Effect of dilution on the degree of dissociation of an electrolyte My book says that on dilution, the degree of dissociation of an electrolyte increases. Consider the reaction:

$$\ce{A(aq) &lt;=&gt; C(aq) + D(aq)}$$

Let $\ce{A}$ be the electrolyte that splits into its ions $\ce{C}$ and $\ce{D}$.
According to my book, on dilution, the concentration of $\ce{C}$ and $\ce{D}$ will decrease. Hence according to the Le Chatelier's principle, the reaction must proceed in forward direction.

Won't the concentration of $\ce{A}$ also decrease? In that sense, shouldn't the reaction proceed both in the forward and backward direction?

Please do let me know where am I going wrong.
If our dummy dissociation degree is $$\alpha=\frac{[\text{C}]}{[\text{A}]+[\text{C}]}=\frac{[\text{D}]}{[\text{A}]+[\text{D}]}$$ and the total concentration of dissociated and undissociated species is $$x=[\text{A}]+[\text{D}]=[\text{A}]+[\text{C}]$$ the equilibrium constant, on which le Chatelier's principle is (well, should be) based is $$K=\frac{[\text{C}][\text{D}]}{[\text{A}]}=\frac{\alpha^2x}{1-\alpha}$$ Now, we'd like to dilute (to decrease) the total concentration $x$. Since the equilibrium constant cannot change, $\alpha$ must increase to keep the constant constant (sorry). And all the concentrations decrease upon dilution, which you can easily prove by checking whether derivatives $\frac{\partial[\text{A}]}{\partial x}$, $\frac{\partial[\text{C}]}{\partial x}$ and $\frac{\partial[\text{D}]}{\partial x}$ are positive.

I really hate them easy "intuitive" principles. In my opinion, resorting to the discussion of the equilibrium constants gives you a clearer picture.

And yes, the reaction proceeds in both forward and reverse directions, if only because it is an equilibrium process.