Why does water weaken ion ion attractions?

Question

My lecturer has told me that water, having a high dielectric value, will "shield" ions and reduce ion-ion attractions.

I really don't see why. Having water molecules surround these two ions doesn't mean that Coloumb's force doesn't exist anymore - nothing in $F=\frac{kQq}{r^2}$ has changed just because water molecules are surrounding them. I recognise that the water molecules will arrange to face the ions in a polarised way (oxygen to the $+$ ion and hydrogens to the $-$ ion). However, I don't see how having a shell of water surrounding you will reduce the attraction. I guess it would prevent them from coming closer due to the sheer physical size of the water shell, but wouldn't there still be attraction?

Can someone explain to me why the ion-ion attraction decreases when water is there?

Answer 1

Electrostatic field caused by ions or any other external source causes partial orientation of molecular electric dipoles along the field vector. Their orientation causes displacements of center of positive and negative charges. This displacement causes induced electrostatic field, that decreases the primary field intensity by the value of relative permittivity.

By other words, you replace the vacuum permittivity in the Coulomb law by the water permittivity.

$$F = \frac {q_1 \cdot q_2}{4 \cdot \pi \cdot \epsilon \cdot r^2}$$

Answer 2

In addition to the answer by Poutnik, here is another perspective:

Although in the force between the ion pair nothing changes, this is not the only force at play here. Water is a highly polar molecule and its dipole (permanent as well as induced by the ions) interacts with the ions. Thus, we have to consider another Coulomb force term for each water-ion pair in the system. These Coulomb force will be smaller individually, but there are many of them. Due to the alignment of the dipole, each water molecule will provide some negative charge to the cation and some positive charge to the anions. Effectively, this will a) shield the ions charges from each other and b) somewhat smear out the charge over a larger area/volume reducing the charge density.

In essence, this is captured in the water permittivity. But a more detailed (and more cumbersome) description would consider explicite Coulomb terms between every water-ion pair.

Answer 3

As another perspective consider the interaction energy between two ions, this is

$$\displaystyle E=\frac{q_1q_2}{4\pi\epsilon_0}\frac{1}{\epsilon r}$$

where $\epsilon$ is the dielectric constant of the solvent, approx 2 for hexane/benzene and 80 for water and $r$ is the ionic separation. ($\epsilon_0$ is the permittivity for free space)

At $1$ nm separation the energy is

$$\displaystyle \frac{-(1.602\cdot 10^{-19})^2}{4\pi \cdot 8.854\cdot 10^{-8)} }\frac{1}{\epsilon 10^{-9}}\approx 2.3\cdot 10^{-19}\frac{1}{\epsilon} \;\mathrm{J}$$

and so for a non polar solvent the energy is $\approx 10^{-19}$ joules which is a lot, thermal energy at room temperature is $k_BT=4.1\cdot 10^{-21}$ joules so thermal energy is no where near enough to separate the ions, it is $28 k_BT$. In a polar solvent such as water at the same distance the energy is $2.8\cdot 10^{-21}$ so comparable to thermal energy so the ions can drift away from one another.