Question

Calculate the pH of a 0.0148 M solution of arginine hydrochloride (arginine·HCl, H2Arg ). Arginine has pKa values of 1.823 (pKa1), 8.991 (pKa2), and 12.01 (pKa3). Calculate the concentration of each species of arginine in the solution. H3Arg+2, H2Arg+, HArg, Arg-

Answer 1

H2Arg+ + H2O <==> HArg + H3O+

let x amount has reacted

pKa = -logKa

Ka2 = [H3O+][HArg]/[H2Arg+]

1.021 x 10^-9 = x^2/0.0148

x = [H3O+] =3.89 x 10^-6 M

pH = -log[H3O+] = 5.41

[H2Arg+] = 0.0148 - 3.89 x 10^-6 = 0.0148 M

HArg + H2O <==> Arg- + H3O+

let x amount reacted

Ka3 = [Arg-][H3O+]/[HArg]

9.77 x 10^-13 = x^2/3.89 x 10^-6

[Arg-] = 9.77 x 10^-13 M

[Harg] = 3.89 x 10^-6 M

H2Arg+ + H2O <==> H3Arg2+ + OH-

let x amount reactied

Kb3 = Kw/Ka1 = [H2Arg2+][OH-]/[H2Arg+]

1 x 10^-14/0.015 = x^2/0.0148

x = [H3Arg2+] = 9.93 x 10^-8 M