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

Especie HClO <------> H+ ClO-

Inicio 0.2 0 0Reacciona ξ ξ ξEquilibrio 0.2-ξ ξ ξ

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:

Tambien se puede determinar utilizando la ecuacion de Henderson-Hasselbalch, sustituyendo las concentraciones al equilibrio de las especies involucradas de la siguiente forma:

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.

( ) 2528

8.3 10[ ][ ] 3.45 10[ ] 0.2 0.199917a

xH ClOK xHClO

ξξ

−+ −−= = = =

−

5

7.4617 8

[ ]log[ ]

[ ]log[ ][ ] [8.3 10 ]log 4.08 log 7.4617[ ] [0.199917]

10 3.45 10

Base conjugadapH pKaAcido

ClOpH pKaHClOClO xpKa pHHClO

Ka x

−

− −

− −

= +

= +

= − = − =

= =

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- <------> NH3 H2O

Inicio 0.05 0 0.1Reacciona ξ ξ ξEquilibrio 0.05-ξ ξ 0.1+ξ

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:

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:

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- <------> NH3 H2O

Inicio 0.05 0.01 0.1Reacciona ξ ξ ξEquilibrio 0.05-ξ 0.01-ξ 0.1+ξ

La constante de equilibrio tiene exactamente el mismo valor (104.8), solamente se modificara la expresion en funcion del avance de reaccion.

4.83 314 14

4 4

4.8

5

1410

[ ] [ ][ ] 10[ ][ ] 10 [ ] 10

0.1 10(0.05 )

[ ] 3.17 1010[ ] 3.15 10

[ ]9.5

R

R

NH NH H KaKNH OH NH

K

OH

HOH

pH

ξξ ξ

ξ

+

+ − − + −

− −

−+ −

−

= = = =

+= =−

= = ×

= = ×

=

3

4

[ ] 0.1log 9.2 log 9.5[ ] 0.05

NHpH pKaNH += + = + =

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+ <------> NH3 H+

Inicio 0.05 0.1 0.01Reacciona ξ ξ ξEquilibrio 0.05-ξ 0.1+ξ 0.01-ξ

Aqui se empleara la expresion de la constante de acidez, por lo que:

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

4.8

5

1410

0.1 10(0.01 )(0.05 )

[ ] 3.65 1010[ ] 2.74 10

[ ]9.56

RK

OH

HOH

pH

ξξ ξ

ξ − −

−+ −

−

+= =− −

= = ×

= = ×

=

9.23

43

10

[ ][ ] (0.1 )(0.01 ) 10[ ] (0.05 )

9.99999962 100.01 log[3.801893963 10 ] 9.42

NH HKaNH

pH

ξ ξξ

ξξ

+−

+

−

−

+ −= = =−

= ×= − = − × =

Resolucion: Supongamos que tenemos 100 mL de disolucion de NH3. 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.

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 acido-base 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.

H3PO4

HSO3 NH2

SO3- NH2

H2PO4-

SH

SH

S

SH

H2PO4-

HPO4 2-

S

SH

S

S

HPO4 2-

PO4 3-

2.2 3.23 7.2 8.85 10.43 12.3pKa

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.

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.

O O

O

O

NH2OH

NH3+OH

OH

NHO

O

NHO

NH2-O

O OH

O

O

NH2OH

aspirinam-aminofenol (acido) paracet

4.37 4.59.5 9.82

m-aminofenol (base)

IntestinoEstomago

pKa8.01.2-3.0

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.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 140.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

α (F

raccio

n m

ol)

pH

N N+

N2+

pH0 1.00E+00 9.99E-12 9.99E-04 9.99E-011 1.00E-01 9.90E-10 9.90E-03 9.90E-012 1.00E-02 9.09E-08 9.09E-02 9.09E-013 1.00E-03 5.00E-06 5.00E-01 5.00E-014 1.00E-04 9.09E-05 9.09E-01 9.09E-025 1.00E-05 9.89E-04 9.89E-01 9.89E-036 1.00E-06 9.89E-03 9.89E-01 9.89E-047 1.00E-07 9.09E-02 9.09E-01 9.09E-058 1.00E-08 5.00E-01 5.00E-01 5.00E-069 1.00E-09 9.09E-01 9.09E-02 9.09E-0810 1.00E-10 9.90E-01 9.90E-03 9.90E-1011 1.00E-11 9.99E-01 9.99E-04 9.99E-1212 1.00E-12 1.00E+00 1.00E-04 1.00E-1313 1.00E-13 1.00E+00 1.00E-05 1.00E-1514 1.00E-14 1.00E+00 1.00E-06 1.00E-17

[H+] αN αN+ αN2+

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500 Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu

The Acid–Base Chemistry of Nicotine:Extensions, Analogies, and a GeneralizationAddison AultDepartment of Chemistry, Cornell College, Mount Vernon, IA 52314; aault@cornell-iowa.edu

The acid–base chemistry of nicotine provides a nice ex-ample of equilibria in aqueous solution where the major formin which the solute is actually present depends upon the pHof the solution. As Summerfield points out (1), nicotine canexist in aqueous solution either as the dication, N2+, themonocation N1+, or the neutral molecule, N.

dication of nicotine

+

+

N

N

H

monocation of nicotine

+

N3

CH3HN

CHH

N2+ N1+

pKa = pK1 = 3 pKa = pK2 = 8

nicotine

N

N

N

CH3

The question is: Which is the predominant form in anaqueous solution at a particular pH? The intuitive answer isthat the dication, N2+, should predominate in strongly acidicsolutions, the neutral molecule, N, should predominate instrongly basic solutions, and the monocation N1+ could pre-dominate in solutions of intermediate pH. More precisely,as we will show next, N2+ will predominate in solutions moreacidic than a pH of 3, the monocation N1+ will predominatein solutions whose pH lies between 3 and 8, and the free base,N, will predominate in solutions that are more basic than apH of 8. Furthermore, at a pH of 3 forms N2+ and N1+ willbe present in equal concentrations, and at a pH of 8 formsN1+ and N will be present in equal concentrations.

Following Summerfield’s lead we will write an expres-sion for the fractional concentration of the free base, N, as afunction of acid ionization constants and hydrogen ion con-centrations, and then extend this treatment to the fractionalconcentrations of N1+ and N2+.

We start by writing eq 1, the equilibrium expression forthe ionization of the dication, N2+, to form N1+ and hydro-nium ion,

K1 =

N1+ H+

N2+(1)

and then eq 2, the equilibrium expression for the ionizationof 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 ratioof the concentrations of N2+ and N1+,

N2+

N1+=

H+

K1(3)

and eq 2 can be rewritten as eq 4 to express the ratio of theconcentrations of N1+ and N.

N1+

N=

H+

K2

(4)

You can then see that the ratio of N2+ to N, the productof the left sides of eqs 3 and 4, equals the product of theright sides of eqs 3 and 4, as shown in eq 5.

N2+

N1+× N1+

N=

N2+

N=

H+

K2

×H+

K1

(5)

The Fractional Concentration of N

We now write in eq 6 the expression for the fractionalconcentration of the free base, N, as the concentration of Nover the sum of the concentrations of N2+, N1+, and N, whereαN is read “fractional concentration of N”.

αN = N

N2+ + N1+ + N(6)

Dividing top and bottom by N gives eq 7.

αN =1

N2+

N+

N1+

N+ 1

(7)

Substituting from eqs 4 and 5 then gives the fractional con-centration of the free base, N, as eq 8.

αN =1

H+

K2

×H+

K1

+H+

K2

+ 1(8)

Equation 8, written in this form, provides the insightfor understanding the variation of the fractional concentrationof the free base, N, as a function of the acidity of the solu-tion. For N to be the major form in solution, its fractionalconcentration must approach unity. The only way in whichthis can happen is for the first two of the three terms of thedenominator of eq 8 to approach 0, so that the denominatorapproaches 0 + 0 + 1 = 1 and the value of the fraction approaches1. This can happen only when [H+] is small; that is, this willbe true only when the solution is somewhat basic. We willsee next that the solution must be more basic than pH 8 forthe free base to be the predominant form of nicotine.

Substituting the values for K1 and K2 for the dication ofnicotine into eq 8 gives eq 9,

αN =1

H+

10�8×

H+

10�3+

H+

10�8+ 1

(9)

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and from eq 9 we can see that when [H+] is exactly 10�8 thefractional concentration of N will be 1/(0.00001 + 1 + 1) or1/2, and that as [H+] becomes substantially less than 10�8 thefractional concentration of N will approach 1.

The Fractional Concentration of N1+

We can think about the fractional concentration of N1+

in the same way. First, we write the fractional concentrationof N1+, αN1+ , as in eq 10.

αN1+ =N1+

N2+ + N1+ + N(10)

Dividing top and bottom by N1+ gives eq 11.

αN1+ =1

N2+

N1++ 1 + N

N1+

(11)

Substituting from eqs 3 and 4 then gives the fractional con-centration of N1+ as eq 12.

αN1+ =1

H+

K1

+ 1 +K2

H+

(12)

You can see from eq 12 that the only conditions underwhich the fractional concentration of N1+ can approach unity(the denominator of eq 12 approaches 1) is for [H+] to besmaller than K1 and larger than K2.

Substituting the values for K1 and K2 for the dication ofnicotine into eq 12 gives eq 13,

αN1+ =1

H+

10�3+ 1 +

10�8

H+

(13)

and from eq 13 we seen that for N1+ to be the major speciesin solution [H+] must be between 10�3 and 10�8, or the pHmust be between 3 and 8.

The Fractional Concentration of N2+

Approaching the fractional concentration of N2+ in the sameway we first write its fractional concentration as in eq 14.

αN2+ =N2+

N2+ + N1+ + N(14)

Dividing top and bottom by N2+ gives eq 15,

αN 2+ =1

1 +N1+

N2++ N

N2+

(15)

and substituting from eqs 3 and 5 then gives the fractionalconcentration of N2+ as eq 16.

αN 2+ =1

1 +K1

H+ +K1

H+ ×K2

H+

(16)

Equation 16 indicates that the dicationic form of nico-tine, N2+, can be the major species when [H+] is larger thanboth K1 and K2; that is, when the solution is strongly acidic.

Substituting the values for K1 and K2 for the dication ofnicotine into eq 16 gives eq 17,

αN 2+ =1

1 +10�3

H+ +10�3

H+ × 10�8

H+

(17)

and from eq 17 we seen that for N2+ to be the major speciesin solution [H+] must be greater than 10�3; the pH of thesolution 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 acidof nicotine; the alternative monocation, N1+′ is also present,but in lower concentration. The reason for this is that thepyridine nitrogen, being sp2 hybridized, is more electronegativeand therefore less basic than the pyrrolidine nitrogen, whichis sp3 hybridized and therefore less electronegative and morebasic than the pyridine nitrogen (2):

+N

H

N1+ N1+'

pKa = 8 pKa = 3

+

N3

N

CHH

N

CH3

Since the pKa values of N1+ and N1+′ differ by 5 powersof ten, their concentrations will always differ by 5 powers often, and the concentration of N1+′ will always be 1 × 10�5

times that of N1+. In solutions whose pH is between 3 and 8,when N1+ is the major form in which nicotine is present, theconcentration of N1+′ will be 1 × 10�5. In solutions moreacidic than a pH of 3 or more basic than a pH of 8, the con-centration of N1+′ will be 1 × 10�5 times that of the lowerfractional concentration of N1+. It is for these reasons thatthe concentration of N1+′ could be ignored in the precedingcalculations.

Summary and Generalization

When a solute can exist in more than one form, say A,B, C, …, the fractional concentration of any one form, sayA, can be expressed by an equation of the form of eq 18:

fractional concentration of A = 1

1 +B

A+

C

A+ …

(18)

When a single equilibrium separates A from anotherform, say B, the relative concentrations of A and B will bedetermined by the ratio of the equilibrium constant that re-lates A and B, and the concentration of another species thatdetermines the relative amounts of A and B. This “other

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502 Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu

species” is typically hydronium ion but could be somethingelse, such as the substrate or the inhibitor of an enzymaticreaction, or anything else that enters into a rapidly reversibleequilibrium with A.

When two equilibria separate A from another form, sayC, the relative concentrations of A and C will be determinedby the product of two ratios of equilibrium constant andconcentration. Both of these possibilities were illustrated inthe nicotine example. You can determine the sense of the ratio,whether it is equilibrium constant over concentration or con-centration over equilibrium constant, either analytically, asin the nicotine example, or through your chemical intuitionand experience.

Application to Amino Acids

The acid–base chemistry of amino acids follows the samepattern. A typical “neutral” amino acid can exist in threeforms in aqueous solution: the cationic or “acidic” form A+,the neutral zwitterionic form A+/�, and the anionic or “basic”form A�.

+

−

+

R C

H

NH H

H

C

O

O H R C

H

NH H

H

C

O

O−

R C

H

H

H

C

O

O

N

A+

pKa = pK1 = 2

A+/- A-

pKa = pK2 = 10

Following the example of nicotine, we expect that A+

will be the predominant form of the amino acid present insolutions more acidic than pH 2, the neutral zwitterionicform A+/� will be the major form present in solutions whosepH lies between 2 and 10, and the basic form, A�, will bethe major form in solutions that are more basic than a pH of10. The major neutral form will be the zwitterionic form A+/�

because the carboxyl proton of A+, having a pKa of 2, is moreacidic than the ammonium proton, with its pKa of 10. Thealternative neutral uncharged form, A0, will never be presentat a high concentration.

+

−R C

H

NH H

H

C

O

O R C

H

H

HO

H

C

O

N

A+/- A0

Since the two pKa values of A+ differ by 8 powers of ten,A0 will always be 10�8 of the concentration of A+/�. You caneasily extend this line of reasoning to amino acids that containeither acidic or basic groups in the side chain. For asparticacid, for example,

aspartic acid; A+; major form at pH = 1

+

HOOH

O

NH3O

pKa = 4pKa = 2

pKa = 10

the significant forms present in aqueous solutions are A+, A+/�,A1�, and A2�:

++

O

NH3O

O −

O

NH3O

A+ A+/-

OHOHHO

+

− OO −

O

NH3O

− OO −

O

NH2O

A1- A2-

In solutions more acidic than pH 2, A+ is the predomi-nant form; in solutions whose pH is between 2 and 4, A+/� isthe predominant form (the isoelectric pH is 3); in solutionswhose pH is between 4 and 10, A1� is the predominant form;and in solutions more basic than pH 10, A2� is the predomi-nant form.

Application to Enzyme Kinetics

We can express the rate of a simple enzyme-catalyzedreaction by eq 19, in which kp is the rate constant for con-version of the enzyme–substrate complex, ES, to product, andE0 is the total enzyme concentration.

rate = kp [E0](fraction of E0 present as ES) (19)

The maximum rate is realized when the fractional con-centration of ES approaches 1, and it is therefore essential tounderstand what factors determine the fractional concentra-tion of ES. In general we can imagine that the enzyme ispresent as free enzyme, E, as the enzyme–substrate complex,ES, as an enzyme–inhibitor complex, IE, and as a complexof the enzyme with both substrate and inhibitor, IES. We cantherefore express in a general way the fractional concentrationof ES as shown in eq 20.

αES =ES

ES + E + IE + IES(20)

Dividing top and bottom by [ES] gives eq 21.

αES =1

1 +E

ES+

IE

ES+

IES

ES

(21)

Proceeding now as in ref 3, we can reexpress this fraction interms of concentrations and equilibrium constants as eq 22

αES =1

1 +KM

S+

KM

S×

I

K2

+I

K3

(22)

where KM, the Michaelis constant, is the substrate concen-tration at which the rate is half maximal in the absence ofinhibitor; K2 is the dissociation constant for the dissociation

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of I from IE; and K3 is the dissociation constant for the lossof I from IES.

In the absence of inhibitor, I, the IE and IES terms ineq 21 are zero. When this is so, high substrate concentrationcan drive the E term to zero and the rate will be maximal.If the inhibitor binds only to free enzyme (the IES term iszero), the rate can still be driven to the original maximumby a high substrate concentration, since the E and IE termsin the denominator of eq 21 can both be driven to zero. This isone of the characteristics of competitive inhibition. If, however,I binds only to ES to give IES (IE does not form), highsubstrate concentration cannot raise the rate to the maximumthat can be achieved in the absence of inhibitor because the

IES term in the denominator of eq 21 cannot be driven to zeroby high [I]; the fractional concentration of ES never becomesequal to 1. This is one of the characteristics of uncompetitiveinhibition. Formation of both IE and IES accounts for thekinetic behavior called noncompetitive inhibition. Reference 3presents 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.

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JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education 1397

An Acid–Base Chemistry Example:Conversion of Nicotine

John H. SummerfieldDepartment of Chemistry, Missouri Southern State College, Joplin, MO 64801; summerfield-j@mail.mssc.edu

Much of the traditional general chemistry class focuseson the nuts and bolts of chemical calculations. As a result itis often difficult to infuse the class with timely chemistrytopics. This Journal has provided some excellent applicationsover the years (1–8).

The current government interest in nicotine conversionby cigarette companies (9) provides an example of acid–basechemistry that can be explained to students in the secondcollege semester of general chemistry. The discussion ofacid–base chemistry tends to coincide with the students’preregistration for the next semester’s organic chemistry class.Organic chemistry has a notorious reputation. Thus studentstend to be anxious about this future class. One result ofthis anxiety is that although the explanation for nicotineconversion relies on organic acid–base chemistry, the studentsare particularly interested because organic chemistry is alreadyon their minds. This example is also suitable for an AP highschool chemistry class, but is probably too advanced for anintroductory high school course.

The weak base ammonia is often added to cigarettetobacco (10). The U.S. Food and Drug Administration (FDA)has argued that this added ammonia enhances the deliveryof nicotine into the smoker’s bloodstream. In contrast, tobaccocompanies argue that it is important to know as much aspossible about nicotine chemistry in order to provide smokersatisfaction, and this knowledge is the underlying reason fortheir interest in ammonia as an additive (11).

The Structure of Nicotine

Nicotine is composed of two ring structures. One is abenzene-like structure in which one carbon is replaced bya nitrogen. This structure is called pyridine, and, in acidicsolution a proton adds to the nitrogen:

H

Pyridine

+

++

NHN (1)

The other portion of nicotine is a five-membered carbon ringwith one carbon replaced by a nitrogen. The nitrogen has amethyl (–CH3) group attached to it. This molecule is calledmethylpyrrolidine. In acidic solution, as with pyridine, aproton adds to the nitrogen.

++

N

H CH3

N

CH3

H +

Methylpyrrolidine

(2)

When pyridine and methylpyrrolidine are bonded togetherthe new molecule is called nicotine. Nicotine occurs in a +2form (1) when the nitrogens in both rings are protonated. Ifonly the methylpyrrolidine takes on a proton, a +1 form (2)results. If both nitrogens are free to act as a base, the free-base form occurs (3).

+

++

H

N

N

H CH3N

N

HH

H CH3

N

N

H

CH3

1 2 3

Why the proton bonds to the methylpyrrolidine ratherthan the pyridine in the +1 form is open to interpretation(12). Rather than go off on this tangent, we simply accept thatthe two nitrogens differ in acidity as indicated by their pKavalues. When protonated, the pyridine ring with a pK1 = 3.1at 20 °C is much more acidic than the methylpyrrolidine ringwith pK2 = 8.0 at 20 °C.

The Nicotine–Ammonia Reaction

We begin with the +2 form of nicotine. As ammonia isadded, the pH increases. With reduced hydrogen ion con-centration, in accordance with LeChâtelier’s principle, thepyridine ring loses its proton, changing the +2 ion to +1. Asthe pH approaches eight, the proton on the methylpyrrolidinering is lost, changing the nicotine to the neutral free-baseform. The fraction of the nicotine that is in free-base form isshown by Freiser (13) to be

α =K1K2

K1K2 + K1 H+ + H+ 2(3)

where K1 is the equilibrium constant for the loss of the firstproton (from the pyridine ring), K2 is the equilibrium constantfor the loss of the second proton (from the methylpyrrolidinering), 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

Applications and Analogiesedited by

Ron DeLorenzoMiddle Georgia College

Cochran, GA 31014

In the Classroom

1398 Journal of Chemical Education • Vol. 76 No. 10 October 1999 • JChemEd.chem.wisc.edu

of eq 3 are divided by K1K2 and then four substitutions aremade:

[H+] = 10{pH

[H+]2 = 10{2pH

K2 = 10{pK2(4)

K1K2 = 10{(pK1+pK2)

Equation 3 is now

α = 1 + 10 { pH

10 { pK 2+ 10 {2pH

10 { pK 1+pK 2

{1

(5)

In Table 1 values for eq 5 are shown for different pHvalues. Also, the ratio of free-base form to +1 form is cal-culated. From Table 1 we can see that the free-base formbecomes increasingly important as the pH increases. At pH =8, a typical pH after ammonia has been added, free-basenicotine accounts for half of the total nicotine content. Thenicotine–ammonia reaction is shown in eq 6.

NH3 NH4++

+N

N

H

H CH3

+N

N

H

CH3

Nicotine

(6)

The free-base form is uncharged. As a result, it is ableto pass through cell membranes more easily than the +1 form(14 ). Since the free-base form is more easily absorbed intothe cell than the +1 form, the conversion to the free-base formusing ammonia improves the delivery of nicotine to thesmoker.

The FDA’s view of the part played by ammonia in tobaccosmoke is analogous to what takes place when cocaine is “free-based”. The hydrochloride salt of cocaine is mixed with aque-ous ammonia. The product is the free-base form of cocaine,which is then smoked (15).

Literature Cited

1. Holme, T. A. J. Chem. Educ. 1994, 71, 919.2. Dhawale, S. W. J. Chem. Educ. 1993, 70, 395.3. Hecht, C. E. J. Chem. Educ. 1992, 69, 645.4. McCullough, T. J. Chem. Educ. 1992, 69, 543.5. Lisenky, G. J. Chem. Educ. 1990, 67, 562.6. Fulkrod, J. E. J. Chem. Educ. 1985, 62, 529.7. Mattice, J. J. Chem. Educ. 1983, 60, 1042.8. Glanville, J.; Rau, E. J. Chem. Educ. 1973, 50, 65.9. Raloff, J. Some Cigarette Makers Manipulate Nicotine. Science

News, July 2, 1994, p 7.10. Kluger, R. Ashes to Ashes; Knopf: New York, 1996; pp 744–745.11. Food and Drug Administration. Fed. Regist. 1995, 21(Aug 11),

801–804.12. March, J. Advanced Organic Chemistry; Wiley Interscience: New

York, 1985; pp 234–235. Katritzky, A. R. Handbook of Heterocy-clic Chemistry; Pergamon: New York, 1985; p 145.

13. Freiser, H. Concepts & Calculations in Analytical Chemistry: ASpreadsheet Approach; CRC: Boca Raton, FL, 1992; pp 60–62.

14. Stryer, L. Biochemistry; Freeman: New York, 1988; p 284.15. Inciardi, J. A. In The Epidemiology of Cocaine Use and Abuse; Res.

Monogr. 110; Schober S.; Schade, C., Eds.; U.S. Department ofHealth and Human Services: Washington, DC, 1991; p 265.

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