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One More Definition
Bronsted-Lowry
Chapter 14:
Acids & Bases
Acid = compound that can donate H+ ions
Base = compound that can accept H+ ions
Lewis
Acid = compound that can accept electron pairs
Base = compound that can donate electron pairs
Part A: Kw and Ka Equilibria
+ H+ NH3
NH4+
NH3 can accept a H+
(B-L base – focus on what it reacts with)
It can do this because N has lone pair electrons
(Lewis base – focus on why it reacts)
Chem 112
Dr. Gentry
Acid & Base Chemistry
= compound that generates H+ ions in solution
= compound that generates OH‒ ions in solution
H
F
HCl
H+ + Cl‒
NaOH Na+ + OH‒
F B
+
N H
••
Acid
Base
Lewis Acids & Bases
(electron pair acceptors and donors)
Arrhenius
F
H
acid
base
F
H
F B N H
F
H
Subset of the general class of dissolution reactions
e.g.
NaCl
Na+
+
Cl‒
Why do we care?
No protons donated or accepted !
Because water is both acid and base
… 75% of earth surface / 65% of human body
H2O
H+ + OH‒
Acid & Base Definitions
Arrhenius
Acid
Base
(oldest of the definitions)
= compound that can donate H+ ions
= compound that can donate OH‒ ions
Acid:
Base:
HCl
H+ + Cl‒
NaOH Na+ + OH‒
Bronsted-Lowry
(better definition)
Acid = compound that can donate H+ ions
Base = compound that can accept H+ ions
Bases:
NaOH + H+ H2O + Na+
NH3 + H+ NH4+
Acid & Base Definitions
Arrhenius
Acid
Base
(oldest of the definitions)
= compound that can donate H+ ions
= compound that can donate OH‒ ions
Bronsted-Lowry
(better definition)
Acid = compound that can donate H+ ions
Base = compound that can accept H+ ions
Lewis
We will use Br-L
for rest of class.
Describes how
interacts with
water
(best, can be applied to non-aqueous systems)
Acid = compound that can accept electron pairs
Base = compound that can donate electron pairs
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A base on one side becomes an acid on other side
… base / conjugate-acid pair
NH3 (base) + H+
acid
NH4+ (conjugate acid)
conjugate base
HCl +
Must Memorize
Acid = can donate H+ ions
Base = can accept H+ ions
Bronsted-Lowry:
Cl‒
NH3
(from 1st Semester)
Strong Acids
HCl
HBr
HI
HNO3
H2SO4
(Hydrochloric acid)
(Hydrobromic acid)
(Hydroiodic acid)
(Nitric acid)
(Sulfuric acid)
Weak Acids (Ka < 1)
+
base
NH4
+
conjugate acid
HC2H3O2
HNO2
H2CO3
HF
(Acetic acid)
(Nitrous acid)
(Carbonic acid)
(Hydrofluoric acid)
Strong Bases
LiOH
NaOH
KOH
Ca(OH)2
Na2O
(Lithium hydroxide)
(Sodium hydroxide)
(Potassium hydroxide)
(Calcium hydroxide)
(Sodium oxide)
Weak Bases (Kb < 1)
NH3
N(CH3)3
(Ammonia)
(Methyl Amine)
An Equilibrium Process
Strong & Weak Acids
Strong Acids: completely dissociate in water
HCl
HNO3
H2SO4
H+ + Cl‒
H+ + NO3‒
2H+ + SO4‒2
(Hydrochloric acid → Chloride ion)
(Nitric acid
→ Nitrate ion)
(Sulfuric acid
→ Sulfate ion)
Weak Acids: only partially dissociate in water
HC2H3O2
(Acetic acid)
H + + C 2H 3O 2‒
(Acetate ion)
H2CO3
(Carbonic acid)
H+ + HCO3‒
(Bicarbonate ion)
HNO2
(Nitrous acid)
H+ +
Predicting Direction of Acid/Base Reaction
Equilibrium system so reaction can go in both directions
Which side dominates?
Strong Acid + Strong Base
Weak Base + Weak Acid
Look to acid/base table
Identify acids and bases
Determine which is strongest base and strongest acid
H2SO4 + 2NH3
SO42‒ + 2NH4+
HCO3‒ + SO42‒
HSO4‒ + CO32‒
NO2‒
(Nitrite ion)
Strong & Weak Bases
Strong Bases: completely dissociate in water
NaOH
KOH
Ca(OH)2
Na+ + OH‒
K+ + OH‒
Ca+2 + 2OH‒
(Sodium hydroxide)
(Potassium hydroxide)
(Calcium hydroxide)
Weak Bases: only partially dissociate in water
NH3 + H2O
(Ammonia)
NH4+ + OH‒
(Ammonium ion)
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Acid = can donate H+ ions
Base = can accept H+ ions
Bronsted-Lowry:
What is the concentration of OH- ions in a HCl solution
whose hydrogen ion concentration is 1.3 M?
A base on one side becomes an acid on other side
… base / conjugate-acid pair
NH3 (base) +
H+
acid
Kw = [H+][OH-] = 1.0 x 10-14
+
NH4 (conjugate acid)
[H+] = 1.3 M
[OH-] =
conjugate base
HCl +
NH3
Cl‒
NH4+
+
base
[OH‒] =
conjugate acid
1 x 10-14
1.3
Kw
[H+]
= 7.7 x 10-15 M
An Equilibrium Process
Water Acid/Base Chemistry
Kw = [H3O+] · [OH‒] = 1.00*10‒14
Water acts as both an acid and a base when reacted with itself
H2O + H2O
H3O+
•
+
OH‒
Hydroxide
Ion
Hydronium
Ion
The concentration of OH– ions in a household ammonia
cleaning solution is 0.0025 M. Calculate the concentration
of H+ ions.
[H+ ] [0.0025M] = 1.00 * 10−14
[H+ ] = 4.00 * 10−12 M
accept proton
acid
H2O + H2O
base
c. base
H3
O+
+
OH‒
•
Calculate the concentration of OH– ions in a vinegar
solution whose hydrogen ion concentration is 4.3*10-5 M
[4.3 * 10−5 M] [OH− ] = 1.00 * 10 −14
[OH− ] = 2.3 * 10 −10 M
c. acid
donate proton
Multiple Strong Acids
Water Acid/Base Chemistry
•
H2O + H2O
[H3O+]’s for strong acids are additive
Each acid completely dissociates
Adds its own moles of H3O+ to give total H3O+
H3O+ + OH‒
•
Water is equilibrium reaction
Forward and reverse reactions in balance
Describe the balance using equilibrium constant
[Solvent water is not included in equation]
Most of the water remains in neutral H2O state since Kc << 1
It is the total H3O+ that will be controlled by Kw
25mL of 0.50M HCl and 48mL of 0.15M HNO3 are combined in
one flask.
- What is the pH of the total system?
- What is the pOH of the total system?
1) Use M1V1 = M2V2 to find [H3O+] from each acid in total flask
M2,HCl
Kw = [H3O+] · [OH‒] = 1.00*10‒14
[x] = concentration in molarity for “x”
= (0.50M HCl
· 25mL) / 73mL
M2,HNO3l = (0.15M HNO3 · 48mL) / 73mL
= 0.17M H3O+ from HCl
= 0.10M H3O+ from HNO3
2) Combine to find total [H3O+] = 0.27M H3O+
3) pH = ‒ log(0.27) = 0.57
pOH = 14 – 0.57 = 13.43
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What About Pure Water?
Kw = [H3O+] · [OH‒] = 10‒14
H3O+ + OH‒
2H2O
Pure Water:
• Nitric acid (HNO3) is used in the production of
fertilizer, dyes, drugs, and explosives. Calculate
the pH of a HNO3 solution having a hydrogen ion
concentration of 0.76 M.
pH = − log(0.76) = 0.11
1) [H3O+] · [OH‒] = 1×10‒14
• The pH of a certain orange juice is 3.33. Calculate
the H+ ion concentration.
2) But same amount of H3O+ and OH‒ from reaction
3) So
[H3O+]
=
[OH‒]
[H+ ] = 10−3.33 = 4.7 × 10−4 M
4) Therefore:
[H3
O+]
=
1×
×10‒7 M,
[OH‒]
=
1×
×10‒7 M
(10‒7 M) · (10‒7 M) = 10‒14
pH ‒ A Measure of Acidity
pH = ‒ log ( [H3O+] )
pOH = ‒ log ( [OH‒] )
pH = ‒ log ( [H+] )
Rule of Logarithm Sums:
For Pure Water:
[H3O+] = 1×10‒7 M,
[OH‒] = 1×10‒7 M
10x · 10y = 10 (x+y)
Log (x· y) = Log (x) + Log (y)
‒ log ( [H3O+] ) = ‒ log (1×10‒7) = 7.0
Since [H3O+] · [OH‒] = 10‒14 for water
pH (pure water) = 7
pH + pOH = +14
Then
Logarithms
If
y = 10x
then
log (y) = x
If 1000 = 103 then log (1000 ) = 3
If 0.01 = 10-2 then log (0.01) = ‒2
A logarithm is just the exponent for 10 X
pH = ‒ log ( [H+] )
[H+] = 1×10‒3 M
pH = 3.0
[H+] = 1×10‒7 M
pH = 7.0
[H+] = 1×10‒8 M
pH = 8.0
[H+] = 4.6×10‒8 M
pH = 7.34
pH = ‒ log ( [H3O+] )
pOH = ‒ log ( [OH‒] )
pH + pOH = +14
If: [H3O+ ] = 1×
×10 –4 M
pH = + 4
⇒
pH = +4
pOH = 14 – pH = +10
[OH–] = 10–10 M
[H3O+]
1×10‒7 M
pH
pOH
[OH–]
+ 7
+ 7
1×10‒7 M
1×10‒3
1×10‒10
+ 3
+10
+11
+ 4
1×10‒11
1×10‒4
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• The concentration of OH– ions in a certain household
ammonia cleaning solution is 0.0025 M.
Calculate the concentration of H+ ions… and pH, pOH
pH Indicators
K w = [H+ ] [OH− ] = 1.00 × 10−14
[OH− ] = 0.0025 M
[H+ ] = 1* 10−14 / 0.0025 = 4.0 * 10−12 M
pH + pOH = 14
pOH = − log(0.0025M) = 2.60
pH = − log(4.0 × 10 −12 ) = 11.40
• Calculate the concentration of OH– ions in a HCl
solution whose hydrogen ion concentration is 1.3 M
[OH‒] = 7.7*10‒15M,
pH = -0.11,
pOH = 14.1
• The OH– ion concentration of a blood sample is
2.5 x 10–7 M.
- What is the pOH of the blood?
- What is the pH of the blood?
- What is the[H+] of the blood?
[OH‒] = 2.5×
×10-7 M,
[H+] = 4.0×
×10-8 M,
• A substance that changes color in a specific pH range.
• Indicators are weak acids and have different colors in
their acid (H In) and conjugate base (In─) forms.
pOH = 6.60
pH = 7.40
• Switch from acid to base forms at characteristic pH’s
• Calculate the concentration of OH– ions in a HCl
solution whose hydrogen ion concentration is 1.3 M
[H+]
= 1.3 M,
[OH‒] = 7*10‒15 M,
Acid-Base Indicators
pH = ‒ 0.11,
pOH = 14.1,
HIn(aq) + H2O(l)
H3O+(aq) + In‒(aq)
Color A
Color B
HIn and In– = Acid and Base forms of Indicator
Bromcresol Green
The pH Scale
HIn
In‒ + H+
HIn and In = Indicator
Basic solution:
pH > 7
‒
[H3O+] < 1*10-7
H+ +
Neutral solution:
pH = 7
[H3O+] = 1*10-7
Acidic solution:
pH < 7
[H3O+] > 1*10-7
yellow
pH < 4.4
[BC− ] [H+ ]
= 1.6 * 10 −5
[HBC]
blue
pH > 4.4
pKa = 4.8
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Calculate the equilibrium pH of a 0.62 M nitrous acid (HNO2) solution.
Bronsted-Lowry Acids in Water
Water as the B-L Acid
H2O (aq) + H2O (l)
H+ + NO2‒
HNO2
(donate a H+)
pKa = 3.35
HNO2
H+
0.62
0
0
+x
+x
Initial
Change
Kw = [H3O+] [OH‒] = 1.0*10‒14
Strong Acid as the B-L Acid
HCl (aq) + H2O (l)
H3O+ (aq) + Cl– (aq)
[H O + ][OH− ]
Ka = 3
[HCl]
Weak Acid as the B-L Acid
HNO2 (aq) + H2O (l)
NO2‒
Equilibrium
K a = 10 − pKa
>> 1
•
0.62 ‒ x
+x
+
−
[ H ] eq [ NO2 ] eq
K a = 4.5 * 10 −4 =
K a = 4.5 * 10−4 =
H3O+ (aq) + A– (aq)
(x)2
(x)2
≈
(0.62 − x) 0.62
K a = 4.5 * 10 −4 =
•
[ H+ ] eq [ NO2 − ] eq
[ HNO2 ] eq
[H3 O ][A ]
[HA]
pK a = − log(K a )
x = [H+] = 0.017M
pH = 1.8
(x)2
(0.036 − x)
=
Assume 0.62 ‒ x ≈ 0.62 since x << 0.62M
[acid]initial
> 1000
Ka
Can make simplifying assumption if:
−
(x)2
(0.62 − x)
Same Problem From Previous Slide
Equilibrium reaction for ionization of weak acid in water
+
=
[ HNO2 ] eq
+x
Assume 0.62 ‒ x ≈ 0.62 since x << 0.62M
< 1
Acid Ionization Constant, Ka
HA (aq) + H2O (l)
‒x
must obey stoichiometry
H3O+ (aq) + NO2– (aq)
[H O+ ][NO2 − ]
Ka = 3
[HNO2 ]
Ka =
+
H3O+ (aq) + OH– (aq)
Describes strength of
the acid
K c = 4.5 * 10 −4 =
[H2O] not included in
definition of equation
•
x = 0.017M
(x)2
(x)2
≈
(0.62 − x ) 0.62
∼ 0.62
= 1.0% error
(0.62 − x)
If [HAcid]initial/Ka < 1000 then must solve using full quadratic formula
For ax2+bx+c=0,
x=
−b ± b2 − 4ac
2a
Calculate the equilibrium pH of a 0.96 M hydrofluoric acid (HF) solution.
Acid Ionization Constants
ACID
CONJ. BASE
Ka
7.1 x 10 –4
F–
1.4 x 10 –11
HNO2
4.5 x 10 –4
NO2 –
2.2 x 10 –11
C9H8O4 (aspirin)
3.0 x 10 –4
C 9H 7O 4 –
3.3 x 10 –11
HCO2H (formic)
1.7 x 10 –4
HCO2 –
5.9 x 10 –11
HF
–5
C 6H 7O 6
–
pKa = 3.15
H+
HF
5
2
x 10 +6
Kb
Cl–
HCl
H + + F‒
HF
+
F‒
x 10 –21
1.3 x 10
–10
C6H8O6 (ascorbic)
8.0 x 10
C6H5CO2H (benzoic)
6.5 x 10 –5
C6H5CO2 –
1.5 x 10 –10
CH3CO2H (acetic)
1.8 x 10 –5
CH3CO2 –
5.6 x 10 –10
HCN
4.9 x 10 –10
CN –
2.0 x 10 –5
C6H5OH (phenol)
1.3 x 10 –10
C 6H 5O –
7.7 x 10 –5
Initial
Change
0.96
0
0
‒x
+x
+x
must obey stoichiometry
Equilibrium
K a = 10
•
− pKa
0.96 ‒ x
+
K a = 7.1* 10 −4 =
+x
−
[ H ] eq [ F ] eq
[ HF] eq
+x
2
=
(x)
(0.96 − x)
Assume 0.96 ‒ x ≈ 0.96 since x = very small
Ka =
(x)2
(x)2
= 7.1* 10−4
≈
(0.96 − x) 0.96
[acid] ?
> 1000
Ka
x = [H+] = 0.026M
pH = 1.58
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Marks End of Material
for 3d Exam
Chapter 13
All Sections
Chapter 14
Sections 1-9
7