.0500 + 0.0502) L = 0.1002 L
pOH = - log(0.02 x10-3 / 0.1002) = 3.70
93
pH = 14.00 – 3.70 = 10.30
‫חומצות ובסיסים‬-12
Titration of Strong Acid with Strong Base
Vtitrant/mL Vexcess/mL Vtotal/mL
94
[OH-]/mol L-1
pH
50.0
0
100.0
0
7.00
50.1
0.1
100.1
0.0001
10.00
50.2
0.2
100.2
0.0002
10.30
51
1
101.0
0.0010
11.00
55
5
105.0
0.0048
11.67
60
10
110.0
0.0091
11.96
70
20
120.0
0.0167
12.22
80
30
130.0
0.0200
12.30
0.1 mL of base
increased the
pH by 3 units!
‫חומצות ובסיסים‬-12
Titration of Strong Acid with Strong Base
95
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
More complicated.
• Weak acid is in equilibrium with its conjugate base.
Example
Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M
NaOH. What is the pH after the following titrant
additions: 0 mL, 40.0 mL, 50.0 mL, and 50.2 mL?
0 mL added
HA + H2O
H3O+ + A-
+][A-]
2
[H
O
x
-5
3
Ka = 1.8 x 10 =
≈
[HA]
(0.100)
96
pH = -log(0.0013) = 2.88
x = 0.0013
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
40.0 mL added
50 mL of
0.100 M
ninitial
nadded
nleft
HA(aq) + OH-(aq)
A-(aq)
+
H2O(ℓ)
0.00500
0.00400
0.00100
Each OH- removes 1 HA…
nleft = 0.00500 - 0.00400
0.00400
…and makes 1 A40 mL of 0.100 M
Vtotal = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L
97
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
40.0 mL added (continued)
Henderson-Hasselbalch:
[A-] = nA-/ V
pH = -log(1.8 x
10-5)
(0.00400/0.0900)
+ log
(0.00100/0.0900)
[HA] = nHA/ V
Note: V cancels (could be omitted)
0.0400
pH = 4.74 + log
= 5.35
0.0100
98
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
(c) 50.0 mL added
Equivalence. All HA is converted to A-.
HA(aq) + OH-(aq)
ninitial
nadded
nleft
A-(aq)
+
H2O(ℓ)
0.00500
0.00500
0
0.00500
Vtotal= 100.0 mL, so
[A-] = 0.00500 mol/0.100 L = 0.0500 M
99
A- is basic!
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
Use Kb to solve for the [OH-] generated by [A-]:
[OH-][HA]
Kb =
[A-]
A+
(0.0500 - x)
Kb = 5.6 x
10-10
pOH = 5.28
100
Kb = Kw / Ka = 5.6 x 10-10
H2O
2
x
≈
0.0500
HA
x
+
OHx
x = 5.3 x 10-6 M
pH = 14 - pOH = 8.72
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
(c) 50.2 mL added
0.2 mL of “extra” OH- dominates pH.
Ignore any contribution from A-.
[OH-] = (0.0002 L x 0.100 mol/L) / 0.1002 L
= 2.0 x 10-4 M
pOH = 3.7
pH = 14.0 – 3.7 = 10.3
101
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
Titrate of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.
102
pH = pKa at 50% titration (25 mL) pKa (acetic acid) = 4.74.
Short vertical section compared to strong acid/strong base.
‫חומצות ובסיסים‬-12
Titration of Weak Acid with Strong Base
103
pH = pKa at 50% titration
Weaker acid = shorter vertical section
‫חומצות ובסיסים‬-12
Titration of Weak Base with Strong Acid
104
50.0 mL of 0.100 M ammonia titrated with 0.100 M HCl.
• pH = pKa at 50% to equivalence (pKa = 9.25).
‫חומצות ובסיסים‬-12
Titration of Polyprotic Acids & Bases
Maleic acid:
HOOC–CH=CH–COOH
(diprotic acid)
105
Sodium carbonate
Na+(aq) + CO32-(aq)
(basic anion)
‫חומצות ובסיסים‬-12