Question

Find the pH and concentration of each species of lysine in a solution of 0.010 0 M lysine*HCl, lysine monohydrochloride. The notation "lysine*HCl" refers to a neutral lysine molecule that has taken on one extra proton by addition of one mole of HCl. A more meaningful notation shows the salt (lysineH+)(Cl-) formed in the reaction.

Answer 1

lysineH+ + H2O <==> lysine + H3O+

let x amount has reacted

Ka2 = [lysine][H3O+]/[lysineH+]

8.71 x 10^-10 = x^2/0.01

x = [H3O+] = 2.95 x 10^-6 M

pH = -log[H3O+] = 5.53

[lysineH+] = 0.01 - 2.95 x 10^-6 = 0.01 M

lysine + H2O <==> lysine- + H3O+

let x amount reacted

Ka3 = 3 x 10^-11 = x^2/2.95 x 10^-6

x = [lysine-] = 3 x 10^-11 M

lysineH+ + H2O <==> lysineH2^2+ + OH-

Kb3 = Kw/Ka1 = [lysineH2^2+][OH-]/[lysineH+]

1 x 10^-14/6.92 x 10^-3 = x^2/0.01

x = [lysineH2^2+] = 1.20 x 10^-7 M