edited by: JOHNJ. ALEXANDER University of Cincinnati Cincinnati. Ohio 45221
exam queztion exchange Buffer pH in Electrophoresis Hugh A. Akers Lamar University Beaumont. TX 77710 T h e following question is appropriate for a n instrumental or biological chemistry course t h a t deals with the theory of electrophoresis. It requires intellectual behavior a t the applications level.
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:
Quesllon For the separation of lysine and arginine by (paper) electrophoresis, what buffer p H should b e selected for the best The equation above describes a gencml relntronship that can be resolution? applied tu uther "pairs" of compounds, eg., fmntr acid-acetic arid. Given Information: Lysine, H O Z C C H ( N H Z ) ( C H ~ ) ~ N H ~ , Iwlic acid-propionicwid, or a ~ p a racid-glutamic t~ acid. pK1, 2.18(COzH), pKz 8.95 ( a NH2), pK3 10.53 ( r NH2); arginine, 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 electrophoresis there must be a difference in the net charge on the compounds. 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-Hasselhalchequation pH
= pK
[A-I + 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:
..
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 Ul =
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 Solving far IH+I:
+ K3d2
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 formula learned early in freshman chemistry (number of neutrons = 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 a s test, the student's understanding of the concepts involved. Part or all of the accompanying 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 t h e neutron and of the proton plus a n electron.) (b) Show t h a 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 a r 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
