Resolucion

Segundo Examen Parcial de Quimica Analitica I (1406)
Tema: Reacciones acido-base.
Profesor: Gustavo Gomez Sosa. Grupo 11. Semestre 2009-2
Resolucion
1. Una disolucion 0.20M de acido hipocloroso (HClO) tiene un pH de 4.08. Con estos datos,
calcule Ka para este acido.
Resolucion: Como primer paso se plantea la reaccion y las condiciones de equilibrio
H+
ClO-
0.2
0
0
Reacciona
ξ
ξ
ξ
Equilibrio
0.2-ξ
ξ
ξ
Especie
HClO
Inicio
<------>
Se sabe que pH = 4.08, por lo que [H+]= ξ = 10-4.08 = 8.3 x 10-5 M
[ClO-] = ξ = 8.3 x 10-5 M
[HClO] = 0.2 M - 8.3 x 10-5 M = 0.199917 M
Sustituyendo en la expresion de Ka:
8.3x10− 5 )
(
[ H + ][ClO − ]
ξ2
Ka =
=
=
= 3.45 x10− 8
[ HClO]
0.2 − ξ
0.199917
2
Tambien se puede determinar utilizando la ecuacion de Henderson-Hasselbalch, sustituyendo las
concentraciones al equilibrio de las especies involucradas de la siguiente forma:
pH = pKa + log
[ Base conjugada]
[ Acido]
pH = pKa + log
[ClO − ]
[ HClO ]
pKa = pH − log
[ClO − ]
[8.3x10− 5 ]
= 4.08 − log
= 7.4617
[ HClO]
[0.199917]
Ka = 10− 7.4617 = 3.45 x10− 8
2. Una muestra de 0.0010 mol de NaOH se agrega a 1.0 L de una disolucion 0.10M de NH3 y
0.050M de NH4Cl. Calcule el pH de la disolucion inicial y el cambio en el pH despues de esta
adicion. Suponga ahora que en lugar de la base, se agregara a la disolucion original 0.0010 mol
de HCl. Calcule el cambio en el pH por esta adicion.
Resolucion: pH de la disolucion original: Como primera forma de resolverlo, se define el equilibrio
en disolucion acuosa de la mezcla NH3 /NH4+, en el que el NH3 es la base conjugada del NH4+, acido
que resulta de la disociacion de la sal NH4Cl. Sabemos que pKa (NH3 /NH4+) = 9.2, por lo que
podemos considerar que la reaccion se da en un medio ligeramente basico, donde predominara la
reaccion con iones hidroxido en lugar de protones:
Especie
NH4+
OH-
<------>
Inicio
0.05
0
0.1
Reacciona
ξ
ξ
ξ
Equilibrio
0.05-ξ
ξ
0.1+ξ
NH3
H2O
Al definir la constante de la reaccion en funcion de la constante de acidez y del avance de reaccion,
obtenemos la concentracion de protones y el pH como se describe a continuacion:
KR =
[ NH 3 ]
[ NH 3 ][ H + ]
Ka
=
= − 14 = 104.8
+
−
− 14
+
[ NH 4 ][OH ] 10 [ NH 4 ] 10
KR =
0.1 + ξ
= 104.8
ξ (0.05 − ξ )
ξ = [OH − ] = 3.17 × 10− 5
10− 14
= 3.15 × 10− 10
[OH − ]
pH = 9.5
[H + ] =
Otra forma de obtener el resultado, es al considerar que mezcla inicial representa una disolucion
amortiguadora, por lo que el pH se puede calcular con la ecuacion de Henderson-Hasselbalch:
pH = pKa + log
[ NH 3 ]
0.1
= 9.2 + log
= 9.5
+
[ NH 4 ]
0.05
Cambio en el pH al agregar NaOH: Si ahora se toma en cuenta que se agrega una base fuerte,
originara un aumento de la concentracion de iones OH- y se puede establecer el siguiente equilibrio:
Especie
NH4+
OH-
<------>
Inicio
0.05
0.01
0.1
Reacciona
ξ
ξ
ξ
Equilibrio
0.05-ξ
0.01-ξ
0.1+ξ
NH3
H2O
La constante de equilibrio tiene exactamente el mismo valor (10 4.8), solamente se modificara la
expresion en funcion del avance de reaccion.
KR =
0.1 + ξ
= 104.8
(0.01 − ξ )(0.05 − ξ )
ξ = [OH − ] = 3.65 × 10− 5
10− 14
[H ] =
= 2.74 × 10− 10
−
[OH ]
pH = 9.56
+
Cambio en el pH al agregar HCl Si ahora se toma en cuenta que se agrega un acido fuerte, el
equilibrio se planteara de otra manera, con exceso de protones al inicio:
Especie
NH4+
Inicio
<------>
NH3
H+
0.05
0.1
0.01
Reacciona
ξ
ξ
ξ
Equilibrio
0.05-ξ
0.1+ξ
0.01-ξ
Aqui se empleara la expresion de la constante de acidez, por lo que:
Ka =
[ NH 3 ][ H + ] (0.1 + ξ )(0.01 − ξ )
=
= 10− 9.2
[ NH 4 + ]
(0.05 − ξ )
ξ = 9.99999962 × 10− 3
pH = 0.01 − ξ = − log[3.801893963 × 10− 10 ] = 9.42
3. La grafica que se ilustra en la figura 1 representa el cambio en el pH de una disolucion de NH3
de concentracion desconocida al valorarse con HCl 0.1 M. Con los datos de esta grafica, calcule
la concentracion de la disolucion de amoniaco. Diga la razon por la cual el punto de
equivalencia no se encuentra a un pH de 7.0
Resolucion: Supongamos que tenemos 100 mL de disolucion de NH 3. El punto de equivalencia se
encuentra en 50 mL de HCl agregado para la valoracion, de esta manera:
VaMa = VbMb
(50 mL)(0.10 M) = (100 mL) Mb
Resolviendo:Mb = 0.05 M
El punto de equivalencia no se encuentra a un pH=7.0, ya que la sal formada en la reaccion no es
neutra: (NH4Cl) es la sal que resulta de la reaccion de un acido fuerte (HCl) y de una base debil (NH3),
y se disocia e hidroliza para formar un acido debil (NH4+), lo que le da un caracter ligeramente acido.
4. Determine el pH que resulta de mezclar acido aminobencensulfonico SO3H(C6H4)NH2 0.2 M,
etano-1,2-ditiol HS-(C2H4)-SH 0.5 M y acido fosforico H3PO4 0.1 M.
Resolucion: Como primer paso se plantea una escala de pKa de las distintas especies presentes en la
disolucion,y se encierran en un recuadro.
HSO3
H2PO4-
H3PO4
2.2
H2PO4SO3-
SH
NH2
HS
SH
S
HPO4 2pKa
3.23
7.2
8.85
10.43
HPO4 2-
SH
S
12.3
PO4 3-
NH2
S
S
Se encuentran en la disolucion originalmente H3PO4, HS-(C2H4)-SH y SO3H(C6H4)NH2 . Todos son
acidos, que no reaccionan entre ellos, por lo que el pH estara determinado por el acido mas fuerte, en
este caso el H3PO4. Por lo tanto, determinando Ka/C0 = 10-2.2/0.1 = 0.063, se trara de un acido debil, por
lo que pH=1/2(pKa-logC0 )= 1/2[2.2-log(0.1)]= 1.6.
5. En el aparato digestivo encontramos una variedad de pH que va desde 1.2 a 3.0 en el estomago,
hasta cerca de 8 en el intestino delgado. Los siguientes farmacos se absorben de diferente
manera en el tracto digestivo y se ilustran en la figura 2: (a) aspirina (acido acetil salicilico), (b)
m-aminofenol y (c) paracetamol (p-hidroxiacetanilida). De acuerdo a las propiedades acidobase de estos medicamentos, prediga cual se absorbera mas eficientemente en el estomago, cual
en el intestino delgado y si alguno de ellos se absorbe con igual eficiencia en ambos organos.
Escriba las reacciones acido-base conjugada en cada caso.
R: Con los datos proporcionados, se puede escribir una escala de pKa con las medicamentos y su
respectiva base conjugada; uno de los farmacos es un anfolito y se delimita el pH del estomago y el del
intestino delgado en la misma escala. Analizando cada organo, tanto la aspirina como el paracetamol en
su forma neutra (sin ionizarse) predominan en un pH menor a sus respectivos pKa (4.5 y 9.5
respectivamente), por lo que la ambos se absorben eficientemente en el estomago, que tiene un pH<3.
Por otra parte el m-aminofenol, una base debil, en medio acido predomina en forma protonada
(ionizada) y no puede metabolizarse por la parte lipida de la mucosa y por tanto no se absorbe
eficientemente en el estomago. Pasando al intestino delgado, el paracetamol se absorbe eficientemente
tambien en este organo, porque solo podria ionizarse a partir de pH= 9.5 y no se dan las condiciones
fisiologicas para que esto ocurra, ya que pH =8.0 y asi continua prevaleciendo en forma neutra. La
aspirina, por otra parte, se encontraria ionizada a pH>4.5 y no podria absorberse por la mucosa lipida
del intestino. En el caso del m-aminofenol, como la especie neutra solo prevalece en el intervalo de
4.37<pH<9.82 , se puede absorber eficientemente en el intestino delgado, ya que pH=8.0.
OH
OH
O
m-aminofenol
(acido)
1.2-3.0
O
Intestino
O
O
NH
paracet
m-aminofenol
(base)
pKa
4.5
NH2
O
aspirina
4.37
HO
NH2
HO
NH3+
HO
Estomago
8.0
O
O
9.82
9.5
-O
NH2
O
O
O
NH
6. La nicotina es un liquido con sabor amargo, soluble en agua, tuvo usos como insecticida hace
algunas decadas, es uno de los alcaloides toxicos del tabaco y es el responsable de la adiccion
en el tabaquismo. Consta de un grupo piridina y otro metilpirrolidona unidos para formar una
base heteroaromatica N (Figura 3-3). Puede protonarse en medio acuoso y existir como dication
N2+ (Figura 3-1) o como monocation N+ (Figura 3-2). Dos profesores publicaron reportes de las
propiedades acido-base de la nicotina como ejemplos para ilustrar en el salon de clase de sus
cursos de licenciatura de los primeros semestres de las carreras de quimica [J. H. Summerfield,
J. Chem. Ed. (1999) 76, 1397; A. Ault, J. Chem. Ed. (2001) 78, 500]. En estos reportes se
deducen expresiones para determinar la fraccion mol de las tres especies de nicotina como
funcion de la concentracion de protones en medio acuoso (ecuaciones 6 a 17 del reporte de A.
Ault). Utilizando estas ecuaciones, trace en una misma grafica la variacion de la fraccion mol
de estas tres especies de nicotina como funcion del pH y compare sus resultados con los que se
reportan en la tabla 1 del articulo de Summerfield. Diga tambien porque solamente se utilizan
dos constantes de acidez para la deduccion de estas ecuaciones. Para facilidad, en la seccion
“Datos” al final del examen, se citan los valores de pKa de cada especie.
Resolucion: Las formulas de cada especie de nicotina y las ecuaciones de la fraccion mol se citan a
continuacion. Como se puede observar, la fraccion mol solo depende de [H+] y de dos Ka porque solo
hay dos equilibrios que involucran a tres especies. Utilizando las formulas se puede trazar la grafica.
pH
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
αN
9.99E-12
9.90E-10
9.09E-08
5.00E-06
9.09E-05
9.89E-04
9.89E-03
9.09E-02
5.00E-01
9.09E-01
9.90E-01
9.99E-01
1.00E+00
1.00E+00
1.00E+00
[H+ ]
1.00E+00
1.00E-01
1.00E-02
1.00E-03
1.00E-04
1.00E-05
1.00E-06
1.00E-07
1.00E-08
1.00E-09
1.00E-10
1.00E-11
1.00E-12
1.00E-13
1.00E-14
αN+
9.99E-04
9.90E-03
9.09E-02
5.00E-01
9.09E-01
9.89E-01
9.89E-01
9.09E-01
5.00E-01
9.09E-02
9.90E-03
9.99E-04
1.00E-04
1.00E-05
1.00E-06
αN2+
9.99E-01
9.90E-01
9.09E-01
5.00E-01
9.09E-02
9.89E-03
9.89E-04
9.09E-05
5.00E-06
9.09E-08
9.90E-10
9.99E-12
1.00E-13
1.00E-15
1.00E-17
1.0
0.9
α (Fraccion mol)
0.8
0.7
N
+
N
2+
N
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
6
7
pH
8
9
10
11
12
13
14
In the Classroom
The Acid–Base Chemistry of Nicotine:
Extensions, Analogies, and a Generalization
Addison Ault
Department of Chemistry, Cornell College, Mount Vernon, IA 52314; [email protected]
The acid–base chemistry of nicotine provides a nice example of equilibria in aqueous solution where the major form
in which the solute is actually present depends upon the pH
of the solution. As Summerfield points out (1), nicotine can
exist in aqueous solution either as the dication, N2+, the
monocation N1+, or the neutral molecule, N.
N+
H
N+
N
N+
CH3
H
N
CH 3
CH3
N
H
dication of nicotine
monocation of nicotine
nicotine
N 2+
pKa = pK1 = 3
N 1+
pKa = pK2 = 8
N
N1+ H+
N2+
(1)
and then eq 2, the equilibrium expression for the ionization
of N1+, to form the free base, N, and hydronium ion.
K2 =
N H+
N1+
(2)
Equation 1 can be rewritten as eq 3 to express the ratio
of the concentrations of N2+ and N1+,
N2+
N1+
500
H+
=
K1
+
N1+ H
=
N
K2
(4)
You can then see that the ratio of N2+ to N, the product
of the left sides of eqs 3 and 4, equals the product of the
right sides of eqs 3 and 4, as shown in eq 5.
N2+
N1+
×
+
H+
N1+ N2+ H
=
=
×
N
N
K2
K1
(5)
The Fractional Concentration of N
The question is: Which is the predominant form in an
aqueous solution at a particular pH? The intuitive answer is
that the dication, N2+, should predominate in strongly acidic
solutions, the neutral molecule, N, should predominate in
strongly basic solutions, and the monocation N1+ could predominate in solutions of intermediate pH. More precisely,
as we will show next, N2+ will predominate in solutions more
acidic than a pH of 3, the monocation N1+ will predominate
in solutions whose pH lies between 3 and 8, and the free base,
N, will predominate in solutions that are more basic than a
pH of 8. Furthermore, at a pH of 3 forms N2+ and N1+ will
be present in equal concentrations, and at a pH of 8 forms
N1+ and N will be present in equal concentrations.
Following Summerfield’s lead we will write an expression for the fractional concentration of the free base, N, as a
function of acid ionization constants and hydrogen ion concentrations, and then extend this treatment to the fractional
concentrations of N1+ and N2+.
We start by writing eq 1, the equilibrium expression for
the ionization of the dication, N2+, to form N1+ and hydronium ion,
K1 =
and eq 2 can be rewritten as eq 4 to express the ratio of the
concentrations of N1+ and N.
We now write in eq 6 the expression for the fractional
concentration of the free base, N, as the concentration of N
over the sum of the concentrations of N2+, N1+, and N, where
α N is read “fractional concentration of N”.
αN =
(6)
Dividing top and bottom by N gives eq 7.
1
αN =
(7)
2+
N
N1+
+
+1
N
N
Substituting from eqs 4 and 5 then gives the fractional concentration of the free base, N, as eq 8.
1
αN =
(8)
+
+
H
H+
H
×
+
+1
K2
K1
K2
Equation 8, written in this form, provides the insight
for understanding the variation of the fractional concentration
of the free base, N, as a function of the acidity of the solution. For N to be the major form in solution, its fractional
concentration must approach unity. The only way in which
this can happen is for the first two of the three terms of the
denominator of eq 8 to approach 0, so that the denominator
approaches 0 + 0 + 1 = 1 and the value of the fraction approaches
1. This can happen only when [H+] is small; that is, this will
be true only when the solution is somewhat basic. We will
see next that the solution must be more basic than pH 8 for
the free base to be the predominant form of nicotine.
Substituting the values for K1 and K2 for the dication of
nicotine into eq 8 gives eq 9,
αN =
(3)
N
N + N1+ + N
2+
H+
108
×
1
H+
103
+
Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu
H+
108
(9)
+1
In the Classroom
and from eq 9 we can see that when [H+] is exactly 108 the
fractional concentration of N will be 1/(0.00001 + 1 + 1) or
1/2, and that as [H+] becomes substantially less than 108 the
fractional concentration of N will approach 1.
Equation 16 indicates that the dicationic form of nicotine, N2+, can be the major species when [H+] is larger than
both K1 and K2; that is, when the solution is strongly acidic.
Substituting the values for K1 and K2 for the dication of
nicotine into eq 16 gives eq 17,
The Fractional Concentration of N1+
N1+
α N1+ =
N2+ + N1+ + N
Dividing top and bottom by N1+ gives eq 11.
1
α N1+ =
2+
N
+1+ N
1+
N
N1+
(10)
(11)
Substituting from eqs 3 and 4 then gives the fractional concentration of N1+ as eq 12.
1
α N 1+ =
(12)
+
K
H
+ 1 + 2+
K1
H
You can see from eq 12 that the only conditions under
which the fractional concentration of N1+ can approach unity
(the denominator of eq 12 approaches 1) is for [H+] to be
smaller than K1 and larger than K2.
Substituting the values for K1 and K2 for the dication of
nicotine into eq 12 gives eq 13,
1
α N 1+ = +
(13)
H
108
+1+ +
H
103
and from eq 13 we seen that for N1+ to be the major species
in solution [H+] must be between 103 and 108, or the pH
must be between 3 and 8.
The Fractional Concentration of N2+
Approaching the fractional concentration of N2+ in the same
way we first write its fractional concentration as in eq 14.
N2+
α N2+ =
N2+ + N1+ + N
Dividing top and bottom by N 2+ gives eq 15,
1
α N 2+ =
1+
N1+
N2+
(14)
(15)
+ N
N2+
α N 2+ =
1+
K1
H+
+
H+
×
K2
H+
(16)
(17)
3
1+
10
103 108
+ + × +
+
H
H
H
and from eq 17 we seen that for N 2+ to be the major species
in solution [H+] must be greater than 103; the pH of the
solution must be on the acidic side of 3. At a pH of exactly 3,
N2+ and N1+ will be present in equal concentrations.
The Other Conjugate Acid of Nicotine
The monocation N1+ is the predominant conjugate acid
of nicotine; the alternative monocation, N1+′ is also present,
but in lower concentration. The reason for this is that the
pyridine nitrogen, being sp2 hybridized, is more electronegative
and therefore less basic than the pyrrolidine nitrogen, which
is sp3 hybridized and therefore less electronegative and more
basic than the pyridine nitrogen (2):
N
N+
H
N
CH 3
N+
CH3
H
N 1+
pKa = 8
N 1+'
pKa = 3
Since the pKa values of N1+ and N1+′ differ by 5 powers
of ten, their concentrations will always differ by 5 powers of
ten, and the concentration of N1+ ′ will always be 1 × 105
times that of N1+. In solutions whose pH is between 3 and 8,
when N1+ is the major form in which nicotine is present, the
concentration of N1+ ′ will be 1 × 105. In solutions more
acidic than a pH of 3 or more basic than a pH of 8, the concentration of N1+′ will be 1 × 105 times that of the lower
fractional concentration of N1+. It is for these reasons that
the concentration of N1+′ could be ignored in the preceding
calculations.
Summary and Generalization
When a solute can exist in more than one form, say A,
B, C, …, the fractional concentration of any one form, say
A, can be expressed by an equation of the form of eq 18:
fractional concentration of A =
and substituting from eqs 3 and 5 then gives the fractional
concentration of N2+ as eq 16.
1
K1
1
α N 2+ =
We can think about the fractional concentration of N1+
in the same way. First, we write the fractional concentration
of N1+, α N 1+ , as in eq 10.
1
B
C
1+
+
+…
A
A
(18)
When a single equilibrium separates A from another
form, say B, the relative concentrations of A and B will be
determined by the ratio of the equilibrium constant that relates A and B, and the concentration of another species that
determines the relative amounts of A and B. This “other
JChemEd.chem.wisc.edu • Vol. 78 No. 4 April 2001 • Journal of Chemical Education
501
In the Classroom
species” is typically hydronium ion but could be something
else, such as the substrate or the inhibitor of an enzymatic
reaction, or anything else that enters into a rapidly reversible
equilibrium with A.
When two equilibria separate A from another form, say
C, the relative concentrations of A and C will be determined
by the product of two ratios of equilibrium constant and
concentration. Both of these possibilities were illustrated in
the nicotine example. You can determine the sense of the ratio,
whether it is equilibrium constant over concentration or concentration over equilibrium constant, either analytically, as
in the nicotine example, or through your chemical intuition
and experience.
Application to Amino Acids
H
O
C
C
H
N
+
O
H
H
H
O
R
C
C
H
N
H
−
O
H
+
H
O
R
C
C
H
N
+/-
A
A
pKa = pK2 = 10
A
pKa = pK1 = 2
−
O
H
H
+
-
Following the example of nicotine, we expect that A+
will be the predominant form of the amino acid present in
solutions more acidic than pH 2, the neutral zwitterionic
form A+/ will be the major form present in solutions whose
pH lies between 2 and 10, and the basic form, A, will be
the major form in solutions that are more basic than a pH of
10. The major neutral form will be the zwitterionic form A+/
because the carboxyl proton of A+, having a pKa of 2, is more
acidic than the ammonium proton, with its pKa of 10. The
alternative neutral uncharged form, A0, will never be present
at a high concentration.
H
O
R
C
C
H
N
H
+
−
O
O
O
H
H
O
O
O
+ NH
A
H
O
R
C
C
H
N
H
O
H
H
A+/-
A
O
O
−
O−
O
+ NH
OH
pKa = 10
aspartic acid; A+; major form at pH = 1
502
3
+/-
O
−
O
O−
O
3
NH2
A2 -
In solutions more acidic than pH 2, A+ is the predominant form; in solutions whose pH is between 2 and 4, A+/ is
the predominant form (the isoelectric pH is 3); in solutions
whose pH is between 4 and 10, A1 is the predominant form;
and in solutions more basic than pH 10, A2 is the predominant form.
Application to Enzyme Kinetics
We can express the rate of a simple enzyme-catalyzed
reaction by eq 19, in which kp is the rate constant for conversion of the enzyme–substrate complex, ES, to product, and
E0 is the total enzyme concentration.
rate = kp [E0](fraction of E0 present as ES)
(19)
The maximum rate is realized when the fractional concentration of ES approaches 1, and it is therefore essential to
understand what factors determine the fractional concentration of ES. In general we can imagine that the enzyme is
present as free enzyme, E, as the enzyme–substrate complex,
ES, as an enzyme–inhibitor complex, IE, and as a complex
of the enzyme with both substrate and inhibitor, IES. We can
therefore express in a general way the fractional concentration
of ES as shown in eq 20.
α ES =
ES
ES + E + IE + IES
α ES =
1
IE
IES
E
+
+
1+
ES
ES
ES
(20)
pKa = 2
(21)
Proceeding now as in ref 3, we can reexpress this fraction in
terms of concentrations and equilibrium constants as eq 22
O
+ NH
3
+ NH
Dividing top and bottom by [ES] gives eq 21.
HO
O
O−
O
3
+
A0
Since the two pKa values of A+ differ by 8 powers of ten,
0
A will always be 108 of the concentration of A+/. You can
easily extend this line of reasoning to amino acids that contain
either acidic or basic groups in the side chain. For aspartic
acid, for example,
pKa = 4
O
H
A 1-
The acid–base chemistry of amino acids follows the same
pattern. A typical “neutral” amino acid can exist in three
forms in aqueous solution: the cationic or “acidic” form A+,
the neutral zwitterionic form A+/, and the anionic or “basic”
form A.
R
the significant forms present in aqueous solutions are A+, A+/,
A1, and A2:
1
α ES =
1+
KM
S
+
KM
S
×
I
I
+
K2 K3
(22)
where KM, the Michaelis constant, is the substrate concentration at which the rate is half maximal in the absence of
inhibitor; K2 is the dissociation constant for the dissociation
Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu
In the Classroom
of I from IE; and K3 is the dissociation constant for the loss
of I from IES.
In the absence of inhibitor, I, the IE and IES terms in
eq 21 are zero. When this is so, high substrate concentration
can drive the E term to zero and the rate will be maximal.
If the inhibitor binds only to free enzyme (the IES term is
zero), the rate can still be driven to the original maximum
by a high substrate concentration, since the E and IE terms
in the denominator of eq 21 can both be driven to zero. This is
one of the characteristics of competitive inhibition. If, however,
I binds only to ES to give IES (IE does not form), high
substrate concentration cannot raise the rate to the maximum
that can be achieved in the absence of inhibitor because the
IES term in the denominator of eq 21 cannot be driven to zero
by high [I]; the fractional concentration of ES never becomes
equal to 1. This is one of the characteristics of uncompetitive
inhibition. Formation of both IE and IES accounts for the
kinetic behavior called noncompetitive inhibition. Reference 3
presents a more detailed analysis of these modes of inhibition.
Literature Cited
1. Summerfield, J. H. J. Chem. Educ. 1999, 76, 1397.
2. Loudon, G. M. Organic Chemistry, 3rd ed.; Benjamin/
Cummings: Redwood City, CA, 1995; pp 664, 1182.
3. Ault, A. J. Chem. Educ. 1974, 51, 381.
JChemEd.chem.wisc.edu • Vol. 78 No. 4 April 2001 • Journal of Chemical Education
503
In the Classroom
edited by
Applications and Analogies
Ron DeLorenzo
Middle Georgia College
Cochran, GA 31014
An Acid–Base Chemistry Example:
Conversion of Nicotine
John H. Summerfield
Department of Chemistry, Missouri Southern State College, Joplin, MO 64801; [email protected]
Much of the traditional general chemistry class focuses
on the nuts and bolts of chemical calculations. As a result it
is often difficult to infuse the class with timely chemistry
topics. This Journal has provided some excellent applications
over the years (1–8).
The current government interest in nicotine conversion
by cigarette companies (9) provides an example of acid–base
chemistry that can be explained to students in the second
college semester of general chemistry. The discussion of
acid–base chemistry tends to coincide with the students’
preregistration for the next semester’s organic chemistry class.
Organic chemistry has a notorious reputation. Thus students
tend to be anxious about this future class. One result of
this anxiety is that although the explanation for nicotine
conversion relies on organic acid–base chemistry, the students
are particularly interested because organic chemistry is already
on their minds. This example is also suitable for an AP high
school chemistry class, but is probably too advanced for an
introductory high school course.
The weak base ammonia is often added to cigarette
tobacco (10). The U.S. Food and Drug Administration (FDA)
has argued that this added ammonia enhances the delivery
of nicotine into the smoker’s bloodstream. In contrast, tobacco
companies argue that it is important to know as much as
possible about nicotine chemistry in order to provide smoker
satisfaction, and this knowledge is the underlying reason for
their interest in ammonia as an additive (11).
The Structure of Nicotine
H+
+
N
H
CH3
(2)
CH3
Methylpyrrolidine
When pyridine and methylpyrrolidine are bonded together
the new molecule is called nicotine. Nicotine occurs in a +2
form (1) when the nitrogens in both rings are protonated. If
only the methylpyrrolidine takes on a proton, a +1 form (2)
results. If both nitrogens are free to act as a base, the freebase form occurs (3).
H
H
H
+
+
N
H
CH3
H
+
N
H
N
N
CH3
CH3
N
N
1
2
3
Why the proton bonds to the methylpyrrolidine rather
than the pyridine in the +1 form is open to interpretation
(12). Rather than go off on this tangent, we simply accept that
the two nitrogens differ in acidity as indicated by their pKa
values. When protonated, the pyridine ring with a pK1 = 3.1
at 20 °C is much more acidic than the methylpyrrolidine ring
with pK2 = 8.0 at 20 °C.
The Nicotine–Ammonia Reaction
Nicotine is composed of two ring structures. One is a
benzene-like structure in which one carbon is replaced by
a nitrogen. This structure is called pyridine, and, in acidic
solution a proton adds to the nitrogen:
+ H+
+
N
H
N
+
N
(1)
We begin with the +2 form of nicotine. As ammonia is
added, the pH increases. With reduced hydrogen ion concentration, in accordance with LeChâtelier’s principle, the
pyridine ring loses its proton, changing the +2 ion to +1. As
the pH approaches eight, the proton on the methylpyrrolidine
ring is lost, changing the nicotine to the neutral free-base
form. The fraction of the nicotine that is in free-base form is
shown by Freiser (13) to be
Pyridine
The other portion of nicotine is a five-membered carbon ring
with one carbon replaced by a nitrogen. The nitrogen has a
methyl (– CH3) group attached to it. This molecule is called
methylpyrrolidine. In acidic solution, as with pyridine, a
proton adds to the nitrogen.
α=
K 1K 2
+
+ 2
(3)
K1K2 + K1 H + H
where K1 is the equilibrium constant for the loss of the first
proton (from the pyridine ring), K2 is the equilibrium constant
for the loss of the second proton (from the methylpyrrolidine
ring), and [H+] is the hydrogen ion concentration.
For our discussion the variables of interest are pH and pKa.
To bring out these quantities the numerator and denominator
JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education
1397
In the Classroom
of eq 3 are divided by K1K2 and then four substitutions are
made:
Nicotine Variable
[H+] = 10{pH
+ 2
Table 1. Free-Base Nicotine as a Function of pH
{2pH
[H ] = 10
(4)
K2 = 10{pK2
pH
6
7
8
α
0.0099
0.091
0.50
Free base/+1 form
0.010
0.10
1.0
9
0.91
10
K1K2 = 10{(pK1+pK2)
Equation 3 is now
{ pH
{ 2pH
α = 1 + 10{ pK + 10
10 2 10 { pK 1 +pK 2
{1
(5)
In Table 1 values for eq 5 are shown for different pH
values. Also, the ratio of free-base form to +1 form is calculated. From Table 1 we can see that the free-base form
becomes increasingly important as the pH increases. At pH =
8, a typical pH after ammonia has been added, free-base
nicotine accounts for half of the total nicotine content. The
nicotine–ammonia reaction is shown in eq 6.
H
H
+
+ NH3
N
H
N
N
CH3
+ NH4+
CH3
N
(6)
Nicotine
The free-base form is uncharged. As a result, it is able
to pass through cell membranes more easily than the +1 form
(14 ). Since the free-base form is more easily absorbed into
the cell than the +1 form, the conversion to the free-base form
using ammonia improves the delivery of nicotine to the
smoker.
1398
The FDA’s view of the part played by ammonia in tobacco
smoke is analogous to what takes place when cocaine is “freebased”. The hydrochloride salt of cocaine is mixed with aqueous ammonia. The product is the free-base form of cocaine,
which is then smoked (15).
Literature Cited
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Holme, T. A. J. Chem. Educ. 1994, 71, 919.
Dhawale, S. W. J. Chem. Educ. 1993, 70, 395.
Hecht, C. E. J. Chem. Educ. 1992, 69, 645.
McCullough, T. J. Chem. Educ. 1992, 69, 543.
Lisenky, G. J. Chem. Educ. 1990, 67, 562.
Fulkrod, J. E. J. Chem. Educ. 1985, 62, 529.
Mattice, J. J. Chem. Educ. 1983, 60, 1042.
Glanville, J.; Rau, E. J. Chem. Educ. 1973, 50, 65.
Raloff, J. Some Cigarette Makers Manipulate Nicotine. Science
News, July 2, 1994, p 7.
Kluger, R. Ashes to Ashes; Knopf: New York, 1996; pp 744–745.
Food and Drug Administration. Fed. Regist. 1995, 21(Aug 11),
801–804.
March, J. Advanced Organic Chemistry; Wiley Interscience: New
York, 1985; pp 234–235. Katritzky, A. R. Handbook of Heterocyclic Chemistry; Pergamon: New York, 1985; p 145.
Freiser, H. Concepts & Calculations in Analytical Chemistry: A
Spreadsheet Approach; CRC: Boca Raton, FL, 1992; pp 60–62.
Stryer, L. Biochemistry; Freeman: New York, 1988; p 284.
Inciardi, J. A. In The Epidemiology of Cocaine Use and Abuse; Res.
Monogr. 110; Schober S.; Schade, C., Eds.; U.S. Department of
Health and Human Services: Washington, DC, 1991; p 265.
Journal of Chemical Education • Vol. 76 No. 10 October 1999 • JChemEd.chem.wisc.edu