Amphoteric Substances
• Amphoteric substances can act as either an
acid or a base
because they have both a transferable H and an
atom with lone pair electrons
• Water acts as base, accepting H+ from HCl
•
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
Water acts as acid, donating H+ to NH3
NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH–(aq)
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Brønsted-Lowry
Acid–Base Reactions
H–A
acid
+
HCHO2
acid
H2O
acid
:B
base
+ H2O
base
+
⇔
⇔
NH3
base
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:A–
+
H–B+
conjugate
conjugate
base
acid
CHO2–
+
H3O+
conjugate
conjugate
base
acid
⇔
2
HO–
+
conjugate
base
NH4+
conjugate
acid
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Conjugate Pairs
In the reaction H2O + NH3 ⇔ HO– + NH4+
H2O and HO– constitute an
acid/conjugate base pair
NH3 and NH4+ constitute
a base/conjugate acid
pair
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Practice – Write the formula for the
conjugate acid of the following
H2O
H3O+
NH3
NH4+
CO32−
HCO3−
H2PO41−
H3PO4
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Practice – Write the formula for the
conjugate base of the following
H2O
HO−
NH3
NH2−
CO32−
because CO32− does not have an H, it
cannot be an acid
H2PO41−
H2PO41−
HPO42−
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Example 15.1a: Identify the Brønsted-Lowry acids
and bases, and their conjugates, in the reaction
H2SO4 + H2O ⇔ HSO4–
+
H3O+
When the H2SO4 becomes HSO4−, it loses an H+ so
H2SO4 must be the acid and HSO4− its conjugate base
When the H2O becomes H3O+, it accepts an H+ so
H2O must be the base and H3O+ its conjugate acid
H2SO4
acid
+ H2O
base
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⇔
6
HSO4–
+
conjugate
base
H3O+
conjugate
acid
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Example 15.1b: Identify the Brønsted-Lowry acids
and bases and their conjugates in the reaction
HCO3– +
H2O
⇔
H2CO3
+
HO–
When the HCO3− becomes H2CO3, it accepts an H+ so
HCO3− must be the base and H2CO3 its conjugate acid
When the H2O becomes OH−, it donats an H+ so H2O
must be the acid and OH− its conjugate base
HCO3– +
base
H2O
acid
⇔
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H2CO3
+
conjugate
acid
7
HO–
conjugate
base
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Practice – Identify the Brønsted-Lowry acid,
base, conjugate acid, and conjugate base in
the following reaction
HSO4−(aq) +
Acid
HCO3−(aq)
⇔
Conjugate
Base
Base
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SO42−(aq) +
8
H2CO3(aq)
Conjugate
Acid
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Practice—Write the equations for the following
reacting with water and acting as a monoprotic
acid & label the conjugate acid and base
HBr
HBr + H2O ⇔ Br− + H3O+
Acid
HSO4−
Base
Conj.
base
Conj.
acid
HSO4− + H2O ⇔ SO42− + H3O+
Acid
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Base
9
Conj.
base
Conj.
acid
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Practice—Write the equations for the following
reacting with water and acting as a
monoprotic-accepting base and label the
conjugate acid and base
I−
I− +
Base
CO32−
H 2O ⇔
Acid
HI +
Conj.
acid
OH−
Conj.
base
CO32− + H2O ⇔ HCO3− + OH−
Base
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Acid
10
Conj.
acid
Conj.
base
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Comparing Arrhenius Theory and
Brønsted-Lowry Theory
• Arrhenius theory
•
HCl(aq) →
H+(aq) + Cl−(aq)
HF(aq) ⇔
H+(aq) + F−(aq)
NaOH(aq) →
Na+(aq) + OH−(aq)
NH4OH(aq) ⇔
NH4+(aq) + OH−(aq)
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Brønsted–Lowry theory
HCl(aq) + H2O(l) →
Cl−(aq) + H3O+(aq)
HF(aq) + H2O(l) ⇔
F−(aq) + H3O+(aq)
NaOH(aq) →
Na+(aq) + OH−(aq)
NH3(aq) + H2O(l) ⇔
NH4+(aq) + OH−(aq)
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Strong & Weak Acids
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Acid Ionization Constant, Ka
• Acid strength measured by the size of the
•
equilibrium constant when reacts with H2O
HAcid + H2O ⇔ Acid− + H3O+
The equilibrium constant for this reaction is
called the acid ionization constant, Ka
larger Ka = stronger acid
[Acid− ] × [H3O+ ]
Ka =
[HAcid]
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Example 15.2b: Calculate the [OH−] at 25 °C when
the [H3O+] = 1.5 x 10−9 M, and determine if the
solution is acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[H3O+] = 1.5 x 10−9 M
[OH−]
[H3O+]
[OH−]
Relationships:
Solution:
Check: the units are correct; the fact that the
[H3O+] < [OH−] means the solution is basic
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Practice – Determine the [H3O+] when the
[OH−] = 2.5 x 10−9 M
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Practice – Determine the [H3O+] when the
[OH−] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[OH−] = 2.5 x 10−9 M
[H3O+]
[H3O+]
[OH−]
Relationships:
Solution:
Check: the units are correct; the fact that the
[H3O+] > [OH−] means the solution is acidic
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Example 15.3b: Calculate the pH at 25 °C when the
[OH−] = 1.3 x 10−2 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[OH−] = 1.3 x 10−2 M
pH
[H3O+]
[OH−]
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH > 7 means
the solution is basic
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Practice – Determine the pH @ 25 ºC of a
solution that has [OH−] = 2.5 x 10−9 M
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Practice – Determine the pH @ 25 ºC of a solution
that has [OH−] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[OH−] = 2.5 x 10−9 M
pH
[H3O+]
[OH−]
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH < 7 means
the solution is acidic
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Practice – Determine the [OH−] of a solution
with a pH of 5.40
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Practice – Determine the [OH−] of a solution with a
pH of 5.40
Given:
Find:
Conceptual
Plan:
pH = 5.40
[OH−], M
[H3O+]
pH
[OH−]
Relationships:
Solution:
Check: because the pH < 7, [OH−] should be less
than 1 x 10−7; and it is
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Example: Calculate the pH at 25 °C when the [OH−]
= 1.3 x 10−2 M, and determine if the solution is
acidic, basic, or neutral
Given:
Find:
Conceptual
Plan:
[OH−] = 1.3 x 10−2 M
pH
[OH−]
pOH
pH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH > 7 means
the solution is basic
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Practice – Determine the pOH @ 25 ºC of a
solution that has [H3O+] = 2.5 x 10−9 M
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Practice – Determine the pOH @ 25 ºC of a solution
that has [H3O+] = 2.5 x 10−9 M
Given:
Find:
Conceptual
Plan:
[H3O+] = 2.5 x 10−9 M
pOH
[H3O+]
pH
pOH
Relationships:
Solution:
Check: pH is unitless; the fact that the pH < 7 means
the solution is acidic
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Buffer Systems
LP#1. What is the pH of a buffer solution that contains 0.050 M formic acid
and 0.060 M sodium formate? The Ka of formic acid is 1.8 x 10-4.
[H 3O + ] = K a
[formic acid]
0.050
= 1.8 x 10 -4
= 1.5 x 10 -4
[formate ion ]
0.060
pH = -log [H3O+]= -log (1.5 x 10-4) = -(-3.82) = 3.82
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Normality
Normality
Normality is defined as the # equivalents = N
L of solution
This is mostly used with acids and bases.
For acids: N = M x #H+
For Bases: N = M x #OHExample:
1M HCl = 1N HCl
1M NaOH = 1N NaOH
1M H2SO4 = 2N H2SO4
1M Ca(OH)2 = 2N Ca(OH)2
1M H3PO4 = 3N H3PO4
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LP#1. What is the normality of a 0.012M solution of H2SO3? Ans: 0.024N
LP#2. What is normality of a solution made by dissolving 3.2g of H3PO4 in water to
a total solution volume of 275mL? The MM of H3PO4=98.0g/mol.
3.2g H3PO4 x 1 mol H3PO4 = 0.03265 mol H3PO4
98.0g H3PO4
0.3265 mol H3PO4 = 0.1187 mol H3PO4
0.275 L solution
L
0.1187 mol H3PO4 x 3 equiv = 0.3562 equiv = 0.36 N
L
mol
mol
OR
0.1187 M H3PO4 x 3 H = 0.36N
H3PO4
LP#3. How many equivalents are in 12 mL of 0.25M H3PO4
0.25M x 3 = 0.75N = 0.75 Equ
12mL x
1 L = 0.012 L
L
1000 mL
0.75 Equ x 0.012L = 0.009 Equ
L
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For Neutralization Reactions: (Nacid)(Vacid) = (Nbase)(Vbase)
LP#4. How many mL of 0.20 N LiOH are required to neutralize 44
mL of 0.30 N H2C2O4?
(Vbase) = (Nacid)(Vacid) = (0.30 N)(44 mL) = 66 mL
(Nbase)
0.20N
What if the concentrations are not given in normality?
LP#5. How many mL of 0.450M Ba(OH)2 will be required to
completely neutralize 122 mL
of 87.0 mL of 0.150M H3PO4?
Ba(OH)2 = (0.45M) x 2 = 0.90N
H3PO4 = (0.15M) x 3 = 0.45N
(Vbase) = (Nacid)(Vacid) = (0.450 N)(87.0 mL) = 43.5 mL
(Nbase)
0.900N
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