Question

How does molarity affect Ka?

Answer 1

No effect.

Explanation:

`K_a` is the equilibrium constant of the dissociation of an acid.

In general, the acid dissociation is written as:

`HA(aq) rightleftharpoons H^+(aq) + A^(-)(aq)`

The expression of `K_a` is: `K_a = ([H^+][A^-])/([HA])`

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`.