How do the ions in electrolytes lead to lower dielectric constants compared to pure water?

I am interested in the salts used in the gel electrophoresis buffers. As I understand it, one of their roles is to change the conductivity of the medium, because otherwise the electric field would be too weak inside the medium to move the negatively charged sample.


How does this work from a mechanistic perspective? I found one lead [here](https://chemistry.stackexchange.com/a/19214/135232), but I don't see how the diagram explains the effect of the rearrangements on the overall electric field. (It seems to imply that the effects on the saline-associated electric field compared to pure water would be net zero to me, since the fields from the charge separation cancel out with the ones from the water.)


Using simplifications like point charges and intuition, is it possible to explain why the dielectric constant becomes lower in saline solutions compared to pure water?




An electric dipole with a dipole moment $\vec{p}$ in an electric field $\vec{E}$ has the energy : $\Delta E = - \vec{p}.\vec{E}$. This energy is minimal if the dipole moment and the field are aligned. If they are aligned by applying a weak field, the compound has a good susceptibility. However, some ionic compounds cannot align their dipoles when a field is applied especially in the case of solids like $\text{NaCl}$ where the dipole moments are antiparallel but the atoms cannot move leading to a low susceptibility and a low dielectric constant.


In the case of $\text{H}\_2\text{O}\_{(aq)}$, the effective dipole moment of each water molecule can be aligned in the direction of the field because water molecules can move as the hydrogen bonds are weaker than covalent bonds.


When $\text{NaCl}$ is in water, all the ions are spherically surrounded by water molecules : $\text{Na}^+$ ions form hydrogen bonds with $O$ and $\text{Cl}^-$ ions with $H$. If a field is applied along one axis in this spherically symmetric system and the water molecules is polarized as in a pure water, the hydrogens with a charge $\delta^+$ trying to move in the direction of the field will create an electrostatic repulsion with $\text{Na}^+$ ions which are also following the same direction, this process is identical for $\text{Cl}^-$ and $O$. This means that for aligning a water molecule when a field is applied with the presence of ions there is an electrostatic energy cost to pay with the energy $-\vec{p}.\vec{E}$. This means that the field $\text{E}$ should be stronger for a more negative $\Delta E$ to achieve the same process than in pure water. Therefore a salted water has dropped the water susceptibility more significantly if more ions are present.


A simple approach when we deal with a mixture is to calculate approximately some properties by applying the rules of mixture at the equilibrium :


$$\epsilon\_r(\text{H}\_2\text{O/NaCl}) \approx \frac{\epsilon\_r(\text{H}\_2\text{O}+\epsilon\_r(\text{NaCl})}{2} \approx 42$$


Aproximately the same result that in your reference. I am not good at drawing but I hope the picture of dipoles in a field and this repulsion is not complex.


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