Sapling Learning macmillan learning Map .e structure of glycine is shown at the

Question

Sapling Learning macmillan learning Map .e structure of glycine is shown at the right along with the pK values of the -carboxyl and -amino groups pK,-9.60 H3N-CH2-C-OH pK,-2.34 Glycine is a diprotic acid which can exist in three ionized forms: H2A, HA°, and A-. (HA° consists primarily of the zwitterionic form with a negatively charged carboxyl group and positively charged amino group, but a very small amount of the completely neutral species does exist) Calculate the fraction (a) of each species present at pH 10.1. Number Number Number HA H,A

Explanation / Answer

Given pK1 and pK2, find K1 and K2 as

K1 = 10^(-pK1) = 0.004571

K2 = 10^(-pK2) = 2.5118*10-10

Also, pH of the solution is 10.1; therefore,

[H+] = 10^(-pH) = 10^(-10.1) = 7.94328*10-11

Use the empirical formula and plug in values to obtain

(H2A+) = [H+]2/[H+]2 + K1[H+] + K1K2 = (7.94328*10-11)2/[(7.94328*10-11)2 + (0.004571).(7.94328*10-11) + (0.004571).(2.511*10-10)] = (6.3*10-21)/(1.51*10-12) = 4.17*10-9 (ans).

(HA0) = K1[H+]/[H+]2 + K1[H+] + K1K2 = (0.004571).(7.94328*10-11)/(1.51*10-12) 0.24 (ans).

(A-) = K1K2/[H+]2 + K1[H+] + K1K2 = (0.004571).(2.511*10-10)/(1.51*10-12) = 0.76 (ans)

The sum will be slightly more than 1 and this is due to rounding off of digits (ans).

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