Download Chemistry 2000 Lecture 19: Organic acids

Chemistry 2000 Lecture 19: Organic acids
Marc R. Roussel
Ka
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The strength of an acid HA is measured by Ka , the
equilibrium constant for the dissociation of the acid into a
proton and its conjugate base:
HA H+ + A−
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Equivalently, we can think of Ka as the equilibrium constant
for the reaction of an acid with water:
HA + H2 O H3 O+ + A−
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Larger Ka =⇒ stronger acid
pKa
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Ka values range over many orders of magnitude, so they are
not very convenient for some purposes (e.g. comparisons
between acids).
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Define
pKa = − log10 Ka
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Smaller pKa =⇒ stronger acid
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Strong acids may have negative pKa values (dissociate
completely in water).
Organic acids
Stronger acid
Weaker acid
Functional group
Carboxylic acids
Phenols
Water
Alcohols
pKa
3–5
1–10
14 (pKw )
15–18
Acidity of carboxylic acids
O
C
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OH
Carboxylic acids are among the most acidic organic
compounds.
Acid strength depends on
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the polarity of the bond to the dissociable hydrogen (in this
case the O-H bond), and
the stability of the conjugate base.
Acidity of carboxylic acids
Polarity of the O-H bond
O
C
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OH
The carbonyl group is electron withdrawing: Because of the
large electronegativity of oxygen, electron density is pulled
away from the carbon atom. This in turn pulls electron
density away from the hydrogen atom, increasing the partial
positive charge on this atom.
Acidity of carboxylic acids
Stability of carboxylate ions
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The carboxylate anion is resonance stabilized:
O
O
−
C
C
O
−
O
The negative charge is therefore spread over a larger region of
the molecule, which tends to stabilize the anion.
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Substituents on the adjacent carbon to the carboxylic acid
functional group can have a substantial effect on the pKa .
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Consider the following series:
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Molecule CH3 COOH FCH2 COOH F2 CHCOOH F3 CCOOH
pKa
4.67
2.66
1.24
0.23
The fluorine atom is very electronegative and creates a partial
positive charge on carbon 2:
H
δ−
F
O
δ+
C
H
C −
O
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The nearness of the partial positive charge on carbon 2 and of
the negative charge of the carboxylate group stabilizes the
latter.
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This is called an inductive effect.
Alcohols
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Unlike carboxylic acids, most alcohols are only about as acidic
as water itself.
Why?
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No electron-withdrawing group (in an ordinary alcohol) to
increase the polarization of the O-H bond
No resonance stabilization of the negative charge of the
conjugate base
Inductive effects can increase the acidity of an alcohol.
Example:
Molecule
pKa
CH3 CH2 OH
15.9
Cl3 CCH2 OH
12.24
Phenols
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Phenols are much more acidic than ordinary alcohols.
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Phenol itself has a pKa of 9.95, vs 17 for cyclohexanol.
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Why? Resonance stabilization of the charge on the phenolate
anion:
.. −
:O:
H
:O:
H
H
:O:
:O:
H
H
H
−
H
H
H
H
H
..
..
−
H
H
H
..
H
H
−
H
H
H
H
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Some phenols are even more acidic because the substituents
can participate in charge delocalization.
Example: p-nitrophenol has a pKa of 7.21.
.. −
:O:
:O:
H
:O:
H
H
H
H
H
H
H
..
−
H
H
H
..
O
..
N
+
..
H
−
+
+
.. −
O:
..
..
O
..
N
.. −
O:
..
..
O
..
:O:
:O:
H
H
H
H
H
H
−
..
H
..
O
..
N
H
+
.. −
O:
..
− ..
:O
..
N
+
.. −
O:
..
N
.. −
O:
..
Acid dissociation equilibria
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If we know the Ka and concentration of an acid, we can
calculate the pH.
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Reminder:
pH = − log10 aH+
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We usually don’t need to take the autoionization of water into
account unless the concentration of protons liberated from the
acid is similar to the concentration of protons generated by
autoionization.
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We often can treat the acid as if it’s mostly undissociated.
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Note that we’ll assume 25◦ C in all of the following
calculations.
Example: pH of a 0.32 M phenol solution
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The pKa of phenol is 9.95.
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pH = 5.22
Example: pH of a 4.2 × 10−5 M ethanoic (acetic) acid
solution
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The pKa of ethanoic acid is 4.76.
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pH = 4.71
Balance of acid and conjugate base at given pH
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Sometimes, we put an acid into a solution of fixed pH (a
buffer) and want to know how much is in the acid and how
much in the conjugate base form.
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This is an easy problem because the pH fixes aH+ , which
immediately gives us the ratio of the conjugate base to the
acid:
Ka
a −
= A
aH+
aHA
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This can easily be converted to percentages of the two forms
if we add the equation
[HA] + [A− ] = 100%
(with a slight abuse of notation).
Example: ethanoic acid at pH 4
Ka
1.74 × 10−5
[CH3 COO− ]
=
=
0.174
=
aH+
10−4
[CH3 COOH]
and
[CH3 COOH] + [CH3 COO− ] = 100%
∴ [CH3 COOH] + 0.174[CH3 COOH] = 100%
∴ [CH3 COOH] = 85%
∴ [CH3 COO− ] = 15%
Distribution curves
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If we repeat the above calculation at a number of different pH
values and plot the results, we obtain distribution curves for
the acid and its conjugate base.
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Note: If pH = pKa , we have
10−pKa
aA−
=
=1
aHA
10−pH
In other words, 50% of the acid is undissociated, and 50% in
the form of the conjugate base at the pKa .
Distribution curve of ethanoic acid
100
acid
base
90
80
70
%
60
50
40
30
20
10
0
0
2
4
6
8
pH
10
12
14
Distribution curves for polyprotic acids
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The calculation is analogous for polyprotic acids except that
there are two (or more) equilibria and three (or more) forms of
the acid to consider.
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When the pKa ’s of a polyprotic acid differ by several units,
the distribution curves look like a simple superposition of
distribution curves for the monoprotic case.
Example: Ethanedioic (oxalic) acid at pH 3
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We want to calculate the fraction (or percentage) of the acid
in each of the three possible ionization states: (COOH)2 ,
HOOCCOO− and (COO)2−
2 .
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The pKa s of the two protons are 1.27 and 4.27.
Answer: [HOOCCOO− ] = 93%, [(COO)2−
2 ] = 5.0% and
[(COOH)2 ] = 1.7%
Distribution curves of ethanedioic acid
100
(COOH)2
HOOCCOO(COO)22
90
80
70
%
60
50
40
30
20
10
0
0
2
4
6
8
pH
10
12
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Take a log, have an equation named after you. . .
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We are now familiar with the equation
Ka =
(aA− )(aH+ )
aHA
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If we take the negative log of this equation, we get
aA−
− log10 Ka = − log10 aH+ − log10
aHA
aA−
∴ pKa = pH − log10
aHA
aA−
or pH = pKa + log10
aHA
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This last equation is called the Henderson-Hasselbach
equation.