Question

What is the strength of the ionic interactions between an ion with 2 positive charges and an ion with 1 negative charges, when the distance between them is 15.5 nanometers and the solvent is benzene, with a dielectric constant of 4.6?

Answer 1

According to the Coulomb's law, the force is given as follows:

$$F = k\dfrac{|q_1q_2|}{\varepsilon r^2}$$

where $k = 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant, $q_1 = +2e$ is the charge of the first ion, $q_2 = -e$ is the charge of the second ion (here $e = 1.6\times 10^{-19}C$ is the elementary charge), $r = 15.5\times 10^{-9}m$ is the distance between ions, and $\varepsilon = 4.6$ is the dielectric constant of benzen.

Thus, obtain:

$$F = k\dfrac{2e^2}{\varepsilon r^2} = 9\cdot 10^9\cdot \dfrac{2\cdot (1.6\times 10^{-19})^2}{4.6\cdot (15.5\times 10^{-9})^2} \approx 4.2\times 10^{-13}N$$

Answer. $4.2\times 10^{-13}N$.