a 0.100M solution ofchloroacetic acid (CiCH2COOH) is 11.0% ionized. Using this information, calculate ClCH2COO, H, ClCH2COOH and Ka for chloroacetic acid. I'm having a real hard time doing this problem give me some help on how to do it.
2007-02-24 14:50:13 · 3 answers · asked by __ 3 in Science & Mathematics ➔ Chemistry
If the Chloroacetic acid is ionized, that means it has dissociated into H+ and ClCH2COO. The other 89% is ClCH2COOH. You can use this and the concentration to figure out [H+] and [ClCH2COO] and [ClCH2COOH}. Then we know that Ka= [H+][ClCH2COO]/[ClCH2COOH]. Knowing this, this problem becomes a "plug and chug" problem. You can use this for most acids.
2007-02-24 15:00:04 · answer #1 · answered by cbh12052000 2 · 0⤊ 0⤋
Didn't you do this a few days ago. Actually, the answers are almost staring at you. Just write out the equation.
ClAC-OH -> ClACO + H+
(acid) (acetate) (H-ion)
Then, since Ka is defined as :
(acetae) (H-ion)/{acid) = Ka
The 11% ionization means
(0.011)(0.011)/(.1-0.011) = Ka about 1.5x10^-3
2007-02-24 22:58:44 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
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2007-02-24 23:08:06 · answer #3 · answered by ApCyuMz_05 2 · 0⤊ 0⤋