Final answer:
To find the pH of a 0.144M ascorbic acid solution, we use the first dissociation constant Ka1 and ignore the second due to its small size. An ICE table is set up to solve for [H+], from which we calculate pH directly. The estimated pH is around 2.97.
Step-by-step explanation:
The question concerns determining the pH of a 0.144M solution of ascorbic acid, which is a diprotic acid with dissociation constants Ka1=8.00 x 10-5 and Ka2=1.60 x 10-12. Because ascorbic acid is a weak acid and its first dissociation constant is much larger than the second, we can assume the contribution to the [H+] from the second dissociation is negligible. Therefore, we can determine the pH by considering only the first dissociation.
Using the acid dissociation constant Ka, the concentration of the acid ([HA] = 0.144M), and assuming that [H+] will be equal to [A-] and much less than the initial concentration of HA, we can set up an ICE table and use the Ka expression to solve for [H+]:
Ka = [H+][A-]/[HA]
[H+]^2 / (0.144 - [H+]) = 8.00 x 10-5
[H+] = √(8.00 x 10-5 * 0.144)
[H+] ≈ 1.08 x 10-3M
Then, the pH is calculated as:
pH = -log(1.08 x 10-3)
We can then estimate the pH to be around 2.97, although the actual calculation would need to be done with a calculator to get an exact number.