For example solution C has 5 M of HA and 10 M of A^{-}, so the dissociation degree is 2/3 or 0.66 (you calculated it as 1/3 by accidentally switching A with HA). Solution E has dissociation degree of 0.75 - so the substance from solution E is dissociated more. Does it mean it is a stronger acid? No! K_{a} for E is 0.9, while K_{a} for C is 20, so it is C that is a much stronger acid.

I still don't agree with you. I admit I'm a bit rusty on the subject but something still feels off. You state at the K

_{a} of E is 0.9 which I assume you got by assuming it was initially a typo and that [HA]=100

**m**M and [A

^{-}]=300

**m**M. If we assume it is not a typo (as ridiculous as that is) we get the value 0.9 for the K

_{a}. Furthermore if we assume [HA]=100μM and [A

^{-}]=300μM we get the value 9×10

^{-6}.

Unless I did my calculations wrong (which I won't rule out) something seems off. There seems to be some sort of of bias in the system due to ionic strength considerations. When comparing the inherit acidicities of two compounds I would argue it is inappropriate to account for ionic strength.

I honestly can't argue with your mathematics - that's how we define K

_{a} and that's generally the metric used to determine acidity of a substance is just... doesn't feel right.