I can't type all of this on one line; lets call chloroacetic acid just HAc(since the Cl plays no role anyway---except of course it makes it a much stronger acid than straight acetic acid).
11% ionized means 0.1M is 0.1 x 0.11 = 0.011M
...........HAc ==> H^+ + Ac^-
.........HAc ==> H^+ + Ac^-
I........0.1M.....0......0
C......-0.011...0.011..0.011
E.....0.1-0.011..0.011..0.011
Substitute those numbers into Ka expression and solve for Ka. I would evaluate equilibrium HAc first. You can insert Cl where ever you need it.
A .1M solution of chloroacetic acid is 11% ionized. Using this information, calculate [ClH2COO-], {H+], [ClCH2COOH] and Ka for cloroacetic acid.
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