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 0.118 M solution of ascorbic acid.

Question

Determine the equilibrium concentrations of all species in the solution:

1. H2C6H6O6
2. HC6H6O6^-
3. C6H6O6^2-

Answer 1

H2C6H6O6 ===> H⁺ + HC6H6O6^- Ka1 = 8.00x10^-5

HC6H6O6^- ===> H+ + C6H6O6^2- Ka2 = 1.60x10^-12

8.00x10^-5 = (x²)/0.118 - x and assuming x is small we can neglect it to facilitate the calculations

x = [H⁺] = 3.1x10^-3 M (this is very small relative to 0.118 M so assumption was valid)

pH = 2.51. Now to find equilibrium concentrations of various species, use an ICE table....

H2C6H6O6 ===> H⁺ + HC6H6O6^-

0.118....................0............0..........Initial

-3.1x10^-3.........+3.1x10^-3....+3.1x10^-3......Change

0.1149............0.0031.........0.0031 .........Equilibrium

So, to summarize. From 1st ionization, the equilibrium concentrations are as follows:

[H2C6H6O6] = 0.115 M

[HC6H6L6^-] = 0.0031 M

NOTE: The second ionization if VERY small relative to the first, so almost all of the dissociation takes place during the first ionization. If you want to calculate the equilibrium concentrations after the second VERY SMALL change, set up another ICE table and repeat the above calculation.

HC6H6O6^- ===> H+ + C6H6O6^2-

0.0031..................0.0031.......0.........Initial

-x....................+x.................+x.........Change

0.0031-x...........0.0031-x.......x.........Equilibrium

Use 1.6x10^-12 = (0.0031-x)(x)/0.0031-x and solve for x and plug back in to the ICE table to find final equilibrium concentrations of HC6H6O6^- and C6H6O6^2-