Question

Although acetic acid is normally regarded as a weak acid, it is about $\mathbf{34}\mathbf{\%}$dissociated in a ${\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{M}$solution $\mathbf{25}\mathbf{°}\mathbf{C}$. It is less than $\mathbf{1}\mathbf{\%}$ dissociated in 1M solution. Discuss this variation in the degree of dissociation with dilution in terms of LeChâtelier's principle, and explain how it is consistent with the supposed constancy of equilibrium constants.

Short answer

As the amount of water is increased, then the equilibrium reaction will favor the formation of ions ${\mathrm{H}}_3{\mathrm{O}}^{+}$and $A^{-}$.

Thus, an increase in dilution increases the degree of dissociation.

But at the same time, the value of K (the equilibrium constant) remains constant due to the decrease in the concentration of${\mathrm{H}}_3{\mathrm{O}}^{+}$ and$A^{-}$ on dilution.

Step-by-step solution

Dissociation of a weak acid

Let us consider a weak acid HA.

The dissociation of the weak acid is as shown below:

$\mathbf{HA}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{\rightleftharpoons }{\mathbf{OH}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{+}{\mathbf{A}}^{\mathbf{-}}$

Application of LeChatlier’s principle

As the amount of water is increased, then the above equilibrium reaction will favor the formation of ions${\mathrm{H}}_3{\mathrm{O}}^{+}$ and ${\mathrm{A}}^{-}$.

Thus, an increase in dilution increases the degree of dissociation.

But at the same time, the value of K (the equilibrium constant) remains constant due to the decrease in the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$and ${\mathrm{A}}^{-}$.