Coulomb force and potential energy in water vs vacuum If you have 2 ions of equal but opposite charge, will the force between them be larger in a vacuum and smaller in water? Would this be because the relative permittivity of water is greater than 1 (around 80)? Also, what does this mean for the interaction energy? Would an interaction be larger in vacuum? I am trying to figure out whether a reaction is more likely to take place in the two different media between the two ions...
Yes, much larger in vacuum. The interaction is proportional to the reciprocal of the dielectric constant (relative permittivity) $1/\epsilon$ so is less in water. Think of it as an attenuation of the electric field around an ion, the larger $\epsilon$ is the more the field is attenuated and is true whatever the charges on the ions are making them attractive or repulsive.  The force at distance r between charges $q_1, q_2$ is  

$$F=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in SI units; $\epsilon_0 $ is the permittivity of free space. The charges are $q=ze$ where e is the charge on the electron and z the ionic valency, $\pm 1, 2 \cdots $ etc. 

The interaction energy is 

$$U=\frac{q_1q_2}{(4\pi\epsilon_0)\cdot\epsilon r} \qquad \mathrm{Joule}$$

The electric field around charge $q_1$ is 

$$E=\frac{q_1}{(4\pi\epsilon_0)\cdot\epsilon r^2}$$

in V/m. The force acting on  a second charge $q_2$ is $F=Eq_2$ 

Its worth plugging in some numbers for the energy for different $\epsilon$ and comparing this to thermal energy $k_BT$.