Chemistry 102: Acids: find pH of solution

Question

14.

Ascorbic acid (vitamin C, C₆H₈O₆) is a diprotic acid (K_a1 = 8.0×10^–5 and K_a2 = 1.6×10^–12). What is the pH of a 0.220 M  solution of ascorbic acid?



Answer 1

Since the 2nd Ka is so much less than the 1st Ka, we can ignore it when calculating pH as it will not affect the concentration of hydronium ions significantly.

C₆H₈O₆ ==> C₆H₇O₆ + H⁺

Ka = [C₆H₇O₆][H⁺] / [C₆H₈O₆]

8.0x10^-5 = (x)(x) / 0.220 - x and assume x is small relative to 0.220 so ignore it in the denominator

8.0x10^-5 = x² / 0.220

x² = 1.76x10^-5

x = [H⁺] = 4.20x10^-3 (which is about 1.9% of 0.220 so above assumption was valid)

pH = -log [H+] = -log 4.20x10^-3

pH = 2.38

Answer 2

Hi Katelyn! I wrote out a response in a white board and posted it here:

https://imgur.com/Okwgdts

Let me know if you have any questions.