# Current and concentration

• gracy
When concentration decreases degree of dissociation increases ,that's why current increases.This is actually incorrect. When you dilute the solution current decreases as expected. But the decrease is slower than one would expect if it was just a matter of dilution. That's because the dissociation degree goes up, partially counteracting effects of the dilution. As I already told you elsewhere, this can be easily explained following properties of the dissociation equilibrium.That's because the dissociation degree goes up, Why?that's what i want to understand.How can we explain this by le chatelier's principle?Let us take Electrolytes in compound form (undissociated form )as reactants and their dissf

#### gracy

Why degree of dissociation increases when concentration decreases?It was written in my textbook that when concentration decreases i.e on dilution there is increase in current because of increase in charge carriers i.e ions,I thought that it could be because on dilution water is added,so now more number of charge carriers because of H+ And OH ions.Am i right?
But in my textbook it is written that when concentration decreases degree of dissociation increases ,that's why current increases.This point is really not clear to me,can anyone help?Why degree of dissociation increases (and in turn current)when concentration decreases?

Concentration of H+ and OH- present in pure water is so low it can be safely neglected when it comes to the discussion of the conductivity. Ultra pure water has a specific resistivity of 18 MΩ·cm - so your idea that they matter is wrong.

when concentration decreases degree of dissociation increases ,that's why current increases

This is actually incorrect. When you dilute the solution current decreases as expected. But the decrease is slower than one would expect if it was just a matter of dilution. That's because the dissociation degree goes up, partially counteracting effects of the dilution. As I already told you elsewhere, this can be easily explained following properties of the dissociation equilibrium.

That's because the dissociation degree goes up,
Why?that's what i want to understand.How can we explain this by le chatelier's principle?Let us take Electrolytes in compound form (undissociated form )as reactants and their dissociated ions as products so increasing water i.e on dilution degree of dissociation increases ,so dissociated ions increases i.e product increases,so does this mean water is reactant here so that when it's concentration increases ,equilibrium shifts towards product?I know it doesn't make any sense ,but i have been told to show some efforts so .that's what my understanding is...Please tell me how adding water favors more dissociation?Is it because as we know in aqueous as well as molten state electrolytes dissociate?I know in aqueous state electrolytes dissociates but is this dissociation directly proportional to water,adding more water more dissociation?

"Why" is not a good question in science, "how" is much better.

But if you really want "why" - think of the dissociation as if it was a reaction between water molecules and dissociating substance molecules. Evey entity in water solution (be it ion or a neutral molecule) is to some extent surrounded by so called "hydration water". Charged ions attract much more water molecules (water is a dipole, so it is attracted to charges). Dissociation is then

AB + nH2O ↔ A(H2O)k+ + B(H2O)l- (n=k+l)

and adding water shifts the equilibrium to the right

Just remember, this is a handwavy argument, not guaranteed to survive exact analysis.

Last edited:
gracy
"Why" is not a good question in science, "how" is much better.
I will always remember this.Thanks.

and adding water shifts the equilibrium to the right
I wouldn't say so. It is a purely entropic phenomenon.

In fact, higher dissociation with reduced concentration also takes case in the gas phase without water, e.g.
## \mathrm{N_2O_4 \leftrightharpoons 2 NO_2}##.
The point is that the entropy of a gas molecule or an ion increases the more space it has available. Hence increasing the volume at fixed particle number (i.e. decreasing concentration) will increase the entropy of the gasses or ions and the reaction will be driven in the direction where more gas molecules / ions are present.

gracy
I was just trying to use LeChetelier's principle which the OP called for.

While I generally agree with the entropic argument, solvation which bonds water molecules lowers entropy - so there are two competing processes here.

I think what you have in mind is a change of the activity of the ions with concentration.
Assuming a sufficiently diluted solution so that Debye Hueckel Theory is applicable, the activity of water is always very nearly 1. The activity of the ions is smaller than the concentration, but the difference becomes smaller with higher concentration.
So yes, this would counteract dissociation to some degree, but would never outweigh the entropic effect.

No, what I mean is that water molecules that get caught in the solvation layers lose their ability to move freely. This both makes the solvation exothermic and "endoentropic". During dissolution on one hand you have increase of entropy (solid becomes dispersed as ions), on the other hand you have decrease of entropy (water molecules become trapped around these ions). Overall effect will depend on the solvation number and I have no idea what the exact numbers are (but I do remember measurements of the solvation entropy being used to determine solvation numbers, so these things are definitely known).

Yes, that's an effect probably relevant at even higher concentrations than what I had in mind. This effect will have a great influence on the dissociation constant, but at lower concentration (when the solvation shells of the ions don't overlapp, statistically), it will hardly vary with concentration.