Transcript CHE 106: General Chemistry
CHE 116: General Chemistry
CHAPTER SIXTEEN Copyright © Tyna L. Heise 2001-2002 All Rights Reserved
Prof. T. L. Heise
CHE 116 1
Acids and Bases: Review
Properties of Acids
sour taste
change with litmus Properties of Bases
bitter taste
change with litmus
Prof. T. L. Heise
CHE 116 2
Acids and Bases: Review
1830 - scientists have recognized that all acids contain hydrogen, but not all hydrogen bearing compounds are acids 3 Svante Arrhenius - linked acid behavior with the presence of an H + and base behavior with the presence of an OH -
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Arrhenius’ definition is useful, but restricts acid base reactions to aqueous conditions 4 Johannes Bronsted and Thomas Lowry proposed a more general definition which involves the transfer of an H + ion from one molecule to another
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
H + ion is simply a proton with no surrounding valence electron.
Small particle interacts strongly with the nonbonding pairs of water molecules to form hydrated hydrogen ions
chemists use H + interchangeably and H 3 O + 5
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Fig 16.1
Demonstrates the interconnections possible between hydrogenated water 6
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Definitions: Acid - any compound which transfers an H+ to another molecule 7 Base - any compound which accepts a transfer of an H+ from another molecule * an acid and base always work together
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Definitions: Amphoteric - some substances can be an acid or a base depending on reaction Conjugate Acid Base pairs - two compounds that differ only in the presence of an H + . The molecule with the extra H is the acid.
Prof. T. L. Heise
CHE 116 8
Bronsted-Lowry Acids and Bases
Fig 16.7, 16.8
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO 3 F PO 4 3 CO 10
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: HSO 3 F PO 4 CO 3 Given a base, bases accept H + , so add an H + to each molecule.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO 3 HSO 3 F PO 4 CO 3 Given a base, bases accept H + , so add an H + to each molecule.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO HF 3 HSO 3 F PO 4 CO 3 Given a base, bases accept H + , so add an H + to each molecule.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO HF HPO 3 4 2 HSO 3 F PO 4 CO 3 Given a base, bases accept H + , so add an H + to each molecule.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: Write the formula for the conjugate acid of each of the following: H 2 SO HF HPO HCO 3 4 2 + HSO 3 F PO 4 CO 3 Given a base, bases accept H + , so add an H + to each molecule.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2 , with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2 , with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
17 O 2 + H 2 O
OH + OH -
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2 , with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
18 O 2 + H 2 O
OH + OH -
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample Exercise: When lithium oxide, Li 2 O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O 2 , with water. Write the reaction that occurs, and identify the conjugate acid base pairs.
acid base O 2 + H 2 O
OH + OH base acid
Prof. T. L. Heise
CHE 116 19
Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:
the more readily a substance donates an H + , the less readily it’s conjugate base will accept one
the more readily a substance accepts an H + , the less readily it’s conjugate acid will donate one
the stronger one of the substances is, the weaker it’s conjugate
Prof. T. L. Heise
CHE 116 20
Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:
strong acids completely transfer their protons to water, leaving no undissociated molecules
weak acids are those that only partly dissociate in aqueous solution and therefore exist in the solution as a mixture of acid molecules and component ions 21
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Relative Strengths of Acids and Bases:
negligible acids are those that have hydrogen but do not donate them at all, their conjugate bases would be extremely strong, reacting with water to complete their octet and leaving OH behind.
In every acid base reaction, the position of the equilibrium favors transfer of H + from stronger side to weaker side
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CHE 116 22
Bronsted-Lowry Acids and Bases
Fig. 16.4
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Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3 (aq)+H 2 O(l)
HPO 4 2 (aq)+OH (aq) b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 24
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3 (aq)+H 2 O(l)
HPO 4 2 (aq)+OH (aq) 2 acids are: H 2 O and HPO 4 2 2 bases are: PO 4 3 and OH 25
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3 (aq)+H 2 O(l)
HPO 4 2 (aq)+OH (aq) 2 acids are: H 2 O and HPO 4 2 2 bases are: PO 4 3 and OH red indicates strength
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CHE 116 26
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3 (aq)+H 2 O(l)
HPO 4 2 (aq)+OH (aq) 2 acids are: H 2 O and HPO 4 2 2 bases are: PO 4 3 and OH reverse reaction favored
Prof. T. L. Heise
CHE 116 27
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: a) PO 4 3 (aq)+H 2 O(l)
HPO 4 2 (aq)+OH (aq) 2 acids are: H 2 O and HPO 4 2 2 bases are: PO 4 3 and OH shifts left
Prof. T. L. Heise
CHE 116 28
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 29
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH 30
Prof. T. L. Heise
CHE 116
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH red indicates strength
Prof. T. L. Heise
CHE 116 31
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH favors forward reaction
Prof. T. L. Heise
CHE 116 32
Bronsted-Lowry Acids and Bases
Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right: b) NH 4 + (aq)+OH (aq)
NH 3 (aq)+H 2 O(l) 2 acids are: NH 4 + and H 2 O 2 bases are: NH 3 and OH shifts right
Prof. T. L. Heise
CHE 116 33
The Autoionization of Water
One of the most important properties of water is its ability to ac as either a Bronsted acid or Bronsted base, depending on circumstances.
One water molecule can donate a proton to another water molecule Fig 16.10
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Prof. T. L. Heise
CHE 116
The Autoionization of Water
The autoionization of water, although rapid and weak, does exist as an equilibrium, and therefore has an equilibrium constant expression: K eq = [H 3 O + ][OH ] [H 2 O] 2 * because water is a liquid, it can be excluded from the equation...
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Prof. T. L. Heise
CHE 116
The Autoionization of Water
K eq [H 2 O] 2 = [H 3 O + ][OH ] K w = [H 3 O + ][OH ] = 1.0 x 10 -14 * this equation is not only applicable to water, but to all aqueous solutions, and it is upon this fact that the pH scale was built.
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Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 b) [OH ] = 3 x 10 -9 c) [OH ] = 1 x 10 -7 37
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 then [OH ] must equal 1.0 x 10 -14 which is 2 x10 -5 [OH ] = 5.0 x 10 -10 38
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: a) [H + ] = 2 x 10 -5 then [OH ] must equal 1.0 x 10 -14 which is 2 x10 -5 [OH ] = 5.0 x 10 -10 [H + ] > [OH ] so acidic 39
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH ] = 3 x 10 -9 then [H + ] must equal 1.0 x 10 -14 which is 3 x10 -9 [H + ] = 3.3 x 10 -6 40
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: b) [OH ] = 3 x 10 -9 then [H + ] must equal 1.0 x 10 -14 which is 3 x10 -9 [H + ] = 3.3 x 10 -6 [H + ] > [OH ] so acidic 41
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH ] = 1 x 10 -7 then [H + ] must equal 1.0 x 10 -14 which is 1 x10 -7 [H + ] = 1.0 x 10 -7 42
Prof. T. L. Heise
CHE 116
The Autoionization of Water
Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic: c) [OH ] = 1 x 10 -7 then [H + ] must equal 1.0 x 10 -14 which is 1 x10 -7 [H + ] = 1.0 x 10 -7 [H + ] = [OH ] so neutral 43
Prof. T. L. Heise
CHE 116
The pH Scale
For convenience, we can use a logarithmic version of concentration to turn the very small concentrations of [H + ] and [OH ] into whole numbers. p(anything) = - log[anything] p(H) = -log[H + ] p(OH) = - log[OH ] ** pH + pOH = 14
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CHE 116 44
The pH Scale
Prof. T. L. Heise
CHE 116 45
The pH Scale
Common household products and their relative pH’s.
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Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH?
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Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH?
48 pH = -log[H + ]
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH?
49 pH = -log[H + ] pH = -log[3.8 x 10 -4 ]
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: In a sample of lemon juice [H + ] is 3.8 x 10 -4 M. What is the pH?
50 pH = -log[H + ] pH = -log[3.8 x 10 -4 ] pH = 3.42
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH?
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Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH?
pH = -log[H + ] 52
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH?
pH = -log[H + ] pH = -log[5.3 x 10 -9 ] 53
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A commonly available window-cleaning solution has a [H + ] is 5.3 x 10 -9 M. What is the pH?
pH = -log[H + ] pH = -log[5.3 x 10 -9 ] pH = 8.28
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CHE 116 54
The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].
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Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].
56 pH = -log[H + ]
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].
57 pH = -log[H + ] 9.18 = -log[H + ]
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].
58 pH = -log[H + ] 9.18 = -log[H + ] 10 -9.18
= [H + ]
Prof. T. L. Heise
CHE 116
The pH Scale
Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ].
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Prof. T. L. Heise
pH = -log[H + ] 9.18 = -log[H + ] 10 -9.18
= [H + ] 6.61 x 10 -10 = [H + ] CHE 116
The pH Scale
Measuring pH: a pH can be measured quickly and accurately using a pH meter.
A pair of electrodes connected to a meter capable of measuring small voltages
a voltage which varies with pH is generated when the electrodes are placed in a solution
calibrated to give pH 60
Prof. T. L. Heise
CHE 116
The pH Scale
Measuring pH: a pH can be measured quickly and accurately using a pH meter.
Electrodes come in a variety of shapes and sizes
a set of electrodes exists that can be placed inside a human cell
acid base indicators can be used, but are much less precise 61
Prof. T. L. Heise
CHE 116
The pH Scale
Fig 16.7
Prof. T. L. Heise
CHE 116 62
Strong Acids and Bases
Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions.
Strong Acids HCl HBr HI HNO 3 HClO 3 HClO 4 H 2 SO 4
Prof. T. L. Heise
monoprotic diprotic CHE 116 63
Strong Acids and Bases
Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions.
64 Calculating the pH of a solution made up entirely of ions means the [H+] is proportional to [acid]
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
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Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
66 pH = -log[H + ]
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
67 pH = -log[H + ] 2.66 = -log[H + ]
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
68 pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66
= [H + ] 2.2 x 10 -3 = [H + ]
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66
= [H + ] 2.2 x 10 -3 2.2 x 10 -3 M H = [H + + ] 1 mole HNO 1 mole H + 3 69
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
An aqueous solution of HNO 3 has a pH of 2.66. What is the concentration of the acid?
pH = -log[H + ] 2.66 = -log[H + ] 10 -2.66
= [H + ] 2.2 x 10 -3 2.2 x 10 -3 M H = [H + + ] 1 mole HNO 1 mole H + 3 2.2 x 10 -3 HNO 3
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CHE 116 70
Strong Acids and Bases
The most soluble common bases are the ionic hydroxides of the alkali and alkaline earth metals.
Due to the complete dissociation of the base into its ion components makes the pH calculation equally straightforward 71
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
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Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 73
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 11.89 = -log[H + ] 74
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89
= [H + ] 75
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89
= [H + ] 1.29 x 10 -12 = [H + ] 76
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89
= [H + ] 1.29 x 10 -12 1.0 x 10 -14 = [OH = [H ] + ] 1.29 x 10 -12 77
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
pH = -log[H + ] 11.89 = -log[H + ] 10 -11.89
= [H + ] 1.29 x 10 -12 1.0 x 10 -14 = [OH = [H + ] ] = 7.8 x 10 -3 1.29 x 10 -12 78
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
1.0 x 10 -14 = [OH ] = 7.8 x 10 -3 1.29 x 10 -12 7.8 x 10 -3 M OH 1 mol KOH 1 mol OH 79
Prof. T. L. Heise
CHE 116
Strong Acids and Bases
Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?
1.0 x 10 -14 = [OH ] = 7.8 x 10 -3 1.29 x 10 -12 7.8 x 10 -3 M OH 1 mol KOH 1 mol OH 7.8 x 10 -3 M KOH 80
Prof. T. L. Heise
CHE 116
Weak Acids
Most acids are weak acids and only partially dissociate in aqueous solution.
The extent to which a weak acid dissociates can be expressed using an equilibrium constant for the ionization reaction HX + H 2 O
Ka = [H 3 O + ][X ] [HX] H 3 O + + X *the larger the K a the stronger the acid
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CHE 116 81
Weak Acids
Calculating K a from pH: a much more complicated calculation is required for the determinations of weak acids, in many cases, due to the extremely small magnitude of the values, some simpler approximations can be made.
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CHE 116
Weak Acids
Sample exercise: Niacin, one of the B vitamins, has the following molecular structure: C O H O N A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?
b) What is the acid-dissociation constant?
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CHE 116 83
Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?
pH = -log[H + ] 3.26 = -log[H + ] e -3.26
= [H + ] 5.5 x 10 -4 = [H + ] 84
Prof. T. L. Heise
CHE 116
Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?
C 6 H 4 NOOH
C 6 H 4 NOO + H + 85 initial 0.020 change -5.5 x 10 -4 0 +5.5 x 10 -4 equil
Prof. T. L. Heise
0.01945
5.5 x 10 -4 0 +5.5 x 10 -4 5.5 x 10 -4 CHE 116
Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
a) What % of the acid is ionized in this solution?
% = part/total x 100 = 5.5 x 10 -4 /0.01945 x 100 = 2.8% 86
Prof. T. L. Heise
CHE 116
Weak Acids
Sample exercise: A 0.020 M solution of niacin has a pH of 3.26
b) What is the acid-dissociation constant?
C 6 H 4 NOOH
C 6 H 4 NOO + H + K a = [C 6 H 4 NOO -
][
H
[
C 6 H 4 NOOH
]
+
] = (
5.5 x 10 -4
)(
5.5 x 10 -4 0.01945
= 1.55 x10 -5
)
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CHE 116 87
Weak Acids
Using K a to calculate pH: Using the value of K a and knowing the initial concentration of the weak acid, we can calculate the concentration of H + (aq).
Example: Calculate the pH of a 0.30 M solution of acetic acid.
HC 2 H 3 O 2 (aq)
H + (aq) + C 2 H 3 O 2 (aq) 88
Prof. T. L. Heise
CHE 116
Weak Acids
Example: Calculate the pH of a 0.30 M solution of acetic acid.
HC 2 H 3 O 2 (aq)
H + (aq) + C 2 H 3 O 2 (aq) From Table 16.2, K a = 1.8 x 10 -5 Ka = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] set up data table of concentrations involved...
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CHE 116
Weak Acids
Ka = 1.8 x 10 -5 = [H + ][C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] set up data table of concentrations involved… [HC 2 H 3 O 2 ] [H + ] [C 2 H 3 O 2 ] Initial 0.30 M 0 0 Change -x +x +x Equilibrium 0.30 - x x x 90
Prof. T. L. Heise
CHE 116
Weak Acids
Input concentrations in formula Ka = 1.8 x 10 -5 = [x][x] [0.30 -x] This will lead to a quadratic equation, but we can simplify the problem a bit. All weak acids dissociate so little that we can assume the initial concentration of the acid remains essentially the same, that means we can rewrite the formula to read...
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CHE 116 91
Weak Acids
Ka = 1.8 x 10 -5 = [x][x] [0.30] 1.8 x 10 -5 (0.30) = x 2 1.8 x 10 -5 (0.30) = x 0.0023 = x 0.0023 = [H + ] pH = -log [H + ] = -log[0.0023] = 2.64
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CHE 116 92
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
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CHE 116
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
C 6 H 4 NOOH
C 6 H 4 NOO + H + K a = 1.5 x10 -5 = [C 6 H 4 NOO ][H + ] [C 6 H 4 NOOH] 94
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
C 6 H 4 NOOH C 6 H 4 NOO H + initial 0.010 0 0 change -x +x +x equilibrium 0.010 -x x x
Prof. T. L. Heise
CHE 116 95
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
C 6 H 4 NOOH
C 6 H 4 NOO + H + K a = 1.5 x10 -5 = [x][x] [0.010] 96
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
1.5 x10 -5 [0.010] = x 2 1.5 x10 -5 [0.010] = x 3.87 x 10 -4 = x
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CHE 116 97
Weak Acids
Sample Exercise: The K a for niacin is 1.5 x 10 -5 . What is the pH of a 0.010 M solution of niacin?
3.87 x 10 -4 = x 3.87 x 10 -4 = [H + ] pH = -log[H + ] pH = -log[3.87 x 10 -4 ] pH = 3.41
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CHE 116 98
Weak Acids
The results seen in the two previous examples are typical of weak acids. The lower number of ions produced during the partial dissociation causes less electrical conductivity and a slower reaction rate with metals.
Percent ionization is a good way to discover the actual conductivity, however...
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Prof. T. L. Heise
CHE 116
Weak Acids
As the concentration of a weak acid increases, the % ionized decreases.
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Prof. T. L. Heise
CHE 116
Weak Acids
As the concentration of a weak acid increases, the % ionized decreases.
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Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise b) a 1.0 x 10-3 M solution 102
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in a) the previous exercise Our approximation was good so % = part x 100 total = 3.87 x 10 -4 x 100 = 3.9% 0.010
Prof. T. L. Heise
CHE 116 103
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution K a = 1.5 x10 -5 = [x][x] [1.0 x 10 -3 ] 1.5 x10 -5 (1.0 x 10 -3 ) = x 2 1.2 x 10 -4 104
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution 105 K a = 1.5 x10 -5 = [x][x] 1.5 x10 -5 [1.0 x 10 -3 ] but, 1.2 x 10 -3 (1.0 x 10 -3 ) = x 2 1.0 x 10 -3 x 100 1.2 x 10 -3
Prof. T. L. Heise
is greater than 5% so use quadratic...
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution K a = 1.5 x10 -5 = [x][x] [1.0 x 10 -3 - x] -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0 106
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0 x = -b ± b 2 - 4ac 2a x =
Prof. T. L. Heise
CHE 116 107
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution -1.5 x 10 -5 (1.0 x10 -3 ) + 1.5 x 10 -5 x + x 2 = 0 x = -b ± b 2 - 4ac 2a x = 1.1 x 10 -4 or -1.3 x 10 -4
Prof. T. L. Heise
CHE 116 108
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10 -4 1.1 x 10 -4 x 100 1.0 x 10 -3 109
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the % of niacin molecules ionized in b) a 1.0 x 10-3 M solution x = 1.1 x 10 -4 1.1 x 10 -4 x 100 = 11% 1.0 x 10 -3 110
Prof. T. L. Heise
CHE 116
Weak Acids
Polyprotic Acids: many acids have more than one ionizable H atom. H 2 SO 3 (aq)
H + (aq) + HSO 3 (aq) K a1 HSO 3 (aq)
H + (aq) + SO 3 -2 (aq) K a2 The Ka are labeled according to which proton is dissociating.
-it is always easier to remove the first proton than the second
Prof. T. L. Heise
CHE 116 111
Weak Acids
Prof. T. L. Heise
CHE 116 112
Weak Acids
If K a K a1 values differ by 10 3 or more, only use to determine calculations.
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Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
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Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
H 2 C 2 O 4
HC 2 O 4 + H + 115 Ka = 5.9 x 10 -2 = [HC 2 O 4 ][H [H 2 C 2 O 4 ] + ]
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
H 2 C 2 O 4
HC 2 O 4 + H + strong acid so 100% dissociation 0.020 M H+ 116
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
HC 2 O 4 -
C 2 O 4 -2 + H + I
D
0.020 0 0.020
-x +x +x E 0.020 - x x 0.020 + x 117
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
Ka = 6.4 x 10 -5 = [0.020 + x ][x] [0.020 - x] use your assumption 118
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
Ka = 6.4 x 10 -5 = [0.020 ][x] [0.020] 6.4 x 10 -5 = x = [C 2 O 4 2 ] 119
Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
[H + ] = 0.020
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Prof. T. L. Heise
CHE 116
Weak Acids
Sample Exercise: Calculate the pH and concentration of oxalate ion, [C 2 O 4 2 ] in a 0.020 M solution of oxalic acid.
pH = -log[H + ] pH = - log[0.020] pH = 1.70
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Prof. T. L. Heise
CHE 116
Weak Bases
Many substances behave as weak bases in water. Such substances react with water, removing protons from water, leaving the OH ion behind.
NH 3 + H 2 O
NH 4 + + OH K b = [NH 4 + ][OH ] [NH 3 ] * K b is the base dissociation constant utilizing the [OH ]
Prof. T. L. Heise
CHE 116 122
Weak Bases
* K b is the base dissociation constant utilizing the [OH ]
bases must contain one or more lone pair to bond with the H + from water.
as before, the larger the K b the base the stronger
stronger base have low pOH, but high pH 123
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
A) pyridine B) methylamine C) nitrous acid 124
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
A) pyridine K b = 1.7 x 10 -9 B) methylamine K b = 4.4 x 10 -4 C) nitrous acid K b = 2.2 x 10 -11 125
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
A) pyridine K b = 1.7 x 10 -9 K b = 1.7 x 10 -9 = [x][x] [0.05] x = [OH ] = 9.2 x 10 -6 pOH = 5.0 so pH = 9.0
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Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
B) methylamine K b = 4.4 x 10 -4 K b = 4.4 x 10 -4 = [x][x] [0.05] x = [OH ] = 4.6 x 10 -3 pOH = 2.32 so pH = 11.68
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Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
C) nitrous acid K b = 2.2 x 10 -11 K b = 2.2 x 10 -11 = [x][x] [0.05] x = [OH ] = 1.0 x 10 -6 pOH = 5.97 so pH = 8.02
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Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?
A) pyridine pH = 9.0
B) methylamine pH = 11.68
C) nitrous acid pH = 8.02
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Prof. T. L. Heise
CHE 116
Weak Bases
Identifying a Weak Base
Neutral substances that have an atom with a nonbonding pair of electrons that can serve as a proton acceptor.
Most of these are nitrogen atoms
Anions of weak acids 130
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
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Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
132 pH = 10.50 so pOH = 3.50
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
pOH = 3.50
NH 3 + H 2 0
NH 4 + + OH [OH ] = 3.16 x 10 -4 133
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
NH 3 + H 2 0
NH 4 + + OH x -3.16 x 10 -4 0 0 +3.16 x 10 -4 +3.16 x 10 -4 x - 3.16 x 10 -4 3.16 x 10 -4 3.16 x 10 -4 134
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
NH 3 + H 2 0
NH 4 + + OH K b = 1.8 x 10 -5 = [3.16 x 10 -4 ][3.16 x 10 -4 ] [x - 3.16 x 10 -4 ] 135
Prof. T. L. Heise
CHE 116
Weak Bases
Sample exercise: A solution of NH 3 in water has a pH of 10.50. What is the molarity of the solution?
NH 3 + H 2 0
NH 4 + + OH K b = 1.8 x 10 -5 = [3.16 x 10 -4 ][3.16 x 10 -4 ] [x - 3.16 x 10 -4 ] 136 x = 0.0058 M
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is equal to the product of the equilibrium constants for the two added reactants.
NH 4 + (aq)
NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l)
NH 4 + (aq) + OH (aq) H 2 O(l)
H + (aq) + OH (aq) 137
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b NH 4 + (aq)
NH 3 (aq) + H 2 O(l)
NH NH 4 + 3 (aq) + H + (aq) + OH (aq) (aq) H 2 O(l)
H + (aq) + OH (aq) 138 K a x K b = K w pK a + pK b = pK w = 14
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: Which of the following anions has the largest base-dissociation constant?
A) NO 2 B) PO 4 3 C) N 3 139
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: Which of the following anions has the largest base-dissociation constant?
A) NO 2 K a = 4.5 x 10 -4 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 2.2 x 10 -11 4.5 x 10 -4 140
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: Which of the following anions has the largest base-dissociation constant?
B) PO 4 3 K a = 4.2 x 10 -13 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 2.4 x 10 -2 4.2 x 10 -13
Prof. T. L. Heise
CHE 116 141
Relationship Between K
a
and K
b Sample exercise: Which of the following anions has the largest base-dissociation constant?
C) N 3 K a = 1.9 x 10 -5 K a x K b = 1.0 x 10 -14 K b = 1.0 x 10 -14 = 5.2 x 10 -10 1.9 x 10 -5 142
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: Which of the following anions has the largest base-dissociation constant?
A) NO 2 K b = 2.2 x 10 -11 B) PO C) N 3 4 3 K b = 2.4 x 10 -2 K b = 5.2 x 10 -10 143
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline?
144
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline?
145 pK a + pK b = 14 4.90 + x = 14 x = 9.1
Prof. T. L. Heise
CHE 116
Relationship Between K
a
and K
b Sample exercise: The base quinoline has a pK a of 4.90. What is the base dissociation constant for quinoline?
146 pK b = 9.1
pK b = -log[K b ] 9.1 = -log[K b ]
Prof. T. L. Heise
[K b ] = 7.9 x 10 -10 CHE 116
Acid Base Properties of Salt Soln’s
Salt solutions have the potential to be acidic or basic.
Hydrolysis of a salt
acid base properties are due to the behavior of their cations and anions
perform the necessary double replacement reaction and examine the products using your strength rules 147
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
If a strong acid and strong base are produced, the resultant solution will be neutral.
If a strong acid and weak base are produced, the resultant solution will be acidic.
If a strong base and a weak acid are produced, the resultant solution will be basic.
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Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
If a weak acid and weak base are produced, the resultant solution will be dependent on the K a values.
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Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO 3 , Fe(NO 3 ) 3 150 NaNO 3 + H 2 O
NaOH + HNO 3 SB SA
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO 3 , Fe(NO 3 ) 3 151 Fe(NO 3 ) 3 + 3H 2 O
Fe(OH) 3 + 3HNO 3 WB SA
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
A) NaNO 3 , Fe(NO 3 ) 3 152 Fe(NO 3 ) 3 + 3H 2 O
Fe(OH) 3 + 3HNO 3 WB SA
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO 153
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO 154 KBr + H 2 O
KOH + HBr SB SA
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
B) KBr, KBrO 155 KBrO + H 2 O
KOH + HBrO SB WA
Prof. T. L. Heise
CHE 116
Acid Base Properties of Salt Soln’s
Sample exercise: In each of following, which salt will form the more acidic solution.
B) KBr , KBrO 156
Prof. T. L. Heise
CHE 116
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidic
a molecule containing H will transfer a proton only if the H-X bond is polarized like H -- X the stronger the bond the weaker the acid and vice versa
Prof. T. L. Heise
CHE 116 157
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidic
oxyacids exist when the H is attached to an oxygen bonded to a central atom if the OH’s attached are equal in number to the O’s present, acid strength increases with electronegativity 158
Prof. T. L. Heise
CHE 116
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidic
oxyacids exist when the H is attached to an oxygen bonded to a central atom the more O’s present compared to the OH’s, the more polarized the OH bond becomes and the stronger the acid is 159
Prof. T. L. Heise
CHE 116
Acid-Base Behavior & Chem Structure
How does the chemical structure determine which of the behaviors will be exhibited?
Acidic
carboxylic acids exist when the functional group COOH is present the strength also increases as the number of electronegative atoms in the molecule increase 160
Prof. T. L. Heise
CHE 116