Ascorbic acid ( H2C6H6O6 ) is a diprotic acid. The acid dissocation constants for H2C6H6O6 are 𝐾a1=8.00×10−5 and 𝐾a2=1.60×10−12. Determine the pH of a 0.135 M solution of ascorbic acid.

Question

Determine the equilibrium concentrations of all species in the solution.

ph

[H2C6H6O6]

[HC6H6O−6]

[C6H6O2−6]

Answer 1

Because the second dissociation constant (1.6x10^-12) is more than 3 orders of magnitude less than the first dissociation constant (8.00x10^-5), it can be ignored as far as making any meaningful contribution to the [H+], and hence has negligible effect on pH.

H2C6H6O6 ==> H⁺ + HC6H6O6^-

Ka1 = 8.00x10^-5 = [H⁺][HC6H6O6^-]/[H2C6H6O6]

8.00x10^-5 = (x)(x)/0.135 - x and assuming x is small compared to 0.135, it can be ignored in the denominator

x² = 1.08x10^-5

x = 3.29x10^-3 M = [H⁺] (assumption above was correct as this is only 2.4% of 0.135 and so is negligible)

pH = -log 3.29x10^-3 = 2.48

Equilibrium concentrations as follows:

[H2C6H6O2] = 0.135 - 0.00329 = 0.132 M

[HC6H6O6^-] = 3.29x10^-3 M

[C6H6O62-] = 7.3x10^-8 M (see below)**

HC6H6O6^- ==> H+ + C6H6O62^-

1.6x10^-12 = (x)(x)/3.29x10^-3

x2 = 5.3x10^-15

x = 7.3x10^-8 M = equilibrium concentration of C6H6O6^2-