# How does molarity affect Ka?

Oct 19, 2015

No effect.

#### Explanation:

${K}_{a}$ is the equilibrium constant of the dissociation of an acid.

In general, the acid dissociation is written as:

$H A \left(a q\right) r i g h t \le f t h a r p \infty n s {H}^{+} \left(a q\right) + {A}^{-} \left(a q\right)$

The expression of ${K}_{a}$ is: ${K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

The concentrations in the expression of ${K}_{a}$ are the equilibrium concentration, they change however, their ratio remains constant, and therefore, the name of constant ${K}_{a}$.

The ONLY factor that affect ${K}_{a}$ is temperature, and therefore there is no influence of concentration change on the value of ${K}_{a}$.