Final answer:
The pH of a 0.270 M solution of ascorbic acid can be calculated using its two dissociation constants, K1 and K2. The first dissociation gives a pH of approximately 4.10, and the second dissociation gives a pH of approximately 12.80. The final pH of the solution will depend on both dissociations and can be calculated using the weighted average formula.
Step-by-step explanation:
The pH of a 0.270 M solution of ascorbic acid (vitamin C, C6H8O6) can be calculated using its two dissociation constants, K1 and K2. Ascorbic acid is a diprotic acid, which means it can donate two protons (H+ ions). The dissociation equations are as follows:
C6H7O6- + H2O ⇌ C6H8O6 + OH-
C6H6O62- + H2O ⇌ C6H7O6- + OH-
First, we can determine the concentration of each species at equilibrium using the initial concentration of ascorbic acid. Let's assume the concentration of H2O is negligible compared to ascorbic acid.
For the first dissociation, the concentration of C6H7O6- (ascorbate ion) is equal to the concentration of H+ ions (since the reaction goes to completion).
C6H7O6- = [H+] = 8.0 x 10-5 M
For the second dissociation, the concentration of C6H6O62- is equal to the concentration of H+ ions.
C6H6O62- = [H+] = 1.6 x 10-12 M
The pH can be calculated using the equation: pH = -log[H+].
For the first dissociation: pH = -log(8.0 x 10-5) ≈ 4.10.
For the second dissociation: pH = -log(1.6 x 10-12) ≈ 12.80.
Since ascorbic acid is diprotic, the pH of the final solution will depend on both dissociations. We need to consider the concentration of each species and calculate the weighted average pH:
pH = (concentration of H+ from 1st dissociation + 2 * concentration of H+ from 2nd dissociation) / 3
To calculate the final pH, substitute the calculated concentrations into the equation and calculate the weighted average.