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Equilibrium
1. Molecular ------------------- Chapter 15
2. Ionic (Weak Acid / Base ) - Chapter 16
3. Ionic (“Insoluble” Salts) - Chapter 17
STRONG Acids & Bases
1. COMPLETELY IONIZED1. COMPLETELY IONIZED1. COMPLETELY IONIZED1. COMPLETELY IONIZED
(a) HCl(aq) ���� H+ (aq) + Cl- (aq)
(b) NaOH(aq) ���� Na+(aq) + OH-(aq)
2. ONLY IONS PRESENT2. ONLY IONS PRESENT2. ONLY IONS PRESENT2. ONLY IONS PRESENT
(a) H+ (aq) & Cl- (aq)
(b) Na+(aq) & OH-(aq)
3. NO EQUILIBRIUM3. NO EQUILIBRIUM3. NO EQUILIBRIUM3. NO EQUILIBRIUM
WEAKWEAKWEAK Acids & Bases
1.1.1.1. NOT COMPLETELY IONIZEDNOT COMPLETELY IONIZEDNOT COMPLETELY IONIZEDNOT COMPLETELY IONIZED
HF(aq) ↔↔↔↔ H+ (aq) + F- (aq)
2. Both IONSIONSIONSIONS and MOLECULESMOLECULESMOLECULESMOLECULES Present
HF(aq) ↔↔↔↔ H+ (aq) + F- (aq)
3. EQUILIBRIUM exists
for Weak Acids & Bases
The [H] of 0.50 M HCN(aq) is ? {Ka = 4.9 x 10–10
What do you do 1st ?
1ST Write & Balance “Reaction”
HCN(aq) ↔ H+ (aq) + CN- (aq)
2nd Write Equilibrium Expression
][
]][[H
HCN
CNKa
−+
=
Initial Change Equilibrium
Find [H] of 0.50 M HCN {Ka = 4.9 x 10–10
HCN(aq) ↔ H+ (aq) + CN- (aq)
I NITIALNITIALNITIALNITIAL ……. 0.5 0 0
C HANGEHANGEHANGEHANGE ……. - x + x + x
E QUILIBRIUMQUILIBRIUMQUILIBRIUMQUILIBRIUM 0.5 – x x + 0 x + 0
Ka = 4.9 x 10-10 ; [HCN] = 0.5 – x ; [H+]=[CN-] = x
x2≅ (4.9 x 10-10) (0.50) ≅ 2.45 x 10-10
[H+]=[CN-] = x ≅ 1.565 x 10-5 ≅ 1.6 x 10-5
][
]][[H
HCN
CNKa
−+
=
0.5
x
x -0.5
x
]5.0[
]][[x 109.4
2210
≈=
−
=×−
x
x
What approximation?
Yes!!
Why?
Since x ≅ 1.6 x 10-5 is so small
0.50 - x = 0.50
0.5
x
x -0.5
x
22
≈
pH – A measure of Acidity
The pH of a solution is defined as the
negative logarithm of the hydrogen ion
concentration (in mol/L).
pH = –log [H+]
pOH = –log [OH-]
pH + pOH = 14
Determine the pH & pOH of 0.50 M
HCN(aq)
pH = - log [H+]
x = [H+] = [CN-] ≅ 1.6 x 10-5
pH = - log [1.6 x 10-5]
pH = 4.8
pOH = 14.0 – 4.8 = 9.2
Definitions of Acids & Bases
1. Arrhenius
2. Bronsted-Lowry
3. Lewis
Acid / Base
ACIDACIDACIDACID: A substance which dissociates to form hydronium ions (H3O
+) in solution :
HA(aq) → H+(aq) + A–(aq)
BASEBASEBASEBASE: A substance that dissociates to form hydroxide ions (OH–) in solution :
MOH(aq) → M+(aq) + OH–(aq)
M = a Metal
-LOWRY Acid / Base
ACID:ACID:ACID:ACID: Substance that can donate H+
for example HF(aq), HCl(aq), HNO3(aq), etc
BASE:BASE:BASE:BASE: Substance that can accept H+
for example OH- , F- , Cl- , Br- , I- , etc
LEWIS LEWIS LEWIS Acid / Base
• Acid is an electron-pair acceptor. These are
generally cations and neutral molecules with
vacant valence orbitals. For example H+ & BF3.
• Base is an electron-pair donor. These are
generally anions and neutral molecules with
available pairs of electrons, such as H2O, NH3, O2–
• The bond formed is called a coordinate bond.
STRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASES
HCl(aq) is a strong acid but Cl- is a “weak” base
Why?
Because Cl- will not react with H+
Why Not?
If it did, HCl(aq) {a strong acid} would form
which does not exist
STRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASES
HF(aq) is a weak acid and F- is a “strong” base
Why?
Because F- will react with H+
Why ?
HF(aq) {a weak acid} would form
which does exist
STRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASESSTRENGTH OF ACIDS AND BASES
HClO4
HI
HBr
HCl
HNO3
HF
HNO2
HCOOH
NH4+
HCN
H2O
ClO4–
I–
Br –
Cl –
NO3 –
F –
NO2 –
HCOO –
NH3
CN –
OH –
ACID BASE ACID BASE
Incre
asin
g A
cid
Str
ength
Str
on
g A
cid
s
–BASE PAIRS
An Acid and a Base that
differ only in the presence
{or absence} of a proton.
Conjugate acid–base pairs
HF + H2O = H3O+ + F-
Conjugate acid–base pairs
–BASE PAIRS
HClO4
HI
HBr
HCl
H2SO4
HNO3
H3O+
HSO4–
HSO4–
HF
HNO2
HCOOH
NH4+
HCN
H2O
NH3
ClO4–
I–
Br –
Cl –
HSO4 –
NO3 –
H2O
SO42–
SO42–
F –
NO2 –
HCOO –
NH3
CN –
OH –
NH2 –
ACID CONJ. BASE ACID CONJ. BASE
What are the conjugate bases of the following
Brønsted–Lowry acids
H2SO4
HClO4
H2SO4
HSO4–
H2O
+ H2O � H3O+ + HSO4
-
+ H2O �
+ H2O �
+ H2O �
+ H2O �
Which one of the conjugate bases of the
following Brønsted-Lowry acids is incorrect?
(a) ClO- for HClO
(b) HS- for H2S
(c) NH3 for NH4+
(d) SO42- for HSO4
-
(e) H2SO4 for HSO4-
Aqueous Equilibria:
WEAK ACIDS
Dissociation (or ionization) of Water
H2O(l) + H2O(l) = H3O+(aq) + OH–(aq)
or simply
H2O(l) = H+(aq) + OH–(aq)
Keq = [H+] [OH–]
This a product (concentration x concentration)
and it is water,
therefore called the Ion Product Constant of water
H2O(l) = H+(aq) + OH–(aq)
The equilibrium gives us the ion product of water.
Kw = Kc = [H+] [OH–] = 1.0 x10–14
If either [H+] or [OH–] is known the other can be
determined
When does the constant Kw change ?
If the concentration of OH– ions in a certain
cleaning solution is 0.0001 M, what is the
concentration of H+ ions?
[H+] [OH–] = 1.0 x10–14
[OH-] = 1.0 x10–4
[H+] = 1.0 x10–14 / 1.0 x10–4
[H+] = 1.0 x10–10
Calculate the concentration of OH– ions in a HCl
solution whose hydrogen ion concentration is 0.1M
[H+] [OH–] = 1.0 x10–14
[H+] = 1.0 x10–1
[OH-] = 1.0 x10–14 / 1.0 x10–1
[OH-] = 1.0 x10–13
pH – A measure of Acidity
The pH of a solution is defined as the
negative logarithm of the hydrogen ion
concentration (in mol/L).
pH = –log [H+]
pOH = –log [OH-]
pH + pOH = 14
PERCENT DISSOCIATIONPERCENT DISSOCIATIONPERCENT DISSOCIATION
• A measure of the strength of an acid.
• Stronger acids have higher percent
dissociation.
• Percent dissociation of a weak acid
decreases as its concentration increases.
Percent Dissociation
For Acids
100%[HA]
][HonDissociati % ×=
+
Calculate the % dissociation of 0.5 M HCN (aq)
From the equilibrium calculation
x ≅ 1.6 x 10-5 = [H+]
100%[HA]
][HonDissociati % ×=
+
3-5
3.2x10100%[0.50]
][1.6x10onDissociati % =×=
−
Ka for 0.5 M HCN (aq)
If % DISSOCIATION = 3.2 x 10-3
[H+] = [CN-] = (% dissociation) ([HA]) = 1.6 x 10-5
][
]][[H
HCN
CNKa
−+
= 100%[HA]
][HonDissociati % ×=
+
1055
101.5]50.0[
]106.1][[1.6x10 −
−−
== xx
Ka
0.1 M AMMONIA SOLUTION, if pH = 11.1
NH3 (aq) = NH4+ (aq) + OH - (aq)
[NH4+ ] = [OH - ]
100%][NH
][OHonDissociati %
3
×=
−
1.34%100%[0.10]
][1.34x10onDissociati %
-3
=×=
Aqueous Equilibria:
WEAK BASES
WEAKWEAKWEAK Bases
NH3(gas) = Ammonia
NH4OH = Ammonium Hydroxide
NH3 (aq) = Aqueous Ammonia ≡ NH4OH
NH4OH = NH4+ (aq) + OH - (aq)
or
NH3 (aq) = NH4+ (aq) + OH - (aq)
Calculate pOH & pH for a 0.10 M aqueous
ammonia solution {Kb = 1.8 x 10-5
NH3 (aq) = NH4+ (aq) + OH - (aq)
x = [OH-] = [NH4+] = 1.34 x 10-3
][NH
]][OH[NH
3
4
−+
=bK
5
3
4 108.1x]-[0.10
[x][x]
][NH
]][OH[NH−
−+
=== xKb
Calculate pOH & pH for a 0.10 M NH3(aq)
[OH-] = 1.34 x 10-3
pOH = -log [OH-] = 2.9
pH = 14 – 2.9 = 11.1
pH of some common fluids
• Gastric juice in stomach 1.0 – 2.0
• Lemon juice 2.4
• Vinegar 3.0
• Grapefruit juice 3.2
• Orange juice 3.5
• Urine 4.8 – 7.5
• Saliva 6.4 – 6.9
• Milk 6.5
• Blood 7.35 – 7.45
• Tears 7.4
• Milk of Magnesia 10.6
• Household Ammonia 11.5
pH + pOH = 14
Neutral solutions: [H+] = [OH-] = 1.0 x 10 –7 M
pH = pOH = 7
Acidic solutions: [H+] > 1.0 x 10 –7 M,
pH < 7.00
Basic solutions: [H+] < 1.0 x 10 –7 M,
pH > 7.00
The OH– ion concentration of a sample of
blood is 2.5x10–7 M. What is its pH
[H+] [OH–] = 1.0 x10–14
[H+] (2.5x10–7) = 1.0 x10–14
[H+] = 4.0 x10–8
pH = –log [H+] = –log (4.0 x10–8)
pH = 7.39794
Really ?
Calculate the pH of a 0.01M HNO3 (aq) solution
pH = –log [H+]
Nitric is a STRONG acid !
100% Ionized: 1 HNO3 → 1 H+ + 1 NO3-
[H+] = [NO3-] = [HNO3]
[HNO3] = 1.0 x 10 –2 M
pH = –log (1.0 x 10 –2) = 2
Acid Ionization Constants
7.1 x 10 –4
4.5 x 10 –4
3.0 x 10 –4
1.7 x 10 –4
8.0 x 10 –5
6.5 x 10 –5
1.8 x 10 –5
4.9 x 10 –10
1.3 x 10 –10
HF
HNO2
C9H8O4 (aspirin)
HCO2H (formic)
C6H8O6 (ascorbic)
C6H5CO2H (benzoic)
CH3CO2H (acetic)
HCN
C6H5OH (Phenol)
F–
NO2 –
C9H7O4 –
HCO2 –
C6H7O6 –
C6H5CO2 –
CH3CO2 –
CN –
C6H5O –
ACID Ka CONJ. BASE Kb
1.4 x 10 –11
2.2 x 10 –11
3.3 x 10 –11
5.9 x 10 –11
1.3 x 10 –10
1.5 x 10 –10
5.6 x 10 –10
2.0 x 10 –5
7.7 x 10 –5
Conjugate acid–base pairs
Relationship between Ka and Kb:
Ka × Kb = Kw
Ka × Kb = Kw
For HF(aq)
Ka x Kb = (7.1 x 10-4)(1.4 x 10-11) = 9.94 x 10-15
For HCN(aq)
Ka x Kb = (4.9 x 10-10)(2.0 x 10-5) = 9.8 x 10-15
For C6H5OH (aq) (Phenol)
Ka x Kb = (1.3 x 10–10)(7.7 x 10–5) = 10.01 x 10-15
Determine the concentration of a HNO3 (aq)
solution with a pH of 2
pH = –log [H+]
pH = 2 therefore
[H+] = 1.0 x 10 –2 M
Nitric is a STRONG acid therefore
[H+] = [NO3-] = [HNO3]
[HNO3] = 1.0 x 10 –2 M
Acids , Bases
&
SALTS
Why does NaOH(aq) react with HCl(aq) ?
HCl(aq) + NaOH(aq) → H2O + NaCl(aq)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → H2O +
Na+(aq) + Cl-(aq)
Na+(aq) & Cl-(aq) are spectator ions
Does NaOH(aq) react with HCl(aq) ?
Yes & the reaction is:
H+(aq) + OH–(aq) = H2O(l)
How Strongly does this reaction take place?
The strong acid strong base reaction
H+(aq) + OH–(aq) = H2O(l)
Is the reverse of the ionization of water
H2O(l) = H+(aq) + OH–(aq)
With Kw = [H+] [OH–] = 1.0 x10–14
The reverse reaction H+(aq) + OH– (aq) = H2O(l)
Would have a K = 1.0 x 10+14
Acid–Base Properties of Salts
SALTS IN WATER EITHER FORM
neutral solutions
basic solutions
or acidic solutions
Acids REACT with Bases
Will NaOH (aq) react with HCl( aq) ?
Yes , 100%
HCl(aq) + NaOH(aq) → H2O + NaCl(aq)
Why does NaOH(aq) react with HCl(aq) ?
Because water is formed
H+(aq) + OH–(aq) = H2O(l)
Will NaCl(aq) react with water?
No
Why Not ?
Because if it did a strong acid HCl(aq) and a
strong base NaOH(aq) would be formed and
neither exist in water
H2O + NaCl(aq) � HCl(aq) + NaOH(aq)
Will sodium ion react with water?
No because a strong base would be formed
What base ?
NaOH
Will potassium ion react with water?
No because a strong base would be formed
What base ?
KOH
Will chloride ion react with water?
No because a strong acid would be formed
What acid ?
HCl(aq)
Will nitrate ion react with water?
No because a strong acid would be formed
What acid ?
HNO3(aq)
What ions will react with water?
The ones that will not form
strong acids or bases
Can you name them?
Salts that produce neutral solutions
are those formed from strong acids
and strong bases.
For example NaCl(aq)
Can you think of others ?
Salts that produce basic solutions are
those formed from weak acids and
strong bases.
For example NaF
Can you think of others?
Salts that produce acidic solutions
are those formed from strong acids
and weak bases.
For Example NH4Cl
Can you think of others?
Acid–Base Properties of Salts
• Salts that produce neutral solutions are those
formed from strong acids and strong bases.
• Salts that produce basic solutions are those
formed from weak acids and strong bases.
• Salts that produce acidic solutions are those
formed from strong acids and weak bases.
Strength of Acids
HClO4
HI
HBr
HCl
H2SO4
HNO3
H3O+
HSO4–
HF
HNO2
HCOOH
NH4+
HCN
ClO4–
I–
Br –
Cl –
HSO4 –
NO3 –
H2O
SO42–
F –
NO2 –
HCOO –
NH3
CN –
ACID CONJ. BASE ACID CONJ. BASE
Predict whether the following solutions will
be acidic, basic or nearly neutral:
(a) NH4I
(b)SrCl2
(c) KCN
(d) Fe(NO3)3
(e) LiClO4
(f) Na3PO4
(a) ______
(b) _______
(c) _______
(d) ______
(e) ______
(f) ______
Last Topic in Chapter 16
The strength of an acid depends
on its tendency to ionize.
Molecular Structure
and Acid Strength
The strength of an acid depends
on its tendency to ionize.
The stronger the bond, the weaker the acid.
Or
The more polar the bond, the stronger the acid.
For Acids in Same Group
Bond strength plays the key roll
Oxoacids of The Halogens
Cl–O–H (aq) Strongest
Br–O–H (aq)
I–O–H (aq) weakest
as electronegativity increases,
acid strength increases
Oxoacids of The Halogens
as electronegativity increases, acid strength
increases
Oxoacids of Chlorine
HClO (aq)
HClO2 (aq)
HClO3 (aq)
HClO4 (aq)
acid strength increases with
increasing oxidation number
Oxoacids of Chlorine:
Predict the relative strengths of the
following groups of oxoacids:
• a) HClO, HBrO, and HIO
• b) HNO3, and HNO2
• c) H3PO3, and H3PO4.
End
Chapter 16