Download - CHAPTER 16 – ACIDS AND BASES
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CHAPTER 16 – ACIDS AND BASES
ACID – A compound the produces hydrogen ions in a water solutionHCl (g) → H+(aq) + Cl-(aq)
BASE – A compound that produces hydroxide ions in a water solutionNaOH (s) → Na+(aq) + OH-(aq)
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1884 SVANTE ARRHENIUSProposed the first definitions of acids and bases
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1923
Expanded the definitions of acids and bases
THOMAS LOWRY
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ACID – A hydrogen ion (or proton) donorBASE – A hydrogen ion (or proton) acceptor
1923
Expanded the definitions of acids and bases
JOHANNES BRØNSTED
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HCl + H2O →
acid base
HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water
+-
Cl- + H3O+
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NH3 + H2O →
base acid
AMPHOTERIC – A substance that can act as an acid or a base
+ -
NH4+ + OH-
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Acids turn into bases, and bases turn into acids
HCl + H2O → Cl- + H3O+
acid base
+-
acid base
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Acids turn into bases, and bases turn into acids
HCl + H2O → Cl- + H3O+
acid base conjugate base
of HCl
conjugate acid
of H2O
+-
acid basebase acid
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NH3 + H2O → NH4+ + OH-
base acid
+ -
base acid
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NH3 + H2O → NH4+ + OH-
base acid conjugate acid
of NH3
conjugate base of H2O
+ -
base acidacid base
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Conjugate base of HClO4
When HClO4 acts as an acid, it becomes: ClO4-
Conjugate base of H2CO3
When H2CO3 acts as an acid, it becomes: HCO3-
Conjugate acid of CF3NH2
When CF3NH2 acts as a base, it becomes: CF3NH3+
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POLYPROTIC ACID – An acid with more than one ionizable hydrogen ion
H2CO3
H3PO4
Hydrogen ions become successively more difficult to ionize
diprotictriprotic
H2SO4 (aq) → H+(aq) + HSO4-(aq)
Strong Acid
HSO4- (aq) → H+(aq) + SO4
2-(aq)
Weaker Acid
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WATER
Water ionizes to a small extent
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WATER
Water ionizes to a small extent
2H2O (l) → H3O+(aq) + OH-(aq)
-+
Makes solutions acidic Makes solutions basic
Equal amounts of H3O+ and OH- make a solution NEUTRAL pure water is neutral
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Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ↔ H3O+(aq) + OH-(aq)
Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant
K = [H3O+][OH-]
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Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)Association: H3O+(aq) + OH-(aq) → 2H2O (l) Equilibrium 2H2O (l) ↔ H3O+(aq) + OH-(aq)
Because of this equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant
Kw = [H3O+][OH-]
ION-PRODUCT CONSTANT FOR WATER (Kw) – The product of the hydronium ion and hydroxide ion molarities in any water solution
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The Kw depends on temperature
10
2540
0.29 x 10-14 M2
1.00 x 10-14 M2
2.92 x 10-14 M2
Temp (ºC) Kw
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Calculate the hydronium ion and hydroxide ion molarities in pure water at 25ºC.
(x)(x)
1.00 x 10-14 M2 =
2H2O (l) → H3O+(aq) + OH-(aq)Molarities before Ionization:Molarities at Equilibrium:
Kw = [H3O+][OH-]
= x2 x
1.00 x 10-7 M =
In pure water, [H+] = [OH-] = 1.00 x 10-7 M
0 0
x x
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Calculate the hydroxide ion molarity in a water solution in at 25ºC in which the [H3O+] = 1.00 x 10-5.
(1.00 x 10-5 M)(x)
1.00 x 10-14 M2 =
Kw = [H3O+][OH-]
1.00 x 10-9 M = x
1.00 x 10-14 M2 _________________
1.00 x 10-5 M
= x
= [OH-]
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Calculate the hydronium ion molarity in a water solution in at 25ºC in which the [OH-] = 8.00 x 10-3.
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[H3O+] = [OH-] :
[H3O+] > [OH-] :
[H3O+] < [OH-] :
the solution is NEUTRALthe solution is ACIDICthe solution is BASIC
All water solutions contain both hydronium and hydroxide ions
For all water solutions: Kw = [H3O+][OH-]
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THE pH SCALE
pH – The negative logarithm of the hydronium ion molarity of a solution
The common logarithm of a number is:the exponent to which 10 must be raised to equal the number
100 1000 0.001
100 = 102 1000 = 103 0.001 = 10-3
200
200 = 102.3
log 100 = 2 log 1000 = 3 log 0.001 = -3 log 200 = 2.3
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[H3O+] log [H3O+] pH
0.1 M0.01 M0.001 M0.02 M
-1.0-2.0-3.0-1.7
1.02.03.01.7
For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)
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Calculate the pH of pure water at 25ºC.
= -log(1.00 x 10-7 M)
pH = -log[H3O+]
For pure water:[H3O+] = 1.00 x 10-7 M
= 7.000
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pH 7 = Neutral
< 7Acidic
> 7Basic
10. M1. M0. 1 M
0.01 M0.001 M0.0001 M0.00001 M0.000001 M0.0000001 M0.00000001 M0.000000001 M0.0000000001 M0.00000000001 M0.000000000001 M0.0000000000001 M0.00000000000001 M
[H3O+] [OH-]
0.000000000000001 M0.00000000000001 M0.0000000000001 M
0.000000000001 M0.00000000001 M0.0000000001 M0.000000001 M0.00000001 M0.0000001 M0.000001 M0.00001 M0.0001 M0.001 M0.01 M0.1 M1 M
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pH 7 = Neutral
< 7Acidic
> 7Basic
Battery Acid
Gastric Fluid
SodaOrange Juice
Tap Water
“Pure” WaterSea WaterBaking Soda
AmmoniaBleachDrain Cleaner
10. M1. M0. 1 M
0.01 M0.001 M0.0001 M0.00001 M0.000001 M0.0000001 M0.00000001 M0.000000001 M0.0000000001 M0.00000000001 M0.000000000001 M0.0000000000001 M0.00000000000001 M
[H3O+] Common Substances
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Calculate the pH of orange juice if it has a hydronium ion concentration of 1.6 x 10-3 M.
= -log(1.6 x 10-3 M)
pH = -log[H3O+]
[H3O+] = 1.6 x 10-3 M
= 2.80
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Calculate the pH of toothpaste that has a hydroxide ion concentration of 5.6 x 10-5 M.
= -log(1.79 x 10-10 M)
pH = -log[H3O+]
= 9.74
(x)(5.6 x 10-5 M)
Kw = [H3O+][OH-]
= 1.79 x 10-10 M1.00 x 10-14 M2 = x _________________
5.6 x 10-5 M
1.00 x 10-14 M2 =
[OH-] = 5.6 x 10-5 M
= [H3O+]
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Calculate the pH of milk if it has a hydroxide ion concentration of 3.2 x 10-9 M.
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Calculate the hydrogen ion concentration in blood, which has a pH of 7.4.
= 0.0000000398 M
pH = -log[H3O+]
-pH = log[H3O+]
antilog (-pH) = [H3O+]
antilog (-7.4) = [H3O+]
For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)
= 3.98 x 10-8 M = 4 x 10-8 M
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Calculate the hydroxide ion concentration in egg yolks, which have a pH of 5.65.
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pH OF STRONG ACID SOLUTIONS
Strong acids completely ionize
HCl (g)
HCl (aq)
H+(aq) + Cl-(aq)
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Calculate the pH of a 0.015 M hydrochloric acid solution.
HCl (aq) + H2O (l) → H3O+(aq) + Cl-(aq)x
1 M
0.015
1 M
= -log(0.015 M)
pH = -log[H3O+]
= 1.82
x 1 M H3O+
____________
1 M HCl
= 0.015 M H3O+ 0.015 M HCl
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Calculate the pH of a 0.0400 M sulfuric acid solution.
H2SO4 (aq) + 2H2O (l) → 2H3O+(aq) + SO4 2-(aq)
x
2 M
0.0400
1 M
= -log(0.0800 M)
pH = -log[H3O+]
= 1.097
x 2 M H3O+
______________
1 M H2SO4
= 0.0800 M H3O+ 0.0400 M H2SO4
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pH OF STRONG BASE SOLUTIONS
Strong bases completely dissociate
NaOH (s)
NaOH (aq)
Na+(aq) + OH-(aq)
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Calculate the pH of a 0.40 M sodium hydroxide solution.
NaOH (aq) → Na+(aq) + OH-(aq) x
1 M0.40
1 M
= -log(2.50 x 10-14 M)
pH = -log[H3O+] = 13.60
(x)(0.40 M)
Kw = [H3O+][OH-]
= 2.50 x 10-14 M1.00 x 10-14 M2 = x _________________
0.40 M
1.00 x 10-14 M2 =
= [H3O+]
x 1 M OH-
______________
1 M NaOH
= 0.40 M OH-0.40 M NaOH not [H3O+]
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BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it
7.00
pH
1.00 L H2O0.10 moles HCl
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BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it
1.00
pH
1.00 L H2O0.10 moles HCl
3.45
pH
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
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BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or base is added to it
1.00
pH
1.00 L H2O0.10 moles HCl
3.37
pH
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
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A solution is a buffer if is contains a weak acid and its conjugate base
Weak acid:Conjugate base:
HFF-
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
If a strong acid is added to the buffer:
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H+ + F- → HF
The strong acid (H+) is reacted away by the conjugate base (F-)
A solution is a buffer if is contains a weak acid and its conjugate base
Weak acid:Conjugate base:
HFF-
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
no more strong acid in the solution
If a strong acid is added to the buffer:
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A solution is a buffer if it contains a weak acid and its conjugate base
Weak acid:Conjugate base:
HFF-
1.00 L 1.0 M HF, 1.0 M F-
If a strong base is added to the buffer:
OH- + HF → H2O + F-
The strong base (OH-) is reacted away by the weak acid (HF)
no more strong base in the solution
0.10 moles OH-
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