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First of all I have learned that $K_\mathrm{a} = \frac{[\ce{H+}] [\ce{A-}]} {[\ce{HA}]}$ . Also I have been informed that $K_\mathrm{a}$ value is independent of concentration.

So let us take an example. Suppose 1 mole of acid $\ce{HA}$ dissociates to form 0.9 moles of $\ce{H+}$ ions and 0.9 mole of $\ce{A-}$ ions and 0.1 mole of undissociated $\ce{HA}$. So, the $K_\mathrm{a}$ value will be $K_\mathrm{a} = \frac{[0.9] [0.9]} {[0.1]}=8.1$.

But if we take 2 moles of same acid then, 1.8 moles of H+ ions and 1.8 moles of A- ions will be formed leaving 0.2 moles of undissociated HA. This time $K_\mathrm{a} = \frac{[1.8] [1.8]} {[0.2]}=16.2$.

So we can see that $K_\mathrm{a}$ has increased with increase in concentration. Can you tell me where my assumption is wrong?

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Before we begin, note that $K_a$ is defined in terms of concentration and not amount. Furthermore, $K_a$ is a constant so it must be the same provided the conditions, other than concentration, are the same.

The nature of equilibrium is such that it does not always give the same absolute reaction (i.e. a fixed percentage of reaction, or the reaction does not proceed to the same "position" each time). This is unlike a fixed forward reaction which always ends at 90% completion. In other words, although the 1 molar solution of HA gave 0.9M of $H^+$ and $A^-$, the 2 molar solution would only give out...

$$ K_a = 8.1 = x^2/(2-x) $$

where $x$ is the number of moles of $H^+$ produced. When you solve this, you would get $x=1.66M$

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  • $\begingroup$ But why Does this happen?Why is 90% dissociation not seen in 2 moles? Can you explain please? $\endgroup$
    – Jajati
    May 21, 2016 at 6:24
  • $\begingroup$ @Jajati you could get 90% dissociation in 2 moles, if the concentration was still 1M (i.e. You had 2 dm3 instead of just 1 dm3). But if you're increasing the concentration by dumping dry HA into the system that is already in equilibrium, you could think of it as the system trying to keep the number of moles down in accordance to ale Chatelier's Principle. $\endgroup$
    – IT Tsoi
    May 21, 2016 at 7:01
  • $\begingroup$ OK.But if there is a solution of 2M . In that case will the % of dissociated HA molecules be less than the % in 1 mole?Does that mean that in high concentration,strength of an acid decreases?Because we see less degree of dissociation? $\endgroup$
    – Jajati
    May 21, 2016 at 9:52
  • $\begingroup$ So yes, % dissociated decreases but that doesn't mean that the acid gets "weaker". So if you look at the 2M vs 1M solution, the 2M solution still has more dissociated protons than the 1M solution. $\endgroup$
    – IT Tsoi
    May 21, 2016 at 10:03
  • $\begingroup$ ok thank you very much for your help. Then last question. If the % dissociation of 2M solution is less than 1M then does that mean that if we obtain 0.5M solution then we will see more % dissociation . If yes then why so?And if not then can you please explain the relation of %dissociation and molar concentration . $\endgroup$
    – Jajati
    May 21, 2016 at 17:37

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