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

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

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.

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  • $\begingroup$ Your question is good one . So let me first correct you your argument "conc. Of both product and reactant will decrease on dilution so the reaction will remain in equilibrium Without any change in no of moles of either product or reactant" First of all Le Chatliers principle talks about combined conc. Of reactant and product so on dilution combined conc will decrease so to increase the combined conc. Reaction will proceed in direction with greater no of moles. I Hope it will help you. $\endgroup$ – Rohan kumar Oct 20 '19 at 15:57
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/72609/… $\endgroup$ – Karsten Theis Oct 21 '19 at 11:04
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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.

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Your book is right, On dilution, the degree of dissociation of an electrolyte increases. But your interpretation is wrong. If you increase the degree of dissociation of A, it means that there is less A, and more B and C in the solution. The degree of dissociation is by definition the fraction of A which are transformed into B and C. This fraction increases by dilution, according to le Chatelier's principle.

Finally, by dilution, all concentrations decrease, whatever their degree of dissociation.

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Water is often described as the universal solvent because most things will dissolve in water if you have enough time and water.

For a compound that is only slightly soluble in water adding more water will indeed increase the number of electrolytes formed.

It may help to think about your reaction in this format:

$$\ce{A + H2O <=> C + D}$$

If copious amounts of water are needed to get the material to dissolve thinking of water as a reactant may help you to understand the equilibrium. Of course, adding water still has an effect on the concentration of reactants and products.

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