How can value of acid dissociation constant is vary with concentration of acid?
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 $x=1.66M$