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Author: Subject: trying to understand Ka
soma
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[*] posted on 16-4-2016 at 02:32
trying to understand Ka


I'm wondering what the Ka of a substance is at 50% dissociation. If Ka=[A-][H+]/AH then e.g. if you have 6 moles of acetic acid and it's 50% dissociated, then you would have 3 moles of H+ and 3 moles of CH3CO2. Would that mean Ka=3x3/6?
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[*] posted on 16-4-2016 at 04:46


No. First, you need to express [HA], [A-], and [H+] as concentrations, not amounts of moles. This is because there's an implicit solvent term in the equilibrium constant.

Assuming 1 liter of an ideal solution, Ka would be (3*3)/3 = 3. This is because there are only 3 moles/liter of HA left after 3 moles/liter have dissociated.

The actual Ka of acetic acid is 1.75*10−5

[Edited on 4-16-2016 by Metacelsus]




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