Calculate the pH of a soln of acetic acid that is 3% ionized, Ka = 1.8 x 10^-5.
Ok, thanks to the help from a very kind person, I realized what I needed to. But now, Im getting the wrong answer over and over again. I get 3.13. But thats not one of the options I have for the answers. If anyone could help me out I would appricate it.
2007-06-30 15:03:27 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
That is the answer look for an option thats rounded up or down?
2007-06-30 15:07:37 · answer #1 · answered by Anonymous · 0⤊ 1⤋
The denominator should have been 0.97 instead of 0.03. It is 97% not ionized.
Try the calculation again
Does pH of 2.38 fit better?
2007-06-30 22:39:59 · answer #2 · answered by reb1240 7 · 1⤊ 0⤋
You're right. Working backwards from your answer:
pH = -log [H+]
3.13 = -log [H+]
[H+] = .00074
[H+] / [HAc] = .03
.00074 / [HAc] = .03
[HAc] = .025
Ka = [H+][Ac-] / [HAc]
1.8 x 10^-5 = .00074^2 / .025
1.8 x 10^-5 is close to 2.2 x 10^-5
2007-06-30 22:36:52 · answer #3 · answered by Anonymous · 0⤊ 0⤋