Map Sapling Learning macmillan learning The structure of phenylalanine is shown
ID: 103653 • Letter: M
Question
Map Sapling Learning macmillan learning The structure of phenylalanine is shown at the right along with the pK values of the -carboxyl and -amino groups pK,-9.13 H3N-CH-C-OH pK,-1.83 Phenylalanine is a diprotic acid which can exist in three ionized forms: H2AT, HA°, and AT. (HA consists primarily of the zwitterionic form witha negatively charged carboxyl group and positively charged amino group, but a very small amount of the completely neutral species does exist) CH2 Calculate the fraction (a) of each species present at pH 1.70. Number Number Number H,A HAExplanation / Answer
Given pK1 and pK2, find K1 and K2 as
K1 = 10^(-pK1) = 0.001479
K2 = 10^(-pK2) = 7.4*10-10
Also, pH of the solution is 1.7; therefore,
[H+] = 10^(-pH) = 10^(-1.7) = 0.01995
Use the empirical formula and plug in values to obtain
(H2A+) = [H+]2/[H+]2 + K1[H+] + K1K2 = (0.01995)2/[(0.01995)2 + (0.001479).(0.01995) + (0.001479).( 7.4*10-10)] = (3.98*10-4)/(4.275*10-4) =0.931 (ans).
(HA0) = K1[H+]/[H+]2 + K1[H+] + K1K2 = (0.001479).(0.01995)/(4.275*10-4) 0.069 (ans).
(A-) = K1K2/[H+]2 + K1[H+] + K1K2 = (0.001479).( 7.4*10-10)/(4.275*10-4) = 2.56*10^-9 (ans)