How does molarity affect Ka?

Question

How does molarity affect Ka?

1 Answer

Impact of this question

16958 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-19T21:11:44

No effect.

Explanation:

$K_{a}$ $K_a$ is the equilibrium constant of the dissociation of an acid.

In general, the acid dissociation is written as:

$H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)$ $HA(aq) rightleftharpoons H^+(aq) + A^(-)(aq)$

The expression of $K_{a}$ $K_a$ is: $K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]}$ $K_a = ([H^+][A^-])/([HA])$

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

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

Science

Math

Humanities

... and beyond

How does molarity affect Ka?

1 Answer

Explanation:

Related questions

Impact of this question