Question

Calculate the pH of a 0.0158 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.

H3Arg2+

H2Arg+

HArg

Arg-

Answer 1

You have to use the next equation

[H] = (Ka1*Ka2*[H₂Arg] + Ka1*Kw) / (Ka₁ + [H₂Arg])

The pka have been provided

Ka = 10^-Pka

ka1	0.01479108
ka2	1.0209E-09
ka3	9.7724E-13

the value have been provided , kw = 1 x10^-14, so the value for H will be

[H] = 2.8 x 10^-6 M

PH = -log [H] = 5.55

Now rewrite the equilibrium expression for the first dissociation

H₃A   →   H⁺ + H₂A^-

Ka = [H⁺][H₂A^-] / [H₃A]

0.015 = 2.8 x 10^-6 M * 0.0158 / [H₃A]

[H₃A] = 2.94 x 10^-6 M

For [HA]  we use the second dissociation

H₂A   →   H⁺ + HA^-

Ka₂ = [H⁺][HA^-] / [H₂A]

1.02x10^-19 = 2.8 x 10^-6 M * [HA] / [0.0158]

[HA] = 5.75 x 10^-6 M

For the [A] we use the third dissociation

HA^-   →   H⁺ + A^-

Ka₃ = [H⁺][A^-] / [HA]

9.77x10^-13 = 2.8 x 10^{-6 *} X / (5.75x10^-6)

X = Arg^- = 2 x 10^-12 M