edited by:

exam queztion exchange JOHN J. ALEXANDER University of Cincinnati Cincinnati. Ohio 45221

Buffer pH in Electrophoresis Hugh A. Akers Lamar University Beaumont. TX 77710

The following question is appropriate for an instrumental or biological chemistry course tha t deals with the theory of electrophoresis. It requires intellectual behavior a t the ap- plications level.

Quesllon

For the separation of lysine and arginine by (paper) elec- trophoresis, what buffer p H should be selected for the best resolution?

Given Information: Lysine, H O Z C C H ( N H Z ) ( C H ~ ) ~ N H ~ , pK1, 2.18(COzH), pKz 8.95 ( a NH2), pK3 10.53 ( r NH2); ar- ginine, H02CCH(NH2)(CH2)3NHC(NH)NH2, pK1, 2.17 (COzH), pK2 9.04 ( a NH2), pK3 12.48 (guanidinium).

Acceptable Solutlon

Graphic Solution: For two substances to be separated by electro- phoresis there must be a difference in the net charge on the com- pounds. An examination of the pK's and predominant ionic forms for lysine and arginine reveals that there is negligible difference between the net charge of lysine and arginine below pH 9; however, above pH 9 a difference in charge is apparent.

Using the Henderson-Hasselhalch equation

[A-I pH = pK + log- [HA1

and considering only the third pK's an inspection of a plot of the net charge, [Ad]I([HA] + [A-I), versus pH from 9.5 to 13.5 for lysine and arginine reveals the greatest difference in net charge around pH .. . 11.5.

Rigorous Solution: The electrophoretic mobility, u, of a substance is a function of the geometry and mass of the substance, the viscosity of the medium. and the electrical ootential aoolied. The net mobilitv. . . , . L ' , is a pn,durt uf theelectn,phu;etir md5lity and the net charge.

Fur example, fur iysine, when the pH is near pKv:

Ul = UIILYS-I = urK31 [Lys0I + [LYs-I IH+I + K31

For a separation we are interested in the difference between the net mobilities of lysine and arginine or

- - urKar - U=K, [H+l + Ksr IHfl + Kao

Combining terms and taking the natural Logarithm, we have:

Differentiating (Ur - U.) with respect to [W], setting equal to zero (to determine the [H+] that produces the maximum difference in net mobilities), and combining terms

urK3r([H+l + K&)% = u.K%([H+l + K3d2

Solving far IH+I:

Since the molecular shapes and masses of lysine and arginine are comparable, ur should be approximately equal to u,, and the second term in the last equation would reduce to zero and:

The equation above describes a gencml relntronship that can be applied tu uther "pairs" of compounds, eg., fmntr acid-acetic arid. Iwlic acid-propionic wid, or a ~ p a r t ~ acid-glutamic acid.

Finding the Number of Neutrons when the Mass Loss Exceeds 1 AMU

Thomas P. Chlrplch Memphis State University Memphis, TN 38152

This question tests the student's ability to reconcile a for- mula learned early in freshman chemistry (number of neu- trons = mass number minus atomic number) witb an apparent conflict arising from a phenomenon described later in nuclear chemistry (loss of mass when protons and neutrons form a nucleus). The goal is to increase, as well as test, the student's understanding of the concepts involved. Part or all of the ac- companying hint may be included, depending on the level of the students.

Question

When protons and neutrons combine t o form a helium nucleus, there is a mass loss of 0.031 amu. When they combine to form the heavy atom, ipBi, the mass lost is 1.7 amu. (a) Show why this loss does not invalidate the rule that the number of neutrons can be found hy subtracting the atomic number from the mass number. (Hint: Calculate the mass loss per nucleon and compare this witb the mass of the neutron and of the proton plus an electron.) (b) Show tha t the rule would not work if there were no mass loss.

Acceptable Solution

(a) The mass loss per nucleon is

1.7 amu = 0.0081 arndnucleon

209 nucleons

Smce the mass uf a neutron is 1 .U087 amu nnd that uf a protun plus an electron is I.Ou79 am", the n n a s s ~ after the maas loss ar t even chraer to I amu and the ttml mass i~ quite rime to tho sum of the number of neutrons and protons.

(b) If there were no mass loss, the rule for finding the number of neutrons would not work for heavier atoms. In the case of z3Bi, the mass of the atom would then be (83 X 1.0079) + (126 X 1.0087) = 210.8 amu (alternative calculation: 209 + 1.7 = 211), which would give an incorrect number for the neutrons (211 - 83 = 128). The heavier the atom, the greater the discrepancy.

1014 Journal of Chemical Education