Question

4. (4 points) Lysine is triprotic amino acid (separately loses three H’s in solution). The pKa's are 2.18, 8.95, and 10.53. a. Write the three ionization equilibrium reactions for lysine (C6H5N2O2). pk = 10.5 NH, CH2 CH2 b. Determine the concentration of each species in a 0.10 M solution. CH2 CH, *H3N-CH- pk = 90 O pk, -22

I have absolutely no idea how to do this question, or where even to start. There is no other information given for the question to solve. kw=1.0x10-14

Answer 1

Part a.

The three ionization equilibrium reactions for lysine can be written as follows.

(i) COOH <----> COO^-; (C-terminus) pK = 2.18

(ii) NH3⁺ <----> NH₂; (Side chain) pK = 10.53

(iii) NH3⁺ <----> NH₂; (N-Terminus) pK = 8.95