Calculate the pH of a 0.0175 M solution of arginine hydrochloride ( arginine·HCl , H2Arg+ ). Arginine has p𝐾a values of 1.823 ( p𝐾a1 ), 8.991 ( p𝐾a2 ), and 12.01 ( p𝐾a3 ).

Question

Calculate the pH of a 0.0175 M solution of arginine hydrochloride (arginine·HCl, H2Arg+). Arginine has p𝐾a values of 1.823 (p𝐾a1), 8.991 (p𝐾a2), and 12.01 (p𝐾a3).

pH=

Calculate the concentration of each species of arginine in the solution.

[H3Arg2+]=

M

[H2Arg+]=

×10

M

[HArg]=

×10

M

[Arg−]=

×10

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