Map Sapling Learning macmillan learning The structure of phenylalanine is shown

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 HA

Explanation / 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)

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