Bronsted Lowry Acid and Base.pptx

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‫ובסיסים‬ ‫חומצות‬
1
Acid-Base Concepts: The Brønsted-Lowry Theory
Arrhenius Acid: A substance that dissociates in water to
produce hydrogen ions, H+
Arrhenius Base: A substance that dissociates in water to
produce hydroxide ions, OH–
M+(aq) + OH–(aq)
MOH(aq)
H+(aq) + A–(aq)
HA(aq)

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Conjugate Acid-Base Pairs: Chemical species whose
formulas differ only by one hydrogen ion, H+
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

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Acid-Dissociation Equilibrium

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Base-Dissociation Equilibrium

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NH3(g) + H2O(ℓ) NH4
+(aq) + OH-(aq)
Base:
H+ acceptor
Acid:
H+ donor
Brønsted-Lowry Acid = Proton donor
Brønsted-Lowry Base = Proton acceptor
Works for non-aqueous solutions and explains why NH3
is basic:
Brønsted-Lowry Concept

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Conjugate acid-base pairs are molecules or ions
related by the loss/gain of one H+:
Conjugate Acid Conjugate Base
H3O+ H2O
CH3COOH CH3COO-
NH4
+ NH3
H2SO4 HSO4
-
HSO4
- SO4
2-
HCl Cl-
donate H+
accept H+
NH4
+ and NH2
- are not conjugate, conversion requires
2 H+.
Conjugate Acid-Base Pairs

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Identify the base conjugate to HF(aq) and the acid
conjugate to HCO3
-(aq).
HCO3
- + H+ H2CO3
HF H+ + F-
( CO3
2- is the base conjugate to HCO3
- )
Conjugate
Acid
Base:
H+ acceptor
Conjugate
Base
Acid:
H+ donor
Conjugate Acid-Base Pairs

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Water acts as a base when an acid dissolves in water:
HBr(aq) + H2O(ℓ) H3O+(aq) + Br-(aq)
acid base acid base
Water is amphiprotic - it can donate or accept a proton
(act as acid or base).
But water acts as an acid for some bases:
H2O(ℓ) + NH3(aq) NH4
+(aq) + OH-(aq)
acid base acid base
Water’s Role as Acid or Base

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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.
A B
..
..
+ A B
Acid
e- pair acceptor
Base
e- pair donor
Coordinate
covalent bond
Coordinate covalent bond: a shared e- pair, with both e-
donated by the same atom.
Lewis Acids & Bases

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Lewis acid Lewis base
+
H O
H
H
H
O
H
H
Examples:
+ N
H
H
H
B
F
F
F
B
F
F
F
N
H
H
H
H+ + H2O H3O+
Lewis acid Lewis base
Lewis Acids & Bases

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Metal ions can act as Lewis acids. They have:
• Missing e-.
• Empty valence orbitals.
• Metal ion + Lewis base → complex ion
Hydroxide ions are Lewis bases.
• O is e- rich (has large electronegativity):
• Can easily donate an e- pair to a bond.
:O–H
:
:
δ+
δ-
-
Ag+ (aq) + 2 :NH3 (aq) [H3N:Ag:NH3]+ (aq)
Positive Metal Ions as Lewis Acids

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Many metal hydroxides are amphoteric – they react
with acids and bases.
Positive Metal Ions as Lewis Acids
Al(OH)3(s) + OH-(aq) [Al(OH)4]-(aq)
Aluminum hydroxide can act as a Lewis acid:
Al(OH)3(s) + 3 H3O+(aq) Al3+(aq) + 6 H2O(ℓ)
or as a Brønsted-Lowry base:

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δ-
Non-metal oxides act as Lewis acids.
• O attracts e- from any multiple bond.
• Leaves the other non-metal e- deficient.
• Susceptible to attack by Lewis base:
S
O O
δ+
δ-
S
O O
δ+
δ-
δ-
O = C = O
δ+ δ-
δ-
Neutral Molecules as Lewis Acids
O = C = O
H O H O
C = O
O
- -

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Acid Strength and Base Strength
Weak Acid: An acid that is only partially dissociated in
water and is thus a weak electrolyte

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Hydrated Protons and Hydronium Ions
H+(aq) + A(aq)
HA(aq)
[H(H2O)n]+
For our purposes, H+ is equivalent to H3O+.
n = 4 H9O4
+
n = 1 H3O+
n = 2 H5O2
+
n = 3 H7O3
+
Due to high reactivity of the hydrogen ion, it is actually
hydrated by one or more water molecules.

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H3O+(aq) + A(aq)
HA(aq) + H2O(l)
Base
Acid
Base
Acid
Weaker acid + Weaker base
Stronger acid + Stronger base
With equal concentrations of reactants and
products, what will be the direction of reaction?

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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.
Weakly ionized, reverse reaction readily occurs.
The conjugate base is relatively strong.
Weak acid + H2O H3O+ + conjugate base
Relative Strength of Acids & Bases

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Acid strength: HF > NH4
+.
(Base strength: NH3 > F-)
HF has greater tendency to ionize
than NH4
+.
Reactant Favored
NH4
+ + F- NH3 + HF
Problem Is the following aqueous reaction product or reactant
favored?
Acid
strength
increasing
Base
strength
increasing
Conj acid. Conj. base
H2SO4
HSO4
-
HBr
Br-
HCl
Cl-
HF
F-
NH4
+
NH3
OH-
O2-
H2
H-
CH4
CH3
-
Relative Strength of Acids & Bases

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Factors That Affect Acid Strength
Bond Polarity

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Bond Strength

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Oxoacids

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Oxoacids

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Organic acids
Contain the carboxylic acid functional group:
Carboxylic Acids & Amines

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• 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:
The carboxylic acid H is acidic (will ionize). All other
protons are non acidic:
CH3 – C
O
O-
CH3 – C
O-
O
+ H2O + H3O+
CH3 – C
O
OH
CH3 – C
O
O-

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Strengths of Carboxylic Acids
R-COOH
Acid Ka
Acetic CH3COOH 1.8 x 10-5
Hexanoic CH3CH2CH2CH2CH2COOH 1.4 x 10-5
Chloroacetic CH2ClCOOH 0.0013
Trichloroacetic CCl3COOH 0.3
Trifluoroacetic CF3COOH 0.6
All hydrocarbon R’s “pull” equally… ~equal strength.
Add a halogen to R = more e- withdrawing
= stronger acid

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Amine groups are basic: R–NH2, R–NHR or R–NR2
..
.. ..
• R = any hydrocarbon.
• Lone-pair on N accepts a proton (like NH3).

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Name R
glycine H
alanine CH3
valine (CH3)2CH
serine HOCH2
… and others
=
−
−
H O
R–C–C–O–H
NH2
Amino Acids & Zwitterions
Amino acids contain acid and amine functional groups.
Alpha
carbon
They are crystalline solids (m.p. > 200°C). Why?
• The solid is a zwitterion (has multiple charges)
=
−
−
H O
R–C–C–O-
NH3
+
The acid H+ has
transferred to the
amine (base)

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Amino Acids & Zwitterions
Amino acid in solution have pH dependent structures.
Alanine:
=
−
−
H O
CH3– C– C–O-
NH3
+
=
−
−
H O
CH3– C– C–O–H
NH3
+
=
−
−
H O
CH3– C– C–O-
NH2
OH-
H+
base
added
acid
added
Positively
charged
Negatively
charged
Zwitterion
(no net charge)

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Dissociation of Water
H3O+(aq) + OH(aq)
2 H2O(l)
Kw = [H3O+][OH]
Ion-Product Constant for Water:
Dissociation of Water:
Kw = (1.0 × 107)(1.0 × 107) = 1.0 × 1014
at 25 oC: [H3O+] = [OH] = 1.0 × 107 M

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Kw = [H3O+][OH] = 1.0 × 1014
[OH] =
[H3O+]
1.0 × 1014
[H3O+] =
[OH]
1.0 × 1014

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Acidic: [H3O+] > [OH]
Neutral: [H3O+] = [OH]
Basic: [H3O+] < [OH]

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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
Autoionization of Water

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The pH Scale
pH = log[H3O+] [H3O+] = 10
pH
Acidic: pH < 7
Neutral: pH = 7
Basic: pH > 7

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The pH Scale
The hydronium ion concentration for lemon juice is
approximately 0.0025 M. What is the pH when [H3O+] =
0.0025 M?
pH = log(0.0025) = 2.60
2 significant figures
2 decimal places

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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)
(Valid in all aq. solns. at 25°C: acidic, neutral or basic)
pKw = pH + pOH = 14.00
The pOH scale

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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
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?
pH Calculations

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H3O+ concentrations can be measured with an:
Electronic pH meter:
• fast and accurate.
• preferred method.
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.
Measuring pH

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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.
H3O+(aq) + In(aq)
HIn(aq) + H2O(l)
Color B
Color A

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Equilibria in Solutions of Weak Acids
Acid-Dissociation Constant:
H3O+(aq) + A(aq)
HA(aq) + H2O(l)
[H3O+][A]
[HA]
Ka =

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Copyright ©2016 Cengage Learning. All Rights Reserved. 12
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Calculation of [H+] in a Water Solution of a Weak Acid
• If
• Note x = [H+]eq and a = [HB]o
• % ionization can be obtained by multiplying 100%
0.05,
x
a x a
a
  
then
+
eq
o
[H ]
=
[HB]
x
a
+
eq
o
[H ]
% = × 100%
[HB]
x
a
If the % ionization is less than 5%, your
assumption is valid. If not, solve for x
using the quadratic equation

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The pH of 0.250 M HF is 2.036. What are the values of Ka and
pKa for hydrofluoric acid?
H3O+(aq) + F(aq)
HF(aq) + H2O(l)
0.250 ≈0 0
x +x +x
0.250  x x x
x = [H3O+] = 102.036 = 0.00920 M
pKa =  log(Ka)

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(0.00920)(0.00920)
0.241
= 3.51 × 104
[HF] = 0.250  x = 0.250  0.00920 = 0.241 M
[F] = [H3O+] = 0.00920 M
[H3O+][F]
[HF]
Ka =
pKa =  log(Ka) = log(3.51 × 104) = 3.455
[H3O+][F]
[HF]
=
Ka =

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Calculating Equilibrium Concentrations of Weak Acids
Calculate the pH of a 0.10 M HCN solution. At 25 oC, Ka = 4.9 × 1010.
H3O+(aq) + CN(aq)
HCN(aq) + H2O(l)
0.10 ≈0 0
x +x +x
0.10  x x x
[H3O+][CN]
[HCN]
Ka =

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Calculating Equilibrium Concentrations of Weak Acids
x2
0.10
≈
4.9 × 1010 =
(x)(x)
(0.10  x)
pH =log([H3O+]) = log(7.0 × 106) = 5.15
x = [H3O+] = 7.0 × 106 M

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Percent Dissociation in Solutions of Weak Acids
Percent dissociation =
[HA]initial
[HA]dissociated
× 100%

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Polyprotic Acids
Some acids “release” multiple H+
Each H+ ionization has a different Ka.
• 1st proton is easiest to remove
• 2nd is harder, etc.
Acid(aq) Name Acidic H
H2S Hydrosulfuric Acid 2
H3PO4 Phosphoric Acid 3
H2CO3 Carbonic Acid 2
HOOC-COOH Oxalic Acid 2
C3H5(COOH)3 Citric Acid 3

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Polyprotic Acids

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Phosphoric acid (H3PO4) has three acidic protons:
H3PO4(aq) + H2O(ℓ) H3O+(aq) + H2PO4
- (aq)
Ka = 7.2 x 10-3
H2PO4
-(aq) + H2O(ℓ) H3O+(aq) + HPO4
2- (aq)
Ka = 6.3 x 10-8
HPO4
2-(aq) + H2O(ℓ) H3O+(aq) + PO4
3- (aq)
Ka = 4.6 x 10-13
Ka Values for Polyprotic Acids
H
+
harder
to
remove

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Equilibria in Solutions of Weak Bases
NH4
+(aq) + OH(aq)
NH3(aq) + H2O(l)
Base
Acid
Acid
Base
[NH4
+][OH]
[NH3]
Kb =
BH+(aq) + OH(aq)
B(aq) + H2O(l)
[BH+][OH]
[B]
Kb =
Base-Dissociation Constant:

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Calculate the pH of a 0.40 M NH3 solution. At 25 oC, Kb = 1.8 × 105.
[NH4
+][OH]
[NH3]
Kb =
NH4
+(aq) + OH(aq)
NH3(aq) + H2O(l)
0.40 0 ≈0
x +x +x
0.40  x x x

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x2
0.40
≈
1.8 × 105 =
(x)(x)
(0.40  x)
pH =log([H3O+]) = log(3.7 × 1012) = 11.43
x = [OH] = 0.0027 M
= 3.7 × 1012 M
0.0027
1.0 × 1014
[H3O+] =

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Relation Between Ka and Kb
H3O+(aq) + OH(aq)
2 H2O(l)
H3O+(aq) + NH3(aq)
NH4
+(aq) + H2O(l)
NH4
+(aq) + OH(aq)
NH3(aq) + H2O(l)
[NH4
+][OH]
[NH3]
[H3O+][NH3]
[NH4
+]
×
Kw
Ka
Kb
= [H3O+][OH] = Kw
= (5.6 × 1010)(1.8 × 105) = 1.0 × 1014
Ka × Kb =

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Relation Between Ka and Kb
pKa + pKb = pKw = 14.00
Ka × Kb = Kw
Kb =
Ka
Kw
Ka =
Kb
Kw
conjugate acid-base pair

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Acid–Base Properties of Salts
• 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

The following ions do not react appreciably with water to
produce either H3O+ or OH ions:
Salts That Yield Neutral Solutions

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Ammonium ion (NH4
+) is the conjugate acid of the weak
base ammonia (NH3) while chloride ion (Cl) is neither
acidic nor basic.
H3O+(aq) + NH3(aq)
NH4
+(aq) + H2O(l)
Salts such as NH4Cl that are derived from a weak base
(NH3) and a strong acid (HCl) yield acidic solutions.
Salts That Yield Acidic Solutions

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Hydrated cations of small, highly charged metal
ions, such as Al3+.

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Salts That Yield Basic Solutions
Cyanide ion (CN) is the conjugate base of the weak acid
hydrocyanic acid (HCN) while sodium ion (Na+) is neither
acidic nor basic.
HCN(aq) + OH(aq)
CN(aq) + H2O(l)
Salts such as NaCN that are derived from a strong base
(NaOH) and a weak acid (HCN) yield basic solutions.

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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?
Salts That Contain Acidic Cations and Basic Anions

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HCO3
(aq) + OH(aq)
CO3
2(aq) + H2O(l) Kb
H3O+(aq) + NH3(aq)
NH4
+(aq) + H2O(l) Ka
(NH4)2CO3:
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).

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= 1.8 × 104
5.6 × 1011
1.0 × 1014
Kb for CO3
2 =
Ka for HCO3

Kw
=
= 5.6 × 1010
1.8 × 105
1.0 × 1014
Ka for NH4
+ =
Kb for NH3
Kw
=
Basic, Ka < Kb
(NH4)2CO3:
HCO3
(aq) + OH(aq)
CO3
2(aq) + H2O(l) Kb
H3O+(aq) + NH3(aq)
NH4
+(aq) + H2O(l) Ka

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CO3
2-(aq) + H2O(ℓ) HCO3
-(aq) + OH-(aq)
[ ]initial 1.50 0 0
[ ]change - x +x +x
[ ]equil 1.50 - x x x
Kb = 2.1 x 10-4 = = ≈
[HCO3
-][OH-]
[CO3
2-]
x2
(1.50 – x)
x2
1.50
x = 1.77 x 10-2 pOH = −log(1.77 x 10-2) = 1.75
pH = 14.00 - 1.75 = 12.25
What is the pH of a 1.50 M Na2CO3(aq)? Kb = 2.1 x 10-4
pH of a Salt Solution

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NH4
+(aq) + H2O(ℓ) NH3(aq) + H3O+(aq)
[ ]initial 0.132 0 0
[ ]change - x +x +x
[ ]equil 0.132 - x x x
x = 8.57 x 10-6 pH = −log(8.57 x 10-6) = 5.07
Ka = 5.6 x 10-10 = = ≈
[NH3][H3O+]
[NH4
+]
x2
(0.132 – x)
x2
0.132
0.132 M NH4Br? Acidic, basic or neutral? Ka(NH4
+) = 5.6 x 10-10.
pH of a Salt Solution
Acidic!

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The Common-Ion Effect
H3O+(aq) + CH3CO2
–(aq)
CH3CO2H(aq) + H2O(l)
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

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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.
H3O+(aq) + CH3CO2
–(aq)
0.10 ≈0 0.10
–x +x +x
0.10 – x x 0.10 + x
[H3O+][CH3CO2
–]
[CH3CO2H]
Ka =

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x(0.10)
0.10
≈
1.8 × 10–5 =
(x)(0.10 + x)
(0.10 – x)
pH = –log([H3O+]) = –log(1.8 × 10–5) = 4.74
x = [H3O+] = 1.8 × 10–5 M

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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.
Le Châtelier’s Principle

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Buffer Solutions
Buffer Solution: A solution that contains a weak acid and
its conjugate base and resists drastic changes in pH
CH3CO2H + CH3CO2
–
HF + F–
NH4
+ + NH3
H2PO4
– + HPO4
2–
Weak acid
+
Conjugate base
For
example:

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Buffer = chemical system that resists changes in pH.
Example
Add 0.010 mol of HCl or NaOH to:
Buffer Solutions
1 L of solution
pH
Initial after HCl after NaOH
Pure water 7.00 2.00 12.00
[CH3COOH] = 0.5 M +
[CH3COONa] = 0.5 M
4.74 4.72 4.76
A buffered solution

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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)
Buffer Action

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Added OH- removed by the acid:
CH3COOH + OH- CH3COO- + H2O
Added H3O+ removed by the conjugate base:
CH3COO- + H3O+ CH3COOH + H2O
Buffer Action

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The equilibrium maintains the acid/base ratio.
CH3COOH + H2O CH3COO- + H3O+
pH remains stable.
Buffer Action

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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)
H2CO3(aq) + H2O(ℓ) H3O+(aq) + HCO3
-(aq)
Buffer Action

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Henderson-Hasselbalch equation
Depends on the [acid]/[base] – not absolute amounts.
[A-]
[HA]
pH = pKa + log With pKa = -log Ka
Note: pH = pKa when [HA] = [A-]
The pH of Buffer Solutions
[H3O+] = Ka
[HA]
[A-]
[HA]
[A-]
log [H3O+] = log Ka + log
[BH+]
[B]
pOH = pKb + log
B: weak base

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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
The pH of Buffer Solutions
What is the HPO4
2-/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([HPO4
2-]/[H2PO4
-])
log([HPO4
2-]/[H2PO4
-]) = 0.192
[HPO4
2-]/[H2PO4
-] = 1.5

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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
10 hydrogen carbonate carbonate ion 4.7 x 10-11 10.33
Common Buffers
Useful buffer range: pH = pKa ±1
(10:1 or 1:10 ratio of [A-]/[HA]).
Buffer Capacity: The amount of
acid (or base) that can be added
without large pH changes.

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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.
Addition of Acid or Base to a Buffer
[A-]
[HA]
pH = pKa + log
pKa = -log(6.3 x 10-8) = 7.20
(a) No base added:
0.25
0.40
pH = 7.20 + log = 7.00
Ka (H2PO4
-) = 6.3 x 10-8

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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:
H2PO4
- + OH- → HPO4
2- + H2O
[A-]
[HA]
pH = pKa + log
0.35
0.30
pH = 7.20 + log = 7.27
ninitial 0.40 0.25
nadded 0.10
nafter (0.40 – 0.10) 0 (0.25 + 0.10)
Use n directly. [ ] = n/ V and V is
the same for both (cancels)
Addition of Acid or Base to a Buffer

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• Standard solution (titrant) is added from a buret.
• Equivalence occurs when a stoichiometric amount of
titrant has been added.
Acid-Base Titrations

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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]

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The acid-Base indicator is a weak acid that changes
color with changes in pH.
Detection of the Equivalence Point
[H3O+][In-]
[HIn]
Ka =
Observed color will vary (depends on the relative
amounts of HIn and In- in solution).
HIn(aq) + H2O(ℓ) H3O+(aq) + In-(aq)
color in acid color in base

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The acidic form dominates when the [HIn] >> [In-]
-pKa = -pH - 1 pH = pKa -1
The basic color dominates when [In-] >> [HIn]
-pKa = 1 - pH pH = pKa +1
[H3O+][In-]
[HIn]
Ka =
[H3O+]
10
=
[HIn]
[In-]
= 10
If
[H3O+][In-]
[HIn]
Ka = = 10 [H3O+]
[In-]
[HIn]
= 10
If

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90

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91
Acid-base titration curve = plot of pH vs Vtitrant added.
Titrate 50.0 mL of 0.100 M HCl with 0.100 M NaOH
nacid = 0.0500 L
= 5.00 x 10-3 mol
0.100 mol
L
Initial pH = -log(0.100) = 1.000 (fully ionized acid)
Titration of Strong Acid with Strong Base

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After 40.0 mL of base added (before equivalence)
Base removes H3O+ : H3O+ + OH-  2 H2O
[H3O+] =
nacid – nbase added
Vacid(ℓ) + Vbase(ℓ) added
[H3O+] =
(5.00 x 10-3 – 4.00 x 10-3) mol
(0.0500 + 0.0400) L
= 0.0111 M
pH = -log(0.0111) = 1.95
At Equivalence (50.0 mL added)
All acid and base react. Neutral salt. pH = 7.00.
Acid concentration becomes:

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93
After Equivalence (50.2 mL added)
All acid consumed.
nOH- added = 0.0502 L = 5.02 x 10-3 mol
0.100 mol
L
original nacid = 0.0500 L = 5.00 x 10-3 mol
0.100 mol
L
Vtotal = (0.0500 + 0.0502) L = 0.1002 L
pOH = - log(0.02 x10-3 / 0.1002) = 3.70
pH = 14.00 – 3.70 = 10.30
nOH- remaining = 0.02 x 10-3 mol

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Vtitrant/mL Vexcess/mL Vtotal/mL [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!

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95

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96
More complicated.
• Weak acid is in equilibrium with its conjugate base.
HA + H2O H3O+ + A-
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?
x2
(0.100)
≈
[H3O+][A-]
[HA]
Ka = 1.8 x 10-5 = x = 0.0013
pH = -log(0.0013) = 2.88
0 mL added
Titration of Weak Acid with Strong Base

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97
HA(aq) + OH-(aq) A-(aq) + H2O(ℓ)
40.0 mL added
ninitial 0.00500
nadded 0.00400
nleft 0.00100 0.00400
Each OH- removes 1 HA…
nleft = 0.00500 - 0.00400
…and makes 1 A-
40 mL of 0.100 M
50 mL of
0.100 M
Vtotal = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L

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98
40.0 mL added (continued)
(0.00400/0.0900)
(0.00100/0.0900)
pH = -log(1.8 x 10-5) + log
0.0400
0.0100
pH = 4.74 + log = 5.35
Henderson-Hasselbalch:
[A-] = nA-/ V
[HA] = nHA/ V
Note: V cancels (could be omitted)

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99
HA(aq) + OH-(aq) A-(aq) + H2O(ℓ)
(c) 50.0 mL added
Equivalence. All HA is converted to A-.
Vtotal= 100.0 mL, so
[A-] = 0.00500 mol/0.100 L = 0.0500 M
ninitial 0.00500
nadded 0.00500
nleft 0 0.00500
A- is basic!

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Use Kb to solve for the [OH-] generated by [A-]:
Kb = 5.6 x 10-10 ≈ x2
0.0500
pOH = 5.28 pH = 14 - pOH = 8.72
[OH-][HA]
[A-]
Kb =
A- + H2O HA + OH-
(0.0500 - x) x x
Kb = Kw / Ka = 5.6 x 10-10
x = 5.3 x 10-6 M

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101
(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

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pH = pKa at 50% titration (25 mL) pKa (acetic acid) = 4.74.
Short vertical section compared to strong acid/strong base.
Titrate of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.

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103
pH = pKa at 50% titration
Weaker acid = shorter vertical section

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104
50.0 mL of 0.100 M ammonia titrated with 0.100 M HCl.
• pH = pKa at 50% to equivalence (pKa = 9.25).
Titration of Weak Base with Strong Acid

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Titration of Polyprotic Acids & Bases
Maleic acid: Sodium carbonate
HOOC–CH=CH–COOH Na+(aq) + CO3
2-(aq)
(diprotic acid) (basic anion)

Bronsted Lowry Acid and Base.pptx

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