I did the experiment to find out the dissociation constant of a weak acid using Henderson equation. However, after going through the theoretical part of the experiment, I am wondering why we need two separate equations for finding out the Acid Dissociation constant and the Equilibrium constant.

Simply put, the expression for dissociation constant and equilibrium constant is identical. As such, when an acid dissociates, we could have explained it using the equilibrium constant equation as well; then, why do we need a new term called Acid Dissociation constant for explaining the same?

One explanation for this question that I can think of right now is - we can use the equation for equilibrium constant only when an equilibrium is achieved, and the concentration of acid, i.e., [Acid] must be present in the denominator of the equilibrium constant equation. However, we can use the equation for acid dissociation constant irrespective of the fact that the reaction reaches an equilibrium or not. So, the presence of the [Acid] term in the denominator is of little significance in case of strong acid since they get dissociated completely, however, it will still have a great significance for dissociation of a weak acid.

I have got an explanation in terms of activity from this answer, but is there a simpler explanation? A more intuitive one will help me understand the concept better.

  • 4
    $\begingroup$ Acid dissociation constant = equilibrium constant of acid dissociation. $\endgroup$
    – Poutnik
    Oct 28 at 16:35
  • $\begingroup$ @Poutnik, Okay. But, why do we need the acid dissociation constant equation to calculate the dissociation constant of the weak acid? Can't it be done using equilibrium constant equation? $\endgroup$
    – Sanu_012
    Oct 28 at 16:43
  • 2
    $\begingroup$ See the first comment. They are synonyms. The same equation using different synonyms is still the same equation. // Provide eventual equations you refer to. $\endgroup$
    – Poutnik
    Oct 28 at 16:53
  • $\begingroup$ The answer by Lautman has provided the required equations. I guess he has made your point clearer. $\endgroup$
    – Sanu_012
    Oct 28 at 17:04
  • 1
    $\begingroup$ It was your job to make your question clear. $\endgroup$
    – Poutnik
    Oct 28 at 17:05

For any system at equilibrium we can define an equilibrium constant as follows:

$$\ce{A(aq) + B(aq) <=> C(aq) + D(aq) }$$

$$K_\text{eq} = \frac{[\ce{C}][\ce{D}]} {[\ce{A}][\ce{B}]}$$

When the solutions are aqueous, we might use the subscript $c$ for the $K$, to indicate concentration. When we are using gas pressure instead of concentration, we use the subscript $p$.

In the case of acid dissociation, it's just another reaction at equilibrium with $\ce{H3O+}$ in the products.

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^-(aq)}$$

$$K_\text{a} = \frac{[\ce{H3O+}][\ce{A^-}]} {[\ce{HA}]}$$

The term "acid dissociation constant" is used instead of "equilibrium constant" for added clarity. In this situation they mean the same thing.

  • $\begingroup$ So, having a hydronium is essential in case of acid dissociation constant equation? We can identify an acid dissociation reaction explicitly by the presence of hydronium ion? $\endgroup$
    – Sanu_012
    Oct 29 at 1:28
  • $\begingroup$ Yes. The presence of hydronium as a product is what makes it an acid dissociation. $\endgroup$ Nov 2 at 15:06

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