Question

What fraction of 1,6-hexanedoic acid (adipic acid) is in each form (H2A, HA−, A2−) at pH 5.00?

Answer 1

Represent hexane-1,6-dioic acid (adipic acid) as H₂A and write down the step-wise dissociation:

H₂A <=====> H⁺ + HA^-; pK₁ = 4.43; ====> K₁ = 3.71*10^-5

HA^- <=====> H⁺ + A^2-; pK₂ = 5.41; ====> K₂ = 3.89*10^-6

Again, set up the equations for HA^- and A²- in terms of K₁ and K₂.

K₁ = [HA^-][H⁺]/[H₂A]

====> [HA^-] = [H₂A].(K₁/[H⁺])

K₂ = [H⁺][A^2-]/[HA^-]

====> [A^2-] = [HA^-](K₂/[H⁺]) = [H₂A]*K₁K₂/[H⁺]²

Mass balance equation:

C = [H₂A] + [HA^-] + [A^2-] = [H₂A] + [H₂A]*K₁/[H⁺] + [H₂A]*K₁K₂/[H⁺]²

= [H₂A](1 + K₁/[H⁺] + K₁K₂/[H⁺]²)

Fraction of H₂A in the mixture, α (H₂A) = [H₂A]/C = 1/(1 + K₁/[H⁺] + K₁K₂/[H⁺]²) = [H⁺]²/([H⁺]² + K₁[H⁺] + K₁K₂)

Given pH = 5.0, [H⁺] = 1.0*10^-5

Plug in the value to obtain

α (H₂A) = (1.0*10^-5)²/[(1.0*10^-5)² + (3.71*10^-5)*(1.0*10^-5) + (3.71*10^-5)*(3.89*10^-6)] = (1.0*10^-10)/(1.0*10^-10 + 3.71*10^-10 + 3.89*10^-11) = 1.0*10^-10/5.099*10^-10 = 0.196 (ans).

α (HA^-) = [HA^-]/C = [H₂A].(K₁/[H⁺])/[H₂A](1 + K₁/[H⁺] + K₁K₂/[H⁺]²) = K₁[H⁺]/([H⁺]² + K₁[H⁺] + K₁K₂) = (3.71*10^-5)*(1.0*10^-5)/[(1.0*10^-5)² + (3.71*10^-5)*(1.0*10^-5) + (3.71*10^-5)*(3.89*10^-6)] = (3.71*10^-10)/(6.153*10^-10) = 0.603 (ans).

α (A^2-) = [A^2-]/C = K₁K₂/([H⁺]² + K₁[H⁺] + K₁K₂) = (3.71*10^-5)*(3.89*10^-6)/[(1.0*10^-5)² + (3.71*10^-5)*(1.0*10^-5) + (3.71*10^-5)*(3.89*10^-6)] = 1.443*10^-10/(6.153*10^-10) = 0.234 (ans).