Calculate the concentration of ethanoic acid with pH 4.53. (Ka = 1.74x10^-5)
No idea how to do this...
Calculate the concentration of weak acid? Given pH and Ka Watch
- Thread Starter
- 05-11-2017 21:33
- 05-11-2017 21:52
pH = 4.53 = -log[H+]
[H+] = 2.951x10^-5
Ka = 1.74x10^-5 = [H+][CH3COO-] / [CH3COOH]
1.74x10^-5 = [2.951x10^-5]^2 / [CH3COOH]
[CH3COOH] = 5.0 x 10^-5
I think that's correct. Not completely sure though.