Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Jules18 on December 09, 2009, 11:18:16 PM
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Hey guys, I think I've done the better half of this question, but I'm stuck at the end:
Calculate the concentration of the acetate ion (M) in a solution prepared by dissolving 1.00×10-3 mol of HCl(g) in 1.00 L of 1.90 M aqueous acetic acid. Ka = 1.7E-5
. CH3COOH --> CH3COO- + H+
initial: 1.90 M 0 0
change: - x +x +x
equlbm: ~1.9 x x
blah blah blah, calculations ... x = 5.68E-3 . So that would be the acetate concentration before you add the 1E-3 mol H+.
And at that point I'm not sure what to do.
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. CH3COOH --> CH3COO- + H+
initial: 1.90 M 0 0
change: - x +x +x
equilibrium: ~1.9 x x
Well before adding HCl is as above. But after that It would be...
. CH3COOH <--> CH3COO- + H+
initial: 1.90 M 0 0
change: - x +x +x
equilibrium: ~1.9 x 10-3+x ~10-3
You know why?
-
. CH3COOH --> CH3COO- + H+
initial: 1.90 M 0 0
change: - x +x +x
equilibrium: ~1.9 x x
Well before adding HCl is as above. But after that It would be...
. CH3COOH <--> CH3COO- + H+
initial: 1.90 M 0 0
change: - x +x +x
equilibrium: ~1.9 x 10-3+x ~10-3
You know why?
Should be
equilibrium: 1.9-x ~1.9 x 10-3+x ~10-3